mm-3189 === Subject: 'and' or 'or' I'm a student teacher with a couple of questions about what my teacher said in class: Given a function with denominator (3x+4)(x-5), the question was to find the restrictions on x. The teacher said the restrictions were 5 and -4/3. A student raised her hand and said, isn't it 5 *or* -4/3? The teacher said, no, it's 'and', because they're both restrictions. I found this explanation confusing and I think the student did too even though she just said oh, okay (as students tend to do). I'm no expert on logic, but isn't it the case that either one or both of the factors, cannot be zero, and so doesn't that mean we should state the restrictions as 5 inclusive or -4/3? --- Another exercise on the handout read: Given f(x) = 3x-2, state the restrictions of x when 4 < f(x) < 10. A lot of the students and myself were confused by this. I thought she meant what values of x are not permitted because the function is undefined at them, which would have been none. But when she took the question up, she said she meant, what values of x correspond to the function values from 4 to 10. Would you --- === Subject: Re: 'and' or 'or' > I'm a student teacher with a couple of questions about what my teacher > said in class: > Given a function with denominator (3x+4)(x-5), the question was to find > the restrictions on x. The teacher said the restrictions were 5 and > -4/3. A student raised her hand and said, isn't it 5 *or* -4/3? The > teacher said, no, it's 'and', because they're both restrictions. I > found this explanation confusing and I think the student did too even > though she just said oh, okay (as students tend to do). I'm no > expert on logic, but isn't it the case that either one or both of the > factors, cannot be zero, and so doesn't that mean we should state the > restrictions as 5 inclusive or -4/3? Whether you say and or or depends on where you place the not. 1/((3x+4)(x-5)) is not defined if and only if x=5 or x=-4/3 equivalently 1/((3x+4)(x-5)) is defined if and only if x<>5 and x<>-4/3 So, I would say either of these is a better answer than 5 and -4/3 or 5 or -4/3 because in these last two cases it is not clear what is meant by it. -- http://www.math.ohio-state.edu/~edgar/ === Subject: Re: 'and' or 'or' > I'm a student teacher with a couple of questions about what my teacher > said in class: > Given a function with denominator (3x+4)(x-5), the question was to find > the restrictions on x. The teacher said the restrictions were 5 and > -4/3. A student raised her hand and said, isn't it 5 *or* -4/3? Neither, I don't think 5 is a restriction, I think it's a number. Here is a legitimate way of expressing the idea using and: x cannot be 5 and x cannot be -4/3. and here is a legitimate way of expressing the idea using or: x cannot be 5 or -4/3. Just to make plain the connection with de Morgan, here they are with brackets: (x cannot be 5) and (x cannot be -4/3). x cannot be (5 or -4/3). It can be a good idea to express mathematical thoughts in full, grammatically correct, sentences just as with any other expression of an idea. -- I don't know who you are Sir, or where you come from, but you've done me a power of good. === Subject: Re: 'and' or 'or' > ... The grammar of or and and can be a bit puzzling. For example, someone who says I don't like tea or coffee. probably means the same as if they had said I don't like tea and I don't like coffee. -- I don't know who you are Sir, or where you come from, but you've done me a power of good. === Subject: Re: 'and' or 'or' >The grammar of or and and can be a bit puzzling. I'll say! Those who had Fire and flood insurance for property in Louisiana now probably wish they had Fire OR flood insurance. For the topic at hand in the thread let me add that the student teacher should take this opportunity to emphasize what a lousy idea it is for students to think of mathematics as symbol pushing. It would be so much better for students to embed the symbols in sentences: If x^2-6x+8=0 then (x-2)(x-4)=0, so either x-2=0 or x-4=0, so either x=2 or x=4. Surely the students would understand instantly that AB=0 does not imply (A=0 AND B=0). As a side benefit, it becomes clear that this line of reasoning is only an if-then statement, not an if-and-only-if. Making this distinction now will help a lot when student go to solve things like x - 2 = sqrt( 2 x - 4 ) or x / (x-1) = x^2 / (x-1) later, not to mention any bit of mathematics with proofs or logic in them. dave PS -- If someone's going to treat you to ice cream, make sure the benefactors are mathematicians. If they say You may have chocolate or vanilla, that doesn't preclude ...or both! === Subject: Re: 'and' or 'or' > I'm a student teacher with a couple of questions about what my teacher > said in class: > Given a function with denominator (3x+4)(x-5), the question was to find > the restrictions on x. The teacher said the restrictions were 5 and > -4/3. A student raised her hand and said, isn't it 5 *or* -4/3? The > teacher said, no, it's 'and', because they're both restrictions. I > found this explanation confusing and I think the student did too even > though she just said oh, okay (as students tend to do). I'm no > expert on logic, but isn't it the case that either one or both of the > factors, cannot be zero, and so doesn't that mean we should state the > restrictions as 5 inclusive or -4/3? How can both factors be 0 at the same time? That would mean x is simultaneously 5 and -4/3. You inclusive or makes no sense either. The teacher was not using and in the logical sense but rather as connection or addition, such as 4 and 2 are 6 as opposed to the logical 4 AND 2 = 0. > --- > Another exercise on the handout read: Given f(x) = 3x-2, state the > restrictions of x when 4 < f(x) < 10. A lot of the students and myself > were confused by this. I thought she meant what values of x are not > permitted because the function is undefined at them, which would have > been none. But when she took the question up, she said she meant, what > values of x correspond to the function values from 4 to 10. Would you The difference is the inclusion of the when clause. Obviously, restriction means more than simply undefined. Note in the first problem you are asked what x CANNOT be. In the second, you are asked what x MUST be. Cannot and must are both restrictions. > --- === Subject: Re: 'and' or 'or' > I'm a student teacher with a couple of questions about what my teacher > said in class: > Given a function with denominator (3x+4)(x-5), the question was to find > the restrictions on x. The teacher said the restrictions were 5 and > -4/3. A student raised her hand and said, isn't it 5 *or* -4/3? The > teacher said, no, it's 'and', because they're both restrictions. I > found this explanation confusing and I think the student did too even > though she just said oh, okay (as students tend to do). I'm no > expert on logic, but isn't it the case that either one or both of the > factors, cannot be zero, and so doesn't that mean we should state the > restrictions as 5 inclusive or -4/3? Randy Poe already gave a good reply. I can add this: Suppose you want to solve the equation (x-5) (x-6) = 0 in the real numbers for instance. Then you can correctly write (x-5) (x-6) = 0 <==> x = 5 or x = 6 and you can correctly say: the solution set for that equation is the set { 5, 6 }, or less formally the equation has two solutions, namely x = 5 and x = 6, or sometimes the equation has two solutions, namely x1 = 5 and x2 = 6, or even the equation has two solutions for x, namely 5 and 6. In the first case you use a formal expression with the correct logical structure. In the second case you use set notation, and in the remaining cases you use everyday language to enumerate the solutions: you have this solution *and* you have that solution and so on... It takes a bit getting used to this, but the teacher was right. > --- > Another exercise on the handout read: Given f(x) = 3x-2, state the > restrictions of x when 4 < f(x) < 10. A lot of the students and myself > were confused by this. I thought she meant what values of x are not > permitted because the function is undefined at them, which would have > been none. But when she took the question up, she said she meant, what > values of x correspond to the function values from 4 to 10. Would you Yes, no problem. The phrase Given f(x) = 3 x - 2 usually implies that x is defined everywhere. Then you set up a string of equivalences starting from the restriction. 4 < f(x) < 10 <==> 4 < f(x) and f(x) < 10 <==> 4 < 3 x - 2 and 3 x - 2 < 10 <==> x > 2/3 and x < 8/3 <==> 2/3 < x < 8/3 Dirk Vdm > --- === Subject: Re: 'and' or 'or' >> I'm a student teacher with a couple of questions about what my teacher >> said in class: >> Given a function with denominator (3x+4)(x-5), the question was to find >> the restrictions on x. The teacher said the restrictions were 5 and >> -4/3. A student raised her hand and said, isn't it 5 *or* -4/3? The >> teacher said, no, it's 'and', because they're both restrictions. I >> found this explanation confusing and I think the student did too even >> though she just said oh, okay (as students tend to do). I'm no >> expert on logic, but isn't it the case that either one or both of the >> factors, cannot be zero, and so doesn't that mean we should state the >> restrictions as 5 inclusive or -4/3? > Randy Poe already gave a good reply. > I can add this: > Suppose you want to solve the equation > (x-5) (x-6) = 0 > in the real numbers for instance. > Then you can correctly write > (x-5) (x-6) = 0 > <== x = 5 or x = 6 > and you can correctly say: > the solution set for that equation is the set { 5, 6 }, > or less formally > the equation has two solutions, namely x = 5 and x = 6, > or sometimes > the equation has two solutions, namely x1 = 5 and x2 = 6, > or even > the equation has two solutions for x, namely 5 and 6. > In the first case you use a formal expression with the correct > logical structure. In the second case you use set notation, > and in the remaining cases you use everyday language to > enumerate the solutions: you have this solution *and* you > have that solution and so on... > It takes a bit getting used to this, but the teacher was right. >> --- >> Another exercise on the handout read: Given f(x) = 3x-2, state the >> restrictions of x when 4 < f(x) < 10. A lot of the students and myself >> were confused by this. I thought she meant what values of x are not >> permitted because the function is undefined at them, which would have >> been none. But when she took the question up, she said she meant, what >> values of x correspond to the function values from 4 to 10. Would you > Yes, no problem. > The phrase > Given f(x) = 3 x - 2 > usually implies that x is defined everywhere. > Then you set up a string of equivalences starting from the restriction. > 4 < f(x) < 10 > <== 4 < f(x) and f(x) < 10 > <== 4 < 3 x - 2 and 3 x - 2 < 10 > <== x > 2/3 and x < 8/3 Typo.............................x<12/3 > <== 2/3 < x < 8/3 > Dirk Vdm >> --- === Subject: Re: 'and' or 'or' > 4 < 3 x - 2 and 3 x - 2 < 10 > <== x > 2/3 and x < 8/3 > Typo.............................x<12/3 Yes, more of a sillo Dirk Vdm === Subject: Re: 'and' or 'or' > I'm a student teacher with a couple of questions about what my teacher > said in class: It is or as x can take on only one value at a time. === Subject: Re: 'and' or 'or' days. My association with the Department is that of an alumnus. >> I'm a student teacher with a couple of questions about what my teacher >> said in class: >It is or as x can take on only one value at a time. That's no good. Then x is different from 5 or different from -4/3 would means that x can take any value, since given any specific value, it will EITHER be different from 5 or it will be different from -4/3, hence will satisfy the condition. The problem is that the statement was not really given. Either and or or will be appropriate, depending on how you phrase it. If you specify the values that x cannot be, then you should use 'and': x different from 5 and different from -4/3; 'or' would be incorrect. On the other hand, you could also say x should not be equal to 5 or to -4/3, in which case 'and' would be inappropriate. It is just an application of De Morgan's Law. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: 'and' or 'or' > I'm a student teacher with a couple of questions about what my teacher > said in class: >>It is or as x can take on only one value at a time. > That's no good. Then x is different from 5 or different from -4/3 > would means that x can take any value, since given any specific value, > it will EITHER be different from 5 or it will be different from -4/3, > hence will satisfy the condition. > The problem is that the statement was not really given. Either and > or or will be appropriate, depending on how you phrase it. If you > specify the values that x cannot be, then you should use 'and': x > different from 5 and different from -4/3; 'or' would be incorrect. On > the other hand, you could also say x should not be equal to 5 or to > -4/3, in which case 'and' would be inappropriate. > It is just an application of De Morgan's Law. x can be any real number except 5 or -4/3. === Subject: Re: 'and' or 'or' > I'm a student teacher with a couple of questions about what my teacher > said in class: > Given a function with denominator (3x+4)(x-5), the question was to find > the restrictions on x. The teacher said the restrictions were 5 and > -4/3. That's stated little sloppily. The restrictions are that x can't take the value 5 or the value -4/3. > A student raised her hand and said, isn't it 5 *or* -4/3? The > teacher said, no, it's 'and', because they're both restrictions. I > found this explanation confusing and I think the student did too even > though she just said oh, okay (as students tend to do). I'm no > expert on logic, but isn't it the case that either one or both of the > factors, cannot be zero, and so doesn't that mean we should state the > restrictions as 5 inclusive or -4/3? No, since 5 isn't a statement that is true or false, and neither is -4/3. If you turn it into a precise statement with a truth value, then you can determine whether and or or is appropriate. Nor would I say 5 is a restriction. I would think that a term restriction refers to a statement about x, a constraint such as x != 5. (I'm using != to mean is not equal to). Then I could say the restrictions on x are: x != 5 AND x != -4/3 Both of these must be true for any valid x, so AND is appropriate. > --- > Another exercise on the handout read: Given f(x) = 3x-2, state the > restrictions of x when 4 < f(x) < 10. I read this as what values must you limit x to such that f(x) satisfies 4 < f(x) < 10? > A lot of the students and myself > were confused by this. I thought she meant what values of x are not > permitted because the function is undefined at them, which would have > been none. But when she took the question up, she said she meant, what > values of x correspond to the function values from 4 to 10. Would you The way you are using restrictions is a little odd, and the answer depends on how that is defined. I think the meaning as this teacher (or her text) is using the word is what conditions must x satisfy? Again, it becomes unambiguous if translated into a full sentence that includes x. correct as a full sentence x is not a member of the set {...}. In the first problem, it would have been just as valid to write that the conditions on x are: x < -4/3 OR -4/3 < x < 5 OR x > 5 That equivalently defines the conditions that x must satisfy. So my advice is: write down a complete mathematical sentence that includes x. - Randy === Subject: Hamming Bound I've previously dealt with the Hamming bound but nothing that looked like THIS. I'm not sure how to prove this. I would appreciate some help. For any (n,k) block code with minimum distance 2t+1 or greater the number of data symbols satisfies n-k >= log_q_[1+ (n 1)(q-1) + (n 2)(q-1)^2+....+(n t) (q-1)^t] (here (n 1) should be read n choose 1, (n 2) as n choose 2 and (n t) as n choose t) Prove this statement which is known as the Hamming bound. Jeff === Subject: Re: Hamming Bound >I've previously dealt with the Hamming bound but nothing that looked >like THIS. I'm not sure how to prove this. I would appreciate some >help. >For any (n,k) block code with minimum distance 2t+1 or greater the >number of data symbols satisfies >n-k >= log q [1+ (n 1)(q-1) + (n 2)(q-1)^2+....+(n t) (q-1)^t] >(here (n 1) should be read n choose 1, (n 2) as n choose 2 and >(n t) as n choose t) >Prove this statement which is known as the Hamming bound. >Jeff You'll have to decide if the following addresses your question adequately. You may well have better references at hand, but, if not, here are a few that I found quickly (via Google): Proof of the Hamming Bound for (n, M, d) block code http://en.wikipedia.org/wiki/Hamming bound Linear code (and notation for same) http://en.wikipedia.org/wiki/Linear code http://planetmath.org/encyclopedia/QuaternaryCode.html I'm guessing your question concerns linear, or [n, k, d]-codes. Toward the bottom of the 'proof' page, you'll see an inequality that reads ( with index k ==> j to avoid confusion with k in [n, k, d] ) M x Sum[j=0..t] (n choose j) (q-1)^j <= q^n Rewriting the inequality as q^n / M >= ..., putting M = q^k, and taking log q of both sides, n-k >= log q Sum[j=0..t] (n choose j) (q-1)^j which is, hopefully, what you wanted. HTH === Subject: 1/89 and the Fibonacci sequence. The mystery of 1/89 and the Fibonacci sequence 1/89 = 01123595505617977528089887640449438202247191/ 01123595505617977528089887640449438202247191/ 01123595505617977528089887640449438202247191/ 01123595505617977528089887640449438202247191/ 01123595505617977528089887640449438202247191/ 01123595505617977528089887640449438202247191/ 01123595505617977528089887640449438202247191/ 01123595505617977528089887640449438202247191/ 01123595505617977528089887640449438202247191/ 01123595505617977528089887640449438202247191/ .. With a continuing decimal expansion that has a period of 44. Then adding the Fibonacci sequence in this manner -- 0112358 +13 +>21 +>>34 +>55 +>>89 +>>144 +>233 Which creates a right +>>377 one step offset. +>610 +>>987 +>>1597 +>2584 +>>4181 +>6765 +>10946 +>>..... etc. ---------------------------------- 01123595505617977528089887640449438202247191... = 1/89? Will this continue repeating the period of 1/89 no matter how many fibonacci numbers are added in this manner? If it does, can it be proved? Others like 1/80 can be represented as -- The sequence of -- 0,1,2,4,8,16,32,64,128... .01248 +>>16 +>32 +>>64 +>128 +>>256 +>512 +>1024 +>>2048 +>4096 +>>8192 +>>16384 +>32768 +>>65536 +>>...... etc. ---------------------------------- =.0124999999999999999999999999999... or .0125 = 1/80 Another is the silver mean sequence where the difference between terms converges on sqrt(2)+1. Found in (OEIS) A000129. 0,1,2,5,12,29,70,169,408,985,2378,5741,13860,33461... Then 1/79 represents the adding together below. 0125 +>>12 +>29 +>>70 +>>169 +>408 +>>985 +>>2378 +>5741 +>13860 +>>33461 +>80782 +>195025 +>>470832 +>>1136689 +>2744210 +>>6625109 +>>15994428 +>........ etc. --------------------------------------------------- 0126582278481012658227848101265822784810126582278481... = 1/79 having a period of 13. 1/69 is this sequence added in the above manner. 0,1,3,10,33,109,360,1189,3927,12970,42837,141481,467280... In OEIS as A006190. 1/59 also in OEIS as sequence A0010076 If all this is true, what is the sequence to be added in the above manner for --- 1/78,1/81,1/82,.. etc.? Dan === Subject: Re: 1/89 and the Fibonacci sequence. >The mystery of 1/89 and the Fibonacci sequence >1/89 = >01123595505617977528089887640449438202247191/ >01123595505617977528089887640449438202247191/ >01123595505617977528089887640449438202247191/ >01123595505617977528089887640449438202247191/ >01123595505617977528089887640449438202247191/ >01123595505617977528089887640449438202247191/ >01123595505617977528089887640449438202247191/ >01123595505617977528089887640449438202247191/ >01123595505617977528089887640449438202247191/ >01123595505617977528089887640449438202247191/ >With a continuing decimal expansion that has >a period of 44. >Then adding the Fibonacci sequence in this manner -- >0112358 > +13 > +>21 > +>>34 > +>55 > +>>89 > +>>144 > +>233 Which creates a right > +>>377 one step offset. > +>610 > +>>987 > +>>1597 > +>2584 > +>>4181 > +>6765 > +>10946 > +>>..... etc. >---------------------------------- >01123595505617977528089887640449438202247191... >= 1/89? >Will this continue repeating the period of 1/89 >no matter how many fibonacci numbers are added >in this manner? Yes. >If it does, can it be proved? Using Binet's form for the Fibonacci numbers, sum_{n=0}^infty F_n b^(-n) = b/(b^2 - b - 1). Your observation is the case b=10. See also my posting from 24 July in the thread Bizarritudes, Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: 1/89 and the Fibonacci sequence. Sorry about the multiple posts but this was a bitch to edit! The mystery of 1/89 and the Fibonacci sequence 1/89 = 01123595505617977528089887640449438202247191/ 01123595505617977528089887640449438202247191/ 01123595505617977528089887640449438202247191/ 01123595505617977528089887640449438202247191/ 01123595505617977528089887640449438202247191/ 01123595505617977528089887640449438202247191/ 01123595505617977528089887640449438202247191/ 01123595505617977528089887640449438202247191/ 01123595505617977528089887640449438202247191/ 01123595505617977528089887640449438202247191/ .. With a continuing decimal expansion that has a period of 44. Then adding the Fibonacci sequence in this manner -- ***OOPs- fixed below*** 0112358 >>+13 >>+>21 >>+>>34 >>+>55 >>+>>89 >>+>>144 >>+>233 Which creates a right >>+>>377 one step offset. >>+>610 >>+>>987 >>+>>1597 >>+>2584 >>+>>4181 >>+>6765 >>+>10946 >>+>>..... etc. ---------------------------------- 01123595505617977528089887640449438202247191... = 1/89? Will this continue repeating the period of 1/89 no matter how many fibonacci numbers are added in this manner? If it does, can it be proved? Others like 1/80 can be represented as -- The sequence of -- 0,1,2,4,8,16,32,64,128... >.01248 >+>>16 >+>32 >+>>64 >+>>128 >+>256 >+>>512 >+>>1024 >+>2048 >+>>4096 >+>8192 >+>16384 >+>>32768 >+>65536 >+>>...... etc. ---------------------------------- =.0124999999999999999999999999999... or .0125 = 1/80 Another is the silver mean sequence where the difference between terms converges on sqrt(2)+1. Found in (OEIS) A000129. 0,1,2,5,12,29,70,169,408,985,2378,5741,13860,33461... Then 1/79 represents the adding together below. 0125 +>>12 +>29 +>>70 +>>169 +>408 +>>985 +>>2378 +>5741 +>13860 +>>33461 +>80782 +>195025 +>>470832 +>>1136689 +>2744210 +>>6625109 +>>15994428 +>........ etc. --------------------------------------------------- 0126582278481012658227848101265822784810126582278481... = 1/79 having a period of 13. 1/69 is this sequence added in the above manner. 0,1,3,10,33,109,360,1189,3927,12970,42837,141481,467280... In OEIS as A006190. 1/59 also in OEIS as sequence A0010076 If all this is true, what is the sequence to be added in the above manner for --- 1/78,1/81,1/82,.. etc.? Dan === Subject: Re: 1/89 and the Fibonacci sequence- Not necessary to stay silent about the 1/89 mystery and related mysteries: Try and explore power series developments like... 1/(1 - x - x^2) = 1 + x + 2x^2 + 3x^3 + 5x^4 + ..., which gives 1/0.89 = 1.1235... and 1/0.9899 = 1.0102030509.... You will see the Fibonacci numbers being generated when you develop the power series, and you will see them to fall into place when you multiply out the identity 1 = (1 - x - x^2)(1 + x + 2x^2 + 3x^3 + 5x^4 + ...) 1/(1 - x) = 1 + x + x^2 + x^3 + ..., which gives 1/0.98 = 1.020408163265.... and 14/0.98 = 100/7 = 14.285714....; All power series developments of reciprocals of polynomials yield recurrent relations for the coefficients. >Sorry about the multiple posts but this was >a bitch to edit! >The mystery of 1/89 and the Fibonacci sequence >1/89 = >01123595505617977528089887640449438202247191/ >01123595505617977528089887640449438202247191/ >01123595505617977528089887640449438202247191/ >01123595505617977528089887640449438202247191/ >01123595505617977528089887640449438202247191/ >01123595505617977528089887640449438202247191/ >01123595505617977528089887640449438202247191/ >01123595505617977528089887640449438202247191/ >01123595505617977528089887640449438202247191/ >01123595505617977528089887640449438202247191/ >.. >With a continuing decimal expansion that has >a period of 44. >Then adding the Fibonacci sequence in this manner -- >***OOPs- fixed below*** >0112358 >+13 >+>21 >+>>34 >+>55 >+>>89 >+>>144 >+>233 Which creates a right >+>>377 one step offset. >+>610 >+>>987 >+>>1597 >+>2584 >+>>4181 >+>6765 >+>10946 >+>>..... etc. >---------------------------------- >01123595505617977528089887640449438202247191... >= 1/89? >Will this continue repeating the period of 1/89 >no matter how many fibonacci numbers are added >in this manner? >If it does, can it be proved? >Others like 1/80 can be represented as -- >The sequence of -- 0,1,2,4,8,16,32,64,128... >>.01248 >>+>>16 >>+>32 >>+>>64 >>+>>128 >>+>256 >>+>>512 >>+>>1024 >>+>2048 >>+>>4096 >>+>8192 >>+>16384 >>+>>32768 >>+>65536 >>+>>...... etc. >---------------------------------- >=.0124999999999999999999999999999... >or .0125 = 1/80 >Another is the silver mean sequence where the difference >between terms converges on sqrt(2)+1. Found in (OEIS) A000129. >0,1,2,5,12,29,70,169,408,985,2378,5741,13860,33461... >Then 1/79 represents the adding together below. >0125 >+>>12 >+>29 >+>>70 >+>>169 >+>408 >+>>985 >+>>2378 >+>5741 >+>13860 >+>>33461 >+>80782 >+>195025 >+>>470832 >+>>1136689 >+>2744210 >+>>6625109 >+>>15994428 >+>........ etc. >--------------------------------------------------- >0126582278481012658227848101265822784810126582278481... >= 1/79 having a period of 13. >1/69 is this sequence added in the above manner. >0,1,3,10,33,109,360,1189,3927,12970,42837,141481,467280... >In OEIS as A006190. >1/59 also in OEIS as sequence A0010076 >If all this is true, what is the sequence to be added >in the above manner for --- >1/78,1/81,1/82,.. etc.? >Dan === Subject: Re: 1/89 and the Fibonacci sequence. The mystery of 1/89 and the Fibonacci sequence 1/89 = 01123595505617977528089887640449438202247191/ 01123595505617977528089887640449438202247191/ 01123595505617977528089887640449438202247191/ 01123595505617977528089887640449438202247191/ 01123595505617977528089887640449438202247191/ 01123595505617977528089887640449438202247191/ 01123595505617977528089887640449438202247191/ 01123595505617977528089887640449438202247191/ 01123595505617977528089887640449438202247191/ 01123595505617977528089887640449438202247191/ .. With a continuing decimal expansion that has a period of 44. Then adding the Fibonacci sequence in this manner -- ***OOPs- fixed below*** 0112358 >>+13 >>+>21 >>+>>34 >>+>55 >>+>>89 >>+>>144 >>+>233 Which creates a right >>+>>377 one step offset. >>+>610 >>+>>987 >>+>>1597 >>+>2584 >>+>>4181 >>+>6765 >>+>10946 >>+>>..... etc. ---------------------------------- 01123595505617977528089887640449438202247191... = 1/89? Will this continue repeating the period of 1/89 no matter how many fibonacci numbers are added in this manner? If it does, can it be proved? Others like 1/80 can be represented as -- The sequence of -- 0,1,2,4,8,16,32,64,128... 01248 >>+>>16 >>+>32 >>+>>64 >>+>>128 >>+>256 >>+>>512 >>+>>1024 >>+>2048 >>+>>4096 >>+>8192 >>+>16384 >>+>>32768 >>+>65536 >>+>>...... etc. ---------------------------------- =.0124999999999999999999999999999... or .0125 = 1/80 Another is the silver mean sequence where the difference between terms converges on sqrt(2)+1. Found in (OEIS) A000129. 0,1,2,5,12,29,70,169,408,985,2378,5741,13860,33461... Then 1/79 represents the adding together below. 0125 >+>>12 >+>29 >+>>70 >+>>169 >+>408 >+>>985 >+>>2378 >+>5741 >+>13860 >+>>33461 >+>80782 >+>195025 >+>>470832 >+>>1136689 >+>2744210 >+>>6625109 >+>>15994428 >+>........ etc. --------------------------------------------------- 0126582278481012658227848101265822784810126582278481... = 1/79 having a period of 13. 1/69 is this sequence added in the above manner. 0,1,3,10,33,109,360,1189,3927,12970,42837,141481,467280... In OEIS as A006190. 1/59 also in OEIS as sequence A0010076 If all this is true, what is the sequence to be added in the above manner for --- 1/78,1/81,1/82,.. etc.? Dan === Subject: Steiner Systems Hi! I need a listing of the 77 Steiner blocks for S(3,6,22) AND a listing of the 132 Steiner blocks for S(5,6,12). (Both are of hextuples) Paul === Subject: Limits and e Limit[(1+1/n)^nx] as n->inf = Limit[(1+x/n)^n] as n->inf Can anyone give me a reference. Both limits are e^nx ; I think it's Euler and saw a demonstration of it years ago. Joe === Subject: Re: Limits and e > Limit[(1+1/n)^nx] as n->inf = Limit[(1+x/n)^n] as n->inf > Can anyone give me a reference. > Both limits are e^nx I think the n in e^nx is a typo. Since n->inf is a variable binding operator, n can't appear free as well. -- I don't know who you are Sir, or where you come from, but you've done me a power of good. === Subject: Re: Limits and e >> Limit[(1+1/n)^nx] as n->inf = Limit[(1+x/n)^n] as n->inf >> Can anyone give me a reference. >> Both limits are e^nx > I think the n in e^nx is a typo. Since n->inf is a variable > binding operator, n can't appear free as well. I don't understand this. I just want to write Limit of the sequence {2^x , (3/2)^2x , (4/3)^3x ,...} Joe > -- > I don't know who you are Sir, or where you come from, > but you've done me a power of good. === Subject: Re: Limits and e > Jim Spriggs > message > .. >> Limit[(1+1/n)^nx] as n->inf = Limit[(1+x/n)^n] as >> n->inf >> Can anyone give me a reference. >> Both limits are e^nx > I think the n in e^nx is a typo. Since > n->inf is a variable > binding operator, n can't appear free as well. > I don't understand this. I just want to write Limit > of the sequence {2^x , > (3/2)^2x , (4/3)^3x ,...} > Joe Note that lim[(1+1/n)^nx] = {lim[(1+1/n)^n]}^x as n --> oo. Thus you can just compute lim[(1+1/n)^n] and raise the result to the x'th power. Jim Spriggs was right about the typo lim[(1+1/n)^nx] =! e^(nx) ..since lim[(1+1/n)^n] = e as n --> oo, implying that lim[(1+1/n)^nx] = lim[(1+1/n)^n]^x = e^x. Usually you will arrive at e^x via lim[(1+x/n)^n] = e^x as n --> oo. Kyle === Subject: Re: Limits and e message >> Jim Spriggs >> message >> .. Limit[(1+1/n)^nx] as n->inf = Limit[(1+x/n)^n] as > n->inf Can anyone give me a reference. > Both limits are e^nx >> I think the n in e^nx is a typo. Since >> n->inf is a variable >> binding operator, n can't appear free as well. >> I don't understand this. I just want to write Limit >> of the sequence {2^x , >> (3/2)^2x , (4/3)^3x ,...} >> Joe > Note that lim[(1+1/n)^nx] = {lim[(1+1/n)^n]}^x as n --> oo. Thus you can > just compute lim[(1+1/n)^n] and raise the result to the x'th power. Jim > Spriggs was right about the typo Sorry. You, Jim right. I was looking at the wrong nx for the typo. Joe > lim[(1+1/n)^nx] =! e^(nx) > ..since lim[(1+1/n)^n] = e as n --> oo, implying that > lim[(1+1/n)^nx] = lim[(1+1/n)^n]^x = e^x. Usually you will arrive at e^x > via > lim[(1+x/n)^n] = e^x as n --> oo. > Kyle === Subject: Re: Limits and e > Narcoleptic Insomniac > message > mathforum.org... >> Jim Spriggs >> in message >> .. Limit[(1+1/n)^nx] as n->inf = Limit[(1+x/n)^n] > as n->inf Can anyone give me a reference. > Both limits are e^nx >> I think the n in e^nx is a typo. Since >> n->inf is a variable >> binding operator, n can't appear free as well. >> I don't understand this. I just want to write >> Limit of the sequence {2^x ,(3/2)^2x , (4/3)^3x ,...} >> Joe > Note that lim[(1+1/n)^nx] = {lim[(1+1/n)^n]}^x as n --> oo. > Thus you can > just compute lim[(1+1/n)^n] and raise the result to > the x'th power. Jim Spriggs was right about the typo > Sorry. You, Jim right. I was looking at the wrong > nx for the typo. > Joe > lim[(1+1/n)^nx] =! e^(nx) > ..since lim[(1+1/n)^n] = e as n --> oo, implying that > lim[(1+1/n)^nx] = lim[(1+1/n)^n]^x = e^x. Usually > you will arrive at e^x via > lim[(1+x/n)^n] = e^x as n --> oo. Just for kicks here's how you do the limit above: Set y = (1 + x/n)^n so that ln(y) = n * ln(1 + x/n), then lim[ln(y)] = lim[n * ln(1 + x/n)] as n --> oo. Notice that as n --> oo, n * ln(1 + x/n) --> oo * 0. This implies that we should rewrite n as... n = 1/(1/n) = 1/(n^(-1)) ..so that we are left with the undeterminant form oo / oo, in which case we can apply l'Hospital's rule to get lim[ln(y)] = lim[1/(n^(-1)) * ln(1 + x/n)] = lim[-1/(n^(-2)) * (-x/(n^2)) / (1 + x/n)] = lim[(-n^2) * (-x/(n^2)) / (1 + x/n)] = x * lim[1 / (1 + x/n)] = x. Since lim[ln(y)] = x and y = (1 + x/n)^n we have e^{lim[ln(y)]} = lim[e^ln(y)] = lim[y] = lim[ln(1 + x/n)^n] = x. Kyle === Subject: Re: Limits and e > Just for kicks here's how you do the limit above: > Set y = (1 + x/n)^n so that ln(y) = n * ln(1 + x/n), > then > lim[ln(y)] = lim[n * ln(1 + x/n)] as n --> oo. > Notice that as n --> oo, n * ln(1 + x/n) --> oo * 0. > This implies that we should rewrite n as... > n = 1/(1/n) = 1/(n^(-1)) > ...so that we are left with the undeterminant form oo > / oo, in which case we can apply l'Hospital's rule to > get > lim[ln(y)] = lim[1/(n^(-1)) * ln(1 + x/n)] > = lim[-1/(n^(-2)) * (-x/(n^2)) / (1 + x/n)] > = lim[(-n^2) * (-x/(n^2)) / (1 + x/n)] > = x * lim[1 / (1 + x/n)] = x. > Since lim[ln(y)] = x and y = (1 + x/n)^n we have > e^{lim[ln(y)]} = lim[e^ln(y)] = lim[y] = lim[ln(1 + x/n)^n] = x. I made a really lame typo in the line above, it should have been e^{lim[ln(y)]} = ... = e^x ..blah, anyways you get the idea. ^_^ > Kyle === Subject: Re: Limits and e > Limit[(1+1/n)^nx] as n->inf = Limit[(1+x/n)^n] as n->inf > Can anyone give me a reference. > Both limits are e^nx ; (1 + x/n)^n tends to e^x as n tends to infinity. > I think it's Euler L Euler, Introductio in analysin infinitorum. English translation: J D Blanton, Introduction to analysis of the infinite, Springer-Verlag. Euler considers (1 + w)^N where w is small and N large. Let x = wN where x is a fixed rational, let N tend to infinity and w to zero so that x remains constant. Then (1 + x/N)^N tends to 1 + x + x^2/2 + x^3/(2*3) + x^4/(2*3*4) + ... ((1 + 1/M)^M)^x tends to e^x where e is the limit of (1 + 1/M)^M. Euler proves that that limit exists by (1 + 1/M)^M = 1 + M/M + M(M - 1)/2 (1/M^2) + M(M - 1)(M - 2)/(2*3) (1/M^3) + ... = 1 + 1 + 1(1 - 1/M)/2 + 1(1 - 1/M)(1 - 2/M)/(2*3) + ... In the limit e = 1 + 1 + 1/2 + 1/(2*3) + 1/(2*3*4) + ... -- I don't know who you are Sir, or where you come from, but you've done me a power of good. === Subject: Re: Limits and e >> Limit[(1+1/n)^nx] as n->inf = Limit[(1+x/n)^n] as n->inf >> Can anyone give me a reference. >> Both limits are e^nx ; > (1 + x/n)^n tends to e^x as n tends to infinity. >> I think it's Euler > L Euler, Introductio in analysin infinitorum. English translation: J > D Blanton, Introduction to analysis of the infinite, Springer-Verlag. > Euler considers > (1 + w)^N > where w is small and N large. Let x = wN where x is a fixed rational, > let N tend to infinity and w to zero so that x remains constant. Then > (1 + x/N)^N tends to 1 + x + x^2/2 + x^3/(2*3) + x^4/(2*3*4) + ... This is what I'm trying to show! > ((1 + 1/M)^M)^x tends to e^x I accept this. I'm trying to get from here to the line above. > where e is the limit of (1 + 1/M)^M. Euler proves that that limit > exists by > (1 + 1/M)^M = 1 + M/M + M(M - 1)/2 (1/M^2) + > M(M - 1)(M - 2)/(2*3) (1/M^3) + ... Can't see why. I'm trying to get from Limit[(1+1/n)^nx] which I accept is e^x to Limit[(1+x/n)^n] also being e^x Joe > = 1 + 1 + 1(1 - 1/M)/2 + 1(1 - 1/M)(1 - 2/M)/(2*3) + ... > In the limit > e = 1 + 1 + 1/2 + 1/(2*3) + 1/(2*3*4) + ... > -- > I don't know who you are Sir, or where you come from, > but you've done me a power of good. === Subject: Re: Limits and e > (1 + 1/M)^M = 1 + M/M + M(M - 1)/2 (1/M^2) + > M(M - 1)(M - 2)/(2*3) (1/M^3) + ... > Can't see why. Use the binomial theorem: (1 + a)^m = 1 + (m choose 1)a + ... + (m choose n)a^n + ... for |a| < 1. I think Newton proved that, so Euler would have known it. Nowadays one would use Taylor's series which one would prove from Rolle's theorem. -- I don't know who you are Sir, or where you come from, but you've done me a power of good. === Subject: Line Graphs What does the following definition mean? Cab anyone explain it? A line graph L(G) (also called an interchange graph) of a graph G is obtained by associating a vertex with each edge of the graph and connecting two vertices with an edge iff the corresponding edges of G meet at one or both endpoints. === Subject: Re: Line Graphs > What does the following definition mean? Cab anyone explain it? > A line graph L(G) (also called an interchange graph) of a graph G is > obtained by associating a vertex with each edge of the graph and > connecting two vertices with an edge iff the corresponding edges of G > meet at one or both endpoints. Yes, I can. (This is from MathWorld, right?) An example will probably help. If you have a graph G with vertices a,b,c,d, and with edges ab, bc, ca, cd, then the line graph L(G) has 4 vertices (because G has 4 edges), which I'll call v1, v2, v3, v4; these represent the edges ab, bc, ca, cd, respectively. The line graph contains the edges v1v2 (since ab and bc share the vertex b), v1v3 (since ab and ca share the vertex a), v2v3, v2v4, and v3v4. The line graph does not contain v1v4, because the edges ab and cd don't share any endpoints. --- Christopher Heckman === Subject: Re: infinity >> Apparently you think that you can increment a value forever and >> ever, and it will always be finite, so why do you think you can add >> 1 element at a time forever and ever to a finite set and get an >> infinite set? This thinking is inconsistent at the most basic level. >There is no add involved. The set of naturals is described by >static properties, not by a process generating its elements, Right, although it might help poor Tony to phrase it a little closer to his way of thinking. We can define lots of sets X_n by X_1 = { 0 } X_2 = { 0, 1 } (i.e add 1 to the previous largest element and X_3 = { 0, 1, 2 } throw that in, too.) etc. Then we can define the set of natural numbers by N = union of all X_n. So there's Tony's dynamic process of creating each of the integers, i.e. of the X_n's, and then the swell foop of combining them all into one set as David suggested. As Tony likes to say, we increment a value forever and ever (it here meaning the maximum value of X_n, so that it is a variable which depends on n, which we can think of as time I guess), and indeed it is indeed finite at all times n . Even more importantly, I should add that the other elements of X_n are also present in X_{n+1} -- they did not increment. But none of the X_n's itself is the set of all integers; making N itself requires the additional step of taking a union of the previous things (or maybe waiting until after the end of time, whatever that means). We're taking the union of infinitely many things; but none of those things is itself infinite, and none of them contains any infinite numbers either. Of course David doesn't need to hear me say this, and Tony won't be helped by hearing me say this, so I don't know why I'm bothering. Beats grading papers I guess. dave === Subject: Re: infinity stephen@nomail.com said: >> ... >> > That was a proof? All it is is a rehashed statement that there is no largest >> > finite integer. Sure, finite F can always be incremented, since finite k can >> > always be incremented. That lack of a largest element, or longest string, >> > doesn't prove infinitude of the set, as far as I'm concerned, so that doesn't >> > prove anything to me. >> How *do* you define finite and infinite? >> However he defines finite, if he claims that >> F = sum S^k for all finite k >> is finite, then he claims that >> F = S^F + (sum S^k for all finite k<>F) >> and it follows that >> F > S^F > Not necessarily. Try replacing finite with not imponderably > enormous. > I think this last line just means that some numbers on the threshold of > imponderability are larger than some other such numbers, even though > those are also threshold values. You see, imponderability is a bit hard > to tie down, and frankly the arguments you mathematicians habitually > use don't work too well on it. > I do not think that matters. Suppose that > F = sum S^k for all not imponderably enormous k > and that F is not imponderably enormous. > Then > F = F + (sum S^k for all not imponderably enormous k <> F) > Presumably if F is not imponderably enormous than we can > safely substract it from both sides. So > 0 = (sum S^k for all not imponderably enormous k <> F) > Well 1,2,3, etc are all not imponderably enormous, and > S^1 + S^2 + S^3 > 0 > for S>0. > Tony is basically claiming that finite numbers exist > that are greater than the sum of all finite numbers, > including themselves. This is going to be problematic > for whatever definition of finite you plug in. > Stephen I am claiming no such thing. You are the one supposing a largest finite in your stuff above, hence the contradiction. I never claimed to have any number for the sum of all finite numbers. -- Smiles, Tony === Subject: Re: infinity > stephen@nomail.com said: > ... >> > That was a proof? All it is is a rehashed statement that there is >> > no largest >> > finite integer. Sure, finite F can always be incremented, since >> > finite k can >> > always be incremented. That lack of a largest element, or longest >> > string, >> > doesn't prove infinitude of the set, as far as I'm concerned, so >> > that doesn't >> > prove anything to me. >> How *do* you define finite and infinite? >> However he defines finite, if he claims that >> F = sum S^k for all finite k >> is finite, then he claims that >> F = S^F + (sum S^k for all finite k<>F) >> and it follows that >> F > S^F Not necessarily. Try replacing finite with not imponderably > enormous. > I think this last line just means that some numbers on the threshold of > imponderability are larger than some other such numbers, even though > those are also threshold values. You see, imponderability is a bit hard > to tie down, and frankly the arguments you mathematicians habitually > use don't work too well on it. I do not think that matters. Suppose that > F = sum S^k for all not imponderably enormous k > and that F is not imponderably enormous. Then > F = F + (sum S^k for all not imponderably enormous k <> F) Presumably if F is not imponderably enormous than we can > safely substract it from both sides. So > 0 = (sum S^k for all not imponderably enormous k <> F) Well 1,2,3, etc are all not imponderably enormous, and > S^1 + S^2 + S^3 > 0 > for S>0. Tony is basically claiming that finite numbers exist > that are greater than the sum of all finite numbers, > including themselves. This is going to be problematic > for whatever definition of finite you plug in. Stephen I am claiming no such thing. You are the one supposing a largest finite in > your > stuff above, hence the contradiction. I never claimed to have any number for > the sum of all finite numbers. To claims that the set of all finite naturals is a finite set. The sum of any finite set of natural numbers is defineable inductively and results in a finite natural number. Clearly that sum must be larger than every member of the set, so that TO's finite set of all finite naturals does not contain all finite naturals. This problem does not arise if the set of all finite naturals were to be infinite, as infinite sums of naturals are not inductively defineable. So TO must choose between having a set of all finite naturals that does not contain all finite naturals or allowing that the set of all finite naturals is not finite. === Subject: Re: infinity Virgil said: > ... > > Yes it does. Say I have 26 letters in my alphabet. I know that the > > longest word in my language is 23 letters long. Then I know, for sure, > > that there cannot be more than 26^23 words in the language. Of course Tony is wrong. He claims there are at most > 350257144982200575261531309080576 words in that language. I claim there > are at most 364267430781488598271992561443798. > Right! 26*(1-26^23)/(1-26) = 364267430781488598271992561443798 words of > at least one and no more than 23 letters. And one more if TO includes a > no-letter word, as he did the other day. Oops, you're right, or at least I erred. No more than 26^23 words of length 23, and no more than sum(x=1->23: 26^x), or 11111111111111111111111 in base 26. In any case, the point is that a finite limit on the lengths of words results in a finite language. -- Smiles, Tony === Subject: Re: infinity > Virgil said: > Yes it does. Say I have 26 letters in my alphabet. I know that > > the longest word in my language is 23 letters long. Then I > > know, for sure, that there cannot be more than 26^23 words in > > the language. Of course Tony is wrong. He claims there are at most > 350257144982200575261531309080576 words in that language. I > claim there are at most 364267430781488598271992561443798. Right! 26*(1-26^23)/(1-26) = 364267430781488598271992561443798 > words of at least one and no more than 23 letters. And one more if > TO includes a no-letter word, as he did the other day. Oops, you're right, or at least I erred. No more than 26^23 words of > length 23, and no more than sum(x=1->23: 26^x), or > 11111111111111111111111 in base 26. In any case, the point is that a > finite limit on the lengths of words results in a finite language. TO claims that the set of ALL finite strings over any alphabet is a finite set. But a concatenation of all those finite strings in that finite set is a finite string that cannot be in that finite set, as it contains every member of that set as a proper substring. SO TO's finite set of all finite strings is not a finite set of all finite strings. OOPS! TO Gooofed Again! === Subject: Re: infinity stephen@nomail.com said: > ... > > That was a proof? All it is is a rehashed statement that there is no largest > > finite integer. Sure, finite F can always be incremented, since finite k can > > always be incremented. That lack of a largest element, or longest string, > > doesn't prove infinitude of the set, as far as I'm concerned, so that doesn't > > prove anything to me. > How *do* you define finite and infinite? > However he defines finite, if he claims that > F = sum S^k for all finite k > is finite, then he claims that > F = S^F + (sum S^k for all finite k<>F) > and it follows that > F > S^F Stephen What are you talking about? Oh yeah, your largest finite. Huyah huyah Ommmmm..... Wake Up! -- Smiles, Tony === Subject: Re: infinity > stephen@nomail.com said: > ... > > That was a proof? All it is is a rehashed statement that there is no > > largest > > finite integer. Sure, finite F can always be incremented, since finite > > k can > > always be incremented. That lack of a largest element, or longest > > string, > > doesn't prove infinitude of the set, as far as I'm concerned, so that > > doesn't > > prove anything to me. How *do* you define finite and infinite? However he defines finite, if he claims that > F = sum S^k for all finite k > is finite, then he claims that > F = S^F + (sum S^k for all finite k<>F) > and it follows that > F > S^F Stephen What are you talking about? Oh yeah, your largest finite. Huyah huyah > Ommmmm..... Wake Up! TO keeps claiming that there is a finite set containing ALL finite naturals. But the finite sum of the elements of such a set must be a finite natural larger than any member of that set, so that TO's set of ALL finite naturals cannot contain all finite naturals. That is what is known as a self-contradiction in TO's claims. So TO is WRONG to claim that the set of all finite naturals must be finite === Subject: Re: infinity David R Tribble said: > Tony Orlow says: >> No, what is disputed is that the values in the set are finite, and >> therefore end. They do not end, but extend into infinite values >> in the infinite set. > TO assumes that not ending requires exending into the infinite, but > that is false. There are all sorts of sets that do not end but do not > extend into the infinite, the set of (finite) naturals being one such > set. The set of real values in the open interval (0,1) for another. > One small but subtle point here; perhaps a better example would be > the infinite set, let's call it H, of rational values in (0,1) with > terminating decimal fractions. Each element is a finite value, > in the sense that it can be represented as a decimal fraction with > a finite number of nonzero digits. If you restrict your values to finite length string representations, then you have restricted your language to a finite size, since you are always using a finite alphabet, or set of numerals. Is this the same size as your set of finite naturals? I doubt it, but I am really not sure at this point. > Otherwise, Tony might object that (0,1) in R contains infinite > values (or infinitely long values), which just confuses the whole > issue. No, I am fully capable of distinguishing between infinite numbers of elements, infinite values for elements, and infinite string representations of elements. You certainly have an infinite set of reals in (0,1), which is why we must allow infinitely long strings to the right of the decimal point, so that we can construct an infinite set of string representations. One the left of the decimal point, you refuse to allow infinitely long strings, so you cannot possibly have an infinite number of whole numbers. In decimal, given n digits, we can have 10^n distinct strings. Can 10^n be infinite if n is finite? If you don't allow infinitely long strings to the left of the point, which represent infinite quantities, then you cannot possibly have an infinite number of string representations of whole numbers. > I'd like to see Tony's explanation of how big set H is. It's obvious > that H contains fewer points than (0,1) in R. It's also obvious that > that H has exactly the same number of members as Q and as N. But it > probably has fewer members than Tony's 'N' (based on our intuition). > Like I said, if you restrict it to finite digits, then it is a finite set, akin to your N, but certainly smaller than mine, as I allow infinite digits. -- Smiles, Tony === Subject: Re: infinity > If you restrict your values to finite length string representations, then you > have restricted your language to a finite size, since you are always using a > finite alphabet, or set of numerals. Is this the same size as your set of > finite naturals? I doubt it, but I am really not sure at this point. TO asserts that there is a finite set of strings that contains every finite string. But that set of strings cannot contain any of the finite strings formed by concatenation of all the strings in the set itself. So that EVERY finite set of finite strings is incomplete, i.e., leaves out some finite strings. Thus TO's assertion that the set of all finite strings is finite leads immediately to a contradiction. So that TO is WRONG! AGAIN! === Subject: Re: infinity decimal point, you refuse to allow infinitely long strings, so you cannot > possibly have an infinite number of whole numbers. In decimal, given n digits, > we can have 10^n distinct strings. Can 10^n be infinite if n is finite? What value of n covers all possible finite strings? - Randy === Subject: Re: infinity Daryl McCullough said: > Tony Orlow says: >> So the real question is this: Is the set of all finite natural >> numbers finite, or infinite? >Finite. Please respond to my inductive proof of this, posted today. > It's nonsense. An inductive proof is a proof of a *universal* statement, > of the form > forall n s.t. n is a finite natural, Phi(n) > The claim The set of all finite natural numbers is finite does > not have this form. You can't prove it by induction. > If S is a set, then the claim S is finite is mathematically > equivalent to > exists n, n is a finite natural, and size(S) < n > The claim S is infinite is mathematically equivalent to > forall n s.t. n is a finite natural, size(S) > n > That has a form that can be proved by induction, by letting > Phi(n) be size(S) > n. > -- > Daryl McCullough > Ithaca, NY Proof: For all n in N, Phi(n) = the set of naturals from 1 up to and including n contains n elements. n=1: {1} has 1 element. n->n+1: A set has n elements with n as a largest element. When we add the successor of n to the set, n+1 is added to the set, incrementing the set size to n+1. So, there are n+1 elements up to and including n+1. Conlcusion: For all n in N, n is the size of the set of naturals from 1 through n. So, if the size of the set is aleph_0, then the set is the set of all naturals from 1 through aleph_0. The size of the set of all naturals through n is n, so if the set is infinite, then n is infinite, and if n is finite, then the set is finite. Your objections to my inductive proofs display a misunderstanding of what they do. Not every inductive proof need be tied to natural numbers, although that is the way peano defined it axiomatically. It would help if you described exactly where you disagree with the logic, because the logic makes sense. -- Smiles, Tony === Subject: Re: infinity ... > Proof: For all n in N, Phi(n) = the set of naturals from 1 up to and > including n contains n elements. > n=1: {1} has 1 element. > n->n+1: A set has n elements with n as a largest element. When we add the > successor of n to the set, n+1 is added to the set, incrementing the set > size to n+1. So, there are n+1 elements up to and including n+1. > Conlcusion: For all n in N, n is the size of the set of naturals from 1 > through n. Right so far. This has never been disputed. > So, if the size of the set is aleph_0, then the set is the set of all > naturals from 1 through aleph_0. Only under the assumption that aleph_0 is in N, something you wish to prove. So your reasoning is circular. If aleph_0 is not in N this conclusion is false (the proof was only for all n in N). What has been argued, repeatedly, is that the size of the set of natural numbers is *not* a natural number, and so aleph_0 is not in N. > The size of the set of all naturals > through n is n, Right. > so if the set is infinite, then n is infinite, Wrong conclusion based on the assumption that the size of N is a natural number and is in N. > and if n > is finite, then the set is finite. Right. > Your objections to my inductive proofs display a misunderstanding of > what they do. The induction you did above was valid, it is the further conclusion that was not justified. > Not every inductive proof need be tied to natural numbers, > although that is the way peano defined it axiomatically. It would help > if you described exactly where you disagree with the logic, because the > logic makes sense. See above. And, every standard mathematical induction refers to induction over the natural numbers. What you need is transfinite induction, but that will fail in this case. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: infinity Tony Orlow says: >Proof: For all n in N, Phi(n) = the set of naturals from 1 up to and >including n contains n elements. >n=1: {1} has 1 element. >n->n+1: A set has n elements with n as a largest element. When we >add the successor of n to the set, n+1 is added to the set, >incrementing the set size to n+1. So, there are n+1 elements up >to and including n+1. >Conlcusion: For all n in N, n is the size of the set of naturals >from 1 through n. Very good. That's a correct proof by induction. >So, if the size of the set is aleph_0, then the set is the set of all >naturals from 1 through aleph_0. But aleph_0 is not a natural number, so it is incorrect to replace n by aleph_0. However, it is true that the size of the set { 1, 2, ... aleph_0 } is aleph_0. >The size of the set of all naturals through n is n, so >if the set is infinite, then n is infinite, and if n >is finite, then the set is finite. We're not talking about sets of the form A_n = { x | x is a natural number and x <= n } we are talking about the set N = { x | x is a (finite) natural number } That is not equal to A_n for any n. So a proof about size(A_n) does not imply any thing about size(N). However, since A_n is a proper subset of N, what you can prove by induction is forall n, size(N) > n -- Daryl McCullough Ithaca, NY === Subject: Re: infinity !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > But aleph_0 is not a natural number, so it is incorrect to replace > n by aleph_0. However, it is true that the size of the set > { 1, 2, ... aleph_0 } > is aleph_0. I think that this is an abuse of notation since it implies a sequence ending in aleph_0. But aleph_0 is a singular disconnected value in that set. It has no predecessor. So I'd rather write it as { aleph_0, 0, 1, 2, ... } Yes, this set has size aleph_0. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: infinity Daryl McCullough said: > Tony Orlow says: >Proof: For all n in N, Phi(n) = the set of naturals from 1 up to and >including n contains n elements. >n=1: {1} has 1 element. >n->n+1: A set has n elements with n as a largest element. When we >add the successor of n to the set, n+1 is added to the set, >incrementing the set size to n+1. So, there are n+1 elements up >to and including n+1. >Conlcusion: For all n in N, n is the size of the set of naturals >from 1 through n. > Very good. That's a correct proof by induction. Glad I could satisfy SOMEBODY around here. >So, if the size of the set is aleph_0, then the set is the set of all >naturals from 1 through aleph_0. > But aleph_0 is not a natural number, so it is incorrect to replace > n by aleph_0. However, it is true that the size of the set > { 1, 2, ... aleph_0 } > is aleph_0. Huh?? What IS that set? It looks like the consecutive whole numbers starting from 1 and forming a set with size aleph_0, but it's not the set of naturals? You see the contradiction here? You have an infinite set, which must have a largest member equal to its size by definition of the set, and yet all members are finite and the set is considered infinite? How can the set be infinite, if it doesn't contain any infinite members, under these circumstances? >The size of the set of all naturals through n is n, so >if the set is infinite, then n is infinite, and if n >is finite, then the set is finite. > We're not talking about sets of the form > A_n = { x | x is a natural number and x <= n } > we are talking about the set > N = { x | x is a (finite) natural number } We are considering under what circumstances the set of naturals can or cannot be infinite, and there is a clear constant relationship between the mnaximum value in the set and the number of elements in the set, which is independent of either. In every case, they are exactly equal. > That is not equal to A_n for any n. So a proof about > size(A_n) does not imply any thing about size(N). However, > since A_n is a proper subset of N, what you can prove by induction > is > forall n, size(N) > n That's all very well. You say there is no n greater than all finite n so we cannot express N as a form of A_n, but the proof demonstrates that in this type of set, the largest member IS the set size. If you claim the set size is aleph_ 0, then that IS, as proven, the largest element of the set. But, we have a contradiction here, since aleph_0 is infinite but the set is said to only have finite elements. We cannot name a largest finite. We all agree on that. As I have shown, since this is the case, we also cannot name a size of the set of finite naturals, since these are the same number. We can name it, but I can name my imaginary purple flying elephant, and it still doesn't make sense. This non-number is a paradoxical little elephant we all have to live with. > -- > Daryl McCullough > Ithaca, NY > -- Smiles, Tony === Subject: Re: infinity > We cannot name a largest finite. We all agree on that. As I have shown, since > this is the case, we also cannot name a size of the set of finite naturals, > since these are the same number. No, they are not the same. Only in your twisted little world. We cannot name a largest finite because a largest finite does not exist. The size of the set of finite naturals does exist, and we can name it. Your set theory is quite pathetic if it cannot determine the size of such a simple set as the set of all finite naturals. Stephen === Subject: Re: infinity > Daryl McCullough said: > Tony Orlow says: Proof: For all n in N, Phi(n) = the set of naturals from 1 up to >and including n contains n elements. n=1: {1} has 1 element. n->n+1: A set has n elements with n as a largest element. When we >add the successor of n to the set, n+1 is added to the set, >incrementing the set size to n+1. So, there are n+1 elements up to >and including n+1. Conlcusion: For all n in N, n is the size of the set of naturals >from 1 through n. Very good. That's a correct proof by induction. > Glad I could satisfy SOMEBODY around here. So, if the size of the set is aleph_0, then the set is the set of >all naturals from 1 through aleph_0. But aleph_0 is not a natural number, so it is incorrect to replace > n by aleph_0. However, it is true that the size of the set { 1, 2, ... aleph_0 } is aleph_0. > Huh?? What IS that set? It looks like the consecutive whole numbers > starting from 1 and forming a set with size aleph_0, but it's not the > set of naturals? The set {1,2,3,...;John Smith} is not a set of naturals either, but is of size aleph_0. What is your problem, TO? > You see the contradiction here? You have an infinite set, which must > have a largest member equal to its size by definition of the set, and > yet all members are finite and the set is considered infinite? How > can the set be infinite, if it doesn't contain any infinite members, > under these circumstances? The same way that the set of all finite naturals cannot be finite. Every finite set of two or more finite naturals has a sum which is a finite natural but is not in the set. Thus no finite set of finite natural numbers can be the set of all finite natural numbers. > We are considering under what circumstances the set of naturals can > or cannot be infinite, and there is a clear constant relationship > between the mnaximum value in the set and the number of elements in > the set, which is independent of either. In every case, they are > exactly equal. If TO continues to assert that the set of all finite natural numbers is finite, he is saying that infinite sets are finite. > We cannot name a largest finite. We all agree on that. As I have > shown, since this is the case, we also cannot name a size of the set > of finite naturals, since these are the same number. We can name it, > but I can name my imaginary purple flying elephant, and it still > doesn't make sense. This non-number is a paradoxical little elephant > we all have to live with. Having a set of all finite natural numbers which does not contain all finite natural numbers is a problem that can be solved only by allowing that set to be infinite (not finite). === Subject: Re: infinity Tony Orlow says: >Proof: For all n in N, Phi(n) = the set of naturals from 1 up to and >including n contains n elements. n=1: {1} has 1 element. n->n+1: A set has n elements with n as a largest element. When we >add the successor of n to the set, n+1 is added to the set, >incrementing the set size to n+1. So, there are n+1 elements up >to and including n+1. Conlcusion: For all n in N, n is the size of the set of naturals >from 1 through n. > Very good. That's a correct proof by induction. > Glad I could satisfy SOMEBODY around here. >So, if the size of the set is aleph_0, then the set is the set of all >naturals from 1 through aleph_0. > But aleph_0 is not a natural number, so it is incorrect to replace > n by aleph_0. However, it is true that the size of the set > { 1, 2, ... aleph_0 } > is aleph_0. > Huh?? What IS that set? It looks like the consecutive whole numbers starting > from 1 and forming a set with size aleph_0, but it's not the set of naturals? > You see the contradiction here? You have an infinite set, which must have a > largest member equal to its size by definition of the set, and yet all members > are finite and the set is considered infinite? How can the set be infinite, if > it doesn't contain any infinite members, under these circumstances? >The size of the set of all naturals through n is n, so >if the set is infinite, then n is infinite, and if n >is finite, then the set is finite. > We're not talking about sets of the form > A_n = { x | x is a natural number and x <= n } > we are talking about the set > N = { x | x is a (finite) natural number } > We are considering under what circumstances the set of naturals can or cannot > be infinite, and there is a clear constant relationship between the mnaximum > value in the set and the number of elements in the set, which is independent of > either. In every case, they are exactly equal. Take the set {10} (the set consisting of the single number 10). The largest element of this set is 10. The number of elements in this set is 1. It is true that the size of the set is equal to the number of elements in the set for sets of the form A_n = { x | x is a natural number and x <= n } However, it is not true that every set is of this form. Nor is it true that if a set is not of this form it cannot have a well defined size. > That is not equal to A_n for any n. So a proof about > size(A_n) does not imply any thing about size(N). However, > since A_n is a proper subset of N, what you can prove by induction > is > forall n, size(N) > n > That's all very well. You say there is no n greater than all finite n so we > cannot express N as a form of A_n, but the proof demonstrates that in this type > of set, the largest member IS the set size. For sets of the form A_n the largest member is the set size. N is not of this form. >If you claim the set size is aleph_ > 0, then that IS, as proven, the largest element of the set. No, your proof has nothing to say about sets of the form of N. > But, we have a > contradiction here, since aleph_0 is infinite but the set is said to only have > finite elements. Yes, we would have a contradiction if your proof said anything about sets of the form of N, but it does not, so there is no contradiction. > We cannot name a largest finite. We all agree on that. As I have shown, since > this is the case, we also cannot name a size of the set of finite naturals, > since these are the same number. No, the size of the set is the largest element only for sets of a very specific form. The set of finite naturals is not of this form. - William Hughes === Subject: Re: infinity Tony Orlow says: >We cannot name a largest finite. We all agree on that. As I have shown, >since this is the case, we also cannot name a size of the set of >finite naturals, since these are the same number. No, they are only the same for sets of the form A_n with n a natural number. The set of all finite natural numbers does not have that form. If U is the set of all finite natural numbers, then size(U) = aleph_0, and aleph_0 is *not* an element of U. -- Daryl McCullough Ithaca, NY === Subject: Re: infinity largest member equal to its size by definition of the set, You're confused. That's the definition for a finite set. > You have an infinite set, and yet all members are finite and the set is > considered infinite? How can the set be infinite, if it doesn't contain > any infinite members, under these circumstances? An infinite set contains an infinite number of members, just like you said. It's a property of the set, not a property of the elements in the set. Consider the reverse: G = { aaa..., bbb..., ccc... } G is a finite set containing three elements, and all three elements are infinite (strings). Set G contains infinite members, but is not itself infinite. === Subject: Re: infinity Tony Orlow says: >> So the real question is this: Is the set of all finite natural >> numbers finite, or infinite? Finite. Please respond to my inductive proof of this, posted today. > It's nonsense. An inductive proof is a proof of a *universal* statement, > of the form > forall n s.t. n is a finite natural, Phi(n) > The claim The set of all finite natural numbers is finite does > not have this form. You can't prove it by induction. > If S is a set, then the claim S is finite is mathematically > equivalent to > exists n, n is a finite natural, and size(S) < n > The claim S is infinite is mathematically equivalent to > forall n s.t. n is a finite natural, size(S) > n > That has a form that can be proved by induction, by letting > Phi(n) be size(S) > n. > -- > Daryl McCullough > Ithaca, NY > Proof: For all n in N, Phi(n) = the set of naturals from 1 up to and including > n contains n elements. > n=1: {1} has 1 element. > n->n+1: A set has n elements with n as a largest element. When we add the > successor of n to the set, n+1 is added to the set, incrementing the set size > to n+1. So, there are n+1 elements up to and including n+1. > Conlcusion: For all n in N, n is the size of the set of naturals from 1 through > n. > So, if the size of the set is aleph_0, then the set is the set of all naturals > from 1 through aleph_0. >The size of the set of all naturals through n is n, so > if the set is infinite, then n is infinite, and if n is finite, then the set is > finite. You missed a set. Let the set of all naturals from 1 through aleph_0 be K. Now let U be the set K minus the element aleph_0. Note there is no n such that U is the set of all naturals through n and U is not K, so U was not considered above. Is U finite or infinite? -William Hughes === Subject: Re: infinity William Hughes said: > Daryl McCullough said: > Tony Orlow says: > So the real question is this: Is the set of all finite natural >> numbers finite, or infinite? Finite. Please respond to my inductive proof of this, posted today. It's nonsense. An inductive proof is a proof of a *universal* statement, > of the form forall n s.t. n is a finite natural, Phi(n) The claim The set of all finite natural numbers is finite does > not have this form. You can't prove it by induction. If S is a set, then the claim S is finite is mathematically > equivalent to exists n, n is a finite natural, and size(S) < n The claim S is infinite is mathematically equivalent to forall n s.t. n is a finite natural, size(S) > n That has a form that can be proved by induction, by letting > Phi(n) be size(S) > n. -- > Daryl McCullough > Ithaca, NY > Proof: For all n in N, Phi(n) = the set of naturals from 1 up to and including > n contains n elements. > n=1: {1} has 1 element. > n->n+1: A set has n elements with n as a largest element. When we add the > successor of n to the set, n+1 is added to the set, incrementing the set size > to n+1. So, there are n+1 elements up to and including n+1. > Conlcusion: For all n in N, n is the size of the set of naturals from 1 through > n. > So, if the size of the set is aleph_0, then the set is the set of all naturals > from 1 through aleph_0. >The size of the set of all naturals through n is n, so > if the set is infinite, then n is infinite, and if n is finite, then the set is > finite. > You missed a set. > Let the set of all naturals from 1 through aleph_0 be K. > Now let U be the set K minus the element aleph_0. > Note there is no n such that U is the set of all naturals > through n and U is not K, so U was not considered above. > Is U finite or infinite? > -William Hughes Of course U is infinite. Aleph_0 would have to be the successor of a number, which would also be infinite. You are not suggesting that the size of the set is now finite? It's aleph_0-1, which is now the largest member of that set. But, all of this leads to contradictions, since there is no largest finite or smallest infinite. Aleph_0 is a concept, an unbounded ill-defined concept, and not any exact number. -- Smiles, Tony === Subject: Re: infinity > William Hughes said: t. Let the set of all naturals from 1 through aleph_0 be K. Now let U > be the set K minus the element aleph_0. Note there is no n such > that U is the set of all naturals through n and U is not K, so U > was not considered above. Is U finite or infinite? -William Hughes > Of course U is infinite. Aleph_0 would have to be the successor of a > number, which would also be infinite. You are not suggesting that the > size of the set is now finite? It's aleph_0-1, which is now the > largest member of that set. But, all of this leads to contradictions, > since there is no largest finite or smallest infinite. Aleph_0 is a > concept, an unbounded ill-defined concept, and not any exact number. TO is hardly in a position to carp. TO claims that the set of finite naturals is finite. The sum of any finite set of two or more finite natural numbers is a finite natural number larger than any member of the set. If, as TO claims, the set of finite naural numbers were actuallly finite, the sum of its members must be a finite natural number larger than every finite natural number, including itself, of course. So the set of all finite natural numbers cannot be finite in any contradiction-free system. AND:If, as TO keeps claiming, the set of all finite strings were truly finite, the concatenation of all those finite strings would still be a finite string, but longer than any string in the *finite* set of *all* finite strings, thus longer than itself. === Subject: Re: infinity >William Hughes said: >> You missed a set. >> Let the set of all naturals from 1 through aleph_0 be K. >> Now let U be the set K minus the element aleph_0. >> Note there is no n such that U is the set of all naturals >> through n and U is not K, so U was not considered above. >> Is U finite or infinite? >Of course U is infinite. Well, U is the set of all finite natural numbers. >Aleph_0 would have to be the successor of a number, No, it isn't. aleph_0 is defined to be the first infinite ordinal. There is no n such that n+1 = aleph_0, because if n < aleph_0, then n+1 < aleph_0. >which would also be infinite. You are not suggesting that the size of the set >is now finite? It's aleph_0-1, which is now the largest member of that set. There is no aleph_0 - 1. Subtraction is not always defined on infinite ordinals. There are two types of ordinals: 1. successor ordinals 2. limit ordinals If x is a successor ordinal, then there is indeed a y such y+1 = x. However, if x is a limit ordinal, you can't reach x by adding 1: forall y, if y < x, then y+1 < x. Now, this is all from the point of view of a set theory that you have rejected. But if you are going to use the symbol aleph_0, then you need to give it some definition, if it's not the standard definition. >But, all of this leads to contradictions, since there is no largest finite or >smallest infinite. There is no largest finite, but there is a smallest infinite, and aleph_0 is its name. >Aleph_0 is a concept, an unbounded ill-defined concept, and >not any exact number. On the contrary, aleph_0 is very precisely defined: aleph_0 is the smallest infinite ordinal -- Daryl McCullough Ithaca, NY === Subject: Re: infinity Daryl McCullough said: > Tony Orlow says: > So the real question is this: Is the set of all finite natural >> numbers finite, or infinite? Finite. Please respond to my inductive proof of this, posted today. It's nonsense. An inductive proof is a proof of a *universal* statement, > of the form forall n s.t. n is a finite natural, Phi(n) The claim The set of all finite natural numbers is finite does > not have this form. You can't prove it by induction. If S is a set, then the claim S is finite is mathematically > equivalent to exists n, n is a finite natural, and size(S) < n The claim S is infinite is mathematically equivalent to forall n s.t. n is a finite natural, size(S) > n That has a form that can be proved by induction, by letting > Phi(n) be size(S) > n. -- > Daryl McCullough > Ithaca, NY > Proof: For all n in N, Phi(n) = the set of naturals from 1 up to and including > n contains n elements. n=1: {1} has 1 element. n->n+1: A set has n elements with n as a largest element. When we add the > successor of n to the set, n+1 is added to the set, incrementing the set size > to n+1. So, there are n+1 elements up to and including n+1. Conlcusion: For all n in N, n is the size of the set of naturals from 1 through > n. So, if the size of the set is aleph_0, then the set is the set of all naturals > from 1 through aleph_0. >The size of the set of all naturals through n is n, so > if the set is infinite, then n is infinite, and if n is finite, then the set is > finite. You missed a set. > Let the set of all naturals from 1 through aleph_0 be K. > Now let U be the set K minus the element aleph_0. > Note there is no n such that U is the set of all naturals > through n and U is not K, so U was not considered above. > Is U finite or infinite? > -William Hughes > Of course U is infinite. Aleph_0 would have to be the successor of a number, > which would also be infinite. OK, define U1 to be the set K minus all the infinite members of K. Note that there is no natural n such that U1 is the set of all naturals 1 through n so your proof by induction does not apply. Is U1 finite or infinite? -William Hughes === Subject: Re: infinity Virgil said: > Daryl McCullough said: > Tony Orlow says: No, what is disputed is that the values in the set are finite, and >therefore end. They do not end, but extend into infinite values >in the infinite set. > TO assumes that not ending requires exending into the infinite, but > that is false. There are all sorts of sets that do not end but do not > extend into the infinite, the set of (finite) naturals being one such > set. The set of real values in the open interval (0,1) for another. We keep switching between two different arguments: 1. Is the set of finite natural numbers a finite set? 2. Does > the set of natural numbers contain infinite numbers? Those are the same argument. > No they are not! The set of finite naturals is Cantor-infinite, and as > TO has no other workable definition of infinite, that makes the set > infinite. All numbers x, 0 The set of finite naturals clearly then is an infinite set that contains > no infinite members. Clearly, any infinite set of whole numbers, or any numbers differing by a constant finite amount, must contain infinite numbers since it needs an infinite range of value to include the infinity of finite intervals in the set. >> So the real question is this: Is the set of all finite natural > numbers finite, or infinite? > Finite. Please respond to my inductive proof of this, posted today. > Then TO must be declaring the set of finite naturals to have a largest > member, since for several weeks now people have been posting unrefuted > proofs that a non-empty ordered set without a largest member is infintie. No, get off it. You may all agree that that is a good criterion for infinitude of a set, but I do not, and have never claimed there is a largest finite number. I don't have to refute your assumptions. I can simply reject them. > A criterion for finiteness of a set that is used commonly is this > one: S is finite <-> exists finite natural number n such that size(S) > < n S is infinite <-> forall finite natural numbers n, size(S) >= n > Which are circular definitions resting on the assumption that all > naturals are finite, which is based on a proof where the rules of > finiteness are never established, except to assume, correctly, that > finite+finite=finite. If all naturals are finite, and you say it is > finite if there is a natural greater than it, this is little better > than saying finite=not infinite. > TO criticizes the definition without providing an alternative. Please comment on the alternative, above. > Try this: > (a) A set is finite if it allows no injection to any proper subset, and > (b) a natural, n, in the set of naturals, N, is finite iff {k in N:k<=n} > is a finite set. > No circularity, and it makes all naturals produced naturally (by a > finite number of successor operations) finite. A finite number of successor operations? Wouldn't that only produce a finite number of naturals in your set? Perhaps you qualify it this way because you are beginning to see that an infinite number of increments produces an infinite sum, and therefore infinite values in the set, but restricting your count of successors to a finite number restricts the number of naturals in your set likewise. This is what I have been trying to explain for months now. You either have a finite set of finite naturals, or an infinite set which includes infinite values. -- Smiles, Tony === Subject: Re: infinity constant finite amount, must contain infinite numbers since it needs an > infinite range of value to include the infinity of finite intervals in > the set. You keep saying that, but you still haven't proved it. I reject your statement. > I don't have to refute your assumptions. I can simply reject them. Sure, and I can reject your assumptions out of hand, too. That's really useful, isn't it? === Subject: Re: infinity > Virgil said: Daryl McCullough said: > Tony Orlow says: No, what is disputed is that the values in the set are finite, and >therefore end. They do not end, but extend into infinite values >in the infinite set. TO assumes that not ending requires exending into the infinite, but > that is false. There are all sorts of sets that do not end but do not > extend into the infinite, the set of (finite) naturals being one such > set. The set of real values in the open interval (0,1) for another. We keep switching between two different arguments: 1. Is the set of finite natural numbers a finite set? 2. Does > the set of natural numbers contain infinite numbers? Those are the same argument. No they are not! The set of finite naturals is Cantor-infinite, and as > TO has no other workable definition of infinite, that makes the set > infinite. > All numbers x, 0 number within zero distance of zero is infinitesimal, and if x is > infinitesimal, 1/x is infinite. >> So the real question is this: Is the set of all finite natural > numbers finite, or infinite? Finite. Please respond to my inductive proof of this, posted today. Then TO must be declaring the set of finite naturals to have a largest > member, since for several weeks now people have been posting unrefuted > proofs that a non-empty ordered set without a largest member is infinite. > No, get off it. You may all agree that that is a good criterion for > infinitude of a set, but I do not, and have never claimed there is a > largest finite number. I don't have to refute your assumptions. I can > simply reject them. TO's rejection valid proofs means that TO's mathematics is self contradictory. > Please comment on the alternative, above. In any ordered set, what TO requires of his infinitesimals is impossible. Try this: > (a) A set is finite if it allows no injection to any proper subset, and (b) a natural, n, in the set of naturals, N, is finite iff {k in N:k<=n} > is a finite set. No circularity, and it makes all naturals produced naturally (by a > finite number of successor operations) finite. > A finite number of successor operations? Wouldn't that only produce a finite > number of naturals in your set? NO! It would only produce finite naturals but an unlimited number of them. For the set of all finite naturals to be finite, the sum of all those finitely many naturals in that finite set would have to be in that set, but in any finite set of naturals with 3 or more members (3 to allow for 0-origin naturals) the sum of all the members is NEVER in the set. So that no finite set of naturals can ever contain all finite naturals. === Subject: Re: infinity Virgil said: > Virgil said: Daryl McCullough said: > Tony Orlow says: No, what is disputed is that the values in the set are finite, and >therefore end. They do not end, but extend into infinite values >in the infinite set. TO assumes that not ending requires exending into the infinite, but > that is false. There are all sorts of sets that do not end but do not > extend into the infinite, the set of (finite) naturals being one such > set. The set of real values in the open interval (0,1) for another. We keep switching between two different arguments: 1. Is the set of finite natural numbers a finite set? 2. Does > the set of natural numbers contain infinite numbers? Those are the same argument. No they are not! The set of finite naturals is Cantor-infinite, and as > TO has no other workable definition of infinite, that makes the set > infinite. > All numbers x, 0 number within zero distance of zero is infinitesimal, and if x is > infinitesimal, 1/x is infinite. >> So the real question is this: Is the set of all finite natural > numbers finite, or infinite? Finite. Please respond to my inductive proof of this, posted today. Then TO must be declaring the set of finite naturals to have a largest > member, since for several weeks now people have been posting unrefuted > proofs that a non-empty ordered set without a largest member is infinite. > No, get off it. You may all agree that that is a good criterion for > infinitude of a set, but I do not, and have never claimed there is a > largest finite number. I don't have to refute your assumptions. I can > simply reject them. > TO's rejection valid proofs means that TO's mathematics is self > contradictory. > Please comment on the alternative, above. > In any ordered set, what TO requires of his infinitesimals is > impossible. Try this: > (a) A set is finite if it allows no injection to any proper subset, and (b) a natural, n, in the set of naturals, N, is finite iff {k in N:k<=n} > is a finite set. No circularity, and it makes all naturals produced naturally (by a > finite number of successor operations) finite. > A finite number of successor operations? Wouldn't that only produce a finite > number of naturals in your set? > NO! It would only produce finite naturals but an unlimited number of > them. Does any iteration of the successor operator produce more than one natural? No. Each time it is applied, one natural is produced. So how do you produce an infinite number of them with finite iterations? That makes no sense. Explanation please. > For the set of all finite naturals to be finite, the sum of all those > finitely many naturals in that finite set would have to be in that set, > but in any finite set of naturals with 3 or more members (3 to allow for > 0-origin naturals) the sum of all the members is NEVER in the set. > So that no finite set of naturals can ever contain all finite naturals. So, you want to add this sum to the set? Go ahead. The set was finite, and now it has one more element. Is it suddenly infinite? No, it's still finite. -- Smiles, Tony === Subject: Re: infinity > Virgil said: > Does any iteration of the successor operator produce more than one > natural? No. Each time it is applied, one natural is produced. So how > do you produce an infinite number of them with finite iterations? > That makes no sense. Explanation please. For the set of all finite naturals to be finite, the sum of all > those finitely many naturals in that finite set would have to be in > that set, but in any finite set of naturals with 3 or more members > (3 to allow for 0-origin naturals) the sum of all the members is > NEVER in the set. So that no finite set of naturals can ever contain all finite > naturals. So, you want to add this sum to the set? It is not just one set, it is EVERY finite set of naturals has this same failing of not containing all finite naturals, so that NO finite set can contain all finite naturals. > Go ahead. The set was finite, and now it has one more element. Is it > suddenly infinite? No, it's still finite. All of them with the added member are still finite and none of them contain ALL finite naturals, since there are ALWAYS missing finite naturals. Since TO asserts some finite set can contain all finite naturals, showing that no finite set can contains all finite naturals makes TO WRONG!. === Subject: Re: infinity Virgil said: >> It would only produce finite naturals but an unlimited number of them. > Does any iteration of the successor operator produce more than one natural? > No. Each time it is applied, one natural is produced. So how do you produce > an infinite number of them with finite iterations? Umm, by applying an infinite number of iterations? If that can't be done, i.e., if we can't apply an infinite number of iterations, then we can only apply a finite number of iterations, which means that there must be a last iteration applied. Which means that there is a last successor, which must be larger than any other successor. But you keep saying (correctly) that there is no largest natural (or whole) number, or equivalently, that there is no largest successor. Therefore we must have an infinite number of iterations. QED. === Subject: Re: infinity !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > Virgil said: > It would only produce finite naturals but an unlimited number of them. >> Does any iteration of the successor operator produce more than one natural? >> No. Each time it is applied, one natural is produced. So how do you produce >> an infinite number of them with finite iterations? > Umm, by applying an infinite number of iterations? > If that can't be done, i.e., if we can't apply an infinite number of > iterations, Which you can't. > then we can only apply a finite number of iterations, which means > that there must be a last iteration applied. Oh, but the finite number of iterations you can apply is not bounded. It really does not serve too much to go on about iterations: the natural numbers' structure does not come into being by them. It comes into being by a set of static properties that are inherently incompatible with iterations. You can _always_ use iterations just up to a point, this point not being fixed in advance. But there is no magic iteration after which all iterations are finished, and so the magic of complete iterations is nothing more than that: magic. Iterations are only fit for modeling subsequences of the naturals. > Therefore we must have an infinite number of iterations. QED. We have an unlimited supply of iterations, so to say. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: infinity > Virgil said: >> It would only produce finite naturals but an unlimited number of them. > Does any iteration of the successor operator produce more than one natural? > No. Each time it is applied, one natural is produced. So how do you produce > an infinite number of them with finite iterations? > Umm, by applying an infinite number of iterations? > If that can't be done, i.e., if we can't apply an infinite number of > iterations, then we can only apply a finite number of iterations, > which means that there must be a last iteration applied. Which means > that there is a last successor, which must be larger than any other > successor. > But you keep saying (correctly) that there is no largest natural > (or whole) number, or equivalently, that there is no largest > successor. > Therefore we must have an infinite number of iterations. QED. But TO cannot, or won't, say which iteration makes the jump from finite to infinite. === Subject: Re: infinity Daryl McCullough said: > Tony Orlow says: No, what is disputed is that the values in the set are finite, and >therefore end. They do not end, but extend into infinite values >in the infinite set. > TO assumes that not ending requires exending into the infinite, but > that is false. There are all sorts of sets that do not end but do not > extend into the infinite, the set of (finite) naturals being one such > set. The set of real values in the open interval (0,1) for another. We keep switching between two different arguments: 1. Is the set of finite natural numbers a finite set? 2. Does > the set of natural numbers contain infinite numbers? Those are the same argument. > No they are not! The set of finite naturals is Cantor-infinite, and as > TO has no other workable definition of infinite, that makes the set > infinite. > All numbers x, 0 number within zero distance of zero is infinitesimal, and if x is > infinitesimal, 1/x is infinite. > The set of finite naturals clearly then is an infinite set that contains > no infinite members. > Clearly, any infinite set of whole numbers, or any numbers differing by a > constant finite amount, must contain infinite numbers since it needs an > infinite range of value to include the infinity of finite intervals in the set. >> So the real question is this: Is the set of all finite natural > numbers finite, or infinite? > Finite. Please respond to my inductive proof of this, posted today. > Then TO must be declaring the set of finite naturals to have a largest > member, since for several weeks now people have been posting unrefuted > proofs that a non-empty ordered set without a largest member is infintie. > No, get off it. You may all agree that that is a good criterion for infinitude > of a set, but I do not, and have never claimed there is a largest finite > number. I don't have to refute your assumptions. I can simply reject them. But you can't reject the fact that given a finite number of finite integers you can take the sum to get a finite integer (you have stated this). So, let K be the set of finite integers. Assume K is finite. Let n be the sum of all the members of K. Then n is a finite integer. But n is not a member of K. Contradiction. Thus K is not finite. Note, this works for *any* definition of infinite, as long as the sum of a finite number of finite integers is finite. It does not use the existence of a largest integer in a finite set. -William Hughes === Subject: Re: infinity Virgil said: > Daryl McCullough said: > What is L? > L is a string length. With an alphabet of size S, we can make S^L > strings of length L. > But what about the set of allowable values of L? If set of allowable > values of L the set of finite naturals, then the set of L's is > Cantor-infinite, and so the set of strings will also be > Cantor-infinite! Sure, Cantor-infinite, but not actually infinite. Cantor's definitions suck. > You claim that the above summation is finite. According to you >> F = sum S^k for all finite k >> is a finite number. >That is correct. > However if F is a finite number, then there are strings of >> length F, and there are S^F strings of length F, which is >> greater than F, the supposed number of finite strings. That's a >> contradiction. Therefore F cannot possibly be a finite number. Darryl just gave a similar proof. When you say for all finite >k then you are saying there is some upper bound to k. No, there is no (finite) upper bound to the set of finite natural > numbers. > No, but k cannot be infinite if it is finite. > But the set of all k can be Cantor-infinite even though each k in that > set is finite. Yes, even though you have a finite number of finite terms, the sum is Cantor- infinite. Doesn't that indicate a problem with Cantor's definitions/ k cannot be infinite. Therefore sum (x=1->k: S^x) is finite. That sum is the number or strings less than or equal to k in length. > It is finite for all finite k, since you are adding k terms of S^x > where x is finite. > But the sum over all finite k is the sum over a Cantor-infinite set of > values, the Cantor-infinite set of finite naturals. Yes, based on the misconception that you can have an infinite set of unique finite naturals you can draw the conclusion that you can have an infinite set of unique finite strings ona finite alphabet. This is your Cantor-infinite. > Until TO faces up to the Cantor-infintieness of the set of finite > naturals, he is being willfully blind to mathematical reality. Cantor-infiniteness isn't at issue. I don't care what Cantor says at this point. There are obviously problems with Cantor's concept of infinity. I am talking about actual logic here, not hocus pocus. Here, now, you say, All the words are finite because if you take >a finite string and add a finite number of symbols you still get a >finite string. No, what we are saying is that if U is a finite set of finite > it follows that U did not contain all finite strings. > The contradiction comes from assuming you have defined the full set > of all finite strings. This is exactly the largest finite argument, > which of course causes contradictions, because you can always add 1. > So what?? I suppose that would seem to indicate infinity, but it > indcates infinity for the values in the set as well and the lengths > of the strings. > But the set of _finite_ naturals has, according to TO, no largest > members, and is, therefore, clearly Cantor-infinite, however TO-finite > TO claims it to be. Cantor may base his claims on such shakey ground as largest member, but there are clearly not an infinite number of unique finite naturals, nor of unique finite strings on any finite alphabet. > If you declare your values and strings finite, though > you give no upper bound, there may be no upper bound to the size of > the set, but it is finite as well, for all the reasons I keep having > to repeat. > And any non-empty ordered set without an upper bound is Cantor-infinite! Think for yourself for a change. Cantor is long gone. Yes. The sum over all finite values of L of S^L is infinite. > So, you can add finite sets of strings to finite sets of strings in > succession, and get an infinite result? > If one has infinitely many such finite terms, yes! There are infinite > series of positive numbers that diverge! First of all, the infinitude of those terms has been in question since the beginning of this conversation, since you are restricting yourself to finite naturals, of which there are only a finite number. More importantly, my whole point here is that you are willing to say that you can add finite numbers of elements at a time, and eventually get an infinite set, but you refuse to consider that, incrementing a value one unit at a time, it can ever reach infinity. The set size IS a number, and you are adding finite quantities to that number at each iteration. If that number can reach infinity through an infinite series of finite additions, then why can't the natural numbers also reach infinity through an infinite series of finite additions? This is totally inconsistent. > But you can't add a finite > number to a number in succession and get an infinite number? Hmmmm... Proof: The set of all finite strings on a finite alphabet is finite. L=0: Given N=S^L, there is S^0=1 string, the null string. L->L+1: Given a finite set of strings of length L or less, we add the > set of strings of length L+1, which has S^(L+1) elements. Since L is > finite, L+1 is finite, and S^(L+1) is finite. So we add this finite > number of strings to the finite number of strings of length L or > less, to get the number of strings of length L+1 or less. A finite > plus a finite is finite. > But we have here an infinite series, S^1 + S^2 + S^3 + ... of positive > terms, which cannot converge unless the terms, S^n, converge to zero. > Does TO wish to state that the sequence f(n) = S^n ha kimit zero? Not if n is allowed to go to infinity, and yet, you have restricted n to finite values, and therefore, the sum is finite. > If you say you > can add the sizes of finite sets, and somehow get to infinity, but > you cannot add other finite numbers and get infinity, then you are > playing a shell game. > The sum of any finite number of positive terms is finite. The sum of any > infinite sequence of positive naturals is a divergent series. Correct. That is why I say that performing the successor operation, or incrementing, an infinite number of times to get an infinite set of whole numbers yields infinite numbers as members of that set, since each natural number is the sum of all the increments up to that point. If you restrict your numbers to the finite, then there are a finite number of them, and if you use these numbers as the lengths of your strings, then you only have finite-length strings, and only have a finite set of them. If you want to have an infinite number of whole numbers, then you need to include infinite whole numbers. > There is no finite upper bound on the lengths of finite strings. > Then there is no reason to conclude that they are all finite. > Since each is only one character longer than some other, all the way > down to one character strings, there is no possible way that any of them > can be infinite any more than the infinite set of finite naturals > contains any infinite naturals. If adding one natural to the set at a time can produce an infinite set by incrementing the set size repeatedly, then incrementing the values in it can just as easily produce an infinite value. Consider each string to be a set of characters. You say adding one character at a time will never produce an infinite string, but addding one natural at a time can produce an infinite set? -- Smiles, Tony === Subject: Re: infinity > Virgil said: > But the set of all k can be Cantor-infinite even though each k in > that set is finite. > Yes, even though you have a finite number of finite terms, the sum is > Cantor- infinite. Doesn't that indicate a problem with Cantor's > definitions/ No one but TO asserts that the number of finite terms is finite. TO apparently cannot tell the size of the terms from the size of the set containing them. The problem is, as usual, internal to TO's understanding, and not with the facts in the case. > But the sum over all finite k is the sum over a Cantor-infinite set > of values, the Cantor-infinite set of finite naturals. > Yes, based on the misconception that you can have an infinite set of > unique finite naturals you can draw the conclusion that you can have > an infinite set of unique finite strings ona finite alphabet. This is > your Cantor-infinite. The misconception is in TO's mind. The Peano axioms guarantee the existence of a set which is Cantor-infinite and such that each subset bounded above by a member of that set is Cantor finite. TO's reiteration ad nauseam of his misconceptions does not make them any less misconceived. Until TO faces up to the Cantor-infiniteness of the set of finite > naturals, he is being willfully blind to mathematical reality. > Cantor-infiniteness isn't at issue. It is TO's refusal to either accept Cantor's definition or come up with a workable alternative that is at issue. So far TO's attempts at a workable alterntive have been laughably inept. > I don't care what Cantor says at this point. There are obviously > problems with Cantor's concept of infinity. I am talking about actual > logic here, not hocus pocus. If TO were talking actual logic here, instead of his own hocus-pocus, he would be getting a more respectful hearing, but the only obvious problem with Cantor, or any of this standard mathematics, is in TO's deliberate misunderstandings and misrepresentations of it. Here, now, you say, All the words are finite because if you >take a finite string and add a finite number of symbols you >still get a finite string. No, what we are saying is that if U is a finite set of finite > that, it follows that U did not contain all finite strings. The contradiction comes from assuming you have defined the full > set of all finite strings. This is exactly the largest finite > argument, which of course causes contradictions, because you can > always add 1. So what?? I suppose that would seem to indicate > infinity, but it indcates infinity for the values in the set as > well and the lengths of the strings. TO is being willfully blind to the obvious again. For any finite set of two or more finite strings of positive lengths, one can create the concatenation of all of them which will be a finite string longer than any of them, so that NO finite set of finite strings can contain all finite strings. In the same way, for finite sets of 3 or more finite naturals, the finite sum of all of the members is not in the set, so there cannot be any finite set that contains all finite naturals. But the set of _finite_ naturals has, according to TO, no largest > members, and is, therefore, clearly Cantor-infinite, however > TO-finite TO claims it to be. > Cantor may base his claims on such shakey ground as largest member, Cantor does not need to say anything about largest members. It is TO's assumptions of the set being finite that requires it. > but there are clearly not an infinite number of unique finite > naturals, nor of unique finite strings on any finite alphabet. Then the finite sum of all allegedly finitely many finite naturals cannot be a finite natural number as it is necessarily larger than any member of the alleged set of all finite natural numbers. And, similarly, the finite concatenation of the allegedly finite set of all finite strings cannot be a finite string. TO's assumptions clearly lead to self-contradiction when finite sums of finite numbers cannot be finite. > If you declare your values and strings finite, though you give no > upper bound, there may be no upper bound to the size of the set, > but it is finite as well, for all the reasons I keep having to > repeat. And those reasons are self-contradictory, as we all keep having to repeat! And any non-empty ordered set without an upper bound is > Cantor-infinite! > Think for yourself for a change. Cantor is long gone. This from someone who cannot think at all with out tripping over his illogic. > First of all, the infinitude of those terms has been in question > since the beginning of this conversation, since you are restricting > yourself to finite naturals, of which there are only a finite number. Since any finite sum of finite naturals is a finite natural,if the set of ALL finite naturals were finite, it would have to contain the sum of all its members, but the sum is clearly larger than of the numbers being summed. OOPS! TO goofed again! > More importantly, my whole point here is that you are willing to say > that you can add finite numbers of elements at a time, and eventually > get an infinite set, but you refuse to consider that, incrementing a > value one unit at a time, it can ever reach infinity. The axioms GIVES us this set, whe do not have to construct it. > But we have here an infinite series, S^1 + S^2 + S^3 + ... of > positive terms, which cannot converge unless the terms, S^n, > converge to zero. Does TO wish to state that the sequence f(n) = S^n has limit zero? > Not if n is allowed to go to infinity, and yet, you have restricted n > to finite values, and therefore, the sum is finite. In finding limits, the variable never actually reaches the target value, it only approaches it. === Subject: Re: infinity But the set of all k can be Cantor-infinite even though each k in that > set is finite. > Yes, even though you have a finite number of finite terms, the sum is Cantor- > infinite. Doesn't that indicate a problem with Cantor's definitions/ Really we are all giving too much credit to Cantor. What Virgil commonly claims to be Cantor-infinite is normally called Dedekind-infinite. I believe Cantor used the more standard definition that set S is finite if there exists a bijection from S to the set {1,2,...,n} for some natural number N. It can be shown that if we assume choice, then both definitions of finite/infinite coincide. (IMO this is one of the best arguments for why choice really is good:) Without choice, it is possible that there are sets which are infinite, but not Dedekind-infinite, because of the lack of some mappings. Note that Cantor still did naive set theory rather then something like ZFC. I really think we should all stop fighting over Cantor. He is not the guy who came up with ALL of set theory. I think it's just easier for cranks to just blame it on one person since then it seems less believable. By today's standards, Cantor's work was non-rigorous and handwavy. Just like many Euler's proofs would be enough to get a D on your analysis homework. Why is Zermelo not such a celebrity? Choice is far less intuitive then anything that Cantor did. I suppose it's fun to talk about Cantor to cranks, just to get their blood boiling. But Cantor-finite is really that all natural numbers are by definition finite (as they can be identified with cardinals, and the finite cardinals are exactly those that are natural numbers), so Tony is out of luck there as there is nothing to prove nor disprove. This as far as I know was the definitions Cantor used. The alternative, that most people here are using is Dedekind-infinite. I suppose because only here is there something to prove. It is obvious, that even though these are the only mathematical definitions of finite/infinite, Tony apparently has some God-given intuitive idea of what it means to be finite/infinite. You will not convince him that he's wrong because 1) He's a total moron (this is the main reason) 2) He doesn't want to be convinced 3) His intuition rather then logic is his main tool and you cannot argue with intuition 4) He's trolling because nobody can be so utterly stupid. Jiri === Subject: Re: infinity set is finite. > Yes, even though you have a finite number of finite terms, the sum is Cantor- > infinite. Doesn't that indicate a problem with Cantor's definitions/ No. What's the problem? As you have repeatedly admitted, there is no end to a list of the values in such a set. Most of us have no problem calling a thing which is unending infinite, and have a great deal of trouble understanding how anybody could apply the adjectives finite and unending to the same thing. > But the sum over all finite k is the sum over a Cantor-infinite set of > values, the Cantor-infinite set of finite naturals. > Yes, based on the misconception that you can have an infinite set of unique > finite naturals Despite your faith-based assertions that you can't, the actual proofs that you can are elementary. > If adding one natural to the set at a time can produce an infinite set by > incrementing the set size repeatedly, It can't. > then incrementing the values in it can > just as easily produce an infinite value. It can't. - Randy === Subject: Re: infinity Randy Poe said: > But the set of all k can be Cantor-infinite even though each k in that > set is finite. > Yes, even though you have a finite number of finite terms, the sum is Cantor- > infinite. Doesn't that indicate a problem with Cantor's definitions/ > No. What's the problem? As you have repeatedly admitted, > there is no end to a list of the values in such a set. > Most of us have no problem calling a thing which is > unending infinite, and have a great deal of trouble > understanding how anybody could apply the adjectives > finite and unending to the same thing. Even though there is no end to the possible values for a natural number, and yet you apply the label finite to natural numbers? This is the root of the problem here. I am only saying the apparently unending set is finite because you are claiming the apparently unending set of values is finite. You allow infinite naturals and this problem goes away. > But the sum over all finite k is the sum over a Cantor-infinite set of > values, the Cantor-infinite set of finite naturals. > Yes, based on the misconception that you can have an infinite set of unique > finite naturals > Despite your faith-based assertions that you can't, the > actual proofs that you can are elementary. Faith-based? I have offered nothing but numerical arguments and caclulations to support my position. You offer gedankens and verbiage, and no math. > If adding one natural to the set at a time can produce an infinite set by > incrementing the set size repeatedly, > It can't. Then the recursive definition of the natural numbers defines a finite set. Be serious. > then incrementing the values in it can > just as easily produce an infinite value. > It can't. Look up infinite series, for god's sake. > - Randy -- Smiles, Tony === Subject: Re: infinity !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > Randy Poe said: >> But the set of all k can be Cantor-infinite even though each k >> in that set is finite. >> Yes, even though you have a finite number of finite terms, the >> sum is Cantor- infinite. Doesn't that indicate a problem with >> Cantor's definitions/ >> No. What's the problem? As you have repeatedly admitted, there is >> no end to a list of the values in such a set. Most of us have no >> problem calling a thing which is unending infinite, and have a >> great deal of trouble understanding how anybody could apply the >> adjectives finite and unending to the same thing. > Even though there is no end to the possible values for a natural > number, and yet you apply the label finite to natural numbers? Every single one of them is finite, and there is an infinite amount of them. > This is the root of the problem here. I am only saying the > apparently unending set is finite because you are claiming the > apparently unending set of values is finite. No. The set of values is infinite. But every single value in that set is finite. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: infinity But the set of all k can be Cantor-infinite even though each k in that > set is finite. > Yes, even though you have a finite number of finite terms, the sum is Cantor- > infinite. Doesn't that indicate a problem with Cantor's definitions/ > No. What's the problem? As you have repeatedly admitted, > there is no end to a list of the values in such a set. > Most of us have no problem calling a thing which is > unending infinite, and have a great deal of trouble > understanding how anybody could apply the adjectives > finite and unending to the same thing. > Even though there is no end to the possible values for a natural number, and > yet you apply the label finite to natural numbers? There is no end to the possible values. I apply the name infinite to the possible values. There is an end to every particular natural number. I apply the label finite to each number, as the list of its digits has an end in every base. Why this confusion between an individual number and the set of all numbers? When you say I apply the label finite to natural numbers, is your choice of language deliberately ambiguous (do you mean any particular natural number or the collection of natural numbers?) or do you really not see the difference between each element and the entire collection? > This is the root of the > problem here. I am only saying the apparently unending set is finite because > you are claiming the apparently unending set of values is finite. No, the unending set of values is not finite. Each individual number doesn't have a set of values, it has a single, finite value. When I say that a given number is finite, I have not made a statement about the set of values. > You allow infinite naturals and this problem goes away. No it doesn't. It just avoids discussing the properties of the collection of finite naturals. It's a distraction. > Despite your faith-based assertions that you can't, the > actual proofs that you can are elementary. > Faith-based? Yes. > I have offered nothing but numerical arguments Each of which eventually ends up with a faith-based statement, a leap beyond what is justified by the logic. > and caclulations incrementing the set size repeatedly, > It can't. > Then the recursive definition of the natural numbers defines a finite set. No, the definition of natural numbers does not require that the collection be built one at a time. There is nothing in the definition that makes such a requirement on the set. It DOES require that each ELEMENT of the set be reachable by adding one to one a finite number of times. Again you have gotten confused between the elements and the set. The recursive definition defines the elements. Once the elements are defined, the set is immediately defined as the collection of everything that meets that definition. This isn't your usual quantor dyslexia, it's some other sort of element-set dyslexia. > Be serious. I'm quite serious. Which part did you think was a joke? > then incrementing the values in it can > just as easily produce an infinite value. > It can't. > Look up infinite series, for god's sake. Be more specific. Tell me what part of infinite series for god's sake implies that one can reach an produce value by a process of incrementing. Perhaps you should look up infinite series for god's sake and familiarize yourself with what it means for a series to diverge. One thing it doesn't mean is that anything ever reaches an infinite value. - Randy === Subject: Re: infinity Randy Poe said: > Randy Poe said: But the set of all k can be Cantor-infinite even though each k in that > set is finite. > Yes, even though you have a finite number of finite terms, the sum is Cantor- > infinite. Doesn't that indicate a problem with Cantor's definitions/ No. What's the problem? As you have repeatedly admitted, > there is no end to a list of the values in such a set. > Most of us have no problem calling a thing which is > unending infinite, and have a great deal of trouble > understanding how anybody could apply the adjectives > finite and unending to the same thing. > Even though there is no end to the possible values for a natural number, and > yet you apply the label finite to natural numbers? > There is no end to the possible values. I apply the name > infinite to the possible values. > There is an end to every particular natural number. I > apply the label finite to each number, as the list of > its digits has an end in every base. Only for finite values, not for infinite values. > Why this confusion between an individual number and the > set of all numbers? When you say I apply the label finite > to natural numbers, is your choice of language deliberately > ambiguous (do you mean any particular natural number or > the collection of natural numbers?) or do you really > not see the difference between each element and the > entire collection? You say each and every natural is finite. So, what are you asking? You know what your position is. > This is the root of the > problem here. I am only saying the apparently unending set is finite because > you are claiming the apparently unending set of values is finite. > No, the unending set of values is not finite. then you have infinite values in the set? > Each individual number doesn't have a set of values, it > has a single, finite value. When I say that a given > number is finite, I have not made a statement about > the set of values. When you say ALL naturals are finite, then you certainly ARE making a statement about the set of values, aren't you? > You allow infinite naturals and this problem goes away. > No it doesn't. It just avoids discussing the properties of > the collection of finite naturals. It's a distraction. No the distraction is all this largest finite nonsense. > Despite your faith-based assertions that you can't, the > actual proofs that you can are elementary. > Faith-based? > Yes. > I have offered nothing but numerical arguments > Each of which eventually ends up with a faith-based > statement, a leap beyond what is justified by the logic. Bull. I am not the one who confuses unbounded with infinite, who claims a constantly growing number of balls inexplicably becomes zero, or that I can cut up a ball and reassemble it into two solid balls of the same size. I am not the one who says you can draw a bijection between an infinite number of kids and their fingers, as if they cut nine off and pass them to infinity and they disappear. I am not the one that claims we can remove half the elements of a set and it remains the same size. Standard set theory is what is full of all sorts of warped leaps of logic, like bijection=equivalence, and an infinite number of increments yields a finite sum. All of my arguments are based in formulas, which set theory seems to entirely avoid for understandable reasons, given the fact that none of them would agree with your results. > and caclulations > Stay away from the white powder. > If adding one natural to the set at a time can produce an infinite set by > incrementing the set size repeatedly, It can't. > Then the recursive definition of the natural numbers defines a finite set. > No, the definition of natural numbers does not require > that the collection be built one at a time. There is > nothing in the definition that makes such a requirement > on the set. The set is recusively defined, each member being derived from the previous through successor(). I don't know what definition you're looking at, or what you think it means. > It DOES require that each ELEMENT of the set be reachable > by adding one to one a finite number of times. Again you > have gotten confused between the elements and the set. > The recursive definition defines the elements. Once the > elements are defined, the set is immediately defined as > the collection of everything that meets that definition. If each element is a finite number of steps from 1, then no element is an infinite number of steps from 1, and the set does not have an infinite number of elements, since each element is a step. > This isn't your usual quantor dyslexia, it's some other > sort of element-set dyslexia. It's Bigulosity. > Be serious. > I'm quite serious. Which part did you think was a joke? You seem to have clipped it, but I was referring to your blind insistence that numbers as elements and numbers OF elements are two very different kinds of numbers. They're not. Building in infinite set one lement at a time is the same as creating an infinite number one increment at a time. > then incrementing the values in it can > just as easily produce an infinite value. It can't. > Look up infinite series, for god's sake. > Be more specific. Tell me what part of infinite series > for god's sake implies that one can reach an produce > value by a process of incrementing. Increment means add one. If you do this an infinite number of times you get sum (x=1->oo: 1). Infinite series says that this sum diverges, since the terms do not have a limit of zero at infinity. Therefore, if you have an infinite number of elements, each one greater than the last, then you will have infinite values as the result of the infinite increments. If you do not allow such infinite sums, then you must not allow infinite increments, which means you do not produce an infinite set. > Perhaps you should look up infinite series for god's > sake and familiarize yourself with what it means for > a series to diverge. One thing it doesn't mean is that > anything ever reaches an infinite value. What??? YOU look it up again. There are three possibilities for an infinite series. It can converge, such that the sum of all terms is a finite number. It can diverge in the sense that the sum over all terms is infinite. Or it can diverge in the sense that the sum is never infinite, but has no specific limit at n=oo. This third form is generally an oscillating sequence. In this case, we are looking at a sum which is clearly infinite. If you add n 1's, the sum is n. If you add an infinite number of 1's, whatever that infinite number is, that will be the sum. If you claim to have an infinite number of naturals, then you have incremented an infinite number of times and have infinite values, including a member equal to the set size. > - Randy -- Smiles, Tony === Subject: Re: infinity > Randy Poe said: There is an end to every particular natural number. I apply the > label finite to each number, as the list of its digits has an end > in every base. > Only for finite values, not for infinite values. And as there are no infinite values IN the set of naturals, at least among those allowed by the Peano axioms, the issue does not arise. If TO want to have anything but finite naturals, he must invent a different word for them as the word naturalis limited to finite ones. Why this confusion between an individual number and the set of all > numbers? When you say I apply the label finite to natural > numbers, is your choice of language deliberately ambiguous (do you > mean any particular natural number or the collection of natural > numbers?) or do you really not see the difference between each > element and the entire collection? > You say each and every natural is finite. So, what are you asking? > You know what your position is. But TO's positions are as ambiguous as he can make them. This is the root of the problem here. I am only saying the > apparently unending set is finite because you are claiming the > apparently unending set of values is finite. No, the unending set of values is not finite. > then you have infinite values in the set? Not needed. The set of all naturals, all of which are finite, is not a finite set: The sum of all members of any finite set of two or more naturals results in a natural larger than any member of the set. If the set of ALL naturals were to be finite, the sum of its members would be a natural strictly greater than all naturals, which is impossible. Thus the set of all naturals cannot be finite. Each individual number doesn't have a set of values, it has a > single, finite value. When I say that a given number is finite, I > have not made a statement about the set of values. > When you say ALL naturals are finite, then you certainly ARE making a > statement about the set of values, aren't you? No! It is a statement about the members, not about the set. You allow infinite naturals and this problem goes away. You allow infinitely many finite naturals and all problems go away! No it doesn't. It just avoids discussing the properties of the > collection of finite naturals. It's a distraction. > No the distraction is all this largest finite nonsense. The only non-sense here is that TO has no sense about what is and is not possible given the Peano axioms. Despite your faith-based assertions that you can't, the actual > proofs that you can are elementary. > Faith-based? Yes. I have offered nothing but numerical arguments Each of which eventually ends up with a faith-based statement, a > leap beyond what is justified by the logic. No, the definition of natural numbers does not require that the > collection be built one at a time. There is nothing in the > definition that makes such a requirement on the set. > The set is recusively defined, each member being derived from the > previous through successor(). I don't know what definition you're > looking at, or what you think it means. When the axioms say there is a set with the following properties, no construction of such a set is required. > If each element is a finite number of steps from 1, then no element > is an infinite number of steps from 1, and the set does not have an > infinite number of elements, since each element is a step. This isn't your usual quantor dyslexia, it's some other sort of > element-set dyslexia. > It's Bigulosity. Another word for garbage. > Perhaps you should look up infinite series for god's sake and > familiarize yourself with what it means for a series to diverge. > One thing it doesn't mean is that anything ever reaches an infinite > value. > What??? YOU look it up again. There are three possibilities for an > infinite series. It can converge, such that the sum of all terms is a > finite number. It can diverge in the sense that the sum over all > terms is infinite. Whereever do you find anything like this? The divergence of an infinite series of positive terms , say a_n, n in N, depends only on the behavior of the sequence of finite sums, say b_n = sum_{1<=k<=n} a_k. That divergence occurs if and only if for every finite epsilon, no matter how large, the set of b_n less than epsilon is finite. So divergence (and convergence) is here based on finiteness not on infinitness. === Subject: Re: infinity No, the definition of natural numbers does not require > that the collection be built one at a time. There is > nothing in the definition that makes such a requirement > on the set. > The set is recusively defined, No, the ELEMENTS are recursively defined. Your dyslexis is acting up again. Let's look at the axioms (as specified at Mathworld, http://mathworld.wolfram.com/PeanosAxioms.html) 1. Zero is a number. Axiom about one particular element. 2. If a is a number, the successor of a is a number. Axiom about each element: the successor of every element is an element. 3. zero is not the successor of a number. Axiom about the special element zero. 4. Two numbers of which the successors are equal are themselves equal Axiom about elements. 5. If a set S of numbers contains zero and also the successor of every number in S, then every number is in S. Only axiom that is about the *set* of naturals, and it is not recursive. It defines the set as the collection of all numbers obeying axioms 1-4. So, not surprisingly, your repeated assertion that the set is defined recursively or has to be created by growing from finite sets one element at a time, is not found anywhere among the Peano Axioms. > each member being derived from the previous > through successor(). Yes, each member is defined by the successor operation. Note the subject of your sentence: each member. That is not talking about sets but, as I said, the elements. The set is merely defined as the collection of members which obey these axioms. > If each element is a finite number of steps from 1, then no element is an > infinite number of steps from 1, Correct. We all have been saying that. > and the set does not have an infinite number > of elements, since each element is a step. Non sequitur and incorrect. The set is the collection of all numbers which can be reached in finite steps. At best it is in dispute (in this thread, not in any discussion where the rules of logic are followed) as to whether that collection is finite or infinite. Again you are getting confused, and simply declaring by faith that the collection of things at finite distance must, must, MUST be infinite (stomp, whine, gnash teeth). > This isn't your usual quantor dyslexia, it's some other > sort of element-set dyslexia. > It's Bigulosity. No, it's element-set confusion. Bigulosity is your wacky theory of a particular property of sets. But again, it's a set property, not an element property. Another manifestation of the element- set dyslexia. > Be more specific. Tell me what part of infinite series > for god's sake implies that one can reach an produce > value by a process of incrementing. > Increment means add one. Yes. And adding one to a finite thing, as we all agree, NEVER gets you an infinite thing. Never. Not ever. > If you do this an infinite number of times That involves going to the end of a process which you have agreed many times has no end. > you get sum (x=1->oo: 1) No, you do not. This sum is defined only in terms of the behavior of the finite partial sums SUM (x=1:n, 1) as n takes on increasing FINITE values. Nobody ever talks about adding infinite terms or getting to infinity. > Infinite series says that this sum diverges Indeed it does. And diverge has a very precise meaning, none of which has anything to do with getting to infinity or adding an infinite number of terms. > Therefore, if you have an infinite number > of elements, each one greater than the last, then you will have infinite values > as the result of the infinite increments. More precisely, if you'd been awake in the class where this was taught, you would realize that you don't have infinite values. Rather, you have finite values which are not bounded. > If you do not allow such infinite sums, Infinite sums are allowed but do not have the meaning you think they do. In particular... > then you must not allow infinite increments, ... this is not true. There is nothing in the definition of an infinite sum which actually implies that an infinite number of terms are being added. > Perhaps you should look up infinite series for god's > sake and familiarize yourself with what it means for > a series to diverge. One thing it doesn't mean is that > anything ever reaches an infinite value. > What??? YOU look it up again. I'm quite clear on the definitions, but I'd be happy to cite them from a text or two. Perhaps tomorrow. Are you willing to find a text bolstering your position? > There are three possibilities for an infinite > series. It can converge, Which does not mean that it ever necessarily actually reaches the value which it converges to. > such that the sum of all terms is a finite number. No, such that the limit of the sequence of partial sums is a finite number. This does not imply that that number is ever reached or that the operation sum of all terms ever actually happens. > It can diverge in the sense that the sum over all terms is > infinite. No, not in that sense. It means that there is no finite upper bound to the finite sequence of partial sums. And what THAT means, more precisely, is that for any FINITE value you pick, there will be at least one FINITE partial sum, a sum over a FINITE number of terms, which has a larger but still FINITE value. Nothing as ill-defined as the sum over all terms ever actually crops up in these definitions. > Or it can diverge in the sense that the sum is never infinite, > but has no specific limit Again this is defined by what happens with the sequence of FINITE partial sums over FINITE numbers of terms, at which they take FINITE values. Nothing like the sum of all terms ever occurs. Indeed, in your mixed-up recollection, you have bits of the actual definition since you talk about the sum is never infinite. There is of course at most one sum of the whole series (defined as a limit). If you talk about more than one value for a sum, what you are recalling in your muddled way is the behavior of the FINITE partial sums. > at n=oo. This third form is generally an oscillating sequence. The PARTIAL sums form a sequence which does not converge. Oscillation is one possibility, not the only one. The convergence criterion is quite precise, and not converge merely means the sequence doesn't match that criterion. > In this case, we > are looking at a sum which is clearly infinite. We are looking at a monotonically increasing sequence of finite values. The LIMIT is infinite because we can't find a finite upper bound, even though every element of the sequence is finite. But there's no such thing as the sum of all the terms. That terminology never actually crops up in formal discussion of series. - Randy === Subject: Re: infinity !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw >> Randy Poe said: >> But the set of all k can be Cantor-infinite even though each >> k in that set is finite. >> Yes, even though you have a finite number of finite terms, the >> sum is Cantor- infinite. Doesn't that indicate a problem with >> Cantor's definitions/ >> No. What's the problem? As you have repeatedly admitted, there is >> no end to a list of the values in such a set. Most of us have no >> problem calling a thing which is unending infinite, and have a >> great deal of trouble understanding how anybody could apply the >> adjectives finite and unending to the same thing. >> Even though there is no end to the possible values for a natural >> number, and yet you apply the label finite to natural numbers? > There is no end to the possible values. I apply the name > infinite to the possible values. Objection. All possible values are finite. Only the set of them is infinite. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: infinity David Kastrup said: >> Randy Poe said: >> But the set of all k can be Cantor-infinite even though each >> k in that set is finite. >> Yes, even though you have a finite number of finite terms, the >> sum is Cantor- infinite. Doesn't that indicate a problem with >> Cantor's definitions/ >> No. What's the problem? As you have repeatedly admitted, there is >> no end to a list of the values in such a set. Most of us have no >> problem calling a thing which is unending infinite, and have a >> great deal of trouble understanding how anybody could apply the >> adjectives finite and unending to the same thing. >> Even though there is no end to the possible values for a natural >> number, and yet you apply the label finite to natural numbers? > There is no end to the possible values. I apply the name > infinite to the possible values. > Objection. All possible values are finite. Only the set of them is > infinite. Objection??? Okay, this IS getting like court, and human law is a pain in the ass. OVERRULED!!! -- Smiles, Tony === Subject: Re: infinity > David Kastrup said: > Randy Poe said: >> But the set of all k can be Cantor-infinite even though each >> k in that set is finite. >> Yes, even though you have a finite number of finite terms, the >> sum is Cantor- infinite. Doesn't that indicate a problem with >> Cantor's definitions/ >> No. What's the problem? As you have repeatedly admitted, there is >> no end to a list of the values in such a set. Most of us have no >> problem calling a thing which is unending infinite, and have a >> great deal of trouble understanding how anybody could apply the >> adjectives finite and unending to the same thing. >> Even though there is no end to the possible values for a natural >> number, and yet you apply the label finite to natural numbers? There is no end to the possible values. I apply the name > infinite to the possible values. Objection. All possible values are finite. Only the set of them is > infinite. > Objection??? Okay, this IS getting like court, and human law is a pain in the > ass. OVERRULED!!! The Supreme Court of logic overrules TO. For any finite set of finite naturals, the sum of all the members is a finite natural, and for sets with more than 1 member, that sum is larger than every member of the set being summed, and thus not a member of the set. Thus EVERY finite set of finite naturals is incomplete, i.e., cannot contain all finite naturals. Since the set of all finite naturals exists but is not finite,... === Subject: Re: infinity Virgil said: > Virgil said: Of course, and I am not saying that a finite language is necessarily > small. A > finite language has no upper bound if the string length has no upper > bound, but > if the string length is strictly finite, then so is the language. That's a contradiction. > No, it directly follows. > Any non-empty ordered set without an upper bound is infinite, at least > in the Cantor sense, because one can easily devise an injection from it > onto a proper subset of itself. TO keeps ignoring this obvious fact in order to persist with his > multifarious mischiefs. I am not ignoring it. I have repeatedly rejected that definition, as it > contradicts any efforts to make real sense of infinite set and number theory. > Since TO has not provided anything to replace the Cantor definitions, > TO-finite and TO-infinite for sets are meaningless. > And as he also has given no definition for TO-finite or TO-infinite > for numbers, those are meaningless, too. Well, I gave this some thought, since it seems we all know what we mean by finite and infinite, and yet, you want a rigorous definition of what I mean by finite. The first thought that occurred to me was akin to Peano. We could say 1 is finite, and if a number x is finite then x-1 and x+1 are also finite. Unfortunately, this would require a caveat like the inductive proof that ALL whole numbers are finite, that we only add or subtract 1 a finite number of times, which would require the definition of finite to refer to itself. This is a problem. So, here is my definition of a finite quantity. All quantities x, such that 0 The Cantor definitions for sets are: > A set is finite if there are no injections from it to any proper subset, > and infinite if there is some such injection. > A natural number is finite if the set of all smaller natural numbers is > finite ( which means that all naturals are finite). You can't use the word finite in the definition of finite. > Until TO comes up with alternate definitions, these rule. -- Smiles, Tony === Subject: Re: infinity > So, here is my definition of a finite quantity. All quantities x, such that 0 <=1, are finite, and if x is finite, then 1/x is finite. Zero and all numbers > within zero units of zero are infinitesimal, and their inverses are infinite. > So, a finite set is one with a finite number of elements. > Can you work with that? I mean, it doesn't define exactly what finite means, > but it does define which quantities are finite. Is there a problem with this > definition? So according to you, there is some non-infinitesimal x such that 0 0 numbers within zero units of zero are infinitesimal, and their inverses > are infinite. Would these infinite quantities be countably or uncountably infinite? That is, would the values of these infinite quantities correspond to the number of natural (whole) numbers, or to something larger? > So, a finite set is one with a finite number of elements. Amazing. There's a jump from quantity to set that I must have missed above. === Subject: Re: infinity > Virgil said: Virgil said: > , Of course, and I am not saying that a finite language is > necessarily small. A finite language has no upper bound > if the string length has no upper bound, but if the > string length is strictly finite, then so is the > language. That's a contradiction. > No, it directly follows. > Any non-empty ordered set without an upper bound is infinite, > at least in the Cantor sense, because one can easily devise an > injection from it onto a proper subset of itself. TO keeps ignoring this obvious fact in order to persist with > his multifarious mischiefs. I am not ignoring it. I have repeatedly rejected that definition, > as it contradicts any efforts to make real sense of infinite set > and number theory. Since TO has not provided anything to replace the Cantor > definitions, TO-finite and TO-infinite for sets are > meaningless. And as he also has given no definition for TO-finite or > TO-infinite for numbers, those are meaningless, too. > Well, I gave this some thought, since it seems we all know what we > mean by finite and infinite, and yet, you want a rigorous definition > of what I mean by finite. The whole point is that those of us who go by the Cantor definition of finite versus infinite sets know precisely what we mean by finite sets and infinite sets, but TO has specifically said that he does not accept that definition, but has not given us any satsfactory alternative. > The first thought that occurred to me was akin to Peano. We could say > 1 is finite, and if a number x is finite then x-1 and x+1 are also > finite. Unfortunately, this would require a caveat like the inductive > proof that ALL whole numbers are finite, that we only add or subtract > 1 a finite number of times, which would require the definition of > finite to refer to itself. This is a problem. > So, here is my definition of a finite quantity. All quantities x, > such that 0 finite. Zero and all numbers within zero units of zero are > infinitesimal, and their inverses are infinite. So, a finite set is > one with a finite number of elements. > Can you work with that? I mean, it doesn't define exactly what finite > means, but it does define which quantities are finite. Is there a > problem with this definition? Yes! According to TO's definition, there are infinitesimal numbers which are zero units from zero, but such numbers violate the rules for order in the standard reals. If not x < 0 and not x > 0, then x = 0 in the standard reals, and there is no room for any numbers that are zero units from zero other than zero itself. To the best of my knowledge, that definition violates all of the nonstandard models, which do allow infinitesimals and infinites, as well. So that if TO's definition is to be workable at all, the set of infintesimals would have to be the empty set. Then, and only then, is it workable, and then it conforms to the Cantor definitions. The Cantor definitions for sets are: A set is finite if there are no injections from it to any proper > subset, and infinite if there is some such injection. A natural number is finite if the set of all smaller natural > numbers is finite ( which means that all naturals are finite). > You can't use the word finite in the definition of finite. Until TO comes up with alternate definitions, these rule. So TO's first attempt is either equivalent to the Cantor definitions or is unworkable. === Subject: Re: infinity Virgil said: > Virgil said: Virgil said: > , Of course, and I am not saying that a finite language is > necessarily small. A finite language has no upper bound > if the string length has no upper bound, but if the > string length is strictly finite, then so is the > language. That's a contradiction. > No, it directly follows. > Any non-empty ordered set without an upper bound is infinite, > at least in the Cantor sense, because one can easily devise an > injection from it onto a proper subset of itself. TO keeps ignoring this obvious fact in order to persist with > his multifarious mischiefs. I am not ignoring it. I have repeatedly rejected that definition, > as it contradicts any efforts to make real sense of infinite set > and number theory. Since TO has not provided anything to replace the Cantor > definitions, TO-finite and TO-infinite for sets are > meaningless. And as he also has given no definition for TO-finite or > TO-infinite for numbers, those are meaningless, too. > Well, I gave this some thought, since it seems we all know what we > mean by finite and infinite, and yet, you want a rigorous definition > of what I mean by finite. > The whole point is that those of us who go by the Cantor definition of > finite versus infinite sets know precisely what we mean by finite sets > and infinite sets, but TO has specifically said that he does not accept > that definition, but has not given us any satsfactory alternative. > The first thought that occurred to me was akin to Peano. We could say > 1 is finite, and if a number x is finite then x-1 and x+1 are also > finite. Unfortunately, this would require a caveat like the inductive > proof that ALL whole numbers are finite, that we only add or subtract > 1 a finite number of times, which would require the definition of > finite to refer to itself. This is a problem. So, here is my definition of a finite quantity. All quantities x, > such that 0 finite. Zero and all numbers within zero units of zero are > infinitesimal, and their inverses are infinite. So, a finite set is > one with a finite number of elements. Can you work with that? I mean, it doesn't define exactly what finite > means, but it does define which quantities are finite. Is there a > problem with this definition? > Yes! > According to TO's definition, there are infinitesimal numbers which > are zero units from zero, but such numbers violate the rules for order > in the standard reals. If not x < 0 and not x > 0, then x = 0 in the > standard reals, and there is no room for any numbers that are zero units > from zero other than zero itself. yes, I understand that part of it is non-standard, but I find it necessary. > To the best of my knowledge, that definition violates all of the > nonstandard models, which do allow infinitesimals and infinites, as well. Infinities and infinitesimals, as I understand it, are part of non-standard analysis, at least some forms of it. > So that if TO's definition is to be workable at all, the set of > infintesimals would have to be the empty set. Then, and only then, is it > workable, and then it conforms to the Cantor definitions. Well, the set of infinitesimals, as you see it, would only include zero, if you reject infinitesimals in general. I am not so sure this conforms to the Cantor definitions. Perhaps in this simple form it would, and what does not agree with Cantor relies on some other axioms. However, this forms the basis for much of my objections, at least as it feels to me right now. Yes, I use a lot of intuition in my explorations, and then check them with logic. It's induction!! The Cantor definitions for sets are: A set is finite if there are no injections from it to any proper > subset, and infinite if there is some such injection. A natural number is finite if the set of all smaller natural > numbers is finite ( which means that all naturals are finite). > You can't use the word finite in the definition of finite. Until TO comes up with alternate definitions, these rule. > So TO's first attempt is either equivalent to the Cantor definitions or > is unworkable. it's workable, and goes beyond Cantor in significant ways, I think. Not too bad for a first attempt. Did you ever try to invent your own axiom system? It's not that easy. I'll try to refine it. -- Smiles, Tony === Subject: Re: infinity > Virgil said: > According to TO's definition, there are infinitesimal numbers > which are zero units from zero, but such numbers violate the > rules for order in the standard reals. If not x < 0 and not x > 0, > then x = 0 in the standard reals, and there is no room for any > numbers that are zero units from zero other than zero itself. > yes, I understand that part of it is non-standard, but I find it > necessary. There are not even non-standard models allowing TO's latest idiocy. There is no ordering in which the same two objects can be both equal and not equal as TO proposes. To the best of my knowledge, that definition violates all of the > nonstandard models, which do allow infinitesimals and infinites, as > well. > Infinities and infinitesimals, as I understand it, are part of > non-standard analysis, at least some forms of it. But idiocies are not. Anything less that everything greater than zero is either zero or less than zero in ANY ordering. So that if TO's definition is to be workable at all, the set of > infintesimals would have to be the empty set. Then, and only then, > is it workable, and then it conforms to the Cantor definitions. > Well, the set of infinitesimals, as you see it, would only include > zero, if you reject infinitesimals in general. It is not that I reject infinitesimals in general but that TO's formulation of them makes them impossible. > I am not so sure this > conforms to the Cantor definitions. If it is to be without self-contradiction it must conform. > Perhaps in this simple form it > would, and what does not agree with Cantor relies on some other > axioms. However, this forms the basis for much of my objections, at > least as it feels to me right now. Yes, I use a lot of intuition in > my explorations, and then check them with logic. It's induction!! Considering the quality of TO's logic (quantifier dyslexia, etc,) seduction seem more appropriate than induction. The Cantor definitions for sets are: A set is finite if there are no injections from it to any > proper subset, and infinite if there is some such injection. A natural number is finite if the set of all smaller natural > numbers is finite ( which means that all naturals are finite). > You can't use the word finite in the definition of finite. Until TO comes up with alternate definitions, these rule. So TO's first attempt is either equivalent to the Cantor > definitions or is unworkable. > it's workable, and goes beyond Cantor in significant ways, I think. The whole trouble is that TO doesn't think. Either he can't or he won't or some combination thereof. But TO maintains his idiocies in the face of overwhelming evidences of his errors. That appears to me to be TO non-thinking. > Not too bad for a first attempt. Did you ever try to invent your own > axiom system? It's not that easy. I'll try to refine it. === Subject: Re: infinity And as he also has given no definition for TO-finite or TO-infinite > for numbers, those are meaningless, too. > Well, I gave this some thought, since it seems we all know what we mean by > finite and infinite, and yet, you want a rigorous definition of what I mean by > finite. Are you including yourself in we? The normal mathematical definition of infinite, as in infinite set for example, is quite clear, and give or take some very minor details, can be expressed in a number of ways. No-one (probably including yourself) is totally clear what _you_ mean by infinite, particularly in infinite number (not something that occurs in carefully worded real maths). But I've hypothesized that in fact you are simply thinking of imponderables - utterly enormous, yet perfectly (normal-math) finite numbers. I think you are going to have a *very* hard time producing a real axiomatic definition that picks out this vague intuition (which is actually totally unmathematical, like interesting numbers). > The first thought that occurred to me was akin to Peano. We could say 1 > is finite, and if a number x is finite then x-1 and x+1 are also finite. > Unfortunately, this would require a caveat like the inductive proof that ALL > whole numbers are finite, that we only add or subtract 1 a finite number of > times, which would require the definition of finite to refer to itself. This is > a problem. Right. You see the circularity problem. Remember that in real maths, it's quite easy to say if n is a pofnat so is n+1, because there is no danger that incrementing too many times will go into the imponderable zone, since we're not interested in such a thing - we want to consider *all* the pofnats. > So, here is my definition of a finite quantity. All quantities x, such that 0 <=1, are finite, and if x is finite, then 1/x is finite. Zero and all numbers > within zero units of zero are infinitesimal, and their inverses are infinite. Hmm. Any idea when the definition of infinitesimal is likely to be forthcoming? This is a no-hoper: what on earth is within zero units of zero except zero? > So, a finite set is one with a finite number of elements. > Can you work with that? I mean, it doesn't define exactly what finite means, > but it does define which quantities are finite. Is there a problem with this > definition? There certainly is. Frankly, it's more vacuous than circular. Brian Chandler http://imaginatorium.org === Subject: Re: infinity imaginatorium@despammed.com said: > Virgil said: > And as he also has given no definition for TO-finite or TO-infinite > for numbers, those are meaningless, too. > Well, I gave this some thought, since it seems we all know what we mean by > finite and infinite, and yet, you want a rigorous definition of what I mean by > finite. > Are you including yourself in we? The normal mathematical definition > of infinite, as in infinite set for example, is quite clear, and > give or take some very minor details, can be expressed in a number of > ways. > No-one (probably including yourself) is totally clear what _you_ mean > by infinite, particularly in infinite number (not something that > occurs in carefully worded real maths). But I've hypothesized that in > fact you are simply thinking of imponderables - utterly enormous, yet > perfectly (normal-math) finite numbers. I think you are going to have a > *very* hard time producing a real axiomatic definition that picks out > this vague intuition (which is actually totally unmathematical, like > interesting numbers). > The first thought that occurred to me was akin to Peano. We could say 1 > is finite, and if a number x is finite then x-1 and x+1 are also finite. > Unfortunately, this would require a caveat like the inductive proof that ALL > whole numbers are finite, that we only add or subtract 1 a finite number of > times, which would require the definition of finite to refer to itself. This is > a problem. > Right. You see the circularity problem. Remember that in real maths, > it's quite easy to say if n is a pofnat so is n+1, because there is > no danger that incrementing too many times will go into the > imponderable zone, since we're not interested in such a thing - we want > to consider *all* the pofnats. Incrementing an infinite number of times produces an infinite sum, which is why the caveat would be necessary as it is for inductive proof. > So, here is my definition of a finite quantity. All quantities x, such that 0 <=1, are finite, and if x is finite, then 1/x is finite. Zero and all numbers > within zero units of zero are infinitesimal, and their inverses are infinite. > Hmm. Any idea when the definition of infinitesimal is likely to be > forthcoming? This is a no-hoper: what on earth is within zero units of > zero except zero? Any complaint with the def of finite? No? Good. If you don't like infinitesimals, you don't have to, but zero (and the points around it) are not finite and should not be included in the finites. Here's another rule added: 1. If x>0 and x<=1, x is finite. 2. If x is finite, then 0-x is finite. 3. If x is finite, then 1/x is finite. This covers all finites, positive and negative, in the reals, does it not? Can we then say that if x is not finite and not zero, then x is infinite? Yes, let's: 4. If x is not finite, and not zero, then x is infinite. 5. If x is infinite, then 0-x is infinite. 6. If x is infinite, then 1/x is infinitesimal. 7. Zero is infinitesimal. 8. If x is infinitesimal, 0-x is infinitesimal. 9. if x is infinitesimal, then 1/x is infinite. > So, a finite set is one with a finite number of elements. > Can you work with that? I mean, it doesn't define exactly what finite means, > but it does define which quantities are finite. Is there a problem with this > definition? > There certainly is. Frankly, it's more vacuous than circular. Does it not define the finite vs. infinite numbers accurately? > Brian Chandler > http://imaginatorium.org -- Smiles, Tony === Subject: Re: infinity The Cantor definitions for sets are: > A set is finite if there are no injections from it to any proper subset, > and infinite if there is some such injection. > A natural number is finite if the set of all smaller natural numbers is > finite ( which means that all naturals are finite). > You can't use the word finite in the definition of finite. Learn to read. Finite natural number above is defined in terms of finite set. Finite set is defined in terms of injection. - Randy === Subject: Re: infinity David R Tribble said: > David R Tribble said: >> But even if N contains infinite whole numbers, you yourself said that >> there is no largest whole number, so the set cannot have a largest >> (or last) member. > No, but an infinite set CAN have a last element. Take all the infinite whole > numbers from 100...000 through 111...111. There are an infinite number of > elements, but there is a first and a last. > If that magical last whole number of N is L = 111...111, why > isn't L+1 in N? Is L+1 not a whole number? Oh, I was thinking of base 10, so that would only be 1/10 of decimal N. And, of course, any whole number, finite or infinite, can be incremented. Infinity never ends. > If you allow N to contain infinite-length whole numbers (an infinite > number of them, in fact), why can't I add one more (L+1) to the set? You can!!! > Or are you saying that there are also an infinite number of > infinite-length whole numbers that are not in N? That N can only > hold some infinite whole numbers but not all of them? N is rather arbitrary. We can have log10(N) digits in decimal for N numbers, or N digits for 10^N numbers. It depends what you're working with. -- Smiles, Tony === Subject: Re: infinity Virgil said: > imaginatorium@despammed.com said: Jiri Lebl said: > By the same token, to say that all strings in the language are finite > is to say > that there is some finite natural number K such that all characters > in any > string can be enumerated by the natural numbers {1,2,...,K}. So, let > S and T be > strings constructed by concatenting K characters: Of course this is nonsense. By another token, to say that all people > have a mother is to say that there is some mother M such that all > people can be enumerated as the children of M. So what's her name? Look, when _we_ say all strings in the language are finite we mean: (1) For all S (string in language), exists n s.t. n > len(S) Why do you insist this is the same as saying: (2) Exists K s.t. for all S (string in language), K > len(S) While claiming not to have QD? Brian Chandler > http://imaginatorium.org > Why do you not respond to the post that I was responding to, from which I > took > the very same language? Oh, that's right, he's on your side. All I did was > turn > his argument back on him, almost verbatim, so if you don't like it, say so > whether it suits your purposes or not. I swear, this is like court. > And the jury is quite prepared to convict TO of mopery, dopery, compound > illogic and quantifier dyslexia in the first degree. > TO has specifically claimed that for m and n in the set of finite > naturals N, > (a) for each n there is an m such that m > n > implies > (b) there is an m such that for each n, m > n > despite that obvious fact that (a) is true and (b) is false. > Proof that (a) is true: for each n choose m = n+1, and m > n. > Proof that (b) is false: for any M choose n = m+1, and not (m > n). Look, if I say m is aleph_0, then all naturals are less than m, right? Now, you would object that aleph_0 is not part of the set, and yet, I have already proven that every set of consecutive naturals starting at 1 includes the set size as an element, so if you have aleph_0 naturals, then aleph_0 IS a natural. You want to declare the set size? Fine, that's your m. -- Smiles, Tony === Subject: Re: infinity > Virgil said: > TO has specifically claimed that for m and n in the set of finite > naturals N, (a) for each n there is an m such that m > n implies (b) there is an m such that for each n, m > n despite that obvious fact that (a) is true and (b) is false. Proof that (a) is true: for each n choose m = n+1, and m > n. Proof that (b) is false: for any M choose n = m+1, and not (m > n). > Look, if I say m is aleph_0, Note that each m and n are specifically required to be members of the set, N, of finite naturals. Thus TO is saying that aleph_0 is a member of the set of finite naturals, in particular, TO is saying that aleph_0 is finite. But then there is certainly no need for any of those infinite numbers, that TO keeps trying to bring in. === Subject: Re: infinity Virgil said: > Virgil said: > TO has specifically claimed that for m and n in the set of finite > naturals N, (a) for each n there is an m such that m > n implies (b) there is an m such that for each n, m > n despite that obvious fact that (a) is true and (b) is false. Proof that (a) is true: for each n choose m = n+1, and m > n. Proof that (b) is false: for any M choose n = m+1, and not (m > n). Look, if I say m is aleph_0, > Note that each m and n are specifically required to be members of the > set, N, of finite naturals. > Thus TO is saying that aleph_0 is a member of the set of finite > naturals, in particular, TO is saying that aleph_0 is finite. > But then there is certainly no need for any of those infinite numbers, > that TO keeps trying to bring in. If the size of a set of all consecutive whole numbers starting at 1 is always a member of that set, and you claim your set of naturals defined this way has a size of aleph_0, then aleph_0 must be a member of your set. Is it finite? -- Smiles, Tony === Subject: Re: infinity > If the size of a set of all consecutive whole numbers starting at 1 > is always a member of that set, It is not. The set of consecutive naturals starting at 1 and BOUNDED by some finite natural may have this property, but not all sets of naturals are bounded. If the set of all finite naturals were finite, then it must contain the sum of all its members, since finite sums of finite naturals will always be finite naturals. But any finite set of two or more natural numbers will NEVER have a sum which is a member of the set. So that the claim that the set of all finite naturals is a finite set leads to a contradiction, that it cannot be the set of ALL finite natural numbers. Note that the problem disappears if the finiteness of the set is not claimed. === Subject: Re: infinity Daryl McCullough said: > Tony Orlow says... >There is no basic contradiction between those two statements, Daryl. Did you >even pay any attention to my reasons for claiming that the set of finite >naturals cannot be infinite? > Did you pay attention to the proof that it *is* infinite? > If A is any finite set of positive finite naturals, with at > least two elements, then the sum of all the elements of A > is again a positive finite natural number that is not in A. > It follows that A does *not* contain all positive finite > naturals. > Thus: no finite set contains every finite natural. > That is logically equivalent to > Any set that contains every finite natural must be an infinite set. > -- > Daryl McCullough > Ithaca, NY That proof rests on identifying a largest element, and hearkens back to the rule that any finite ordered set must have a largest element. Your reasoning is that, if it cannot have a largest element, then it cannot be finite, but I have already rejected that rule. Sorry. -- Smiles, Tony === Subject: Re: infinity > Daryl McCullough said: > Tony Orlow says... There is no basic contradiction between those two statements, >Daryl. Did you even pay any attention to my reasons for claiming >that the set of finite naturals cannot be infinite? Did you pay attention to the proof that it *is* infinite? If A is > any finite set of positive finite naturals, with at least two > elements, then the sum of all the elements of A is again a positive > finite natural number that is not in A. It follows that A does > *not* contain all positive finite naturals. Thus: no finite set contains every finite natural. That is logically equivalent to Any set that contains every finite natural must be an infinite > set. -- Daryl McCullough Ithaca, NY > That proof rests on identifying a largest element, and hearkens back > to the rule that any finite ordered set must have a largest element. Not at all. It rest only on the fact that the sum of two or more positive finite naturals is larger than any one of them, which can be proved inductively with no reference to largest elements. > Your reasoning is that, if it cannot have a largest element, then it > cannot be finite, but I have already rejected that rule. Sorry. That largest element is not just a rule, it is a theorem valid in any standard set theory which includes the equivalent of the Peano axioms and the Cantor criterion for finiteness. If TO rejects that, he rejects all set theory. Sorry, TO, but you lose! === Subject: Re: infinity Tony Orlow says... >Daryl McCullough said: >> If A is any finite set of positive finite naturals, with at >> least two elements, then the sum of all the elements of A >> is again a positive finite natural number that is not in A. >> It follows that A does *not* contain all positive finite >> naturals. >> Thus: no finite set contains every finite natural. >> That is logically equivalent to >> Any set that contains every finite natural must be an infinite set. >That proof rests on identifying a largest element, No, it doesn't. It rests on the fact that if you add up all the elements of a set of finite positive numbers, then you get a number that is larger than any element in the set. -- Daryl McCullough Ithaca, NY === Subject: Re: infinity Daryl McCullough said: >Daryl McCullough said: >>stephen@nomail.com said: > I have no idea what L is in your S^L. You are aware that there > is more than one string length, so picking a single L does not > make any sense. It almost makes sense if you think that L is > the maximum string length, i.e. the largest finite natural number. > Of course you also deny that there is a maximum string > length, so I have no idea what S^L is supposed to mean. >>Given any string length and alphabet, that is the maximum number of unique >>srings in the language. >> What is L? >L is a string length. > The length of *which* string? '0' has length 1. '00' has length 2. > '000' has length 3. '0000' has length 4. If U = the set of all > finite sequences of 0s and 1s, then what is L? >With an alphabet of size S, we can make S^L strings of >length L. > You haven't said what L is. Is L a constant? Is it a variable? > Is it a natural number? What is L? >> No, what we are saying is that if U is a finite set of finite >> that, it follows that U did not contain all finite strings. >The contradiction comes from assuming you have defined the full set of all >finite strings. > Didn't you claim that the set of all finite naturals is a finite set? >This is exactly the largest finite argument, which of course >causes contradictions > That's right: If you assume that the set of all finite numbers > is a finite set, that leads to a contradiction: > Let A = any finite set of finite naturals that includes 1 and 2. > Let m = the sum of all elements of A. > Then m is larger than any element of A. > Then m is not an element of A. > Then there exists a finite natural number that is not in A. > What that proves is > If A is a finite set of finite naturals, then A does not > contain all finite naturals. > So the collection of all finite naturals cannot be a finite set. QED Of course, given any finite number we can name a larger one. All this proves is that you cannot name a largest one, or sum of all of them, without causing a contradiction. >because you can always add 1. So what?? > So the collection of all finite naturals is not a finite set, > contrary to your claims. Then how do you explain that we have a set of natural numbers, each occupying 1 unit of the number line, and we have an infinite number, but no two are more than a finite distance apart? How do you fit an infinite number of unit dustances in a finite space? >> Yes. The sum over all finite values of L of S^L is infinite. >So, you can add finite sets of strings to finite sets of strings in >succession, and get an infinite result? > To say that a sum of positive terms t_j is infinite is to say that > given any finite real number R, there exists a finite natural number > n(R) (note: n is a *function* of R) such that > t_0 + t_1 + ... + t_n(R) > R > That's clearly the case for the sum > S^0 + S^1 + ... > Given any finite real number R, we can find a number n(R) such that > S^0 + S^1 + ... + S_n(R) > R So what? Given any finite number, you can find a larger one. What does that prove? That the larger one is infinite? Did you find the largest finite, add one, and get an infinite? No. That proves nothing. Being larger than a finite does not make the sum infinite. >But you can't add a finite number to a number in >succession and get an infinite number? Hmmmm... > The numbers grow without bound, but each number is finite. Likewise, then, you should believe that the set grows without bound, but never becomes infinite. After all, you are adding one element at a time to a finite set. How can it ever achieve infinity? >Proof: The set of all finite strings on a finite alphabet is finite. >L=0: Given N=S^L, there is S^0=1 string, the null string. >L->L+1: Given a finite set of strings of length L or less, we add the set of >strings of length L+1, which has S^(L+1) elements. Since L is finite, L+1 is >finite, and S^(L+1) is finite. So we add this finite number of strings to the >finite number of strings of length L or less, to get the number of strings of >length L+1 or less. A finite plus a finite is finite. >Therefore: The set of all finite strings on a finite alphabet is finite. > That doesn't follow. What follows from your inductive proof is this: > forall L, the set of all strings of length less than or equal to L > is finite > If we let f(L) be the number of strings of length L or less, then > you can prove by induction: > forall L, f(L) is finite. > To prove that the set of *all* finite strings is finite, you need to > show something different: > exists K, forall L, f(L) < K No, that is not required at all. It suffices to say that it is true for all n in N that, if n is finite, then the number of strings less than or equal to n in length is finite. In other words, there is no n in N for which the set of all strings up to and including that length is infinite. sum(x=0->L: S^x) is finite for finite S and L. After all, if you want to prove all n in N are finite do you need to show that exists K, forall n, n> It can't be finite, because for any finite set of strings, there >> is a finite string not in that set. >For any finite set that you define, you can define a larger one, > It doesn't matter whether I can define it or not. If U is > a finite set of finite strings, then the result of concatenating > all the strings in U is a finite string that is not in U. So? You never showed in any way that U was the set of all finite strings. There is no way to have such a set in any definite way. >but you cannot define the entire set of naturals this way. > That's because it's *not* a finite set. The set deifnition is recursive and therefore generates an infinite set, but the values become infinite too. The set of finites, which has no discernible end, is nevertheless finite. The limit on its quantities is the limit on its size. >If you say the set size is infinite because it goes on forever > That's what *you* have said. You've said that infinite means without end. >then I can say the values in the set are infinite for the same reason. > In the case where A = the set of all *finite* naturals, then > by definition of A, no element of A is infinite. You aren't making any > sense. And in that case, the set is finite. You see no end to the set, and no end to the values. You say the values cannot go on forever, so neither can the set. There is no discernible end to the set, but that's because there is no discernible end to the finite numbers. >If you say you can add the sizes of finite sets, and >somehow get to infinity, > I said that if you have a *finite* set of finite naturals, > then you can add all the elements of the set and get a *finite* > result. No, you said that S^0+S^1+S^2+..... would give an infinite result for the size of the language, which implies that adding finite numbers to finite numbers can create infinite sums. But, that contradicts the inductive proof that all naturals are finite. Each time you are adding a natural number as a term, and you are adding a natural number of terms. Any natural times any natural is a natural, if they are all finite. >> There is no finite upper bound on the lengths of finite strings. >Then there is no reason to conclude that they are all finite. > Are you saying that there might be a finite string that is infinite? No, I am saying that there is no reason to exclude infinite strings, expecially when the formula for the number of strings given the number of characters in each string shows that the infinite language requires them. > -- > Daryl McCullough > Ithaca, NY -- Smiles, Tony === Subject: Re: infinity > So the collection of all finite naturals is not a finite set, > contrary to your claims. > Then how do you explain that we have a set of natural numbers, each > occupying 1 unit of the number line, and we have an infinite number, > but no two are more than a finite distance apart? How do you fit an > infinite number of unit dustances in a finite space? The space is not finite unless there is a last number, as has been shown repeatedly. So TO's wits are wandering. Given any finite real number R, we can find a number n(R) such that S^0 + S^1 + ... + S_n(R) > R > So what? That proves the diameter of the set is infinite. > Given any finite number, you can find a larger one. What > does that prove? That the set of numbers under consideration is not a finite set. The numbers grow without bound, but each number is finite. > Likewise, then, you should believe that the set grows without bound, > but never becomes infinite. The set that is without finite bound on its members is infinite but its members need not be. TO does not seem to be able to tell the difference between the properties of a set and the properties of its members. > After all, you are adding one element at > a time to a finite set. How can it ever achieve infinity? It is given to us already infinite by the axioms. The set does not have to achieve anything it does not start with. > No, that is not required at all. It suffices to say that it is true > for all n in N that, if n is finite, then the number of strings less > than or equal to n in length is finite. In other words, there is no n > in N for which the set of all strings up to and including that length > is infinite. But if one puts no such bound on struing length, the set of all finite strings of finite length but with no finite bound on those lengths, is NOT a finite set. I said that if you have a *finite* set of finite naturals, then you > can add all the elements of the set and get a *finite* result. > No, you said that S^0+S^1+S^2+..... would give an infinite result for > the size of the language, which implies that adding finite numbers to > finite numbers can create infinite sums. But, that contradicts the > inductive proof that all naturals are finite. NO! It just means that an infinite series of positive naturals diverges. > No, I am saying that there is no reason to exclude infinite strings Nor is there any reason to require them, as an infinite set of strings can be achieved without having any infinite strings in it at all. === Subject: Re: infinity >Daryl McCullough said: >> Let A = any finite set of finite naturals that includes 1 and 2. >> Let m = the sum of all elements of A. >> Then m is larger than any element of A. >> Then m is not an element of A. >> Then there exists a finite natural number that is not in A. >> What that proves is >> If A is a finite set of finite naturals, then A does not >> contain all finite naturals. >> So the collection of all finite naturals cannot be a finite set. QED >Of course, given any finite number we can name a larger one. That isn't what I said. I said that if you add up all the elements in a finite set of nonnegative numbers (containing at least 1 and 2), you get a new number that is greater than any element in the set. -- Daryl McCullough Ithaca, NY === Subject: Re: infinity Daryl McCullough said: >Daryl McCullough said: >> Let A = any finite set of finite naturals that includes 1 and 2. >> Let m = the sum of all elements of A. >> Then m is larger than any element of A. >> Then m is not an element of A. >> Then there exists a finite natural number that is not in A. >> What that proves is >> If A is a finite set of finite naturals, then A does not >> contain all finite naturals. >> So the collection of all finite naturals cannot be a finite set. QED >Of course, given any finite number we can name a larger one. > That isn't what I said. I said that if you add up all the > elements in a finite set of nonnegative numbers (containing > at least 1 and 2), you get a new number that is greater than > any element in the set. > -- > Daryl McCullough > Ithaca, NY And, how can you add up ALL the finite numbers? Where did you stop? Obviously, before you got to F, so obviously you DIDN'T add up all the finite numbers. That's the crux of your contradiction. There is no largest finite, therefore there is no sum of all finites. -- Smiles, Tony === Subject: Re: infinity > And, how can you add up ALL the finite numbers? If there are only finitely many, start with a sum of zero then add one at a time to the previous sum. > Where did you stop? When all of TO's finitely many finite naturals have all been added in. Of course, the result is of necessity a finite natural number not in TO's set which supposedly contains all of them. === Subject: Re: infinity >> That isn't what I said. I said that if you add up all the >> elements in a finite set of nonnegative numbers (containing >> at least 1 and 2), you get a new number that is greater than >> any element in the set. >And, how can you add up ALL the finite numbers? Where did you stop? So you are saying that some sets of natural numbers are so large that it is impossible to add up all the elements in the set. Those sets have no end. Normal people call such sets infinite sets. But for some reason, you want to call them finite. In that case, we need a new word to mean to mean a finite set that is so large that you can never get to the end of it. How about unbounded? Is that a good word? If not, pick a different word. In that case, it's a matter of terminology: what you call an unbounded set, everyone else calls an infinite set. So, we have: The collection of all finite naturals is an unbounded set. The collection of all strings of finite length is an unbounded set. -- Daryl McCullough Ithaca, NY === Subject: Re: infinity Daryl McCullough said: >> That isn't what I said. I said that if you add up all the >> elements in a finite set of nonnegative numbers (containing >> at least 1 and 2), you get a new number that is greater than >> any element in the set. >And, how can you add up ALL the finite numbers? Where did you stop? > So you are saying that some sets of natural numbers are so large > that it is impossible to add up all the elements in the set. Those > sets have no end. > Normal people call such sets infinite sets. But for some reason, > you want to call them finite. In that case, we need a new word to > mean to mean a finite set that is so large that you can never get > to the end of it. How about unbounded? Is that a good word? > If not, pick a different word. Yes, the finite naturals, and the set of finite strings on a finite alphabet, are both unbounded, but necessarily finite given the restrictions of finitude on the elements. We can never discuss anything fruitfully about infinite sets without discussing properties of the elements in the set. If you specify additional properties of the elements, then you are possibly changing the size of the set. If you limit naturals and strings to finite values and lengths, then you restrict the size of the set accordingly. > In that case, it's a matter of terminology: what you call an > unbounded set, everyone else calls an infinite set. Well, it seems that what people call countably infinite is really unbounded, since every instance of those types of sets seems to be restricted to finite elements. Your uncountably infinite set seem to be actually infinite, such as the reals in [0,1]. Even your set of rationals is limited to finite numerators and denominators, so the set is ultimately finite, although unbounded. There really is no reason, however, why you can't have a truly infinite set of whole numbers or strings. You just have to allow each element to be infinite. > So, we have: > The collection of all finite naturals is an unbounded set. > The collection of all strings of finite length is an unbounded set. That's pretty much it! But, the set of all whole numbers is infinite, and the set of all strings on a finite alphabet is infinite, if those sets are allowed to include infinite members. > -- > Daryl McCullough > Ithaca, NY -- Smiles, Tony === Subject: Re: infinity > Yes, the finite naturals, and the set of finite strings on a finite alphabet, > are both unbounded, but necessarily finite given the restrictions of finitude > on the elements. The sum of any finite set of two or more finite natural numbers is a finite natural number larger than any member of the set. If, as TO claims, the set of finite naural numbers were actuallly finite, the sum of its members must be a finite natural number larger than every finite natural number, including itself, of course. So the set of all finite natural numbers cannot be finite in any contradiction-free system. The sum of any finite set of two or more finite natural numbers is a finite natural number larger than any member of the set. If, as TO claims, the set of finite naural numbers were actuallly finite, the sum of its members must be a finite natural number larger than every finite natural number, including itself, of course. So the set of all finite natural numbers cannot be finite in any contradiction-free system. === Subject: Re: infinity Randy Poe said: > Daryl McCullough said: > K is not a *constant*, it is a variable. It doesn't make any sense > to say S^K is the maximum size of the language. > Yes it does. Say I have 26 letters in my alphabet. I know that the longest word > in my language is 23 letters long. Then I know, for sure, that there cannot be > more than 26^23 words in the language. > You mean if I define A = the set of finite strings then > you know for sure that A has no strings longer than > 23 letters? > What in that definition rules out 24-letter strings? > If I define A = the set of finite strings, can you tell > me what value of K you will use to say for sure that > there are only S^K words in A? No, I am saying that given any finite limit on the lengths of strings, you have a finite limit on the number of strings. If all strings are finite, then the set of strings is finite. > Why can't strings be infinite if there is no upper bound on their lengths? > This, one of your favorite distractions, is just that. Yes > there *can* be infinite strings. But discussing them > adds nothing to the discussion of whether the set of > finite strings is finite. It certainly does. Any finite limit on the lengths of strings results in a finite limit on the number of strings. You don't state a specific limit, but to say they are finite is to say the set of strings is also. > You seem to not want infinite strings, like everyone rejected infinite numbers, > but that's not really the issue here is it? The problem is that you seem to > think you can declare all of the strings to be finite, and accept that the > number of strings possible increases with the maximum length of the string > according to sum(x=1->k:S^x), and that all k in N are finite, but that you can > still have a sum of these finite terms be infinite, because you believe you can > have an infinite set of finite whole numbers. > That's all reasonably accurate. > Do you think the sum 1+2+3+... over the finite numbers is > a finite value, say F? sum(x=1->k:x)=(k^2+k)/2, which is finite for any finite k. Do you think there is a finite k s.t. (k^2+k)/2 is infinite? > If F is finite, isn't it already in the sum 1+2+3+...? Sure, and 1+2+3+...+F is finite also. You can keep increasing finite values with no discernible end. So what? > Okay, say you only have one character in your alphabet. Can you have an > infinite number of them, and have them all be finite? Doesn't the 100th have > 100 characaters, the millionth have a million, and the ooth have oo? > There is no ooth. Do you have an infinite number of them or not? Are there not balls with an an infinite number of predecessors? When you have an infinite number of balls, what do the labels look like? > There are an infinite number of words. You have said yourself > that the process of adding one letter to a finite word, > to make another finite word, never ends. Sure, it never ends, and if it goes on forever, then you have elongated your words forever and they have become infinitely long. > Do you understand that this means the list of finite words > doesn't stop? It has no discernible end, but the finiteness of the lengths of the words implies finiteness for the set of words. > Do you understand that if F is any finite number, an unending > list has more than F elements in it? Of course, but the indexes on the elements of an infinite set include infinite indexes. Each natural can be thought of as its own index, its own position, in the set of naturals. > - Randy -- Smiles, Tony === Subject: Re: infinity > Randy Poe said: > If I define A = the set of finite strings, can you tell me what > value of K you will use to say for sure that there are only S^K > words in A? > No, I am saying that given any finite limit on the lengths of > strings, you have a finite limit on the number of strings. The above is correct, but what follows does not follow. > If all strings are finite, then the set of strings is finite. The set of lengths of all finite strings does not impose any finite limit on the allowable length of finite strings. Since that set has no finite upper bound on its members, it is infinite. > There is no ooth. > Do you have an infinite number of them or not? Are there not balls > with an an infinite number of predecessors? No! Every natural, and the ball with which it is labeled, has finitely many predecessors, but there is no finite limit on how many predecessors. > When you have an infinite > number of balls, what do the labels look like? The members of the infinite set of finite naturals. TO just cannot allow that an endless sequence is just that, endless. TO has to stick something irrelevant on the end of it. === Subject: Re: infinity Do you think the sum 1+2+3+... over the finite numbers is > a finite value, say F? > sum(x=1->k:x)=(k^2+k)/2, which is finite for any finite k. Do you think there > is a finite k s.t. (k^2+k)/2 is infinite? Nope. Your turn. Do you think there's a finite value of k such that sum(x=1->k:x) represents a sum of all the natural numbers? What value of k is that, if so? > There is no ooth. > Do you have an infinite number of them or not? Yes, one for each natural number. As you must know by now, I and virtually everyone else in this world who's ever taken a college-level math course understands that there are an infinite number of natural numbers, and none of them has the value oo. > Are there not balls with an an > infinite number of predecessors? There are not. > When you have an infinite number of balls, > what do the labels look like? Each one of them looks like a finite natural number, with a finite number of digits and a ball next to it with a higher number, also finite. > There are an infinite number of words. You have said yourself > that the process of adding one letter to a finite word, > to make another finite word, never ends. > Sure, it never ends, and if it goes on forever, then you have elongated your > words forever Oops. Got confused between the sets and the elements again. No, you have no elongated any particular word, either forever or by a single character. Every time you add a character, your last (shorter) word is still there and you have a new word, which is also finite. The process of getting a new finite word goes on forever. I can never stop getting a new finite word. You also did your think of agreeing a process has no end and then talking about what happens at the end. I haven't elongated any word forever at the end of my infinite process because there is no end to the infinite process. - Randy === Subject: Re: infinity imaginatorium@despammed.com said: > Daryl McCullough said: > Tony Orlow says... >David R Tribble said: > and therefore have an infinite >difference in value. If two values have an infinite difference, There are no two values that have an infinite difference. > So, in your infinitely long line of units from 1 to N, no two points are > infinitely far apart? Just close your eyes and picture what you're talking > about,... > I suppose you can do this, Tony? Do tell us what the picture looks > like? Does your mind's eye really reach from the left end (the one > that's there) to the right end (the one that isn't)? I picture each point being labeled with a number that denotes its distance from zero. Either there are infinite numbers infinitely far from zero, or there aren't. All finite numbers are a finite distance from zero. If you want an infinite number of numbers you need infinite distance, since they each occupy one unit of dustance on the number line. > Brian Chandler > http://imaginatorium.org -- Smiles, Tony === Subject: Re: infinity > imaginatorium@despammed.com said: Daryl McCullough said: > Tony Orlow says... >David R Tribble said: >and therefore have an infinite >difference in value. If two values have an infinite difference, There are no two values that have an infinite difference. > So, in your infinitely long line of units from 1 to N, no two points are > infinitely far apart? Just close your eyes and picture what you're > talking > about,... I suppose you can do this, Tony? Do tell us what the picture looks > like? Does your mind's eye really reach from the left end (the one > that's there) to the right end (the one that isn't)? > I picture each point being labeled with a number that denotes its distance > from > zero. Either there are infinite numbers infinitely far from zero, or there > aren't. All finite numbers are a finite distance from zero. Can TO establish any finite upper bound on those finite numbers? If not then the set of such finite numbers has, by definition, infinite diameter. > If you want an infinite number of numbers you need infinite distance One needs an infinite diameter to the set of natural numbers but that does not require anything more than a lack of any finite upper bound on distances between natural numbers. And that, as has been adequately proved time and again, follows from the properties of the set of finite natural numbers. === Subject: Re: infinity Virgil said: 1) The difference between numbers A and B is abs(A-B) 2) For finite A and B, abs(A-B) is finite. 3) If all elements are finite, then any two of them is finite. 4) If a number is finite, then it is not infinite. I can't imagine WHICH of these you could possibly argue with. Which is it? > It is the assumptions that TO has carefuly NOT stated that we disagree > with. > One of which is that non-empty ordered finite sets can exist without > largest elements. > Such sets are always Cantor-infinite. > TO has yet to explain how Cantor-infinite sets can be TO-finite. > It is an issue that he has been carefully avoiding for some time now. Well, I have certainly explained my reasoning regarding your infinite sets of finite whole numbers and finite strings on finite alphabets, and why I don't consider them infinite. I don't have a definition based in surjections. I'll have to think about exactly how to define finiteness in my scheme. But my reasoning is not based on anything with which you seem to disagree, so I don't know why you argue with it. -- Smiles, Tony === Subject: Re: infinity > Virgil said: > It is the assumptions that TO has carefuly NOT stated that we > disagree with. One of which is that non-empty ordered finite sets can exist > without largest elements. Such sets are always Cantor-infinite. TO has yet to explain how Cantor-infinite sets can be TO-finite. It is an issue that he has been carefully avoiding for some time > now. Well, I have certainly explained my reasoning regarding your infinite > sets of finite whole numbers and finite strings on finite alphabets, > and why I don't consider them infinite. We undeersatnd WHAT it is that you believe, but cannot understand WHY it is that you believe it in the face of so many sound proofs of its being in error. > I don't have a definition based in surjections. I'll have to think > about exactly how to define finiteness in my scheme. Without some adequate and precise definition of finiteness versus infiniteness of sets, easily applied to arbitrary sets, how can TO be so sure that his notions of it are correct? TO is merely handwaving in the absence of such definitions, since no proofs of anything are possible without such definitions. With our definition (the Cantor criterion), which is both adequate and precise, we can prove our claims and TO cannot do anything in opposition butr wave his hands in oppposition. > But my reasoning is not based on anything with which you seem to > disagree, so I don't know why you argue with it. That TO does not understand what parts of his delusions we disagree with only illustrates his inability to read what has been clearly stated repeatedly. We argue with it because, given the set theories that we are working with, TO is WRONG! === Subject: Re: infinity Daryl McCullough said: > Tony Orlow says... >Daryl McCullough said: >> Use quantifiers! That clarifies things. To say an infinite >> number of whole numbers must have an infinite range means >> (Let C be an infinite set of natural numbers) >> forall finite ranges r, exists x in C, exists y in C >> |x-y| > r >> You keep getting this confused with the statement >> exists x in C, exists y in C, forall finite ranges r, >> |x-y| > r >> The first statement says that the range of C is infinite. >The first statement says that for every finite number, there is a pair of >numbers that differs by more than that amount. > That's what people mean when they say that the range is infinite. > It means that there is no *finite* range that includes all the > elements. >That doesn't necessarily mean the range of the set is infinite. > Yes, it does. No, it doesn't. For any finite number, we can name a larger finite number, right? So, for any finite distance, we can name a greater distance, but that doesn't mean that greater distance is infinite. >It could contain only finites, but finites >never have an end, and you can always find one greater than any >you specify. > That's what people mean when they say that the range is infinite: > given any finite bound, there exists elements outside that bound. > That's why people say that there are infinitely many finite numbers. But, to say that you can always find a bigger finite number doesn't mean that that bigger number is infinite. >No, my logic is not a matter of switching quantifiers, but applying >geometrical understanding of the set, using infinite series, >combinatorics, and other math. > Well, I'm familiar with all of that math, too. It doesn't imply > that there are infinite naturals. It doesn't imply that there > are only finitely many finite naturals. It doesn't imply that > every infinite set of naturals must contain a natural that is > infinitely big. Yes it does. >Not everything is best discussed axiomatically. > You can leave off the formality of quantifiers if it is clear > to everyone what you are talking about. If there is confusion > about what you are talking about, then using quantifiers is the > way to clear up that confusion. That's why quantifiers were > *invented*, to clear up exactly the sort of ambiguity that > appears in natural language when people talk about all > some, etc. >I am trying to get you to picutre all the increments of Peano's >lined up as unit intervals, and tell me if what you see, an >infinite string of units, is really finitely long, or >whether there have to be some infinite distances between some >pairs of points. > If you take the entire real number line and divide it up into > unit intervals, and label each interval with a natural number, > then (1) There are infinitely many intervals, and (2) For any > two pairs of points, the distance between them is finite. So, the distance between any two points on the infinite real line is finite? Come on! You can't possibly believe that! >> No, there won't be. The statement there are points that are infinitely >> far apart is formalized as >Jesus! Try picturing things in your head, > In the picture in my head, there are infinitely many intervals, > but the distance between any two points is finite. You can't picture two points infinitely far apart on an infinite line? What is that line made of besides points, and what makes it infinite besides distance? This is nonsense. > Look, take your mental picture of the set of all intervals. Now, > imagine *removing* all points that are infinitely far away from > the points in interval number 0. What do you have left? An set > of intervals without end such that no two points are infinitely > far apart. That's what *I* picture in my head when I imagine > the naturals as intervals. Yes, but if no two points are infinitely far apart, then you have an overall finite distance, and if each number occupies 1 unit of space on that line, then you can only fit a finite number of them into that finite distance. This is exactly why I say there are a finite number of finite whole numbers, because there can only be a finite number of them between any other two. > -- > Daryl McCullough > Ithaca, NY -- Smiles, Tony === Subject: Re: infinity > Daryl McCullough said: > Tony Orlow says... Daryl McCullough said: > Use quantifiers! That clarifies things. To say an infinite >> number of whole numbers must have an infinite range means >> (Let C be an infinite set of natural numbers) >> forall finite ranges r, exists x in C, exists y in C >> |x-y| > r >> You keep getting this confused with the statement >> exists x in C, exists y in C, forall finite ranges r, >> |x-y| > r >> The first statement says that the range of C is infinite. >The first statement says that for every finite number, there is a pair of >numbers that differs by more than that amount. That's what people mean when they say that the range is infinite. > It means that there is no *finite* range that includes all the > elements. That doesn't necessarily mean the range of the set is infinite. Yes, it does. > No, it doesn't. For any finite number, we can name a larger finite number, > right? So, for any finite distance, we can name a greater distance, but that > doesn't mean that greater distance is infinite. It could contain only finites, but finites >never have an end, and you can always find one greater than any >you specify. That's what people mean when they say that the range is infinite: > given any finite bound, there exists elements outside that bound. > That's why people say that there are infinitely many finite numbers. > But, to say that you can always find a bigger finite number doesn't mean that > that bigger number is infinite. But to have an infinite diameter to a set of numbers one doesn't need infinitely distant numbers, one only neneds a lack of any finite upper bound on distances. If range requires infinite distsnces between numbers, then that are sets with infinite diameters but undefined ranges. Well, I'm familiar with all of that math, too. It doesn't imply > that there are infinite naturals. It doesn't imply that there > are only finitely many finite naturals. It doesn't imply that > every infinite set of naturals must contain a natural that is > infinitely big. > Yes it does. Not in any standard set theory. Only in such slef-contradictory theories as TO is trying to promulgate. Not everything is best discussed axiomatically. You can leave off the formality of quantifiers if it is clear > to everyone what you are talking about. If there is confusion > about what you are talking about, then using quantifiers is the > way to clear up that confusion. That's why quantifiers were > *invented*, to clear up exactly the sort of ambiguity that > appears in natural language when people talk about all > some, etc. I am trying to get you to picutre all the increments of Peano's >lined up as unit intervals, and tell me if what you see, an >infinite string of units, is really finitely long, or >whether there have to be some infinite distances between some >pairs of points. If you take the entire real number line and divide it up into > unit intervals, and label each interval with a natural number, > then (1) There are infinitely many intervals, and (2) For any > two pairs of points, the distance between them is finite. > So, the distance between any two points on the infinite real line is finite? > Come on! You can't possibly believe that! WE can't possibly believe anything else. What TO can believe would shame the Red Queen. > In the picture in my head, there are infinitely many intervals, > but the distance between any two points is finite. > You can't picture two points infinitely far apart on an infinite line? NO! > What is that line made of besides points, and what makes it infinite > besides distance? What makes it infinite is not that any points are actually infinitely far apart, but that there is no finite limit to how far apart points can be. Look, take your mental picture of the set of all intervals. Now, > imagine *removing* all points that are infinitely far away from > the points in interval number 0. What do you have left? An set > of intervals without end such that no two points are infinitely > far apart. That's what *I* picture in my head when I imagine > the naturals as intervals. > Yes, but if no two points are infinitely far apart, then you have an overall > finite distance I still have infinitely many finite distances, and the set of those distances has infinite diameter though it does not have any range at all. > and if each number occupies 1 unit of space on that line, > then > you can only fit a finite number of them into that finite distance. Which finite distance? TO goes on as if there were only one largest finite distance instead of infinitely many of finite distances of unboundedly increasing size. This is > exactly why I say there are a finite number of finite whole numbers, because > there can only be a finite number of them between any other two. But if there are only finitely many of them then there must be a largest one (since non-existence of a largest finite natural implies that there are infinitely many of them). === Subject: Re: infinity Tony Orlow says... >Daryl McCullough said: >> That's what people mean when they say that the range is infinite. >> It means that there is no *finite* range that includes all the >> elements. >>That doesn't necessarily mean the range of the set is infinite. >> Yes, it does. >No, it doesn't. For any finite number, we can name a larger finite number, >right? Right. So the set of all finite numbers is an infinite set, and thus has an infinite range. -- Daryl McCullough Ithaca, NY === Subject: Re: infinity > Tony Orlow says... >Daryl McCullough said: >> That's what people mean when they say that the range is infinite. >> It means that there is no *finite* range that includes all the >> elements. >>That doesn't necessarily mean the range of the set is infinite. >> Yes, it does. >No, it doesn't. For any finite number, we can name a larger finite number, >right? > Right. So the set of all finite numbers is an infinite set, > and thus has an infinite range. Remark on definitions: The RANGE of a set of numbers is, according to TO, the maximum difference between members, even when no such maximum exists. Which means that a range will not exist when no actual maximum exists. The diameter of a set of numbers (or set of points in an arbitrary metric space) is the least upper bound of the distances between members, if there is a finite least upper bound, and is otherwise said to be infinite. Which means that a diameter always exists for such sets of points, whether or not ther is any maximum distance between points. Where both exist, they are equal, but in many of the critical places where TO tries to use ranges, ranges, as least as TO defines them, do not exist. === Subject: Re: infinity imaginatorium@despammed.com said: > Daryl McCullough said: > Tony Orlow says: > Except that, as you just pointed out, for any set of consecutive naturals > starting from 1, the set size IS an element of the set. Therefore, if the set > is infinite, then it contains an infinite element. This is a contradiction. So > how do you resolve this? > Very simply. Extremely simply, though obviously beyond you. We observe > that any element of the set that is equal to the set size must be the > largest element in the set. We deduce that in a set with no largest > element, there is no largest element that could be equal to the set > size. Okay, then what is the size of the set? The proof shows also that the set size IS a member of the set. So, if the set size is infinite, then how can that number NOT be part of the set of whole numbers starting from 1? For any such set, the size IS an element. Do you disagree? > This is what I am trying to show you. > No, you can't _show_ us anything, mathematically speaking, because you > have simply no idea at all what a mathematical argument is. Do tell me: > consider a binary tree in which there are only branching nodes - how > many leaf nodes are there? Use Tinduction, or any other favourite tool. What is the purpose of this question? The obvious answer, using plain logic, is that a tree with no leaf nodes has zero leaf nodes. However, an infinite binary tree really has infinite numbers of leaf nodes. For level n, starting with zero at the root node, there are 2^n nodes. If a tree is x levels deep, it has 2^x leaf nodes. If x is infinite, then 2^x is infinite as well. By pretending that somehow at x=oo all the leaf nodes disappear is inconsistent. But of course, you will say that there are infinite levels so we never get any leaf nodes. I am not sure how you reconcile this with the inductive argument that the number of leaf nodes of a tree with n levels is 2^n. > Brian Chandler > http://imaginatorium.org -- Smiles, Tony === Subject: Re: infinity > What is the purpose of this question? The obvious answer, using plain > logic, is that a tree with no leaf nodes has zero leaf nodes. > However, an infinite binary tree really has infinite numbers of leaf > nodes. And the obvious answer is, as so often happens, wrong. A *maximal* binary tree is one for which NO path has a leaf node, since if there were any leaf nodes, the paths of which they are ends would not be maximal, but could be extended further. Every path must be infinite in length with no last node. === Subject: Re: infinity Except that, as you just pointed out, for any set of consecutive naturals > starting from 1, the set size IS an element of the set. Therefore, if the set > is infinite, then it contains an infinite element. This is a contradiction. So > how do you resolve this? > Very simply. Extremely simply, though obviously beyond you. We observe > that any element of the set that is equal to the set size must be the > largest element in the set. We deduce that in a set with no largest > element, there is no largest element that could be equal to the set > size. > Okay, then what is the size of the set? The proof shows also that the set size > IS a member of the set. I missed this bit of the proof. _How_ does it prove that the set size IS a member of the set? (Yeah, yeah, I guess because for any finite sequence, the maximum member exists and is equal to the set size. In the limit, the maximum member doesn't exist, but somehow must be equal to the set size, so while nonexistent, must creep back into the set disguised as an 'infinite number'. Ho hum.) Anyway, if you care to show this bit of the proof *carefully*, you might like in parallel to provide a proof that sqrt(2) is rational (it's the limit of a sequence, every member of which is rational), and show the difference - why your proof of the set membership of something which doesn't on the face of it exist is valid, while the squrt(2) proof is (I suppose you might agree?) invalid. > So, if the set size is infinite, then how can that > number NOT be part of the set of whole numbers starting from 1? For any such > set, the size IS an element. Do you disagree? Obviously I disagree. We know that for any not-quite-such set (i.e. any *finite* initial segment of the naturals) the largest: (a) exists, (b) is a member, and (c) equals the set size. We also can see that for the complete set of naturals, this largest (a) does not exist, and therefore (b) is not a member, so everyone except you is baffled as to how it can (c) equal the set size. > Do tell me: > consider a binary tree in which there are only branching nodes - how > many leaf nodes are there? Use Tinduction, or any other favourite tool. > What is the purpose of this question? The obvious answer, using plain logic, is > that a tree with no leaf nodes has zero leaf nodes. Good. Sounds about right. > However, an infinite binary > tree really has infinite numbers of leaf nodes. Hmm, and this contradiction doesn't bother you much? You don't think that if something has zero leaf nodes, it can't really have an infinite number of them as well? For level n, starting with zero > at the root node, there are 2^n nodes. If a tree is x levels deep, it has 2^x > leaf nodes. If x is infinite, then 2^x is infinite as well. By pretending that > somehow at x=oo all the leaf nodes disappear is inconsistent. Of course. No mathematician thinks that at infinity anything disappears. You are stuck with your maundering finite imponderable numbers, which is where you think things reach infinity. Mathematicians, though, say unending meaning, er, really, really, really, not having an end. No end. Just endless. Not even an end labelled infinity. But of course, > you will say that there are infinite levels so we never get any leaf nodes. I > am not sure how you reconcile this with the inductive argument that the number > of leaf nodes of a tree with n levels is 2^n. At least you seem to see there is a problem. Brian Chandler http://imaginatorium.org === Subject: Re: infinity imaginatorium@despammed.com said: > imaginatorium@despammed.com said: Except that, as you just pointed out, for any set of consecutive naturals > starting from 1, the set size IS an element of the set. Therefore, if the set > is infinite, then it contains an infinite element. This is a contradiction. So > how do you resolve this? Very simply. Extremely simply, though obviously beyond you. We observe > that any element of the set that is equal to the set size must be the > largest element in the set. We deduce that in a set with no largest > element, there is no largest element that could be equal to the set > size. > Okay, then what is the size of the set? The proof shows also that the set size > IS a member of the set. > I missed this bit of the proof. _How_ does it prove that the set size > IS a member of the set? (Yeah, yeah, I guess because for any finite > sequence, the maximum member exists and is equal to the set size. In > the limit, the maximum member doesn't exist, but somehow must be equal > to the set size, so while nonexistent, must creep back into the set > disguised as an 'infinite number'. Ho hum.) Ho hum? The set of naturals is the set of all consecutive finite whole numbers starting from 1. The proof shows that for any such set, the size of the set is a member of the set. Do you know what the largest finite natural is? No. Do you know what the size of the set of finite naturals is? No. But, you know those two numbers are the same number, so one cannot be finite while the other is infinite. > Anyway, if you care to show this bit of the proof *carefully*, you > might like in parallel to provide a proof that sqrt(2) is rational > (it's the limit of a sequence, every member of which is rational), and > show the difference - why your proof of the set membership of > something which doesn't on the face of it exist is valid, while the > squrt(2) proof is (I suppose you might agree?) invalid. > So, if the set size is infinite, then how can that > number NOT be part of the set of whole numbers starting from 1? For any such > set, the size IS an element. Do you disagree? > Obviously I disagree. We know that for any not-quite-such set (i.e. any > *finite* initial segment of the naturals) the largest: (a) exists, (b) > is a member, and (c) equals the set size. Right, for all n in N. > We also can see that for the > complete set of naturals, this largest (a) does not exist, and > therefore (b) is not a member, so everyone except you is baffled as to > how it can (c) equal the set size. Again, with the no largest finite argument. Do you honestly think that you can just dismiss obvious facts for the entire set because there is no last element? That is entirely irrelvant. If all elements are finite, then this fact holds for all elements in the set. To put it another way, for any such set of size n, n is an element of the set. So, how big is your set of naturals, again? > Do tell me: > consider a binary tree in which there are only branching nodes - how > many leaf nodes are there? Use Tinduction, or any other favourite tool. > What is the purpose of this question? The obvious answer, using plain logic, is > that a tree with no leaf nodes has zero leaf nodes. > Good. Sounds about right. > However, an infinite binary > tree really has infinite numbers of leaf nodes. > Hmm, and this contradiction doesn't bother you much? You don't think > that if something has zero leaf nodes, it can't really have an infinite > number of them as well? It depends how you imagine it. I don't think you can simply pretend there are definitely no leaf nodes when the tree is infinite. This is just another form of the largest finite mantra. > For level n, starting with zero > at the root node, there are 2^n nodes. If a tree is x levels deep, it has 2^x > leaf nodes. If x is infinite, then 2^x is infinite as well. By pretending that > somehow at x=oo all the leaf nodes disappear is inconsistent. > Of course. No mathematician thinks that at infinity anything > disappears. You are stuck with your maundering finite imponderable > numbers, which is where you think things reach infinity. No, you are stuck in yet another misconception about what I am saying. But, do go on..... > Mathematicians, though, say unending meaning, er, really, really, > really, not having an end. No end. Just endless. Not even an end > labelled infinity. And yet, the projectively extended real numbers do have infinity at a point on the circle, opposite zero, the way the numbers SHOULD be viewed from the infinite perspective. It is perfectly valid to view infinity as a point infinitely far from zero. > But of course, > you will say that there are infinite levels so we never get any leaf nodes. I > am not sure how you reconcile this with the inductive argument that the number > of leaf nodes of a tree with n levels is 2^n. > At least you seem to see there is a problem. I tend to think the problem there comes from pretending the leaf nodes disappear at infinity. I prefer to say that if there are n branches, there are n/2 paths and leaf nodes, whether n is finite or infinite. That is more consistent. > Brian Chandler > http://imaginatorium.org -- Smiles, Tony === Subject: Re: infinity > Do you know what the largest finite natural is? I know that there is no such thing, which TO sometimes admits and sometimes denies. === Subject: Re: infinity !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > Ho hum? The set of naturals is the set of all consecutive finite > whole numbers starting from 1. The proof shows that for any such > set, Hand-waving hogwash again. There is only _one_ set of all consecutive finite whole numbers starting from 1. But there is an infinite number of finite set of consecutive finite whole numbers starting with 1. The set of _all_ consecutive finite whole numbers is not one of them. > the size of the set is a member of the set. Of all finite sets of consecutive whole numbers starting with 1. But the set of _all_ consecutive whole numbers starting with 1 is not one of them. > Do you know what the largest finite natural is? No. It is trivial to show that such a number does not exist. > Do you know what the size of the set of finite naturals is? It is trivial to show by induction that it can't be a finite natural. > No. But, you know those two numbers are the same number, Guffaw. > so one cannot be finite while the other is infinite. One does not exist, and the other is not a finite number. There are frameworks in which it is a particular infinite number, though, that labels the equivalence class containing the set of natural numbers as one as its elements. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: infinity stephen@nomail.com said: > stephen@nomail.com said: >> stephen@nomail.com said: > I am not assuming that there is a longest word. A longest word > implies a largest natural number. You always deny that there > is a largest natural number, but once again you are using an > argument that depends on there being a largest natural number. >> I said longest WORDS, not longest word. >> There are no longest words either, so I am not assuming >> anything about them. There is no maximum finite word length. > > I have no idea what L is in your S^L. You are aware that there > is more than one string length, so picking a single L does not > make any sense. It almost makes sense if you think that L is > the maximum string length, i.e. the largest finite natural number. > Of course you also deny that there is a maximum string > length, so I have no idea what S^L is supposed to mean. >> Given any string length and alphabet, that is the maximum number of unique >> srings in the language. >> I asked what L is. L is not the maximum number of unique >> strings in the language. > You said you had no idea what S^L is, even though we have discussed this > before. L is any string length as I said. What do you not understand? > Again, I do not understand what L is. Is L the length of any > string? Well, then lets look at the language of strings on > the alphabet {0,1} of length 100 or less. Is the size of this > language 2^55? 55 afterall is a length less than 100. Or is the size of > the language 2^23? What is the value of L I should plug into > 2^L to determine the size of this language? All of them up to 100. sum(x=0->100: 2^x) > > You should have a summation: > S^0 + S^1 + S^2 + S^3 + .... S^k + ... > for all finite k. That is the number of finite length strings > over an alphabet with size S. >> Yes, and you are summing a finite number of terms, each of which is finite in >> value. Is the sum infinite? No. >> But I am not summing a finite number of terms. I am summing >> all of the finite k. There are an infinite number of finite k. >> You cannot assume there are only a finite number of finite >> k when trying to prove that there are only a finite number >> of finite k. > Well, I have proven, at least to my own staisfaction, that you cannot have an > infinite set of finite whole numbers, so when you say finite k, I say you are > counting a finite number of times and summing a finite number of terms, each of > which is finite. You say I cannot assume that there are a finite number of > finite naturals in trying to prove that there are a finite number of finite > strings on a finite alphabet. > You cannot assume what you are trying to prove. I didn't. You did. > Can you assume there are an infinite number of > finite naturals, when trying to prove that there are an infinite number of > finite strings using a finite alphabet? > No, but I do not need to. I already gave you three proofs > in another post. Here is the simplest of them again. > Suppose that the number of positive finite whole numbers is finite. > Lets call this number F. If F is finite, then F+1 is also > finite, and F+1 <> F. The set {1, 2, 3, .... F, F+1} > contains F+1 finite whole numbers. But this contradicts > the fact that there are only F positive finite whole numbers. > Therefore, there cannot be a finite number of positive finite > whole numbers. That only proves there is no largest whole number. So what? Does the lack of a largest finite mean that there can be no infinites? What is your point? > Nowhere in that proof do I assume that are an infinite number > of finite naturals. In fact, I assume the opposite, and > derive a contradiction. If you do not agree with the above, > please point out the error. The contradiction is not in the idea of a finite set but in naming the size F as the largest finite. > You claim that the above summation is finite. According to you F = sum S^k for all finite k is a finite number. >> That is correct. > However if F is a finite number, then > there are strings of length F, and there are S^F strings > of length F, which is greater than F, the supposed number > of finite strings. That's a contradiction. Therefore > F cannot possibly be a finite number. >> Darryl just gave a similar proof. When you say for all finite k then you >> are saying there is some upper bound to k. >> No, I am saying for all finite k. There is no finite upper >> bound for k. I noticed that you have conveniently ignored my proof >> above. > That was a proof? All it is is a rehashed statement that there is no largest > finite integer. Sure, finite F can always be incremented, since finite k can > always be incremented. That lack of a largest element, or longest string, > doesn't prove infinitude of the set, as far as I'm concerned, so that doesn't > prove anything to me. > Until you define what you mean by 'infinitude' no one will know > what you are talking about. I mean larger than any finite, which apparently doesn't hold for you. >> What about the strings of length F? Which of the following >> do you disagree with: >> 1. F is a finite number >> 2. There are strings of length F >> 3. There are S^F strings of length F >> 4. S^F > F >> 5. There are more than F finite strings. > This is exactly the same as the idea that there is no largest finite. SO > WHAT???? > You claim that there are only F finite strings. But there > are clearly S^F strings with the finite length F. If > S^F is greater than F, how can you claim there are only > F strings? I claim that there are a finite number of finite strings. I never claimed there were F of them. You are again trying to shove the largest finite in my mouth, and it's irrelevant. Whatever finite F you give me, S^F is finite, and sum(x= 0->F:S^x) is finite. > Do you really not see the contradiction? How can there > be only F finite strings, if there exist finite strings with length > F, and S^F is greater than F? Just answer that. Huyah huyah Ommmmm...... Largest Finite. I never said that. >> You apparently disagree with 5, because you insist >> there are only F finite strings. You definitely agree >> with 1, and you seem pretty certain about 3 and 4. So >> I guess 2 is the one you must disagree with, but >> I do not see how you can claim that the set of all >> finite strings does not include strings of length F >> if F is a finite number. > This is all based on trying to pinpoint that largest finite F, which is a > waste of all our time. It doesn't exist. What is your longest finite string, or > largest finite natural? It doesn't exist. The nonexistence of a largest finite > is irrelevant. There is no F s.t. no finite language can be larger. SO WHAT??? > So what???? Do you really not see the contradiciton? Yeah, really not in anything I said. You concocted a contraidiction, but it was irrelevant. > Here is another one for you. Let > F = sum of all finite k > 0 > i.e. > F = 1 + 2 + 3 + 4 + .... > You of course claim that F is finite. If that is the > case, then F will appear on both sides of the above equation. > F = F + (sum of all finite k>0 and k<>F) Are you claiming there is a alrgest finite number? That would appear to be what you are doing. Why would you do that? > Given that F is finite, we can safely subtract it from > both sides of the equation, giving us: > 0 = (sum of all finite k>0 and k<>F) > So according to you, the sum of all finite k>0 and not > equal to F equals 0. So either 1+2+3+ ... adds up to 0, > or there are no finite numbers greater than 0 other than F, > or some other equally bizarre case. I know you will > say 'So what?', but I find it hard to believe you really > cannot see the contradiction. No, according to you that is the case. This has nothing to do with anything that I have said. When did I make any such argument. You cannot deny that for any finite L sum(x=0->L: S^x) is finite. > I don't know what restrictions you put on the meaning of language, but a > language is simply a set of strings constructed from a set of symbols. > I have no idea what you are talking about. A language is a set of > strings. I have never said otherwise. >> L = a*(ba+ab)*b >> is a language containing an infinite number of strings >> (all of which are finite). I do not create this language >> a string at a time. I define that language with some >> finite structure, such as a regular expression, and I am done. > That is one way to construct a language. BTW your language isn't infinite > unless you allow the strings to become infinite. > My language is infinite, even though all the strings are finite. > I can even prove it. Of course you are using your weird > private definition of 'infinite', but I do not care about > your definition. So, for which L does sum(x=0->L:S^x) become infinite? > And that is the only way to construct an infinite language. > It does not have to be a regular expression, but it has > to be defined with some finite structure. You cannot > list all the elements of an infinite language. No kidding. > > The set of all finite length strings over the alphabet {0,1} > is an infinite set. There is no longest string in this > set, and there is no L to plug into your S^L formula. >> If the string length cannot be infinite, then the language cannot be infinite. >> So you keep saying without proof. That may be your definition >> of infinite, but your definition of infinite does not >> include unending sets that any reasonable person would call >> infinite. > Apprently, I have a non-standard understanding of infinity, but nothing I have > heard here convinces me that I am in the least bit wrong. > You do have a non-standard understanding of infinity. If > you would actually share your definition of 'infinity' perhaps > someone could make sense of what you are talking about, but > I doubt you will as that will probably just reveal more > contradictions. If the fact that noone else in the world > agrees with you does not convince you that you are in the > least bit wrong, then I suppose nothing will. Hey, if you > decide that the word 'cat' should really mean 'dog', then > you probably would not think you were in the least bit wrong either. I do not judge my thinking in terms of whether others will like it or agree. I jusdge it in terms of the consistency of the conclusions it draws and how well they mesh with reality and other thinking. I am obviously not the only one who objects to the bizarre reasoning in this area, so I don't feel incredibly alone anyway. But, even if I were the only one on the the planet who thought like I did, I still would not see that as a areason to believe I am just wrong. Everything has to start somewhere. > Stephen -- Smiles, Tony === Subject: Re: infinity > stephen@nomail.com said: >> stephen@nomail.com said: > stephen@nomail.com said: >> I am not assuming that there is a longest word. A longest word >> implies a largest natural number. You always deny that there >> is a largest natural number, but once again you are using an >> argument that depends on there being a largest natural number. > I said longest WORDS, not longest word. There are no longest words either, so I am not assuming > anything about them. There is no maximum finite word length. > I have no idea what L is in your S^L. You are aware that there >> is more than one string length, so picking a single L does not >> make any sense. It almost makes sense if you think that L is >> the maximum string length, i.e. the largest finite natural number. >> Of course you also deny that there is a maximum string >> length, so I have no idea what S^L is supposed to mean. > Given any string length and alphabet, that is the maximum number of unique > srings in the language. I asked what L is. L is not the maximum number of unique > strings in the language. >> You said you had no idea what S^L is, even though we have discussed this >> before. L is any string length as I said. What do you not understand? >> Again, I do not understand what L is. Is L the length of any >> string? Well, then lets look at the language of strings on >> the alphabet {0,1} of length 100 or less. Is the size of this >> language 2^55? 55 afterall is a length less than 100. Or is the size of >> the language 2^23? What is the value of L I should plug into >> 2^L to determine the size of this language? > All of them up to 100. > sum(x=0->100: 2^x) So when I plug in all the finite L, I get an infinite sum of finite values, i.e. sum (x>=0 : 2^x) Note, I did not plug in a single value of L like you kept insisting. There are an infinite number of finite values for x. > There are an infinite number of finite k. > You cannot assume there are only a finite number of finite > k when trying to prove that there are only a finite number > of finite k. >> Well, I have proven, at least to my own staisfaction, that you cannot have an >> infinite set of finite whole numbers, so when you say finite k, I say you are >> counting a finite number of times and summing a finite number of terms, each of >> which is finite. You say I cannot assume that there are a finite number of >> finite naturals in trying to prove that there are a finite number of finite >> strings on a finite alphabet. >> You cannot assume what you are trying to prove. > I didn't. You did. Very mature response. Your entire S^L proof that there are only a finite number of values for L is based entirely on the assumption again that there are only a finite number of values for L. Explain again the relevance of S^L without assuming there are only a finite number of finite values. >> Can you assume there are an infinite number of >> finite naturals, when trying to prove that there are an infinite number of >> finite strings using a finite alphabet? >> No, but I do not need to. I already gave you three proofs >> in another post. Here is the simplest of them again. >> Suppose that the number of positive finite whole numbers is finite. >> Lets call this number F. If F is finite, then F+1 is also >> finite, and F+1 <> F. The set {1, 2, 3, .... F, F+1} >> contains F+1 finite whole numbers. But this contradicts >> the fact that there are only F positive finite whole numbers. >> Therefore, there cannot be a finite number of positive finite >> whole numbers. > That only proves there is no largest whole number. So what? Does the lack of a > largest finite mean that there can be no infinites? What is your point? My point is that if I assume that there are only a finite number of finite numbers that it leads to a contradiction. >> Nowhere in that proof do I assume that are an infinite number >> of finite naturals. In fact, I assume the opposite, and >> derive a contradiction. If you do not agree with the above, >> please point out the error. > The contradiction is not in the idea of a finite set but in naming the size F > as the largest finite. What does it matter what we name it? You claim there is a finite number that is equal to the number of finite numbers. It does not matter if we call it F, or X, or Tony or whatever. You claim this number exists. If it exists, then that number plus one exists, and that number plus one is also finite. Until you define what you mean by 'infinitude' no one will know >> what you are talking about. > I mean larger than any finite, which apparently doesn't hold for you. But you claim that the sum of all finite numbers, which clearly should be larger than any individual finite number, is a finite number. So no, I have know idea what you mean by 'infinitude'. After all, you claim that the sum 1+2+3+..... is not larger than every finite number, and is in fact equal to some finite number. > What about the strings of length F? Which of the following > do you disagree with: > 1. F is a finite number > 2. There are strings of length F > 3. There are S^F strings of length F > 4. S^F > F > 5. There are more than F finite strings. >> This is exactly the same as the idea that there is no largest finite. SO >> WHAT???? >> You claim that there are only F finite strings. But there >> are clearly S^F strings with the finite length F. If >> S^F is greater than F, how can you claim there are only >> F strings? > I claim that there are a finite number of finite strings. I never claimed there > were F of them. You are again trying to shove the largest finite in my mouth, > and it's irrelevant. Whatever finite F you give me, S^F is finite, and sum(x= > 0->F:S^x) is finite. You claim there are a finite number of strings. We can call that finite number F, or X, or Tony, or whatever, if it exists as you said it does. You cannot claim that there are a finite number of finite numbers, but then claim that we cannot call that number F. You are just becoming increasingly dense in your effort to not see the simple contradictions your ideas lead to. >> Do you really not see the contradiction? How can there >> be only F finite strings, if there exist finite strings with length >> F, and S^F is greater than F? Just answer that. > Huyah huyah Ommmmm...... Largest Finite. I never said that. You are the one who keeps invoking 'Largest Finite'. All your arguments implicitly rely on a largest finite. Once again, tell me why 1 + 2 + 3 + .... is a finite number? Here is another one for you. Let >> F = sum of all finite k > 0 >> i.e. >> F = 1 + 2 + 3 + 4 + .... >> You of course claim that F is finite. If that is the >> case, then F will appear on both sides of the above equation. >> F = F + (sum of all finite k>0 and k<>F) > Are you claiming there is a alrgest finite number? That would appear to be what > you are doing. Why would you do that? I never mentioned anything about a largest finite number. There are obviously numbers larger than F, such as F+1. But according to you, F = 1 + 2 + 3 + ... + F + F+1 + F+2 + ... because it equals the sum of all finite numbers, and it to is a finite number. You are the one who claims that 1 + 2 + 3 + 4 + .... is a finite number. Accept the implications of your claim or drop it. Stop trying to pretend that other people are claiming these absurdities. >> Given that F is finite, we can safely subtract it from >> both sides of the equation, giving us: >> 0 = (sum of all finite k>0 and k<>F) >> So according to you, the sum of all finite k>0 and not >> equal to F equals 0. So either 1+2+3+ ... adds up to 0, >> or there are no finite numbers greater than 0 other than F, >> or some other equally bizarre case. I know you will >> say 'So what?', but I find it hard to believe you really >> cannot see the contradiction. > No, according to you that is the case. This has nothing to do with anything > that I have said. When did I make any such argument. You cannot deny that for > any finite L sum(x=0->L: S^x) is finite. Of course I cannot deny that. But you claim that sum (x>=0 : S^x) is finite. And you claim that sum (x>=0 : x) is finite. Why can't you talk about those claims? Why do you always try to deflect the discussion to the irrelevant bounded cases? >> I don't know what restrictions you put on the meaning of language, but a >> language is simply a set of strings constructed from a set of symbols. >> I have no idea what you are talking about. A language is a set of >> strings. I have never said otherwise. L = a*(ba+ab)*b is a language containing an infinite number of strings > (all of which are finite). I do not create this language > a string at a time. I define that language with some > finite structure, such as a regular expression, and I am done. >> That is one way to construct a language. BTW your language isn't infinite >> unless you allow the strings to become infinite. >> My language is infinite, even though all the strings are finite. >> I can even prove it. Of course you are using your weird >> private definition of 'infinite', but I do not care about >> your definition. > So, for which L does sum(x=0->L:S^x) become infinite? For none, which you have been repeatedly told. The language is infinite the moment I define it. It does not start out finite and 'become infinite'. Once again you are making up ridiculous distractions that have nothing to do with the argument. >> And that is the only way to construct an infinite language. >> It does not have to be a regular expression, but it has >> to be defined with some finite structure. You cannot >> list all the elements of an infinite language. > No kidding. Well, you are the one who claimed that there was some other way to construct an infinite language. Apprently, I have a non-standard understanding of infinity, but nothing I have >> heard here convinces me that I am in the least bit wrong. >> You do have a non-standard understanding of infinity. If >> you would actually share your definition of 'infinity' perhaps >> someone could make sense of what you are talking about, but >> I doubt you will as that will probably just reveal more >> contradictions. If the fact that noone else in the world >> agrees with you does not convince you that you are in the >> least bit wrong, then I suppose nothing will. Hey, if you >> decide that the word 'cat' should really mean 'dog', then >> you probably would not think you were in the least bit wrong either. > I do not judge my thinking in terms of whether others will like it or agree. I > jusdge it in terms of the consistency of the conclusions it draws and how well > they mesh with reality and other thinking. I am obviously not the only one who > objects to the bizarre reasoning in this area, so I don't feel incredibly alone > anyway. But, even if I were the only one on the the planet who thought like I > did, I still would not see that as a areason to believe I am just wrong. > Everything has to start somewhere. Yeah, and those cats might really be dogs. Word definitions really cannot be wrong. Language is a communal thing. If you insist on using words contrary to the usage of the community, then you will just be misunderstood. Insisting that the community's definitions are 'wrong' is like saying that the French are wrong because the do not speak English. Stephen === Subject: Re: infinity > That only proves there is no largest whole number. So what? Does the > lack of a largest finite mean that there can be no infinites? What is > your point? It proves that the set of finite whole numbers can be infinite without needing any infinite whole numbers. === Subject: Re: infinity were F of them. Right. So let me guess - there are a *finite number* of them, but this finite number (in asterisks) is somehow not like most finite numbers, because it can't be named? Can't even be called 'X'? Because otherwise we would add one to X and get a contradiction. Hmm, do I need to add a notion of unnamable number to my existing notion of imponderable? (I'll use this as shorthand for imponderably enormous) Well, no, I don't. The whole point of considering an axiomatically defined (somewhat arbitrary) reference imponderable (your 'N') is that there can't be a clearcut threshold between ponderable and imponderable numbers. Therefore, if we decide to consider the set of ponderably long strings, and get a value of 'F' for how many there are, then if someone produces the expression 'F+1' (for example), we of course can ponder it, since it's only one more than something else that's ponderable. Any incipient contradiction is due to the inappropriateness of tools that were designed for doing proper maths. Brian Chandler http://imaginatorium.org === Subject: Re: infinity imaginatorium@despammed.com said: > I claim that there are a finite number of finite strings. I never claimed there > were F of them. > Right. So let me guess - there are a *finite number* of them, but this > finite number (in asterisks) is somehow not like most finite numbers, > because it can't be named? Can't even be called 'X'? Because otherwise > we would add one to X and get a contradiction. Hmm, do I need to add a > notion of unnamable number to my existing notion of imponderable? > (I'll use this as shorthand for imponderably enormous) Yeah, and you can throw your largest finite natural in there too. You folks seem to like numbers like that, so you might as well start a collection. > Well, no, I don't. The whole point of considering an axiomatically > defined (somewhat arbitrary) reference imponderable (your 'N') is that > there can't be a clearcut threshold between ponderable and imponderable > numbers. Right. This is what we had referred to as the Twilight Zone between finite and infinite, that uncountable chasm. > Therefore, if we decide to consider the set of ponderably long > strings, and get a value of 'F' for how many there are, then if someone > produces the expression 'F+1' (for example), we of course can ponder > it, since it's only one more than something else that's ponderable. Any > incipient contradiction is due to the inappropriateness of tools that > were designed for doing proper maths. (sigh) > Brian Chandler > http://imaginatorium.org -- Smiles, Tony === Subject: Re: infinity No, but I do not need to. I already gave you three proofs > in another post. Here is the simplest of them again. > Suppose that the number of positive finite whole numbers is finite. > Lets call this number F. If F is finite, then F+1 is also > finite, and F+1 <> F. The set {1, 2, 3, .... F, F+1} > contains F+1 finite whole numbers. But this contradicts > the fact that there are only F positive finite whole numbers. > Therefore, there cannot be a finite number of positive finite > whole numbers. > That only proves there is no largest whole number. It proves that there are more whole numbers than any finite number. > So what? Does the lack of a > largest finite mean that there can be no infinites? What is your point? The existence of more than finite finite-valued whole numbers means that there are infinitely many. The number of finite whole numbers is larger than any finite number. What do you call a number which is larger than any finite value? > You of course claim that F is finite. If that is the > case, then F will appear on both sides of the above equation. > F = F + (sum of all finite k>0 and k<>F) > Are you claiming there is a alrgest finite number? He's claiming that the sum of all the finite numbers is not a finite number. > That would appear to be what > you are doing. Why would you do that? Of course, that's not what he's doing. You claim that the sum of all the finite numbers is finite. It isn't. It can't be. If F is the sum of all the finite numbers, and F is finite, then it's on the right hand side of that equation as well as the left. So if the sum of all the finite numbers is finite, then it's also larger than itself. Therefore the sum of all the finite numbers is not finite. - Randy === Subject: Re: infinity Randy Poe said: > stephen@nomail.com said: > No, but I do not need to. I already gave you three proofs > in another post. Here is the simplest of them again. Suppose that the number of positive finite whole numbers is finite. > Lets call this number F. If F is finite, then F+1 is also > finite, and F+1 <> F. The set {1, 2, 3, .... F, F+1} > contains F+1 finite whole numbers. But this contradicts > the fact that there are only F positive finite whole numbers. > Therefore, there cannot be a finite number of positive finite > whole numbers. > That only proves there is no largest whole number. > It proves that there are more whole numbers than any > finite number. > So what? Does the lack of a > largest finite mean that there can be no infinites? What is your point? > The existence of more than finite finite-valued whole > numbers means that there are infinitely many. The number > of finite whole numbers is larger than any finite number. > What do you call a number which is larger than any > finite value? > You of course claim that F is finite. If that is the > case, then F will appear on both sides of the above equation. F = F + (sum of all finite k>0 and k<>F) > Are you claiming there is a alrgest finite number? > He's claiming that the sum of all the finite numbers is > not a finite number. > That would appear to be what > you are doing. Why would you do that? > Of course, that's not what he's doing. > You claim that the sum of all the finite numbers is > finite. > It isn't. It can't be. If F is the sum of all the finite > numbers, and F is finite, then it's on the right hand > side of that equation as well as the left. > So if the sum of all the finite numbers is finite, > then it's also larger than itself. > Therefore the sum of all the finite numbers is not finite. > - Randy You are doing exactly what Stephen was. You surmise some finite whole number which is the sum of all whole numbers, and then show that it must include itself in its sum, which causes a contradiction. But, the contradiction comes not from assuming that the sum is finite, but from assuming you can possibly name any such sum. You can't name that any more than you can name the largest finite. This proof is of the same ilk as all your other largest finite arguments, causing a contradiction by specifying some limit on the finites which doesn't exist. All you have proven is there is no largest finite natural, which isn't at issue. I don't talk about the largest finite. If ALL strings in the set are finite, and ALL sets which just include finite strings are finite, then the set is finite. -- Smiles, Tony === Subject: Re: infinity > Randy Poe said: stephen@nomail.com said: No, but I do not need to. I already gave you three proofs in > another post. Here is the simplest of them again. Suppose that the number of positive finite whole numbers is > finite. Lets call this number F. If F is finite, then F+1 is > also finite, and F+1 <> F. The set {1, 2, 3, .... F, F+1} > contains F+1 finite whole numbers. But this contradicts the > fact that there are only F positive finite whole numbers. > Therefore, there cannot be a finite number of positive finite > whole numbers. > That only proves there is no largest whole number. It proves that there are more whole numbers than any finite number. So what? Does the lack of a largest finite mean that there can be > no infinites? What is your point? The existence of more than finite finite-valued whole numbers > means that there are infinitely many. The number of finite whole > numbers is larger than any finite number. What do you call a number which is larger than any finite value? You of course claim that F is finite. If that is the case, > then F will appear on both sides of the above equation. F = F + (sum of all finite k>0 and k<>F) > Are you claiming there is a alrgest finite number? He's claiming that the sum of all the finite numbers is not a > finite number. That would appear to be what you are doing. Why would you do > that? Of course, that's not what he's doing. You claim that the sum of all the finite numbers is finite. It isn't. It can't be. If F is the sum of all the finite numbers, > and F is finite, then it's on the right hand side of that equation > as well as the left. So if the sum of all the finite numbers is finite, then it's also > larger than itself. Therefore the sum of all the finite numbers is not finite. - Randy > You are doing exactly what Stephen was. You surmise some finite whole > number which is the sum of all whole numbers, and then show that it > must include itself in its sum, which causes a contradiction. But, > the contradiction comes not from assuming that the sum is finite, but > from assuming you can possibly name any such sum. You can't name that > any more than you can name the largest finite. This proof is of the > same ilk as all your other largest finite arguments, causing a > contradiction by specifying some limit on the finites which doesn't > exist. All you have proven is there is no largest finite natural, > which isn't at issue. It can be proven inductively that the sum of any finite number of finite natural numbers is a finite natural number, and that the sum of three or more distinct finite naturals is a finite natural larger than any of them. TO claims existence of a finite set containing all finite natural numbers. The sum of all such members must be a finite natural not in that set. OOPS! TO goofed again! > I don't talk about the largest finite. If ALL strings in the set are > finite, and ALL sets which just include finite strings are finite, > then the set is finite. === Subject: Re: infinity which is the sum of all whole numbers, and then show that it must include > itself in its sum, which causes a contradiction. But, the contradiction comes > not from assuming that the sum is finite, but from assuming you can possibly > name any such sum. So you really are claiming now that there are unnameable numbers? There is such a thing as the finite sum of all finite whole numbers but I can't say let F be the finite sum of all finite whole numbers? > You can't name that any more than you can name the largest > finite. If something exists and is finite, what in the world prevents me from giving it a name? - Randy === Subject: Re: infinity Randy Poe said: > You are doing exactly what Stephen was. You surmise some finite whole number > which is the sum of all whole numbers, and then show that it must include > itself in its sum, which causes a contradiction. But, the contradiction comes > not from assuming that the sum is finite, but from assuming you can possibly > name any such sum. > So you really are claiming now that there are unnameable > numbers? There is such a thing as the finite sum of all > finite whole numbers but I can't say let F be the finite > sum of all finite whole numbers? Why don't you say F is the largest finite number, and see what happens? Duh! > You can't name that any more than you can name the largest > finite. > If something exists and is finite, what in the world prevents > me from giving it a name? When you name the largest finite, then I will name the sum of all finites, but then that will be larger than the largest finite, won't it? This is the same old rehashed circular nonsense that Cantorians waste so much time on. Why don't you try explaining at which finite L sum(x=0->L: S^x) becomes infinite? You can't? Well, then........ > - Randy -- Smiles, Tony === Subject: Re: infinity So you really are claiming now that there are unnameable > numbers? There is such a thing as the finite sum of all > finite whole numbers but I can't say let F be the finite > sum of all finite whole numbers? > Why don't you say F is the largest finite number, and see what happens? Duh! There's a difference. I don't claim that the largest finite number exists, so I can't give it a name. There is also a difference in that if I hypothesize for the purpose of a proof that there *is* a largest finite number, I can indeed call it F. And then the one line proof follows immediately: F+1 is a finite number, and is larger. Hence F is not the largest finite. That means there is no largest finite, not that it exists but can't be named. On the other hand, you claim that the sum of all finites does exist, is finite, and yet has no name. > You can't name that any more than you can name the largest > finite. > If something exists and is finite, what in the world prevents > me from giving it a name? > When you name the largest finite, then I will name the sum of all finites, If something EXISTS and IS FINITE, what prevents me from giving it a name? I don't claim the largest finite exists. > but > then that will be larger than the largest finite, won't it? No, it will be larger than any finite, without the need to talk about a largest. > Why don't > you try explaining at which finite L sum(x=0->L: S^x) becomes infinite? It never does. But then I don't claim that there's some finite L at which this sum represents all finite strings either. You do. At which L does that happen? - Randy === Subject: Re: infinity Randy Poe said: > Randy Poe said: > So you really are claiming now that there are unnameable > numbers? There is such a thing as the finite sum of all > finite whole numbers but I can't say let F be the finite > sum of all finite whole numbers? > Why don't you say F is the largest finite number, and see what happens? Duh! > There's a difference. I don't claim that the largest finite > number exists, so I can't give it a name. > There is also a difference in that if I hypothesize for > the purpose of a proof that there *is* a largest finite > number, I can indeed call it F. > And then the one line proof follows immediately: F+1 is > a finite number, and is larger. Hence F is not the largest > finite. > That means there is no largest finite, not that it exists > but can't be named. > On the other hand, you claim that the sum of all finites > does exist, is finite, and yet has no name. The sum of all finite strings IS the size of the set. Are you saying the set has no size? Of course it does. It has elements. The question is whether this size is infinite or not. > You can't name that any more than you can name the largest > finite. If something exists and is finite, what in the world prevents > me from giving it a name? > When you name the largest finite, then I will name the sum of all finites, > If something EXISTS and IS FINITE, what prevents me from giving > it a name? I don't claim the largest finite exists. By putting a size like aleph_0 on the set, you do, since the last element of any set of consecutive whole numbers starting from 1 is equal to the set size. By declaring aleph_0 to be your set size, you have declared aleph_0 to be your largest finite, except that you claim it is infinite, which is self- contradictory. In order to claim there is a specific number which is the sum of all finite (positive) whole numbers is to claim that there is a last element, and that we have added all elements from the first to the last, so this claim runs into the same contradictions that the largest finite does. > but > then that will be larger than the largest finite, won't it? > No, it will be larger than any finite, without the need > to talk about a largest. Fine, but if it's larger than any finite, then it's larger than the largest, right? (sigh) > Why don't > you try explaining at which finite L sum(x=0->L: S^x) becomes infinite? > It never does. But then I don't claim that there's some > finite L at which this sum represents all finite strings > either. So, you claim there are infinite-length strings in your set of finite strings? Do you remember what L is supposed to represent? > You do. At which L does that happen? At infinite L, like I have been saying. You need infinite strings to have an infinite set on a finite alphabet. > - Randy -- Smiles, Tony === Subject: Re: infinity > You need infinite strings to have an > infinite set on a finite alphabet. No one but TO does. TO mistakenly claims that the set of all finite strings over a finite alphabet is a finite set. To concatenate a finite number of finite strings just means to string them all together to form a finite superstring. When one concatenates any finite number of finite strings the result is a finite string. When one concatenates two or more finite strings of length at least 1 each, the result is a string longer than any of its component strings. TO claims that there is a finite set which can contain ALL finite strings. But any concatenation of all the strings in a finite set of strings will produce a finite string longer than any in that. Thus EVERY finite set of finite strings is incomplete, i.e., there cannot be any finite set that contains every finite string. Of course, an infinite set of strings could contain all finite strings without any problems. === Subject: Re: infinity >The sum of all finite strings IS the size of the set. Uh, that didn't make any sense. Did you mean to say The sum of all finite naturals IS the size of the set. In that case, the size of the set of all finite naturals is not itself a finite natural (since it is clearly *larger* than any finite natural). >Are you saying the set has no size? Yes, but its size is not a finite natural. That's why its an infinite set. >By putting a size like aleph_0 on the set, you do, since the last >element of any set of consecutive whole numbers starting from 1 >is equal to the set size. No. What you have proved is that for each finite n, size(A_n) = n (where A_n = the set of all numbers greater than or equal to 1 and less than or equal to n). The set of all naturals is not of that form. >By declaring aleph_0 to be your set size, you have declared >aleph_0 to be your largest finite No, aleph_0 is the smallest infinite, not the largest finite. >In order to claim there is a specific number which is the sum of >all finite (positive) whole numbers is to claim that there is a >last element No. Some sets don't have a last element, but every set has a size. In particular, the set of all finite naturals has no last element, but it has a size. And that size is aleph_0, the first infinite cardinal. >> No, it will be larger than any finite, without the need >> to talk about a largest. >Fine, but if it's larger than any finite, then it's larger than the >largest, right? (sigh) Yes, if U = the set of all finite naturals, then size(U) is greater than any element in U. That's not a contradiction unless you assume that U is finite. >So, you claim there are infinite-length strings in your set of >finite strings? Why do you keep asking that question, when it's been answered many, many times? If A = the set of all finite strings, then size(A) is greater than any finite natural number, but A does not contain any infinite strings (by definition). >Do you remember what L is supposed to represent? No, you never made that clear. >At infinite L, like I have been saying. You need infinite strings to >have an infinite set on a finite alphabet. That's provably false. If A = the set of all finite strings, then size(A) is greater than any finite natural number. Therefore, size(A) is infinite. -- Daryl McCullough Ithaca, NY === Subject: Re: infinity Virgil said: > Virgil said: I mapped paths from the same tree to sets of naturals so as > to establish a bijection between them and the power set of > N. Now you are trying to change things, but fine, let's examine > what you are now proposing. You claim to have branches mapped > to all the naturals in a 1-1 correspondence, and therefore > countable. Then you claim that the set of paths is the > powerset of that set, and therefore uncountable? You really > should draw pictures before writing words. Then you won't > waste 1,000 of them sounding stupid. Here's your tree, with > each number being a branch/node: . > 1 2 > 3 4 5 6 > 7 8 9 10 11 12 13 14 > This form of labelling has nothing to do with me. > The only labels I use are on branches: Left branch and Right > branch, and the construction of my bijections are based entirely on > those labels and nothing else. Bull. You are constructing a bijection with the elements of the tree and the elements of the naturals and of the powerset of the naturals. > The disrepancy is between what I described and what TO > misunderstood me to have described. If he could actually read, > perhaps such discrepancies would not occur, at least so often. Each path in an infinite binary tree consists of a sequence of > branches which can be uniquely numbered with the infinitely many > finite naturals, so that all paths have a branch 1 and a branch 2 > and a branch 3 and so on without end. Each path determines a set by > including the numbers for each right branch and excluding the > numbers for each left branch (the reverse would work equally well). This clearly bijects paths with subsets of N (the infinite set of > finite naturals). Each node in the same tree can be matched with a unique natural in > binary notation as follow: The root node is 1. For each left branch > on the finite path from the root node to a given node, append 0, > and for each right branch append 1. So the children of the root > will be numbered 10 and 11, the grandchildren 100, 101, 110, and > 111, and so on. the number of binary digits will be one more than > the number of branches between the root and the node which matches > the binary number. This clearly bijects the set of nodes of that same maximal binary > tree with N (the infinite set of finite natural numbers). So that unless TO can construct a bijection between N (the infinite > set of finite natural numbers) and P(N), there are more (in the > sense of Cantor) paths than nodes. Grrrrr.....you moron! That's exactly the way I described your proof > to begin with, which you denied. > If TO thinks that what he said and what I said are the same, he has > even worse dyslexia than previusly thought. No, it's exactly as I remembered and described, and you are spewing nonsense to defend your nonsense. > You have one tree where you label each node with a natural number > I never label any nodes with anything in my construction. The only > labeling is done to branches, left or right. You never mentioned left and right. You mentioned bijections between the paths and the elements of the power set, and between the nodes or branches and the natural numbers. In the first, each branch denotes inclusion of a natural, but the entire row of branches corresponds to that natural, not a signle branch. > , and then another tree where you label each node with a bit > representing the membership of a natural in a set which is a member > of the powerset. > Same tree, same labels on branches, no labels on nodes. No, you have two different trees. > In your first tree, where each node is a natural, > That is TO's tree, not mine, as I did not label any nodes at all. > In the second, each path represents a unique > set of naturals, and the tree represents the power set of the > naturals, but each node does not represent a natural number. > But each branch determines whether a particular natural is in, or not > in, the set represented by that path. Each set of branches at a given level correspond to a natural, not each branch. > Each ROW > represents a natural number, and for row n, all of the 2^n nodes in > the row correspond to that natural. So, you did do exactly what I > said you did, which is change your interpretation of the tree > mid-proof. > Even if I had changed the labeling, which I did not, unless I also > changed the underlying tree, both constructions remain valid FOR THAT > TREE. There is no rule that one cannot have more than one label for a > thing if it is convenient to do so. Convenience being the operative motivation. > Since the tree is the same for both, however labeled, TO's objection > fails. Binary trees are binary trees, but they can be interpreted in a number of ways. Perhaps the most common is as analog to the binary number system, with each branch being one of two choices for a next digit, and each path representing a nuique whole number. If we allow infinite binary strings in our whole numbers, then we can construct a bijection between the paths of the tree and the natural numbers. We can also construct a bijection between the paths of the binary tree and the powerset of the natural numbers, as you suggested. Since, as you said, bijection is transitive, this amounts to a bijection between the whole numbers and the power set of the whole numbers, which are clearly different sizes. Let us try inductive proofs regarding the relationship between > branches and paths. > Proof: In a maximal binary tree, number of paths is half the number > of branches, plus 1. False for the tree consisting of only a root node, and false where the > number of branches is odd. The root node has one path, one end. That's half of the nonexistent branches, plus 1. Oh, and maximal means with all leaf nodes at the same level, so there could never be an odd number of branches. Sorry. > And false for binary trees with infinitely many branches. No, proven true, inductively. Give it up. You're wrong. -- Smiles, Tony === Subject: Re: infinity > Binary trees are binary trees, but they can be interpreted in a > number of ways. Perhaps the most common is as analog to the binary > number system, with each branch being one of two choices for a next > digit, and each path representing a nuique whole number. If we allow > infinite binary strings in our whole numbers, then we can construct a > bijection between the paths of the tree and the natural numbers. TO is off his rocker again on including his irrelevant infinite strings. If one includes such infinite strings, there is NO bijection with the set of finite naturals, which is what I am using. There is a bijection, as I have several times described, between the Cantor-infinite set of nodes and the Cantor-infinite set of finite naturals in binary notation and limited to finite strings of digits. There is a bijection , as I have several times described, between the Cantor-infinite set of (complete) paths and the power set of the infinite set of finite naturals. There is no bijection between the set of nodes and the set of paths. === Subject: Re: infinity > Virgil said: > Same tree, same labels on branches, no labels on nodes. No, you have two different trees. Since all maximal binary trees are tree-isomorphic, any difference would make no difference. So the issue is irrelevant. === Subject: Re: infinity Virgil said: > Virgil said: > Same tree, same labels on branches, no labels on nodes. No, you have two different trees. > Since all maximal binary trees are tree-isomorphic, any difference would > make no difference. So the issue is irrelevant. Aren't all balls ball-isomorphic? -- Smiles, Tony === Subject: Re: infinity > Virgil said: Virgil said: Same tree, same labels on branches, no labels on nodes. No, you have two different trees. Since all maximal binary trees are tree-isomorphic, any difference would > make no difference. So the issue is irrelevant. Aren't all balls ball-isomorphic? TO's balls are not in any way isomorphic to anything real. === Subject: Re: infinity > 1 2 > 3 4 5 6 > 7 8 9 10 11 12 13 14 This form of labelling has nothing to do with me. The only labels I use are on branches: Left branch and Right > branch, and the construction of my bijections are based entirely on > those labels and nothing else. > Bull. > You are constructing a bijection with the elements of the tree and the > elements of the naturals and of the powerset of the naturals. True, but irrelevant to my remark. A path in a maximal binary tree is uniquely determined by an infinite sequence left-branch/right-branch options. A node is uniquely determined by a finite sequence of such options. I show that each such infinite sequence may be made to correspond uniquely to a subset of N, and each finite sequence to a member of N, though by differing bijections. === Subject: Re: infinity Virgil said: > 1 2 > 3 4 5 6 > 7 8 9 10 11 12 13 14 This form of labelling has nothing to do with me. The only labels I use are on branches: Left branch and Right > branch, and the construction of my bijections are based entirely on > those labels and nothing else. > Bull. > You are constructing a bijection with the elements of the tree and the > elements of the naturals and of the powerset of the naturals. > True, but irrelevant to my remark. A path in a maximal binary tree is > uniquely determined by an infinite sequence left-branch/right-branch > options. A node is uniquely determined by a finite sequence of such > options. I show that each such infinite sequence may be made to > correspond uniquely to a subset of N, and each finite sequence to a > member of N, though by differing bijections. An infinite binary tree has infinite depth, otherwise what's infinite about it? If it has infinite depth, that is, infinitely long paths, then each node is NOT uniquely determined by some FINITE sequence, but by an INFINITE sequence. So, basically, what you are doing is allowing infinite bit strings in one and calling it uncountable, and limiting the other tree to fnite bit strings and calling it countable. It would appear that countably infinite really means unboundedly finite. -- Smiles, Tony === Subject: Re: infinity > So, basically, what you are doing is allowing infinite bit strings in one and > calling it uncountable, and limiting the other tree to fnite bit strings and > calling it countable. It would appear that countably infinite really means > unboundedly finite. The set of infinite bit strings IS uncountable and the set of finite bitstrings IS countable. Countably infinite is possible, as in the infinite set of finite naturals, but unboundedly finite is not possible, except in TO's dreamworld. === Subject: Re: infinity imaginatorium@despammed.com said: > imaginatorium@despammed.com said: Randy Poe said: > The size of set of natural numbers is a natural unit infinity to use, but one > can have a double set, depending on one's construction. > 2. But N is not the largest natural. N+1, N+2, ..., 2N, 2N+1, 2N+2 > are all in the naturals. > They are all whole numbers relative to the N under investigation. > 3. The range of a set is the maximum difference between elements, > so there are no elements of the natural numbers farther apart than > N. > not in the basic set of natural numbers. > 4. There is no maximum element of the naturals, so there is no > maximum difference between elements. But, if all elements are declared to be finite, no difference between them can > ever be infinite, so the maximum difference in the set is some finite value. Uh, so if you're counting these natural numbers with a ditty, the > ditty, which (I _think_ we agree) need never stop, stops at this last > mentioned finite value? This maximum difference of course might be the > difference between 0 and the largest natural, and the nonexistence of > this largest natural doesn't impair its functioning to stop the > unstoppable ditty? This will never end. Brian Chandler > http://imaginatorium.org > to the fair. My statement is clear. You say all elements are finite. Which do > you disagree with? > 1) The difference between numbers A and B is abs(A-B) > Sounds right. > 2) For finite A and B, abs(A-B) is finite. > Sounds right. > 3) If all elements are finite, then any two of them is finite. > No. Any two of them _are_ finite. > 4) If a number is finite, then it is not infinite. > I don't know that I can agree with *your* statement of this, since you > have never clearly defined what you mean by either term. In the > mathematical statement that looks like the same string of words (ah, > yes, which I see you fervently believe is what you are talking about) > this is true, yes. > I can't imagine WHICH of these you could possibly argue with. Which is it? > None, evidently. How about your next one, which will be something like: > 5) Since for any pair of numbers you give me, I can give you a (finite) > number larger than the difference between them, it follows [joke, Tony, > OK?] that you can give a (finite) number larger than the difference > between any pair of numbers I give you. No, I cannot give any finite number that is larger than all finite numbers, but I CAN say that any infinite number is larger than any finite number, and that if you restrict your set to finite numbers, then there cannot be any infinite numbers in it. Since all numbers are finite, the difference between any pair of numbers is finite, and therefore there can be no infinite difference, and the range of the set is finite. > Brian Chandler > http://imaginatorium.org > **** Sorry about the finger trouble above !! **** -- Smiles, Tony === Subject: Re: infinity > No, I cannot give any finite number that is larger than all finite > numbers, but I CAN say that any infinite number is larger than any > finite number, and that if you restrict your set to finite numbers, > then there cannot be any infinite numbers in it. But restricting a set to finite numbers does not restrict it to being a finite set. > Since all numbers > are finite, the difference between any pair of numbers is finite, and > therefore there can be no infinite difference, and the range of the > set is finite. The range as TO defines it requires existence of a minimal and maximal member to the set, and is undefined if either of them is missing. So that by TO's definition, the set of finite naturals does not have any range. But it does have an infinte diameter. Definition: The *diameter* of a non-empty subset of the reals is the (finite) least upper bound, if it exists, of distances between members, and if the set of distances has no finite upper bound, the diameter is said to be infinite. Note that, according to TO's defnition of range, open intervals have no range, but they have diameters equal to their lenghts. === Subject: Re: infinity Virgil said: > No, I cannot give any finite number that is larger than all finite > numbers, but I CAN say that any infinite number is larger than any > finite number, and that if you restrict your set to finite numbers, > then there cannot be any infinite numbers in it. > But restricting a set to finite numbers does not restrict it to being a > finite set. > Since all numbers > are finite, the difference between any pair of numbers is finite, and > therefore there can be no infinite difference, and the range of the > set is finite. > The range as TO defines it requires existence of a minimal and maximal > member to the set, and is undefined if either of them is missing. Wherever you have two or more values, you have a value range. Just because you have an unbounded set does not mean it has no range. It means you can't firgure out the range. It's one less than the size of the set, one less than the largest element. That's the range of the set of naturals, or any initial segment thereof. > So that by TO's definition, the set of finite naturals does not have any > range. But it does have an infinte diameter. > Definition: The *diameter* of a non-empty subset of the reals is the > (finite) least upper bound, if it exists, of distances between members, > and if the set of distances has no finite upper bound, the diameter is > said to be infinite. Again, you equate unboundedness with infinity. This needs to be properly distinguished. > Note that, according to TO's defnition of range, open intervals have no > range, but they have diameters equal to their lenghts. -- Smiles, Tony === Subject: Re: infinity > Virgil said: > No, I cannot give any finite number that is larger than all > finite numbers, but I CAN say that any infinite number is larger > than any finite number, and that if you restrict your set to > finite numbers, then there cannot be any infinite numbers in it. But restricting a set to finite numbers does not restrict it to > being a finite set. > Since all numbers are finite, the difference between any pair of > numbers is finite, and therefore there can be no infinite > difference, and the range of the set is finite. The range as TO defines it requires existence of a minimal and > maximal member to the set, and is undefined if either of them is > missing. > Wherever you have two or more values, you have a value range. Just > because you have an unbounded set does not mean it has no range. It > means you can't firgure out the range. It's one less than the size of > the set, one less than the largest element. That's the range of the > set of naturals, or any initial segment thereof. Every set of numbers has a diameter that one can figure out, it is the least upper bound of the distances between members, so why is it that ranges can't be figured out? So that by TO's definition, the set of finite naturals does not > have any range. But it does have an infinte diameter. Definition: The *diameter* of a non-empty subset of the reals is > the (finite) least upper bound, if it exists, of distances between > members, and if the set of distances has no finite upper bound, the > diameter is said to be infinite. > Again, you equate unboundedness with infinity. This needs to be > properly distinguished. The DEFINITION of infinite diameter for a set is that the set of distances between its members has no finite upper bound. Thus it is quite proper to speak of sets which have no infinite members as having infinite diameters. That is the difference between ranges and diameters, every set of numbers has a diameter, but unbounded sets of only finite numbers do not have ranges. > Note that, according to TO's defnition of range, open intervals > have no range, but they have diameters equal to their lenghts. === Subject: Re: infinity imaginatorium@despammed.com said: > Virgil said: > ...[?] > I mapped paths from the same tree ... > That's enough. I'm generally avoiding both tree and vase arguments, > since they are far too complex. But let's just try a little... > Let us try inductive proofs regarding the relationship between branches and > paths. > Uh-oh. OK, I'll try as well. I'll investigate how many ends there are > in an endless tree (that's one in which every node branches into two > child nodes, and this never never never never ends). Intuitively, I'd > expect that an endless tree would not have any ends, since I've defined > it that way, but let's see. > Proof: In a maximal binary tree, number of paths is half the number of > branches, plus 1. > 1. For an empty tree consisting of the root node, there is one path in the > tree, of length zero, and no branches. 1 path = 0 branches/2 + 1. > For an empty [not quite the right word, is it] tree, there is one > end. Uh, yeah, one path, like I said. > 2. If there are n branches, and n/2+1 paths, on level x, and we add level x+1, > each of the n/2+1 leaf nodes corresponding to a path will be extended by one > level by adding two branches, and each of the n/2+1 paths will be divided into > two paths, doubling the number of paths. So, our original n branches are > increased by 2*(n/2+1), and our original n/2+1 paths is doubled. This gives n+2 > *(n/2+1) = n+n+2 = 2n+2 branches, and 2*(n/2+1) = n+2 paths. 1/2(2n+2)+1 = n+2. > I think you mean that considering the tree that stops at level x, there > are 2^x [plus/minus epsilon*] ends. Yes, we are proving something true for a maximal binary tree of depth corresponding to the natural numbers starting from zero, as a property of that number. > * Your notation is so woolly I can't be bothered to work it out > exactly. You can't understand the inductive part of the inductive proof? Did you try? If n is the number of branches and n/2+1 is the number of paths, we add twice the number of paths to the branches and double the paths at each step, so if b(n) is the branches on level n, and p(n) is the number of paths on level n, and p (n)=b(n)/2+1, then p(n+1)=b(n+1)/2+1, and the realtionship is maintained from n to n+1. > 3. Therefore, each additional level added to the binary tree maintains this > relationship between branches and paths, and the number of paths is half the > number of branches, plus 1. > Notice how the word Therefore is completely wrong. The statement is > true, but is not a logical deduction from the previous one. Anyway, > turtles all the way down. Reread. In any case, you don't seem to disagree with the conclusion, so why don't you explain it to Virgil? > Proof: In a maximal binary tree with paths of length n, or depth of n, there > are p(n)=2^n paths and b(n) = 2(2^n-1) = 2(p(n)-1) branches. > What do you mean by Proof? Isn't this still part of the same > argument? You would get more respect if you took the trouble to > understand elementary notational conventions. This was a second proof, using a slightly different statement. > Anyway, never mind: to finish off *my* proof... By induction, the > number of ends of an endless tree is infinity. Da-dah!! yes, you are very clever...... > Brian Chandler > http://imaginatorium.org -- Smiles, Tony === Subject: Re: infinity > You can't understand the inductive part of the inductive proof? Did you try? > If > n is the number of branches and n/2+1 is the number of paths, we add twice > the > number of paths to the branches and double the paths at each step, so if b(n) > is the branches on level n, and p(n) is the number of paths on level n, and p > (n)=b(n)/2+1, then p(n+1)=b(n+1)/2+1, and the realtionship is maintained from > n > to n+1. TO's argument is like saying that because (n-1)/n < 1 for all naturals, that the limit of (n-1)/n as n goes towards infinity must be less than 1. What holds for all finite values need not hold in the limit. === Subject: Re: infinity Virgil said: > ...[?] > I mapped paths from the same tree ... > That's enough. I'm generally avoiding both tree and vase arguments, > since they are far too complex. But let's just try a little... > Let us try inductive proofs regarding the relationship between branches and > paths. > Uh-oh. OK, I'll try as well. I'll investigate how many ends there are > in an endless tree (that's one in which every node branches into two > child nodes, and this never never never never ends). Intuitively, I'd > expect that an endless tree would not have any ends, since I've defined > it that way, but let's see. > Proof: In a maximal binary tree, number of paths is half the number of > branches, plus 1. 1. For an empty tree consisting of the root node, there is one path in the > tree, of length zero, and no branches. 1 path = 0 branches/2 + 1. > For an empty [not quite the right word, is it] tree, there is one > end. > Uh, yeah, one path, like I said. Look, you have written your proof - I'm interlining my proof, a different proof, of something different. (What do you mean by path? Is it a 'way through the tree'?) are p(n)=2^n paths and b(n) = 2(2^n-1) = 2(p(n)-1) branches. > What do you mean by Proof? Isn't this still part of the same > argument? You would get more respect if you took the trouble to > understand elementary notational conventions. > This was a second proof, using a slightly different statement. Oh, OK. In a tree that terminates at depth n, p(n) is the number of terminal nodes, and b(n) is the number of 'branches', where a branch is a link from one node to its child. Yeah, OK, I agree with the 'numbers'. > Anyway, never mind: to finish off *my* proof... By induction, the > number of ends of an endless tree is infinity. Da-dah!! > yes, you are very clever...... Well, I'd like you to address the substantive issue. Seems to me that if one were to consider an unending tree, each path through it would be unending, so there would be some infinite number of branches, but there would also be zero leaf nodes. Do you agree? Brian Chandler http://imaginatorium.org === Subject: Re: infinity imaginatorium@despammed.com said: > imaginatorium@despammed.com said: > Virgil said: > ...[?] > I mapped paths from the same tree ... That's enough. I'm generally avoiding both tree and vase arguments, > since they are far too complex. But let's just try a little... Let us try inductive proofs regarding the relationship between branches and > paths. Uh-oh. OK, I'll try as well. I'll investigate how many ends there are > in an endless tree (that's one in which every node branches into two > child nodes, and this never never never never ends). Intuitively, I'd > expect that an endless tree would not have any ends, since I've defined > it that way, but let's see. Proof: In a maximal binary tree, number of paths is half the number of > branches, plus 1. 1. For an empty tree consisting of the root node, there is one path in the > tree, of length zero, and no branches. 1 path = 0 branches/2 + 1. For an empty [not quite the right word, is it] tree, there is one > end. > Uh, yeah, one path, like I said. > Look, you have written your proof - I'm interlining my proof, a > different proof, of something different. (What do you mean by path? > Is it a 'way through the tree'?) Yes a path is a succession of branches from the root, through child nodes, to a leaf, or possibly infinitely long with no discernible leaf. > are p(n)=2^n paths and b(n) = 2(2^n-1) = 2(p(n)-1) branches. What do you mean by Proof? Isn't this still part of the same > argument? You would get more respect if you took the trouble to > understand elementary notational conventions. > This was a second proof, using a slightly different statement. > Oh, OK. In a tree that terminates at depth n, p(n) is the number of > terminal nodes, and b(n) is the number of 'branches', where a branch is > a link from one node to its child. Yeah, OK, I agree with the > 'numbers'. Yay!!! > Anyway, never mind: to finish off *my* proof... By induction, the > number of ends of an endless tree is infinity. Da-dah!! > yes, you are very clever...... > Well, I'd like you to address the substantive issue. Seems to me that > if one were to consider an unending tree, each path through it would be > unending, so there would be some infinite number of branches, but there > would also be zero leaf nodes. Do you agree? If one imagines a tree that goes forever and never ends, one might imagine that there are no leaf nodes. If we look at the tree as representing the binary numbers, each path corresponds to a single infinite string of bits, each branch corresponds to a bit, and the nodes are just the spaces between digits. The root node corresponds to the digital point. So, no leaf nodes means that all bit strings are infinite. I suppose I can live with an image with no leaf nodes, but that doesn't affect the relationship between branches and paths, does it? Aren't there still essentially twice as menay branches as paths, and if so, can there possibly be countably many branches, and uncountably many paths? > Brian Chandler > http://imaginatorium.org -- Smiles, Tony === Subject: Re: infinity > imaginatorium@despammed.com said: > imaginatorium@despammed.com said: > Virgil said: > ...[?] > I mapped paths from the same tree ... That's enough. I'm generally avoiding both tree and vase arguments, > since they are far too complex. But let's just try a little... Let us try inductive proofs regarding the relationship between > branches and > paths. Uh-oh. OK, I'll try as well. I'll investigate how many ends there are > in an endless tree (that's one in which every node branches into two > child nodes, and this never never never never ends). Intuitively, I'd > expect that an endless tree would not have any ends, since I've defined > it that way, but let's see. Proof: In a maximal binary tree, number of paths is half the number > of > branches, plus 1. 1. For an empty tree consisting of the root node, there is one path > in the > tree, of length zero, and no branches. 1 path = 0 branches/2 + 1. For an empty [not quite the right word, is it] tree, there is one > end. > Uh, yeah, one path, like I said. Look, you have written your proof - I'm interlining my proof, a > different proof, of something different. (What do you mean by path? > Is it a 'way through the tree'?) > Yes a path is a succession of branches from the root, through child nodes, to > a > leaf, or possibly infinitely long with no discernible leaf. Proof: In a maximal binary tree with paths of length n, or depth of > n, there > are p(n)=2^n paths and b(n) = 2(2^n-1) = 2(p(n)-1) branches. What do you mean by Proof? Isn't this still part of the same > argument? You would get more respect if you took the trouble to > understand elementary notational conventions. > This was a second proof, using a slightly different statement. Oh, OK. In a tree that terminates at depth n, p(n) is the number of > terminal nodes, and b(n) is the number of 'branches', where a branch is > a link from one node to its child. Yeah, OK, I agree with the > 'numbers'. > Yay!!! > Anyway, never mind: to finish off *my* proof... By induction, the > number of ends of an endless tree is infinity. Da-dah!! > yes, you are very clever...... Well, I'd like you to address the substantive issue. Seems to me that > if one were to consider an unending tree, each path through it would be > unending, so there would be some infinite number of branches, but there > would also be zero leaf nodes. Do you agree? > If one imagines a tree that goes forever and never ends, one might imagine > that > there are no leaf nodes. If we look at the tree as representing the binary > numbers, each path corresponds to a single infinite string of bits, each > branch > corresponds to a bit, and the nodes are just the spaces between digits. The > root node corresponds to the digital point. So, no leaf nodes means that all > bit strings are infinite. I suppose I can live with an image with no leaf > nodes, but that doesn't affect the relationship between branches and paths, > does it? Yes it does! In trees in which every path has a leaf node, there is an obvious bijection between paths and leaf nodes, but in a maximal binary tree there are infinitely many paths and zero leaf nodes. If that correspondence fails so dramatically in the lomiting case, what is TO's evidence that any others still hold? === Subject: Re: infinity Randy Poe said: > Do you really think there is some specific QUANTITY > of natural numbers? > I would think that somebody using words like set A has > twice as many elements as set B thinks that there is > a specific quantity of elements in set A or set B. > Are you saying there aren't? Over any given range that relationship holds, but one can talk about different infinities realtive to each other, and apply those different infinities as ranges to get different size sets with the same pattern of numbers. > 4. There is no maximum element of the naturals, so there is no > maximum difference between elements. > But, if all elements are declared to be finite, no difference between them can > ever be infinite, so the maximum difference in the set is some finite value. Why does there have to be a maximum difference? > The maximum difference is the range. > So there is a maximum difference? > So there is a maximum natural number? No, but given all elements are finite, no number, and therefore no difference between numbers, can ever be infinite. Therefore, the set cannot have an infinite range of values, and an infinite number of unit differences cannot exist within the set, so the set cannot have an infinite number of whole numbers. > No, I introduced it as the definition of value range, since no one seems to > be able to decipher that cryptic term. Oy. > So what is the maximum natural number? There is no exact maximum number, but all numbers are less than infinity, accoding to you. > And if there isn't one, how can there be a maximum difference > or a range? There is no exact range, but there is no infinite range either, unless you have some infinite difference between some pair of values in your set. If the range of values is not infinite, then it is finite, since we have a range of vlaues wherever we have more than one value. If you cannot name the maximum element then you cannot name the range, but if all elements are finite, then the range is finite. > - Randy -- Smiles, Tony === Subject: Re: infinity > Randy Poe said: > I would think that somebody using words like set A has twice as > many elements as set B thinks that there is a specific quantity of > elements in set A or set B. Are you saying there aren't? > Over any given range that relationship holds, but one can talk about > different infinities realtive to each other, and apply those > different infinities as ranges to get different size sets with the > same pattern of numbers. In other words, set size is not constant in TO's set theory, but is whatever he wants it to be. 4. There is no maximum element of the naturals, so there is > no maximum difference between elements. > But, if all elements are declared to be finite, no difference > between them can ever be infinite, so the maximum difference > in the set is some finite value. Non-sequitur, unless TO is assuming a maximal member. Unless there are maximal and minimal members, there cannot be a maximal difference so that such sets do not have any range at all. On the other hand they can, and do, have infinite diameters. Why does there have to be a maximum difference? > The maximum difference is the range. So there is a maximum difference? So there is a maximum natural number? > No, but given all elements are finite, no number, and therefore no > difference between numbers, can ever be infinite. Therefore, the set > cannot have an infinite range of values Lack of an infinite range just means lack of members members, but that does not prohibit an infinite diameter. > So what is the maximum natural number? > There is no exact maximum number, but all numbers are less than > infinity, accoding to you. Right! If the range of values is not infinite, then it is finite The range, as TO has defined it, can fail to exist at all, and must fail to exist for sets like the set of all finite neatural numbers. The diameter, on the other hand, is defined for all sets of finite numbers, and is infinite for unbounded sets of finite numbers, even though such unbounded sets do not contain any infinite numbers. === Subject: Re: infinity Randy Poe said: > stephen@nomail.com said: > stephen@nomail.com said: >> I am not assuming that there is a longest word. A longest word >> implies a largest natural number. You always deny that there >> is a largest natural number, but once again you are using an >> argument that depends on there being a largest natural number. > I said longest WORDS, not longest word. There are no longest words either, so I am not assuming > anything about them. There is no maximum finite word length. > I have no idea what L is in your S^L. You are aware that there >> is more than one string length, so picking a single L does not >> make any sense. It almost makes sense if you think that L is >> the maximum string length, i.e. the largest finite natural number. >> Of course you also deny that there is a maximum string >> length, so I have no idea what S^L is supposed to mean. > Given any string length and alphabet, that is the maximum number of unique > srings in the language. I asked what L is. L is not the maximum number of unique > strings in the language. > You said you had no idea what S^L is, even though we have discussed this > before. L is any string length as I said. What do you not understand? > So L is a variable, which stands for the length of any > particular string in the language? > What we don't understand is why you use it as if it is > a constant for the language as a whole. > So if I have a language which has strings of length 1, > strings of length 2, strings of length 3, .... > strings of length n for all finite n, then what > value of L do you have in mind when you tell me there > are S^L words in my [according to you] finite language? All of them. The total number of words of length L or less is sum(x=1->L: S^x). > L couldn't be the maximum finite number could it? Because > you don't think there is one of those, do you? No, but if all string lengths L are finite, then sum(x=1->L: S^x) is also finite for all L. > But L couldn't mean 1 or 2 or 3, could it, even though > you said L is the length of any string in the language? > L couldn't be 10, because it wouldn't make sense to > say my language has only S^10 words in it. Clearly > there are words of length 11, and 12, and 13... As L approaches infinity, so does sum(x=1->L: S^x). > So are there only S^L words in the entire language? For > what value of L? S^L words of length L given alphabet of size S. > - Randy -- Smiles, Tony === Subject: Re: infinity > Randy Poe said: > So if I have a language which has strings of length 1, > strings of length 2, strings of length 3, .... > strings of length n for all finite n, then what > value of L do you have in mind when you tell me there > are S^L words in my [according to you] finite language? > All of them. The total number of words of length L or less is sum(x=1->L: > S^x). Then TO is claiming that the set of finite naturals is finite despite the successor operation mapping it to a proper subset of itself. Thus TO is claiming that sets that are Cantor-infinite can be TO-finite. Given a choice between Cantor and TO, the sane will always chose Cantor. === Subject: Re: infinity === Subject: Re: a linear algebra question Originator: bergv@math.uiuc.edu (Maarten Bergvelt) oops. ok then. But I think I can prove the claim if V is finite dimensional. The proof runs by induction on the dimension of V. If V=0 there is nothing to prove. Likewise, if V(j)=V for every j, there is nothing to prove. So pick a k with V(k) different from V. Since we know how to solve it if V(k) is zero, we have a solution modulo V(k). We can pull this solution back to V to obtain a family (x(i)) with x(j)-x(i) = a(i,j) mod V(i)+V(j)+V(k) In other words, there are c(i,j) in V(k) with x(j)-x(i) = a(i,j) + c(i,j) mod V(i)+V(j) Now this implies that the family c(i,j) satisfies the same axiom and lies in the smaller space V(k). By induction hypothesis there is a family y(i) with y(j)-y(i) = c(i,j) mod V(i)+V(j). Therefore the family z(i)=x(i)-y(i) will do the trick. === Subject: Re: a linear algebra question Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Wonderful! Here is a related question. Let F be a flabby sheaf of abelian groups having the following property. Any section vanishing on the intersection of two open sets can be written as a section vanishing on the first open set plus a section vanishing on the second open set. (Here and in the sequel ``section'' means ``global section''. For instance we can take for F the sheaf of discontinuous sections of any given sheaf of abelian groups.) Let A be the group of sections, (U_i) an open cover and and A_i the subgroup of sections vanishing on U_i. Our question can be phrased as follows: does the 1-cohomology of this cover with values in F vanish? The answer is yes by Corollary 1 to Theorem 3.8.1 in Grothendieck, Alexander. Sur quelques points d'alg.8fbre homologique. T.99hoku Math. J. (2), 9, 1957, 119--221. I suppose that flabby sheaves of abelian groups don't have the above property in general, but I have no argument to prove this. === Subject: Re: real points on an algebraic variety Originator: bergv@math.uiuc.edu (Maarten Bergvelt) I think the answer is affirmative. Proof as follows. (All citations are from the book Bochnak et al: Real Algebraic Geometry) V(R)subset R^n is an algebraic set. Theorem 2.8.3 implies that V(R) is a finite union of irreducible algebraic sets. Hence for our claim we can assume that V(R) is irreducible. By Prop. 3.3.14, V(R) contains a regular point and by Prop. 3.3.11 it contains a manifold of dimension equal to dim(V(R)). Baire's category Theorem implies that there is j such that V_j(R) has the same dimension as V(R). Since V(R) is irreducible, it follows V(R) = V_j(R).