mm-2503 === >Yep. Absolutely. That's why people who have outgrown your mediaeval >imponderables never place ranges nor declare finished... > Is Tony an idiot, or is he pulling our legs? It's hard for > me to figure it out. He ends his post with Smiles, which > might be a clue that it's all a big joke with him. It is hard to know. Trouble is that endlessly repeating the same confusions, even when they are wrenched into daylight, isn't exactly high in the humour rankings. I think the bottom line is that Tony has some ideas - basically he has studied some finite combinatorics and stuff, but simply assumes he knows what infinity means. He is much more coherent than most of the cranks, often lasting several sentences without collapsing (I mean, Ross Finlayson can't even do verbs and their objects yet). The problem is basically a combination of a massive misconception (that mathematics is a branch of experimental computing, roughly) with a strangely inflated confidence level. I think if I had never met another crank I would have assumed that Tony is just trying to pull our legs, but now I'm happy to accept his (presumed) assurance that he isn't. I also wasted a bit of time the other day reading some of the Phil threads - I had forgotten quite how similar the arguments were. There's always eventually this wriggling bit while sets expand to fill the requirements of the latest objection. Also, originally Tony appeared with this bigulosity idea, trying to assign sizes to infinite sets. One big problem IMO is that many people write, carelessly, as though cardinality (under the standard definition) is The Only Correct Way to Assign a Size to a set. And this is plainly not true: just the simple subset relation as a partial ordering is - at least in some sense - a sort of sizing. It looked as though if the Cardinality Good Anything Else Bad droners could be kept out of the way, Tony might have been guided to seeing how to formalise what he was thinking of, and even (!!) possibly to have seen why it isn't as satisfactory as he first thought. But then things went downhill, and we've been through several ridiculous episodes where he simply argues blindly against some elementary result, which he plainly doesn't properly understand. (I really _don't_ believe anyone could properly understand the A->P(A) bijection proof and still argue against it.) Brian Chandler http://imaginatorium.org === Subject: Re: infinity ... >> Do you consider the following a bijection from S={a,b,c} to its >> power set? >> [...] >> The above function is a bijection from {a,b,c,d,e,g,h,i} >> to P({a,b,c}). It is not a bijection from {a,b,c} to P({a,b,c}). > But, if this set went on forever, then i WOULD be in the set, and for any > subset, you could identify an x such that it mapped to that set. It is true > that no bijection is possible in the finite case. In the proof, a last > element of the set is assumed when we assume there is some string > representing the entire set. Again with the last element red herring. Just let it go, Tony. Consider the set of naturals N = {0, 1, 2, 3, ...} which obviously has no last element. I can map an infinite binary natural to it, so that f(x) = N: x = 2^0 + 2^1 + 2^2 + 2^3 + ... So I've managed to identify an x such that it mapped to the entire (infinite) set N. Granted, x must be an infinite natural, but it still works. And all without resorting to any last element in either the set or in the series. > However, to whatever extent we have considered > the set, say to N elements, we can always consider it to N+1, or 2^N > elements. If this is the set of all naturals, for instance, then N in the > set implies 2^N in the set. There is no end to the bijection. The proof > assumes one. If there is no end to the bijection, then the bijection is not complete, and it therefore not a bijection. If we can't account for all of the mappings of numbers to subsets, then we can't say the mapping is a bijection. None of the standard proofs assume a last element. Just let it go. === Subject: Re: infinity ... David R Tribble said: > Stephen said: >> Do you consider the following a bijection from S={a,b,c} to its >> power set? >> [...] >> The above function is a bijection from {a,b,c,d,e,g,h,i} >> to P({a,b,c}). It is not a bijection from {a,b,c} to P({a,b,c}). > But, if this set went on forever, then i WOULD be in the set, and for any > subset, you could identify an x such that it mapped to that set. It is true > that no bijection is possible in the finite case. In the proof, a last > element of the set is assumed when we assume there is some string > representing the entire set. > Again with the last element red herring. Just let it go, Tony. > Consider the set of naturals > N = {0, 1, 2, 3, ...} > which obviously has no last element. > I can map an infinite binary natural to it, so that f(x) = N: > x = 2^0 + 2^1 + 2^2 + 2^3 + ... Yes, that's the same one can do with *N. ....11111 maps to the entire set. > So I've managed to identify an x such that it mapped to the entire > (infinite) set N. Granted, x must be an infinite natural, but it > still works. And all without resorting to any last element in > either the set or in the series. Actually, since I consider the set of finite naturals to be finite, that number woudl have a finite number of bits and a finite value, but anyway... > However, to whatever extent we have considered > the set, say to N elements, we can always consider it to N+1, or 2^N > elements. If this is the set of all naturals, for instance, then N in the > set implies 2^N in the set. There is no end to the bijection. The proof > assumes one. > If there is no end to the bijection, then the bijection is not > complete, and it therefore not a bijection. If we can't account for > all of the mappings of numbers to subsets, then we can't say the > mapping is a bijection. Oh. What is the end of the mapping between naturals and evens? What is the even corresponding to the last natural, or for that matter, to any of the last half of the naturals? > None of the standard proofs assume a last element. Just let it go. It assumes a completed set with a completed string and a natural that equals that string, corresponding to the completed set. For any completion you consider, than number is beyond it. So, don't assume a completion to the set. -- Smiles, Tony === Subject: Re: infinity ... > Actually, since I consider the set of finite naturals to be finite, > that number woudl have a finite number of bits and a finite value, > but anyway... But if an TO-finite set can be endless (TO-finite but unbounded ordered sets are endless as sequences), then a TO-finite string of bits is equally endless, and can be Dedekind infinite. TO continues to use finite and infinite in two different senses, sometime the Dedekind sense and sometimes the TO sense, without specifying which. Such deliberate ambiguity vitiates any meaning to statements by TO involving use of either finite or infinite without specific explanation of which sense ic meant. > None of the standard proofs assume a last element. Just let it go. > It assumes a completed set with a completed string and a natural that > equals that string, corresponding to the completed set. For any > completion you consider, than number is beyond it. So, don't assume a > completion to the set. There is no such thing as an uncompleted set. the membership of any set is static. If a set exists at all it immutable. Any change in membership creates a new and different set. Thus, if x is not a member of set A, then A / {x}, where '/' means union, is a different set. Similarly, if x is a member of set b, then B{x}, where '' means relative difference, is a different set. Where does To get this TOmatic notion that consants vary? === Subject: Re: infinity ... (Yes, Randy is right...) that string, corresponding to the completed set. For any completion you > consider, than number is beyond it. So, don't assume a completion to the set. Pretty muddled, yet pretty clearly a rejection of the axiom of infinity. So you don't have infinite sets in the sense meant by mathematicians. Back to the middle ages, Tony... Brian Chandler http://imaginatorium.org === Subject: Re: infinity ... <85oe5liuuk.fsf@lola.goethe.zz> William Hughes said: >> Well, as you have not even defined 2^N for infinite TO naturals, you >> have some work to do. > Defined 2^N? It's 2 to the Nth power. What do you need to know? In binary, > it's a 1 followed by N 0's. Please state your complaint in the form of a > question. It is entirely reasonable to ask you to define 2^N. In the past, you've defined 'N' as the sum of an infinite series: N = 1 + 1 + 1 + 1 + ... or perhaps: N = 1 + 2 + 3 + 4 +... or perhaps: N = 1 + 2 + 4 + 8 + ... Most of the time, you simply say that 'N' is an arbitrary infinite number. For 2^N, a 1 followed by N zeros doesn't cut it, if for no other reason than because your 'N' is arbitrary. Perhaps you should define N a bit more precisely before you go throwing it around so freely. Defining your infinite naturals in more concrete terms, perhaps sets and successors, would be a step forward in making sense of your number theories. === Subject: Re: infinity ... David R Tribble said: > William Hughes said: >> Well, as you have not even defined 2^N for infinite TO naturals, you >> have some work to do. > Defined 2^N? It's 2 to the Nth power. What do you need to know? In binary, > it's a 1 followed by N 0's. Please state your complaint in the form of a > question. > It is entirely reasonable to ask you to define 2^N. > In the past, you've defined 'N' as the sum of an infinite series: > N = 1 + 1 + 1 + 1 + ... That is standard N. > or perhaps: > N = 1 + 2 + 3 + 4 +... That is (N^2+N)/2 > or perhaps: > N = 1 + 2 + 4 + 8 + ... That is 2^N-1 > Most of the time, you simply say that 'N' is an arbitrary infinite > number. Yes, or it can be the standard unit infinity. It's the unit we happen to be using for comaprison. > For 2^N, a 1 followed by N zeros doesn't cut it, if for no other > reason than because your 'N' is arbitrary. Perhaps you should define > N a bit more precisely before you go throwing it around so freely. That definition holds for finite or infinite N. It's fine. > Defining your infinite naturals in more concrete terms, perhaps > sets and successors, would be a step forward in making sense of your > number theories. Yes. I wopuld like to be able to express the sum of the harmonic series, or of the primes, for one thing. Things like log2(N), however, seem only describable in terms of some number of zeroes. -- Smiles, Tony === Subject: Re: infinity ... > stevendaryl3016@yahoo.com said: > Let A be any set whatsoever, finite or infinite, it doesn't matter. > > Let f be any function from A to P(A). > > Let w = { x in A | x is not an element of f(x) }. > > Let x = any set in A. > > Let u = f(x). We prove that u is not equal to w. > > By definition of w, we have x in w <-> x is not an element of f(x). > > So x in w <-> x is not an element of u. That means that there are > > two cases: Case 1: x in w, and x is not in u. In that case, u cannot > > equal w. Case 2: x is not in w, and x is in u. In that case, u cannot > > equal w. > > So what we have proved is that forall x, w is not equal to f(x). So > > w is not in the image of f. So f is not a bijection between A and P(A). > > There's no induction. There's no assumption that A is finite. > But there is an assumption that y is in S. If you are assuming you have the > complete set of naturals, that you have identified the last, and can therefore > identify the element that maps to the entire set, then you indeed run into a > contradiction. ... > Goodness you are dim. The axiom of infinity says (in effect) The > naturals are a set. This means we talk about the set of naturals, > letting all the other axioms apply to it. Why do we need to have an > _axiom_ to let us talk about the naturals? Because it is an infinite > set. It goes on forever, and never ends. There is no last natural. > There is no end to them. The end is not merely unspecified, > unidentified, tenuous, or any such, it is *nonexistent*. > (Remembering that nonexistence, like existence, is not a predicate.) > Having proved that there is no last natural, we nonetheless talk about > the complete, entire, total, set of all naturals. All naturals. There > is no natural anywhere in any of your nonsensical doubling and > thinning operations that does not already belong to the mathematical > set of all naturals. That's what all means. > Sorry, mustn't go on - it's pointless anyway, since after thousands of > posts it's pretty unlikely you will ever grasp any of it. But anyway, > to go back to your paragraph: Yes, your reiterations of the standard nonsense don't go very far with me. Sorry. Having noted that there is no last natural, and that the size of the set to which any natural is successor is itself, you nonetheless assign a set size while declaring the nonexistence of a largest element. Here, you again declare you have the entire set, derive a contradiction from that, and deflect it at bijections. The bijection will fail with any finally declared set. Where the sets don't end, it never fails. > Yes, we have the complete set of naturals. > No, we have not identified the last, because there isn't one. Then how do you know the size, or how many bits you need in y? You have, in effect, declared N as the size of the set, and noted that y=2^N and is thus outside the set. The set doesn't end. There is no natural outside of it. If y is a natural, 2^y is a natural. When is that not the case? > Can therefore identify the element that maps to the entire set makes > no sense. We are considering any mapping from a set A to its power set > P(A). In some mappings there is an element mapped to the complete set - > e.g. (a crank favourite) > 0 -> {} > 1 -> {0, 1, 2, 3, ... } // the complete set of naturals > 2 -> {0} > 3 -> {1, 2, 3, 4, ... } // all naturals except 0 > 4 -> {1} > ... That looks like one of Albrecht's. It reminds me of my adjustment to the Peano axioms to include both finite and infinite. It could have some application. But, it's not what I offered. > ... However, both the infinite set of naturals and the infinite > power set go on forever, so you never run out of naturals to map to subsets, > nor subsets to map to naturals. > Right: well done!! You got something right. So, what about that situation is not a bijection? > ... Despite the fact that, within any range up to > S, you cannot map every subset to an element within that range, the lack of a > largest element makes it so there DOES exist an element that maps to all n > through S, but it is more than S. This is a prime example of where the value > range matters in the bijection. > No it isn't. There is no value range in a bijection. Value ranges > are only used in TOmatics, remember. But you are clawing your way > towards grasping what was the natural assumption (that if two sets go > on forever it is never possible to say there can't be a bijection > between them) before Cantor pointed out that it is wrong. What is wrong with my bijection? No specific rule has been broken, has it? > Here's a question for you. Obviously if the TOnats (T) are a set, > within the framework of conventional set theory, the proof applies to > them, and there can be no bijection T <-> P(T). But are they a set? Is > it possible to have the complete, final, total, can never be extended > in response to the next question, set of TOnats? It is never possible to place any exact number on it. Infinite sets can only be compared with each other over any given range. We can speak of the entire set, but as soon as we speak of any specific size, we will encounter contradictions. > Brian Chandler > http://imaginatorium.org -- Smiles, Tony === Subject: Re: infinity ... > > stevendaryl3016@yahoo.com said: > Let A be any set whatsoever, finite or infinite, it doesn't matter. > > Let f be any function from A to P(A). > > Let w = { x in A | x is not an element of f(x) }. > > Let x = any set in A. > > Let u = f(x). We prove that u is not equal to w. > > By definition of w, we have x in w <-> x is not an element of f(x). > > So x in w <-> x is not an element of u. That means that there are > > two cases: Case 1: x in w, and x is not in u. In that case, u cannot > > equal w. Case 2: x is not in w, and x is in u. In that case, u > > cannot > > equal w. > > So what we have proved is that forall x, w is not equal to f(x). So > > w is not in the image of f. So f is not a bijection between A and P(A). > > There's no induction. There's no assumption that A is finite. > > > But there is an assumption that y is in S. If you are assuming you have > > the > > complete set of naturals, that you have identified the last, and can > > therefore > > identify the element that maps to the entire set, then you indeed run > > into a > > contradiction. ... > > Goodness you are dim. The axiom of infinity says (in effect) The > naturals are a set. This means we talk about the set of naturals, > letting all the other axioms apply to it. Why do we need to have an > _axiom_ to let us talk about the naturals? Because it is an infinite > set. It goes on forever, and never ends. There is no last natural. Then why does TO keep making assumptions that require a last naturals? TO seems to want to have things both ways, no largest and having a largest simultaneously. Fortunately for his delusion, in his world of TOmatics both can be true and both false simultaneously. > There is no end to them. The end is not merely unspecified, > unidentified, tenuous, or any such, it is *nonexistent*. > (Remembering that nonexistence, like existence, is not a predicate.) > > Having proved that there is no last natural, we nonetheless talk about > the complete, entire, total, set of all naturals. All naturals. There > is no natural anywhere in any of your nonsensical doubling and > thinning operations that does not already belong to the mathematical > set of all naturals. That's what all means. > > Sorry, mustn't go on - it's pointless anyway, since after thousands of > posts it's pretty unlikely you will ever grasp any of it. But anyway, > to go back to your paragraph: > Yes, your reiterations of the standard nonsense don't go very far with me. TO has his own brand of nonsense which he much prefers, since in his one can be on both sides of every fence simultaneoulsy. > Sorry. Having noted that there is no last natural, and that the size > of the set to which any natural is successor is itself, you > nonetheless assign a set size while declaring the nonexistence of a > largest element. Here, you again declare you have the entire set, > derive a contradiction from that, and deflect it at bijections. The > bijection will fail with any finally declared set. Where the sets > don't end, it never fails. Then let TO produce any bijection between the endless set of finite naturals and its power set. Or does TO still claim that the Dedekind infinite set of finite naturals ends? > > Yes, we have the complete set of naturals. > No, we have not identified the last, because there isn't one. > Then how do you know the size, or how many bits you need in y? We do not need to have a size in any TOish sense to do all that needs doing. Cardinality serves us well, and we do not need any of TO's artificial additions to standard math. > You have, in effect, declared N as the size of the set We have, in effect, not decalred a size at all, other than a cardinality. > > ... However, both the infinite set of naturals and the infinite > > power set go on forever, so you never run out of naturals to map to > > subsets, > > nor subsets to map to naturals. > > Right: well done!! You got something right. > So, what about that situation is not a bijection? NO! > > This is a prime example of where the value range matters in the > > bijection. > > No it isn't. There is no value range in a bijection. Value ranges > are only used in TOmatics, remember. But you are clawing your way > towards grasping what was the natural assumption (that if two sets go > on forever it is never possible to say there can't be a bijection > between them) before Cantor pointed out that it is wrong. > What is wrong with my bijection? No specific rule has been broken, has it? Yes, the one that says that every member of the codomain must be in the image. > > Here's a question for you. Obviously if the TOnats (T) are a set, > within the framework of conventional set theory, the proof applies to > them, and there can be no bijection T <-> P(T). But are they a set? Is > it possible to have the complete, final, total, can never be extended > in response to the next question, set of TOnats? > It is never possible to place any exact number on it. Infinite sets > can only be compared with each other over any given range. At least inside TOmatics, though not ouside its idiocy. > We can > speak of the entire set, but as soon as we speak of any specific > size, we will encounter contradictions. So outside of TOmatics, one merely avoids speaking of size and uses only cardinality. > > Brian Chandler http://imaginatorium.org > === Subject: Re: infinity ... Brian Chandler: >> Yes, we have the complete set of naturals. >> No, we have not identified the last, because there isn't one. > Then how do you know the size, or how many bits you need in y? You have, in > effect, declared N as the size of the set, and noted that y=2^N and is thus > outside the set. The set doesn't end. There is no natural outside of it. If y > is a natural, 2^y is a natural. When is that not the case? It's always the case that for any finite natural y in N, 2^y also exists as a finite natural in N. (You just said so.) Which proves that there is no last (or largest) finite member in N, because for any finite natural y in N, you can always find a larger finite 2^y in N. So you just proved that there is no largest finite natural, and that N has no last finite member, and that N is an infinite set. Or do you still maintain that N is a finite set, contradicting what you just said above? === Subject: Re: infinity ... stevendaryl3016@yahoo.com said: > Let A be any set whatsoever, finite or infinite, it doesn't matter. > > Let f be any function from A to P(A). > > Let w = { x in A | x is not an element of f(x) }. > > Let x = any set in A. > > Let u = f(x). We prove that u is not equal to w. > > By definition of w, we have x in w <-> x is not an element of f(x). > > So x in w <-> x is not an element of u. That means that there are > > two cases: Case 1: x in w, and x is not in u. In that case, u cannot > > equal w. Case 2: x is not in w, and x is in u. In that case, u cannot > > equal w. > > So what we have proved is that forall x, w is not equal to f(x). So > > w is not in the image of f. So f is not a bijection between A and P(A). > > There's no induction. There's no assumption that A is finite. > > But there is an assumption that y is in S. If you are assuming you have the > > complete set of naturals, that you have identified the last, and can therefore > > identify the element that maps to the entire set, then you indeed run into a > > contradiction. ... > Goodness you are dim. The axiom of infinity says (in effect) The > naturals are a set. This means we talk about the set of naturals, > letting all the other axioms apply to it. Why do we need to have an > _axiom_ to let us talk about the naturals? Because it is an infinite > set. It goes on forever, and never ends. There is no last natural. > There is no end to them. The end is not merely unspecified, > unidentified, tenuous, or any such, it is *nonexistent*. > (Remembering that nonexistence, like existence, is not a predicate.) > Having proved that there is no last natural, we nonetheless talk about > the complete, entire, total, set of all naturals. All naturals. There > is no natural anywhere in any of your nonsensical doubling and > thinning operations that does not already belong to the mathematical > set of all naturals. That's what all means. > Sorry, mustn't go on - it's pointless anyway, since after thousands of > posts it's pretty unlikely you will ever grasp any of it. But anyway, > to go back to your paragraph: > Yes, your reiterations of the standard nonsense don't go very far with me. Right. Fairly soon, I think we should all give in. Accept that we just can't understand what Teacher Tony is Telling us. (OK guys? Let's give in. Tony's right. Anything that's been repeated at least 4000 times just has to be true, no?) Do share with us what your next step is? Are you going to write a text book? At least a web page? Convert a university department? Or just tell your grandchildren that the people on Usenet said you were right, so all the universities just must be wrong? > Sorry. Having noted that there is no last natural, and that the size of the set > to which any natural is successor is itself, you nonetheless assign a set size I assign a set size? Sorry, Tony, my brain is getting old and tired. Kindly fish out for me the place in my proof where I assign a set size... Just for the hell of it, here's the proof, with boxes [ ], so you can just put a tick where I assign the size. Note that the proof does not include the words finite or infinite: it depends only on the simplest properties of all sets - basically that they have members, and something either is or is not a member. Consider an arbitrary set A. [ ] (Thus any entity in the universe [proper class] of all entities either is or is not a member of A [ ].) Consider the power set of A. [ ] (The set of all subsets of A, including A itself; thus for any particular one of these subsets B, any element of A either is or is not a member of B [ ].) Now consider any mapping f from A into P(A) [ ]. (At least one such mapping exists [ ]: the canonical map that takes any element x e A [e means 'element of' OK?] to the subset {x} e P(A). So no worries about existence.) We want to show that whatever mapping f is chosen, there will be at least one element q of P(A) (q is a subset of A, right?) which the mapping f does not map anything onto. [ ] Consider an arbitrary element x e A [ ], and suppose f(x) = y e P(A). [ ] Well, y is a subset of A, so x either is or is not an element of y. [ ] Thus we can divide the elements of A into two classes: [ ] CLASS I: those which map to a subset of which they are a member [ ] CLASS II: those which map to a subset of which they are not a member [ ] Call w the subset of A which includes exactly those elements in CLASS II. [ ] [-----end prologue-----] Now I assert that the particular mapping f we are considering at the moment does not map any element of A to the subset w in P(A). [ ] Every element p inside w is mapped to a subset that does not include p, so it cannot be mapped to w. [ ] Every element p' outside w is mapped to a subset that _does_ include p', which thus includes an element outside w, and thus p' cannot be mapped to w. [ ] Therefore the mapping f is not a surjection. (We called that onto when I was doing this stuff.) Thus there is no bijection. But anyway... > while declaring the nonexistence of a largest element. Here, you again declare > you have the entire set, derive a contradiction from that, and deflect it at > bijections. The bijection will fail with any finally declared set. Where the > sets don't end, it never fails. Crapola. You are at least getting to the point of reiterating what would have been the natural view before Cantor pointed out it was wrong. But again you imply that I would go around finally declaring sets. I deny this very strongly. I have never wittingly finally declared a set in my life, and I demand an apology. Sorry - deleting the digital fixation babble. Brian Chandler http://imaginatorium.org === Subject: Re: infinity ... >Right. Fairly soon, I think we should all give in. Accept that we just >can't understand what Teacher Tony is Telling us. (OK guys? Let's give >in. Tony's right. Anything that's been repeated at least 4000 times >just has to be true, no?) Do share with us what your next step is? Are >you going to write a text book? At least a web page? Convert a >university department? Or just tell your grandchildren that the people >on Usenet said you were right, so all the universities just must be >wrong? You're right. People are only arguing with Tony because they have jobs that depend on the blind acceptance of establishment mathematics. Or in some cases, because they have spent a lot of money and time learning material that is nonsense, and they can't admit to themselves that they fell so hard for such a scam. After Tony mathematics has replaced the standard mathematics in schools and textbooks, will we be able to practice pre-revolution mathematics in our own homes, in our spare time, or will that be outlawed? -- Daryl McCullough Ithaca, NY === Subject: Re: infinity ... [proper class] of all entities either is or is not a member of A [ ].) > Consider the power set of A. [ ] (The set of all subsets of A, > including A itself; thus for any particular one of these subsets B, any > element of A either is or is not a member of B [ ].) > Now consider any mapping f from A into P(A) [ ]. (At least one such > mapping exists [ ]: the canonical map that takes any element x e A [e > means 'element of' OK?] to the subset {x} e P(A). So no worries about > existence.) > We want to show that whatever mapping f is chosen, there will be at > least one element q of P(A) (q is a subset of A, right?) which the > mapping f does not map anything onto. [ ] > Consider an arbitrary element x e A [ ], and suppose f(x) = y e P(A). [ > Well, y is a subset of A, so x either is or is not an element of y. [ ] > Thus we can divide the elements of A into two classes: [ ] > CLASS I: those which map to a subset of which they are a member [ ] > CLASS II: those which map to a subset of which they are not a member [ > Call w the subset of A which includes exactly those elements in CLASS > II. [ ] > [-----end prologue-----] > Now I assert that the particular mapping f we are considering at the > moment does not map any element of A to the subset w in P(A). [ ] > Every element p inside w is mapped to a subset that does not include p, > so it cannot be mapped to w. [ ] > Every element p' outside w is mapped to a subset that _does_ include > p', which thus includes an element outside w, and thus p' cannot be > mapped to w. [ ] > Therefore the mapping f is not a surjection. (We called that onto > when I was doing this stuff.) > Thus there is no bijection. Forgot to add in: not only does this proof not mention finite or infinite, it doesn't mention empty or non-empty either. So it works for the empty set too (how many 'bits' is that, I wonder)... The empty set E has no elements and one subset. Therefore the only mapping from E -> P(E) is the null mapping (doesn't map anything to anything). This does not therefore map anything to the only element of P(E) = {{}}, which is the empty set. Brian Chandler http://imaginatorium.org === Subject: Re: infinity ... Goodness you are dim. The axiom of infinity says (in effect) The > > naturals are a set. This means we talk about the set of naturals, > > letting all the other axioms apply to it. Why do we need to have an > > _axiom_ to let us talk about the naturals? Because it is an infinite > > set. It goes on forever, and never ends. There is no last natural. > > There is no end to them. The end is not merely unspecified, > > unidentified, tenuous, or any such, it is *nonexistent*. > > (Remembering that nonexistence, like existence, is not a predicate.) > Yes, your reiterations of the standard nonsense don't go very far with me. > Right. Fairly soon, I think we should all give in. Accept that we just > can't understand what Teacher Tony is Telling us. (OK guys? Let's give > in. Tony's right. Anything that's been repeated at least 4000 times > just has to be true, no?) Do share with us what your next step is? Are > you going to write a text book? At least a web page? Convert a > university department? Or just tell your grandchildren that the people > on Usenet said you were right, so all the universities just must be > wrong? I think, though he is unable to articulate it, that what Tony is rejecting is the Axiom of Infinity. So he's grasping toward a system without one, and replacing the set of natural numbers with some vague, ever shifting thing whose membership is never the same twice and always has just enough members for whatever you're using it for at the moment. At any rate, since we work with a system in which AofI is included and Tony rejects it, that seems to be pretty much the end of the story. Of course there's no hope of convincing Tony of anything, least of all that he contradicts himself sometimes within the space of a single sentence. For at least 4000 posts the game instead has been: - see how long this thread can go, - try to goad TO into ever more ridiculous proclamations - try to figure out what TO's axioms are - Randy === Subject: Re: infinity ... Tony Orlow says... >Yes, your reiterations of the standard nonsense don't go very far with me. That's because you are an idiot. You are incapable of formulating or understanding a mathematical argument. Worse, you believe yourself to be much more competent than you actually are. That's the point, really, of rigor and formalization. If you are speaking loosely, it's very easy to fool yourself into thinking you are making sense. But if you are forced to write down your reasoning in a careful, detailed, rigorous way, you will discover where your ideas are full of hot air. Standard mathematics *has* gone through this process. Your junk has not. That's why standard mathematics is much higher quality than your bull. It's not because mathematicians have a higher IQ, but because they have a much stricter quality control process than you do. You have no quality control AT ALL. Even when you are confronted with an out-and-out contradiction in your ideas, it hardly fazes you---you don't change your ideas in the face of evidence that they are wrong, you just make up new weasel words like tenuous existence, unidentifiable naturals. The fact that you are not stupid makes your behavior all the worse: your idiocy is a *choice* on your part. You would rather be lazy and pretend to be doing mathematics than to work hard and *actually* do mathematics. -- Daryl McCullough Ithaca, NY === Subject: Re: infinity ... Daryl McCullough said: > Tony Orlow says... >Yes, your reiterations of the standard nonsense don't go very far with me. > That's because you are an idiot. You are incapable of formulating > or understanding a mathematical argument. Worse, you believe yourself > to be much more competent than you actually are. So, you wanna go out for a beer, or what? LOL > That's the point, really, of rigor and formalization. If you are > speaking loosely, it's very easy to fool yourself into thinking > you are making sense. But if you are forced to write down your > reasoning in a careful, detailed, rigorous way, you will discover > where your ideas are full of hot air. Then perhaps they will fly. > Standard mathematics *has* gone through this process. Your junk > has not. That's why standard mathematics is much higher quality > than your bull. It's not because mathematicians have a higher > IQ, but because they have a much stricter quality control process > than you do. You have no quality control AT ALL. Even when you > are confronted with an out-and-out contradiction in your ideas, > it hardly fazes you---you don't change your ideas in the face of > evidence that they are wrong, you just make up new weasel words > like tenuous existence, unidentifiable naturals. few paradoxical facts into general rules that defy reason. I don't expect you to see that. I also won't call you an idiot for not seeing what is obvious to people like Martin, Zuhair, Ross, myself and others. > The fact that you are not stupid makes your behavior all the > worse: your idiocy is a *choice* on your part. You would rather > be lazy and pretend to be doing mathematics than to work hard > and *actually* do mathematics. Do you think it is lazy to try to construct a better foundation for mathematics? Well, it is a lot easier than regurgitating what I learned in books. It's so much easier to paint than go to the museum. Only lazy people make anything new. The real work is in defending the status quo against the crackpots. Oh, the hordes at the gate! Woe to those lazy barbarians! > -- > Daryl McCullough > Ithaca, NY -- Smiles, Tony === Subject: Re: infinity ... > Daryl McCullough said: > Tony Orlow says... > > >Yes, your reiterations of the standard nonsense don't go very far > >with me. > > That's because you are an idiot. You are incapable of formulating > or understanding a mathematical argument. Worse, you believe > yourself to be much more competent than you actually are. > So, you wanna go out for a beer, or what? LOL Didn't know they had bars in TOmatica. perhaps it is the bars in TOmatica that TO never leaves, and that is his problem. > > That's the point, really, of rigor and formalization. If you are > speaking loosely, it's very easy to fool yourself into thinking you > are making sense. But if you are forced to write down your > reasoning in a careful, detailed, rigorous way, you will discover > where your ideas are full of hot air. > Then perhaps they will fly. Not far, as such bubbles always burst quickly. > > Standard mathematics *has* gone through this process. Your junk has > not. That's why standard mathematics is much higher quality than > your bull. It's not because mathematicians have a higher IQ, > but because they have a much stricter quality control process than > you do. You have no quality control AT ALL. Even when you are > confronted with an out-and-out contradiction in your ideas, it > hardly fazes you---you don't change your ideas in the face of > evidence that they are wrong, you just make up new weasel words > like tenuous existence, unidentifiable naturals. > I express things the way I see them. Then TO needs a vision upgrade. Without an axiom system things cannot be airtight, that is the point of starting with an axiom system. We have a number of different axiom systems in all of which we have a Dedekind infinite set of (Dedekind finite) naturals from which we can build the usual number systems, calculus, etc., on which almost all modern mathematics, science and engineering is based. TO has no system of axioms, only some mutually- and self-contradictory intuitions which he can not make work in any system that requires adherence to the standard rules of logic. > very few paradoxical facts into general rules that defy reason. TO's reason defies logic. Most people, given the choice, will choose to defy TO and his reason before defying logic. > I don't expect you to see that. I also won't call you an idiot for > not seeing what is obvious to people like Martin, Zuhair, Ross, > myself and others. The difficulty with that group is that while they may publicly disagree with standard logic and mathematics, they do not entirely agree with each other, and occasionally even disagree with themselves. > > The fact that you are not stupid makes your behavior all the worse: > your idiocy is a *choice* on your part. You would rather be lazy > and pretend to be doing mathematics than to work hard and > *actually* do mathematics. > Do you think it is lazy to try to construct a better foundation for > mathematics? If TO were trying to do that, he would not defy logic nor ignore logical arguments. Any mathematics in such defiance of logic is doomed to end up in a midden. Where it belongs. === Subject: Re: infinity ... Tony Orlow says... >Do you think it is lazy to try to construct a better foundation for >mathematics? You haven't done that. You've just posted ill-considered nonsense. You haven't done any actual mathematics. >Well, it is a lot easier than regurgitating what I learned in >books. Nobody has been regurgitating what they read in books. The people who have been arguing with you have been trying to make sense of *your* statements. We've tried to understand what you were saying about bigulosity, we've tried to understand what you've been saying about infinite naturals. People have been working *much* harder at making sense of your nonsense than you ever worked at making sense of actual mathematics. >It's so much easier to paint than go to the museum. Only >lazy people make anything new. The real work is in defending >the status quo against the crackpots. As I've pointed out to you before, nobody has any trouble accepting different points of view about mathematics. I personally am somewhat familiar with recursion theory, ZF set theory, Quine's New Foundations set theory, Aczel's non-well-founded set theory, constructive type theory, the lambda calculus, category theory, nonstandard analysis, combinator theory, Conway's surreal numbers. These are all *drastically* different ways of looking at the foundations of mathematics. I don't reject any of them just because they are different from the status quo. I'm always on the look out for new theories, new ways to look at things. But the difference between all these subjects and Tony Orlow mathematics is that they are all *rigorous*. The people that developed them put *thought* into them. They actually tried to get clear what their basic assumptions were, and they tried to rigorously work out the consequences of those assumptions. You flatter yourself to think that we are rejecting what you say because it goes against the status quo. People aren't rejecting your ideas because they are challenging, or because they are new, or because they attack the status quo. People are rejecting your ideas because they are idiotic. They are ill-thought-out. They are nonsensical. They are a complete mess. -- 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 > Daryl McCullough said: >> Tony Orlow says... >>Yes, your reiterations of the standard nonsense don't go very far with me. >> That's because you are an idiot. You are incapable of formulating >> or understanding a mathematical argument. Worse, you believe yourself >> to be much more competent than you actually are. > So, you wanna go out for a beer, or what? LOL >> That's the point, really, of rigor and formalization. If you are >> speaking loosely, it's very easy to fool yourself into thinking >> you are making sense. But if you are forced to write down your >> reasoning in a careful, detailed, rigorous way, you will discover >> where your ideas are full of hot air. > Then perhaps they will fly. Sure. But pray open the window instead of blowing them here. They stink. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: infinity ... So, what about that situation is not a bijection? You claim to have a bijection f from N to P(N). One element of P(N) is the set N itself. You admit that there is no y in N such that f(y) = N. Therefore f is not a bijection from N to P(N). > What is wrong with my bijection? No specific rule has been broken, has it? The definition of bijection, that it be onto P(N), that every element of P(N) be mapped by some element of N. - Randy === Subject: Re: infinity ... Randy Poe says... >You claim to have a bijection f from N to P(N). >One element of P(N) is the set N itself. >You admit that there is no y in N such that f(y) = N. >Therefore f is not a bijection from N to P(N). Actually, Tony is not claiming that there is a bijection between the *finite* naturals and the power set of the finite naturals. He is claiming that there is a bijection between *N and P(*N), where *N is some kind of nonstandard model of the naturals that includes infinite naturals. Not that that makes any difference to Cantor's proof... -- Daryl McCullough Ithaca, NY === Subject: Re: infinity ... of them. I can give you that, when you're ready, which will probably never > be. No, by all means, please do. I'm ready. Oh, but let me guess. The first point in the real line is: 0:0.000...001, and the next point is 0:0.000...002, and so on. But then how do you explain the points between them (you know, 0:0.000...000:000...001 0:0.000...000:000...002 and so on), if the real line is continuous? You do know what continuous and infinitely divisible mean, right? === Subject: Re: infinity ... Sorry for the length of this post. Normally, I would only respond to the most egregious errors, but obviously you (Tony) did not understand anything I said in my post. Tony Orlow: >> No, it was specifically a bijection between two sets of infinite binary >> strings representing, on the one hand, the whole numbers in *N starting >> from 0, both finite and infinite, in normal binary format, and on the >> other hand, the specification of each subset of whole numbers in *N, >> where each bit which, in the binary number, represents 2^n denotes >> membership of n in the subset. This is a bijection between the whole >> numbers in *N and P(*N), using an intermediate bijection with a common >> set of infinite binary strings. David R Tribble: >> We can show a mapping (surjection) from the finite naturals in N >> to the finite subsets of N (the finite members of P(N)) using your >> mapping strategy. But then you don't have any more naturals left >> to map to the infinite subsets in P(N). Tony Orlow: > No more FINITE naturals. Does *N include infinite values or not? Yes, *N contains all the finite and all the infinite naturals. I'm the one who starting using the *N notation in this thread in the first place, after all, to distinguish the Tony set of naturals from the standard set. But you obviously are not paying attention. Here I'm talking about a bijection between N and P(N). We already agree that a bijection between *N and P(N) is easy, so I don't even bother mentioning it. The next paragraph is where I talk about a bijection between *N and P(*N). David R Tribble: >> Pretty much the same thing for *N - using your binary mapping >> stragegy, we can show a mapping between the finite naturals in *N >> to the finite subsets of *N (which are the finite subsets in P(*N)), >> and a mapping between the infinite naturals in *N to some of the >> infinite subsets in P(*N). But after we've run out of members in >> *N for mapping, there are still more infinite subsets remaining >> in P(*N) that we can't map. Tony Orlow: > Uh, when do we run out of the unending list of infinite natural numbers? > Which is the last one? For every one, there is always another after it, > right? So, when does the bijection break down? After the 2^Nth element of > the set of N naturals? We all know that there is no last member of *N; this has been mentioned hundreds of times already in this thread. Just let it go. It's obvious that you don't understand. We run out of members of *N, finite and infinite, because we use them all to map to subsets of *N. But there remain subsets of *N that still are not mapped, and we've used up all the members of *N in our mapping process, so there are more subsets of *N than members in *N that can map to them. You can't use a member of *N to map to more than one subset; once you've mapped a natural to a subset, it's used up by the mapping process, and you have to choose a different natural to map the next subset. That's my whole point, that there are subsets of *N that cannot be mapped by any members of *N, which is why your strategy is not a proper bijection. David R Tribble: >> Consider the subsets S(n) for any n in *N, using your binary index >> mapping scheme. If n is finite, the subset S(n) is also finite; >> for example, >> S(42) = S(101010(2)) = {1,3,5}. >> If n is infinite, S(n) is also infinite; for example, >> S(...999) = S(...111(2)) = {0,1,2,3,...}. >> But there are still more subsets left in P(*N), and there are not >> enough left over members in *N, finite or infinite, to map to them. >> These subsets contain at least one infinite natural, and that >> natural has no bit index corresponding to a member of *N. Tony Orlow: > What? Every element of *N has an infinite number of bits, even if the string > represents a finite value. Just because an element has an infinite value, and > is represented by an equally infinite bit position, that is not a problem, > since infinite values in binary REQUIRE 1 bits in infinite positions in the > string. It's obvious that you don't understand. A bit position number must be longer than the bitstrings containing its corresponding bit value. Bit position j corresponds to the bitstring value 2^j, and 2^j > j, right? If you're going to use index j in a subset, then there must be a natural k >= 2^j, right? But how do you deal with an infinite natural j, which appears in an infinite number of subsets of *N, if you've already previously mapped all the finite and infinite naturals in *N to the other subsets containing only finite naturals? David R Tribble: >> For example, subset B = {0, ...999} has no corresponding n in *N >> that can map to it, because there is no natural bit index available >> to represent the second member. Tony Orlow: > You are assuming a finite natural number of bits, but that is simply not the > case with *N, or it would ony contain finite values. Are we talking about the > same set? You obviously don't understand what I'm saying. You're trying to create a straw-man argument out of what you *think* I'm saying, them saying it's false, so therefore your logic wins. But it's obvious I'm not saying anything of the sort; I said nothing about a last element nor a last bit, because we all know there are no such things. We're talking about infinite sets and infinite naturals, right? Just let it go. Tony Orlow: > Lets say that, since we are using binary, you mean 0:111...111 with N digits. > This number is 2^N-1. That's the number that corresponds to the subset > containing the first N elements. It's not IN those first N elements, but it's > in the infinite set. And it's not in any of the sets that can be mapped by any member of *N, because all of those members are used to map to other subsets that do not contain any infinite naturals. Try it for yourself. List all the subset mappings using your strategy, and show us where any natural in *N, finite or infinite, maps to any subset containing an infinite natural. There are none. David R Tribble: >> This example is just one of an >> infinite number of finite and infinite subsets of *N that cannot >> be mapped with the finite and infinite members of *N. Tony Orlow: > Ah, but they can, if you avoid assuming in some way that you have identified > the last of them. That's what infinite bijections are all about. You simply > push all differences down the line, until theyre infinitely far away and you > can't see them any more. At lease, that's how transfinite cardinality works. Tony, Tony, Tony. You obviously can't follow what I'm saying. Just let it go. David R Tribble: >> The conclusion is that there does not exist a surjection from the >> members of *N to the members of P(*N), and therefore no bijection >> between them is possible. Tony Orlow: > But that conclusion assumes a last element, whence it draws its > contradictions. Obviously, you're just ignoring what I said. I state quite clearly and explictly that your strategy leaves some subsets of *N unmapped by any members of *N, so even though it is an injection from *N to P(*N), it is not a surjection, so it is not a bijection. It is incumbent upon you to show otherwise, and so far you have not done that. There is no red herring straw-man last element. Just let it go. === Subject: Re: infinity ... Virgil said: > I don't understand why an inductive proof of an equality relationship > does not extend to infinite numbers. > Informally, the Peano axioms may be stated as follows: > 1 There is a natural number 0. > 1 Every natural number a has a successor, denoted by S(a). > 3 There is no natural number whose successor is 0. > 4 Distinct natural numbers have distinct successors: > if not(a = b), then not(S(a) <> S(b)). > 5 If a property is possessed by 0 and also by the successor of > every natural number which possesses it, then it is possessed > by all natural numbers. (This axiom ensures that the proof > technique of mathematical induction is valid.) Says nothing about finiteness there...... > More formally, we define a Dedekind-Peano structure to be an ordered > triple (X, x, f), satisfying the following properties: > 1 X is a set, x is an element of X, and f is a map from X to itself. > 2 x is not in the range of f. > 3 f is injective. > 4 If A is a subset of X satisfying: > 5 If [x is in A, and > (If a is in A, then f(a) is in A)] > then A = X. Nope, still no finiteness...... > A member of the set of such objects so defined is defined as finite if > the set of such objects of which it is ultimately a successor is a > finite set by the Dedekind definition of finite sets. So, all finite numbers are successors to finite sets. Then, which finite is a successor to any infinite portion of the infinite set of finite naturals? None? Hmmmm..... That doesn't seem to work very well. > With that definition of finiteness of such objects, all such objects are > provably finite. Only the ones that are successors to a finite set. The ones that are successors to infinite-size sets are infinite in value. So, you assume there are only finite naturals, and this is your reasoning for why induction doesn't work in the infinite case? You aren't actually capable of thinking about what your axioms mean, are you? You have given no reason why induction only works in the finite case. -- Smiles, Tony === Subject: Re: infinity ... > Virgil said: > > > I don't understand why an inductive proof of an equality relationship > > does not extend to infinite numbers. > > Informally, the Peano axioms may be stated as follows: > 1 There is a natural number 0. > 1 Every natural number a has a successor, denoted by S(a). > 3 There is no natural number whose successor is 0. > 4 Distinct natural numbers have distinct successors: > if not(a = b), then not(S(a) <> S(b)). > 5 If a property is possessed by 0 and also by the successor of > every natural number which possesses it, then it is possessed > by all natural numbers. (This axiom ensures that the proof > technique of mathematical induction is valid.) > Says nothing about finiteness there...... > > More formally, we define a Dedekind-Peano structure to be an ordered > triple (X, x, f), satisfying the following properties: > 1 X is a set, x is an element of X, and f is a map from X to itself. > 2 x is not in the range of f. > 3 f is injective. > 4 If A is a subset of X satisfying: > 5 If [x is in A, and > (If a is in A, then f(a) is in A)] > then A = X. > Nope, still no finiteness...... But when one constrins one self to finiteness, one loses absolutely nothing of what Peano provides, so no infinitenes is needed, or possible. > > A member of the set of such objects so defined is defined as finite if > the set of such objects of which it is ultimately a successor is a > finite set by the Dedekind definition of finite sets. > So, all finite numbers are successors to finite sets. That is not what was said, TO. Learn to read. > Then, which finite is a > successor to any infinite portion of the infinite set of finite naturals? > None? > Hmmmm..... That doesn't seem to work very well. It works absolutly wonderfully everywhere outside TOmatics. And what goes on inside TOmatics is irrelevant to that. > > With that definition of finiteness of such objects, all such objects are > provably finite. > Only the ones that are successors to a finite set. The ones that are > successors > to infinite-size sets are infinite in value. The way things are defined, sets need not have successors, only (finite) naturals need have them. > So, you assume there are only finite naturals NO! We define things so that all Peano naturals correspond to Dedekind finite sets, and are thus definably Dedekind finite. === 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: >> I don't understand why an inductive proof of an equality relationship >> does not extend to infinite numbers. >> Informally, the Peano axioms may be stated as follows: >> 1 There is a natural number 0. >> 1 Every natural number a has a successor, denoted by S(a). >> 3 There is no natural number whose successor is 0. >> 4 Distinct natural numbers have distinct successors: >> if not(a = b), then not(S(a) <> S(b)). >> 5 If a property is possessed by 0 and also by the successor of >> every natural number which possesses it, then it is possessed >> by all natural numbers. (This axiom ensures that the proof >> technique of mathematical induction is valid.) > Says nothing about finiteness there...... Finiteness happens to be a property possessed by 0, and possessed by the successor of every finite number. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: infinity ... stephen@nomail.com said: > stephen@nomail.com said: >> stephen@nomail.com said: stephen@nomail.com said: > David R Tribble said: >> >> But you have not provided a mapping between any set and its powerset, >> infinite or otherwise. > Have too. > > No you have not Tony. Have, too! > The proof that there does not exist > a bijection between a set and its power set is quite short. Then it shouldn't take too much looking to see where it goes wrong.... > > Let f be a function from S to P(S). Our proposed mapping bijection.... > > Define the set w as follows: > > w= { x : x in S and x is not in f(x) } So, w is the set of all elements which are not members of the subsets which they map to through f(x).... > > Clearly w is a subset of S, and so w is an element of P(S). Clearly...but properly? > > We now show that w is not in the image of f. That is, > there does not exist a y such that f(y)=w. So, there can be no y such that it maps to the subset of all elements which do not map to subsets containing themselves? We'll see..... > > Suppose such a y exists. If such a y exists, it must > either be an element of w, or not. One or the other. I'll accept the excluded middle.... > > If y is an element of w, then y is in f(y), which means > it is not an element of w. If y is in w, this means y is a member of the set of elements which map to subsets which do not contain themselves. This means that y is in S but not in f (y). So, indeed, it IS a member of w. There is no reason why both x and y cannot map to subsets which do not contain themselves. If y is in w, then y is in f(y), because w=f(y). Remember, we are assuming there exists a y such that f(y)=w. However w is defined such that y can only be in w, if y is not in f(y). >> Okay, I hit this one at the end of the day, and got confused halfway through it >> and forgot that you're assuming y is mapped to w. Sorry about that. It is clear >> that no element in the set maps to a subset that contains itself, as I >> illustrated below. If f(y)=w, then y can't be in w, but then that means y IS in >> f(y), which means it's in w. Got it. >> >> No, that is not clear at all. It is entirely possible that an >> element maps to a set that contains itself. However no element >> maps to w. You still do not get it. > Well, it's really not possible, given the general bijection I have offered > between an infinite ordered set and its power set. No natural is a member of > the subset of naturals which it denotes, as I demonstrated. > What is really not possible? It really is possible for y to > be a member of f(y). What is really not possible is that > f is a bijection from S to P(S). Not in the bijection I offered. y is never in f(y). >> >> Here is a simple mapping from N to P(N). >> >> 1 -> {1} >> 2 -> {1,2} >> 3 -> {1,2,3} >> 4 -> {1,2,3,4} >> ... >> >> Every element is contained in the subset that it is mapped >> to. For this mapping, w={}, and no element is mapped to {}. > And not every subset is included, so this is not a bijection. > Of course. There is no bijection from S to P(S). Yes, there is. I gave you one. You have yet to point out anything wrong with it, except to say it's impossible so it doesn't exist. One solid counterexample is all that is required to refute a proof of impossibility. >> >> >> That certainly causes a bit of a contradiction, based on the largest-finite >> kind of argument, since you are noting that whatever subset you choose, it >> never contains the natural that maps to it, >> >> No. Try again. > No need. I gave my general bijection, and this fact is true for it. I see no > other bijection between a set and its power set, but this one holds, and that > fact si true. No natural maps to a set which contains it. Any set mapped by a > natural n has a maximal element no larger than log2(n). > Your bijection is not a bijection. Sorry. Why, exactly? Hand waving and denial are not valid logical arguments. What rule of bijection have I violated, specifically? > > Neither of those possibilities causes a contradiction. You are getting confused with your double negatives. If w is the set of all elements which do not map to subsets containing themselves, then being a member of w means simply that y does not map to a subset containing y, which is perfectly possible. Not being a member of w means that an element maps to a subset which DOES contain itself. Where is the contradiction? The contradiction is that if y is in w, then it is not in w, and if y is not in w, then it is in w. That is a pretty obvious contradiction. >> Well, y is in w, as are all the elements. y does not map to w. For any given >> set, there is no element in the set which maps this way to the set itself. And >> yet, when you have infinite sets such as this, can't I just map element 2^S-1 >> to subset S? >> >> What is element 2^S-1? S is a set. Element 2^S-1 means >> nothing to me. > If set S has S elements, from number 0 through S-1, then the element which maps > to the entire set is a string of 1's which is S long, which corresponds to a > value of 2^S-1, which would be element number 2^S (not 2^S-1, sorry - damned > error of 1!). > So what is element 2^S-1? If there are only |S| elements in S, > then there is no element 2^S-1. Remember, you are mapping elements > of S to subsets of S. There is no element 2^S-1 to be mapped to S. > So what you are saying makes no sense. It makes total sense, and it is consistent with what I have been trying to teach you for months. There is no exact number you can pin on these infinite sets. Assumption of any such largest element leads to a contradiction. This is all you have done with this proof. The fact remains that for any n in N, there is a corresponding subset denoted by that bitstring, and for any subset of N, there is a natural in N whose bitstring denotes that subset. If you claim this is not so, then please show me a natural or a subset which has no corresponding element. You have two infinite sets that go on forever. When does the correspondence end? > We can try again. >> f(a) = { a } >> f(b) = { b } >> f(c) = { a, b, c } >> Now w={}, which is not in the range of f. >> How about >> f(a) = { } >> f(b) = { a } >> f(c) = { a, b } >> Now w={a,b,c}. We can keep playing this game, but w will >> never be in the image of f. > Yes, well, besides, those are relatively random mappings, anyway. > Yes, they are random. That is the point. No matter what > you pick for f, it is not a bijection. Except the bijection I offered, which works without flaw, despite your denials. In absence of any valid objection to the construction, my bijection stands. What is your specific objection? > Let's move on > to the infinite case. >> >> Of course in the finite case it is obvious that there >> cannot be a bijection from S to P(S) because P(S) has >> 2^|S| elements, and when |S| is finite it is obvious that >> 2^|S| > |S|. > Yup. Agreed. >> >> However the above proof makes no mention or use >> of the set being finite, and it applies equally well >> to finite and infinite sets. > It demonstrates that the power set is larger, but what does it say about > bijections? > It says that a bijection does not exist. Are you really incapable > of understanding that? I am quite capable of seeing that it is yet another of your proofs by contradiction that derives from the assumption of a largest element and conflates the result to claim that something unrelated is impossible or doesn't exist. While it is true that within any GIVEN range of S there is no element within S that maps to S, it is NOT true that over the entire unending range there is any point where any element does not have a unique corresponding element in the other set. It is impossible to form a bijection between any finite set and its power set, but as you do with all sorts of unequally infinite sets, it is quite simple, as shown, to form a bijection between any infinite ordered set and its power set. Show me where my bijection fails, specifically. Denials don't cut the mustard. >> >> Lets look at the natural numbers and the bijection >> albstorz proposed: >> >> 1 -> N >> 2 -> {} >> 3 -> {1} >> 4 -> N{1} >> 5 -> {2} >> 6 -> N{2} >> 7 -> {1,2} >> 8 -> N{1,2} >> 9 -> {3} >> 10 -> N{3} >> 11 -> {1,3} >> 12 -> N{1,3} >> 13 -> {4} >> 14 -> N{4} >> 15 -> {1,4} >> 16 -> N{1,4} >> 17 -> {2,3} >> 18 -> N{2,3} >> 19 -> {5} >> 20 -> N{5} >> ... >> >> In this case >> w={ 2,3,5,7,9,11,13,15,17,19 .... } >> >> Where does this show up in teh above list? If you >> claim it shows up in position y, then is y in w or not? > Yes, well, in this case you have some elements that map to subsets which > contain them, since Albrecht decided to entertain both ends of the set at once, > so as to avoid the Twilight Zone. Now, you are asking where 1010.....0101010110 > exists? Somewhere beyond N. It's essentially (2^(N+1)/3)+1. In Albrecht's > notation, it's N{1,4,6,8,10,.....} over a range of 2^N. > Somewhere beyond N? What does that mean. Is Somewhere beyond N > in w or not? Not if you assign a specific range N to w. As soon as you limit the set by declaring some actual size N, then it is essentially finitized, and the bijection fails, because we note that the element mapping to it is at 2^N, and outside the stated range. Without this value range, the bijection continues forever and is a valid bijection. It doesn't prove the set is the same size as the power set. It proves bijections alone don't mean equality of set size for infinite sets. > Stephen -- Smiles, Tony === Subject: Re: infinity ... > > What is really not possible is that f is a bijection from S to > P(S). > Not in the bijection I offered. y is never in f(y). Irrelevant. What is relevant to the existence of surjections (including bijections) from a set, S, to its power set., P(S), is that H ={x in S:x not in f(x)} is a well-definied set for ANY mapping, f, from ANY set, S, to its powerset, P(S), but that, by very construction of one can never have f(s) = H for any s in S. > > >> > >> Here is a simple mapping from N to P(N). > >> > >> 1 -> {1} 2 -> {1,2} 3 -> {1,2,3} 4 -> {1,2,3,4} > >> ... > >> > >> Every element is contained in the subset that it is mapped to. > >> For this mapping, w={}, and no element is mapped to {}. > > And not every subset is included, so this is not a bijection. > > Of course. There is no bijection from S to P(S). > Yes, there is. I gave you one. What may be regarded by TO as valid in TOmatics, need not be valid anywhere else, and TO's alleged bijection cannot be one until TO can inject the set of all infinite binary strings to the set of Dedekind finite binary strings. > You have yet to point out anything wrong with it, Several people have pointed out several things wrong with it, but TO simply ignores what he doesn't want to see. > Your bijection is not a bijection. Sorry. > Why, exactly? For any of a multitude of reasons that have been ponted out to TO but carefully ignored by him. > Hand waving and denial are not valid logical arguments. What rule of > bijection have I violated, specifically? Specifically TO has violated the rule that to have a bijection, or even a surjection, each member of the codomain (target set) must be the image of some member of the domain (source set). > So what is element 2^S-1? If there are only |S| elements in S, > then there is no element 2^S-1. Remember, you are mapping elements > of S to subsets of S. There is no element 2^S-1 to be mapped to S. > So what you are saying makes no sense. > It makes total sense, and it is consistent with what I have been > trying to teach you for months. There is no exact number you can pin > on these infinite sets. We are to trying to pin numbers on anything, we merely say that in order to have a set at all, one must be able to say of any object whether it belongs to the set or not, The membership question must be always answerable. What TO is trying to do is to say that there are TO-sets for which the membership question cannot be answered, of which *N and P(*N) are apparently examples, and that for such TO-sets one CAN have bijections *because* the membership question cannot be answered. > Assumption of any such largest element leads to a contradiction. This > is all you have done with this proof. The fact remains that for any n > in N, there is a corresponding subset denoted by that bitstring, and > for any subset of N, there is a natural in N whose bitstring denotes > that subset. If you claim this is not so, then please show me a > natural or a subset which has no corresponding element. The subsets in P(S) which have no source in S depends on the f. Give us an f and we will give you a member of P(S) not in the image of THAT f. But changing f, changes everything else. > Except the bijection I offered, which works without flaw, despite > your denials. Only in the twilight zone of TOmatics. Elsewhere if fails miseerably. > In absence of any valid objection to the construction, > my bijection stands. What is your specific objection? Give us a specific and detailed description of any function from *N to P(*N) and we will give you a specific member of P(*N) not in the image of that function. It will be of form {x in *N: x not in f(x)}. > It says that a bijection does not exist. Are you really incapable > of understanding that? > I am quite capable of seeing that it is yet another of your proofs by > contradiction that derives from the assumption of a largest element > and conflates the result to claim that something unrelated is > impossible or doesn't exist. Where does {x in *N: x not in f(x)} assume the existence of anything like a largest element of anything? TO is losing touch with reality even more thoroughly that his pervius distant contact wthit. > While it is true that within any GIVEN > range of S there is no element within S that maps to S, it is NOT > true that over the entire unending range there is any point where any > element does not have a unique corresponding element in the other > set. TO's range has nothing to do with anything here. The issue is whether TO has a function which for each member of *N produces a subset of *N. If TO has no such function, the issue is moot. If TO says he does have such a function, he has not described in sufficient detail to allow anyone to calculate any single value for any single member of *N, so it thus far remains purely mythical. > It is impossible to form a bijection between any finite set and > its power set, but as you do with all sorts of unequally infinite > sets, it is quite simple, as shown, Not shown. In order to show any such thing one must produce an algorithm which allows calculation of values for such a function. NO such algorithm has been presented. > > Somewhere beyond N? What does that mean. Is Somewhere beyond N > in w or not? > Not if you assign a specific range N to w. Thus TO has sets whose membership is up for grabs. One cannot define a function on a set whose membership is unknown and unknowable, so that TO's alleged functins are not functions. > As soon as you limit the > set by declaring some actual size N, then it is essentially > finitized, and the bijection fails, because we note that the element > mapping to it is at 2^N, and outside the stated range. Then bijection always fails, as we said a long time ago. In order to define a function on any set, the membership of that set must be fixed and determined and not variable, as it is in TOmatics. > Without this value range The *myths* of value ranges do not apply outside TOmatics. > It doesn't prove the set is the same size as the power set. Only TO gives a rat's ass about set sizes in this context. The issue is whether a surjection can exist from any set to its power set. Everywhere except in TOmatics, the answer is unequivocally NO. In TOmatics, every yes-no question has two correct answers and two wrong answers, so no one except TO cares. > It proves bijections alone don't mean equality of set size for > infinite sets. Only in TOmatics does 'set size', other than cardinality, enter into the questin. === Subject: Re: infinity ... > stephen@nomail.com said: >> Well, it's really not possible, given the general bijection I have offered >> between an infinite ordered set and its power set. No natural is a member of >> the subset of naturals which it denotes, as I demonstrated. >> What is really not possible? It really is possible for y to >> be a member of f(y). What is really not possible is that >> f is a bijection from S to P(S). > Not in the bijection I offered. y is never in f(y). Then it is not a bijection. If y is never in f(y), then there does not exist a y such that f(y)=S. S is a member of P(S), and in order for f to be a bijection, there must be a y in S such that f(y) =S. You plainly state that such a y does not exist, so f is not a bijection. Stephen === Subject: Re: infinity ... Randy Poe said: > stephen@nomail.com said: > > What is element 2^S-1? S is a set. Element 2^S-1 means > > nothing to me. > If set S has S elements, from number 0 through S-1, then the element which maps > to the entire set is a string of 1's which is S long, which corresponds to a > value of 2^S-1, which would be element number 2^S (not 2^S-1, sorry - damned > error of 1!). > So you would say that no element x is in f(x), correct? Not given any particular x, no. > Therefore for this mapping, the set > w = {x in S: x not in f(x)} > contains every element of S, right? That is, w = S. Yes. > Now let's consider every element y of S. Obviously y is > in w, since w = S. But then y is not in f(y), since that's > what it means for y to be in w. Then f(y) can't be S. That's true, and this is an interesting argument. You certainly cannot identify any natural which maps to the entire set of naturals. That would require identifying some last element and and establishing a particular number of bits N for the subset representation, so as to say w maps to (2^N)-1. Of course, since w only includes N elements, it doesn't include y, and y is not in w, within the original range of N. This is why one needs infinite sets for their bijections. Since neither set ends, there is always a corresponding element for each element, even though the numbers in the subsets are always smaller than the numbers representing the subsets. That doesn't matter. There is always another one larger, so you never run out of corresponding elements. At least, this is the way your bijections normally work. What rule am I breaking with this bijection? Where does the correspondence break down? At element N+1? > So this means that no matter what y you pick, f(y) can't > be S. So S is not mapped by your bijection. Not within the range of S, no, but over the infinite range, yes. > What's wrong with your proof? Simple. You have f(z) = S, > where z is NOT a member of your original set S. You don't > have f(y) = S for any y IN your set S. That is true. For any given upper bound you set on S, y is greater. You seem to think y is too larger to be in the boundless infinite S. So what is this upper bound which y exceeds, in order to be larger than all natural numbers, finite and infinite? It is larger than whatever value range you assign to S, but that is why you don't try to assign actual sizes to discrete infinite sets. There is no actual size, because there is no end to the process of generating elements. Without an endpoint, it cannot be measured in absolute terms, but must be measured by comparison with a discrete unit infinity, such as N, the identity function between element count and element value. In this bijection, for any given natural n, the subset has a largest element no greater than log2(n), but n is an element of subset 2^n, and where n exists, so does 2^n. None of this precludes bijection. > - Randy -- Smiles, Tony === Subject: Re: infinity ... > Randy Poe said: > > > stephen@nomail.com said: > > What is element 2^S-1? S is a set. Element 2^S-1 means > > nothing to me. > > If set S has S elements, from number 0 through S-1, then the element > > which maps > > to the entire set is a string of 1's which is S long, which corresponds > > to a > > value of 2^S-1, which would be element number 2^S (not 2^S-1, sorry - > > damned > > error of 1!). > > So you would say that no element x is in f(x), correct? > Not given any particular x, no. > > Therefore for this mapping, the set > > w = {x in S: x not in f(x)} > > contains every element of S, right? That is, w = S. > Yes. > > Now let's consider every element y of S. Obviously y is > in w, since w = S. But then y is not in f(y), since that's > what it means for y to be in w. Then f(y) can't be S. > That's true, and this is an interesting argument. You certainly cannot > identify > any natural which maps to the entire set of naturals. That would require > identifying some last element and and establishing a particular number of > bits > N for the subset representation, so as to say w maps to (2^N)-1. Of course, > since w only includes N elements, it doesn't include y, and y is not in w, > within the original range of N. This is why one needs infinite sets for their > bijections. Since neither set ends, there is always a corresponding element > for > each element, even though the numbers in the subsets are always smaller than > the numbers representing the subsets. That doesn't matter. There is always > another one larger, so you never run out of corresponding elements. At least, > this is the way your bijections normally work. > What rule am I breaking with this bijection? The rule that for any function, f, from any set, S, to its power set, P(S), there is no member, s, of S such that f(s) = {x in S:x not in f(x)} > > So this means that no matter what y you pick, f(y) can't > be S. So S is not mapped by your bijection. > Not within the range of S, no, but over the infinite range, yes. So TO is saying that there are elements OUTSIDE of the domain of the function that map onto all those subsets that are not images of elements of the domain. But that does not establish anything like a bijection from the domain set to anything. > > What's wrong with your proof? Simple. You have f(z) = S, > where z is NOT a member of your original set S. You don't > have f(y) = S for any y IN your set S. > That is true. Then TO does not have his claimed bijection fron S to anything. === Subject: Re: infinity ... Tony Orlow says... >> Therefore for this mapping, the set >> w = {x in S: x not in f(x)} >> contains every element of S, right? That is, w = S. >Yes. >> Now let's consider every element y of S. Obviously y is >> in w, since w = S. But then y is not in f(y), since that's >> what it means for y to be in w. Then f(y) can't be S. >That's true, and this is an interesting argument. You certainly >cannot identify any natural which maps to the entire set of naturals. That's why people say There is no bijection between S and the the power set of S. They mean that for all maps f from S to P(S), there is a corresponding w in P(S) such that nothing gets mapped to w. (You would say nothing identifiable gets mapped to w). >What rule am I breaking with this bijection? It's not a bijection. >> So this means that no matter what y you pick, f(y) can't >> be S. So S is not mapped by your bijection. >Not within the range of S, no, Then you don't have a bijection. By definition, f is a bijection only if there is some y in S that maps to w. >> What's wrong with your proof? Simple. You have f(z) = S, >> where z is NOT a member of your original set S. You don't >> have f(y) = S for any y IN your set S. >That is true. That means that f is not a bijection. -- Daryl McCullough Ithaca, NY === Subject: Re: infinity ... > stephen@nomail.com said: > > What is element 2^S-1? S is a set. Element 2^S-1 means > > nothing to me. > > If set S has S elements, from number 0 through S-1, then the element which maps > > to the entire set is a string of 1's which is S long, which corresponds to a > > value of 2^S-1, which would be element number 2^S (not 2^S-1, sorry - damned > > error of 1!). > So you would say that no element x is in f(x), correct? > Not given any particular x, no. Uhoh, we're hinting at specifiable again. Every element of S is a particular x. There are no unparticular elements of S. > Therefore for this mapping, the set > w = {x in S: x not in f(x)} > contains every element of S, right? That is, w = S. > Yes. Good. > Now let's consider every element y of S. Obviously y is > in w, since w = S. But then y is not in f(y), since that's > what it means for y to be in w. Then f(y) can't be S. > That's true, and this is an interesting argument. I'm just repeating the Cantor argument for your particular mapping. I don't think it's easy to get an intuitive grasp for that argument without seeing how it works for a couple of examples. You have provided one example, but the proof is general. > You certainly cannot identify > any natural which maps to the entire set of naturals. Forget identify. What you just agreed to is that there does not EXIST a natural y such that f(y) = S. > That would require > identifying some last element No it wouldn't. Where did last element come into it? All that came into it was agreeing that w = S, whether S has a last element or not. y is in w whether S has a last element or not. Therefore y is not in f(y), whether S has a last element or not. Therefore f(y) can not be S, whether S has a last element or not. > What rule am I breaking with this bijection? We just agreed that it is impossible under this bijection for S to appear on the right hand side. Therefore there is at least one subset of S (S itself) which does not appear on the right hand side. Therefore your bijection breaks the property of mapping to every subset of P(S). > Where does the correspondence break down? At element N+1? I don't know what other elements are not mapped. This is your obsession with finding the end again. We don't need to find the end. We just considered EVERY SINGLE ELEMENT and proved that f(y) can't be S FOR ANY ELEMENT y OF S. That rules out everything except y's which are not elements of S. But bijection S -> P(S) means that for every element s of P(S), there is y which IS AN ELEMENT OF S such that f(y) = s. S is an element of P(S). f(y) does not equal S for any y in S. So it's not a bijection. - Randy === Subject: Re: infinity Is there any value of x for which x+1 is not an element of the original set? ----------------- Yes w-1 in set 0,1,2,3,.......... has 1-1 relation with w in set 1,2,3,4..........., and w is not present in set 0,1,2,3,............. === Subject: Re: infinity element of the original set? > ----------------- > Yes w-1 in set 0,1,2,3,.......... w-1 is not in set 0, 1, 2, 3, ... Let me clarify. What I mean by {0, 1, 2, 3, ...} is 0, plus any number which can be generated by adding 1 to a finite natural number (in other words, the Peano set). What I mean by {1, 2, 3, ..} is the set of numbers which are successors to 0 or other finite natural numbers. > has 1-1 relation with w in set > 1,2,3,4..........., and w is not present in set 0,1,2,3,............. Nor is w in {1, 2, 3, ...} - Randy === Subject: Re: infinity Randy: I thought you were asking about bijection between sets 0,1,2,3,........ and set 1,2,3,......... , bijection doesn't know Peano, it never heared of him. The set 0,1,2,3,...... is followed by w-1 at it's end and w is the end of 1,2,3,4,...... and bijective functions bijects sets to their ends. To tell me that Peano's numbers are afraid from making the final jump , and they are running on a slippery ground that makes them unable to make the final jump, is the problem of these numbers and not the bijective funtions. asked to compare between set A =1,2,3,.......,50 and set B formed by f(x)=2x , the man reaches comparison at number 25 were he finished writing 25 numbers in set A and their 25 counterparts in set B, and then gets tired and stop and say that set A is equivalent to set B since they have 1-1 correspondance up to 25. Your Peano respecting bijection works like that man , since it is obvious that any finite subset of A is well below the half of set A if set A is infinite since any finite number is well below half it's infinity, because half the infinite is infinite and any infinite is bigger than any finite. Randy if bijection is to be fare then it should be carried on to the end. If it is not then how do you know that one of the sets doesn't contain a different member from the other set at it's end. You can't say yes and you can't say No. and so your equality by bijection is not clear weather it leads to an infinite set being identified with a proper infinite subset of it or just another equivalentely sized infinite set, containing a different number at the far end. Think of it deeply. Even if I cannot name a number in set 1,2,3,........ that has bijection to set 0,1,2,3,........ which doesn't exist in 0,1,2,3,...... that doesn't mean that you proved the converse ie the non existance of this number.You are just playing in the darkness, giving certainity to things that you didn't reach. It is better to find better definitions of the infinite than that of reflexion, with it's all contradictive appearance to common sense. Think of it deeply. Zuhair === 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: I thought you were asking about bijection between sets > 0,1,2,3,........ > and set 1,2,3,......... , bijection doesn't know Peano, it never heared > of him. > The set 0,1,2,3,...... is followed by w-1 at it's end and w is the end > of 1,2,3,4,...... Neither set has an end. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: infinity <85br1i22q1.fsf@lola.goethe.zz> Neither set has an end. ................. both has infinite natural ends w-1 and w. this is equivalent to saying that they don't have a natural finite end. Zuhair === 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 > Neither set has an end. > ................. > both has infinite natural ends w-1 and w. this is equivalent to saying > that they don't have a natural finite end. The sets contain only natural numbers, and the definition of natural numbers does not allow for infinite natural numbers. So the sets don't have an end. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: infinity 0,1,2,3,........ > and set 1,2,3,......... , I was. I'm clarifying what I mean by those sets. The first is the set of natural numbers starting with 0. > bijection doesn't know Peano, it never heared > of him. Yes, but these sets are sets of natural numbers. I define natural number by the Peano axioms. If you want another definition of natural number and therefore the sets we're talking about, then give me alternate axioms. > The set 0,1,2,3,...... is followed by w-1 at it's end No it isn't. Not if those are the natural numbers. > and w is the end of 1,2,3,4,...... No it isn't. Not if we started from natural numbers. Also, since I meant the second set to be a proper subset of the first, then it's impossible for 1, 2, 3, ... to have an element that was not present in 0, 1, 2, ... Anyway, I'm talking about the finite natural numbers. So let me restate. Consider the mapping f(x) = x+1 where x is taken from the finite natural numbers {0,1,2,...}. This is a mapping from the finite values {0,1,2, ...} to the finite values {1,2,...}. Is this mapping not a bijection? Why not? - Randy === Subject: Re: infinity This is a mapping from the finite values {0,1,2, ...} to the finite values {1,2,...}. Is this mapping not a bijection? Why not? ------------------------------------ It is a bijection of coarse, but set 1,2,3,4,....... is not a subset of 0,1,2,3,.......... because the first set contains a number which is not present in the second, you can see that if you biject to the end of each set, if you do not do that of coarse you cannot see that and you will think that the first set is a subset of the other. Zuhair === Subject: Re: infinity > This is a mapping from the finite values {0,1,2, ...} to the > finite values {1,2,...}. Is this mapping not a bijection? > Why not? > ------------------------------------ > It is a bijection of coarse, but set 1,2,3,4,....... is not a subset of > 0,1,2,3,.......... > because the first set contains a number which is not present in the > second, you > can see that if you biject to the end of each set, if you do not do > that of coarse you cannot see that and you will think that the first > set is a subset of the other. What does Zuhair think an ellipsis, ..., is? What member of {1,2,3,4,...} does Zuhair find absent from {0,1,2,3,...} ? === Subject: Re: infinity finite values {1,2,...}. Is this mapping not a bijection? > Why not? > ------------------------------------ > It is a bijection of coarse, but set 1,2,3,4,....... is not a subset of > 0,1,2,3,.......... Even if what I mean by {1,2,3,...} is take the set {0,1,2,3...} and remove 0 from it? > because the first set contains a number which is not present in the > second, No it doesn't. Take the first set and remove 0. That's the second set. No chance for another element to sneak in. - Randy === Subject: Re: infinity Randy: many times in that group I said and said it again the set 1,2,3,4,.............. if defined by what is left after removal of zero from the set 0,1,2,3,4,............. is something different from the set 1,2,3,4,.......... defined as the codomain of bijection to the set 0,1,2,3,........... , these two sets are different, but you cannot see that simply because you are not seeing what is happening in the direction of infinity since you are never reaching a finite natural ending number and you are not extending the set to its end then you will never discover the difference, and you will always think they are the same sets. The end of that set is the natural number defined by me as w-1 for the first set and w for the second set.but these numbers are infinite naturals . Without developing a theory of induction like that of Peano's which can extend inductive qualities to involve a number like w, then of coarse you will never be convinced about what I am saying. Zuhair === Subject: Re: infinity > Without developing a theory of induction like that of Peano's which can > extend inductive qualities to involve a number like w, then of coarse > you will never be convinced about what I am saying. > Zuhair We will not in any case be convinced of he falsehoods Zuhair is saying. === Subject: Re: infinity 1,2,3,4,.............. > if defined by what is left after removal of zero from the set > 0,1,2,3,4,............. Good. Let's work with those two sets. Now I want to ask a question about those two sets. Let N = {0,1,2,3...} Let N' = N{0}, the set obtained by removing 0 from N. Consider the map f:N->N' with f(x) = x+1 Is this a bijection from N to N'? > is something different from the set 1,2,3,4,.......... defined as the > codomain Then I don't care about whatever set you are talking about. I want to talk about the set obtained by removing 0 from N. The Dedekind definition you hate talks about bijections between a set and a proper subset. If you don't have a proper subset, then it doesn't apply to discussions of the Dedekind definition of infinite set. So consider my map f, between N and a proper subset of N. Is f a bijection? - Randy === Subject: Re: infinity Is this a bijection from N to N'? No of coarse not! It is a surjection of N to N' . Zuhair === Subject: Re: infinity > Is this a bijection from N to N'? > No of coarse not! > It is a surjection of N to N' . > Zuhair Let f represent the first element of N then there is a natural bijection from N to N{f}. It is callsed the successor operation. === Subject: Re: infinity or bijection from N to N. Zuhair === Subject: Re: infinity No of coarse not! > It is a surjection of N to N' . Oh good. Finally we get an agreement on what question is being asked. OK, for which finite value x in N is there no x+1 in N'? - Randy === Subject: Re: infinity Ok, Randy. Their is no finite value x in N that do not have x+1 in N' , the value x that belong to N but do not have x+1 in N' is w-1 , since this should have w as its image, but w is not a member of N' since N' ends by w-1 . Please Randy review some of the posts I made earlier, You are doing like the man who get tired when he reaches to number 50 ( review earlier posts by me) . Review my defintion of infinite sets. Zuhair === Subject: Re: infinity > Ok, Randy. > Their is no finite value x in N that do not have x+1 in N' , the value > x that belong to N but > do not have x+1 in N' is w-1 , since this should have w as its image, > but w is not a member of N' since N' ends by w-1 . Neither N nor N' ends, if by that is meant that there is any member of either that does not have a successor in that set itself. > Please Randy review some of the posts I made earlier, You are doing > like the man who get tired when he reaches to number 50 ( review > earlier posts by me) . Review my defintion of infinite sets. You definition does not trump Dedekind's or Cantor's. === Subject: Re: infinity > Ok, Randy. > Their is no finite value x in N that do not have x+1 in N' , the value > x that belong to N but > do not have x+1 in N' is w-1 Oh dear. We don't seem to have agreed on the question after all. Consider the set N of finite natural numbers, beginning at 0. Consider N' = N{0}, N with element 0 removed. Consider the mapping f:N->N', f(x) = x+1. Is f(x) a bijection? Why not? - Randy === Subject: Re: infinity Consider the set N of finite natural numbers, beginning at 0. You mean of all finite natural numbers. because if you mean of a finite subset of all finite natural numbers then it is clear that 0,1,2,3,4,......,n will have a bijection to 1,2,3,4,........,n+1 and n+1 is not present in the first set. So you meant the set of all finite natural numbers which an infinite set this set has w-1 as a ending number in it so it is 0,1,2,3,.............,w-1 This is what we mean when we say the set of all finite natural numbers, it is strange that this set differes from other sets in that it contains an infinite number at its end yet it is a member of it. and again I will say that this cannot have bijection with N'= 1,2,3,.........., w-1 because the last member w-1 is the one which cannot be bijected to a number in N' Zuhair === Subject: Re: infinity > Consider the set N of finite natural numbers, beginning at 0. > You mean of all finite natural numbers. Yes I do. I said beginning at 0 because there is a choice of convention as to whether you consider the natural numbers to begin at 1 or 0. I am saying N begins at 0 and includes all finite natural numbers. > So you meant the set of all finite natural numbers which an infinite > set > this set has w-1 as a ending number in it so it is > 0,1,2,3,.............,w-1 No, the set of all finite nautral numbers will not have an end, and especially not an infinite number at the nonexistent end. The set of all finite natural numbers contains only finite natural numbers. > This is what we mean when we say the set of all finite natural numbers, No it isn't. When we talk about the set of all finite natural numbers we do not mean a set with an end, nor do we mean a set that contains infinite quantities. > it is strange that > this set differes from other sets in that it contains an infinite > number at its end yet it is a member of it. It does not. Let us confine our discussion to the set of finite natural numbers. A set of finite things contains only finite things. > and again I will say that this cannot have bijection with N'= > 1,2,3,.........., w-1 > because the last member w-1 is the one which cannot be bijected to a > number in N' There is no last member in the set I am considering. Didn't you already admit that a bijection is possible between the FINITE numbers beginning at 0 and the FINITE numbers beginning at 1? - Randy === Subject: Re: infinity Ok Randy : I will clarify my claims. John said to Jonathan : I will marry Susan Jonathan: Why? John: because I admire her and because she is virgin Jonathan: Since Susan is not virgin then you John are going to marry the non virgin Susan John: No, Since Susan is not virgin I am not going to marry her!!!! Let me tell how I think exactly so that you know my strange claims from were they come from. To me I define an infinite number as the measure of multiplicity of members in an infinite set. Or simply the answer to How many members in that infinite set are their? Now if w is the number which describes the multiplicity of members in for example the set N=1,2,3,4,....................... Now even numbers that belong to N are a subset of N since every even number should belong to N but not every number in N is even . For the same reason odd numbers are a subset of N. Since N members are either even or odds and since their is bijection between even and odds , then the subset of evens is equal in size to the subset of odds. these two subsets of N are half the multiplicity of N. according to the above the number which should describe the multiplicity of evens or odds is the same and it equals w/2 .period. Their is no way to make me yield from this kind of reasoning!!!!! . Now the above is a logical kind of reasoning that I call a common sense fact! it is of the strenght of 1+1=2. You can say it is an axiom to me. Now to me any kind of reasoning that gives a result against the above is to be refuted as erronous if I can prove that, doubtful if I cannot prove its error , or insufficient if it has grounds for it. Furthermore I will try to view any such kind of insufficient reasoning by the backround of logical common sense reasoning above.And thus modefy the insufficient to suite the sane reasoning. Now I said that peano's infinite set of finite naturals should end by an infinite integer that is a member of it which is w-1. ie peano's set= 0,1,2,3,..........,w-1 if you don't put w-1 at the end then this is a finite subset of peano's set. To me to say this is an infinite peano's set and not to include w-1 as it's final member is Paradoxical( although Peano himself axiomatized this paradoxical assumption ). However I am not going to argue with you about that. since proving the above formally is more difficult than proving 1+1=2 by formal logic. Now let me assume the traditional and standard interpretation of peano's set of finite natural numbers.( that it should not include any type of numbers other than finite naturals). According to that their can be a bijection between an infinite peano's set and a proper infinite subset of it. Accordingly the whole can equal the part at infinite level. And so the kind of reasoning that I hold as very strong and common sense would be likely erronous. so the set of all evens is w and all odds is w and all naturals is w. Now you should understand me when I say that since bijection made us reach to a kind of reasoning against my common sense derived one above then I will reject bijection itself as a standared for equality between infinite sets except in certain conditions. In reality I have been long wondering about that subject ie using mapping by bijective functions as a standard for equality between infinite sets. Of coarse it is the standard at finite sets, but at the infinite? I don't think it is always successful in doing so. So you should understand me now when I say that I will only accept bijection as a standard of equality between infinite sets iff it agrees with the kind of logical common sense reasoning I illustrated above. Accordingly not any bijective function between infinite sets can determine their equality. Using your bijection with Peanos strict definitions you will only see the shadow of the infinite and not the infinite itself !!! Now I can have bijection with my shadow at the moring , evening , and noon, does that mean that I have different lenghts at these timings or does it mean that my shadows have the same length at these timings. Of coarse both are false. Only at a certain time My shadow really reflect my real lenght, and only this bijection is successful in determing equality between the shadow and me in lenght.( of coarse the shadow and me are infinite sets of points). Same can be said about bijection between the diagonal of a square and one of its sides, it is a failure of determing equality, while bijection between parallel sides does determine equality. and of coarse these entities are infinite sets of points. I think that a bijective function that determines equality between two infinite sets should have a certain criteria other than just mapping from each set into the other by one -one and onto. That's why I propose an identity bijective function as the standard for equality between any two infinite sets. So an infinite set A has equal number to set B iff an identity bijective function exists between them. Define: the counter i of an infinite set is the number determining it's multiplicity for example set B= 1/2, 1/4,1/8,.................. the counter is the index on base 2 , in each member 1/(2^i) the counter set of B is 1,2,3,........... Now identity bijective function between set N= 1,2,3,.............. and B. is f:N--->N i=f(x)=x. and it causes bijection here. and so N=B in size. Now lets go back to your example of comparing between set N=0,1,2,3,............. and set N'= N/ 0 ( / for except)= ,1,2,3,......... Now what is the counter in N' ? N' = 0 +1 , 0+2 , 0+3 ,............... = 0+i = so the counter of set N' = 1,2,3,4,....... ( see that N' is not 1+0 , 1+1, 1+2 , 1+3 ,............... = 1+i , since here the counter of that set would be 0,1,2,3,............) using the bijective identity function f:N--->N' , i=f(x)=x we get the following 0,1,2,3,......................... 1,2,3,........................ the relation between N and N' using f(x)=x is not bijection. So N>N' . I hope my heiarchy of priorities is now very obvious to you. Also I have a question to you Randy. Why sets cannot deal with dynamic situations? I see it deals with it at finite level , so why it cannot deal with it at infinite level. Example: if Virgil's wirtting speed is one number per second, Zuhair's two numbers per second and TO's three numbers per second. Suppose they are asked to wright numbers from 1,2,3,........ . Suppose the contest is finished by 10 seconds. the traks will be TO = 1,2,3,4,5,6...............,30 Zuhair= 1,2,3,4...........,20 Virgil= 1,2,.......,10 Now what if the race continues to infinity, the three sets are still distict. So why set theory cannot make us see that comparative difference. For that purpose I advocated to include w-1 in the set of 0,1,2,3,........ and w in the set of 1,2,3,......... to illustrate comparative differences, because infinity is a DYNMIC pricess. Because adhering to Peano's definitions will cast away alot of comparative information at the dynamic infinite level, and it is imprecise approach towards the infinite, I do really think that these axioms should be modefined. And I will repeat it for the last time I do really have a strong intuitive sense ( this is not common sense of coarse) that when Peano says the set of all finite natural numbers N are infinite ( no last finite natural number) , this saith is equivalent to saying that N= 0,1,2,3,............,w-1 , with w-1 as a member of that set. So I think Peano's axioms should be modefined to deal with the infinite in a full comparative manner. I hope the message above is clear. Zuhair === Subject: Re: infinity > Their is no way to make me yield from this kind of reasoning!!!!! . There is no way that kind of reasoning will convince anyone any more mathematically compentent than TO. === 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 > To me I define an infinite number as the measure of multiplicity of > members in an infinite set. Which set? There are more than one kind of infinite set. The idea of cardinalities could be reasonably called multiplicity of members. > Or simply the answer to How many members in that infinite set are > their? > Now if w is the number which describes the multiplicity of members in > for example the set > N=1,2,3,4,....................... > Now even numbers that belong to N are a subset of N since every even > number should belong to N but not every number in N is even . For the > same reason odd numbers are a subset of N. Since N members are either > even or odds and since their is bijection between even and odds , then > the subset of evens is equal in size to the subset of odds. Fine so far. > these two subsets of N are half the multiplicity of N. Not so fine. w is a name for the multiplicity of a set. You have not defined what it means to halve a set, much less what it means to halve a cardinality. > according to the above the number which should describe the > multiplicity of evens or odds > is the same and it equals w/2 .period. > Their is no way to make me yield from this kind of reasoning!!!!! . Too bad. You start doing arithmetic with a set property without establishing first that this is possible or reasonable. > Now the above is a logical kind of reasoning that I call a common > sense fact! Facts are facts and don't depend on sense, common or not. You are starting wild handwaving for the purpose of glossing over lacking fundations. > it is of the strenght of 1+1=2. You can say it is an axiom to me. More handwaving. You know that an axiom is the _opposite_ of a fact? It is a declaration that the axiom is _not_ a fact, but an arbitrary starting point which could, with equal validity, be chosen differently. So if you say it is an axiom to me, you declare that you think it can't be proven and it does not logically follow from anything else. > Now to me any kind of reasoning that gives a result against the > above is to be refuted as erronous if I can prove that, doubtful if > I cannot prove its error , or insufficient if it has grounds for it. In short, you will not accept being shown wrong. An interesting stance for a beginner. > Now I said that peano's infinite set of finite naturals should end > by an infinite integer that is a member of it which is w-1. Completely unrelated hog wash. Peano's set has no end, since every member is followed by another distinct member. > ie peano's set= 0,1,2,3,..........,w-1 > if you don't put w-1 at the end then this is a finite subset of peano's > set. There is no end to Peano's set, and so the above is nonsense. > To me to say this is an infinite peano's set and not to include w-1 > as it's final member is Paradoxical( although Peano himself > axiomatized this paradoxical assumption ). To call something infinite and demand that it ends at the same time is more paradoxical to me. > However I am not going to argue with you about that. since proving > the above formally is more difficult than proving 1+1=2 by formal > logic. To prove, you have to start from axioms. And if you purport to prove something about the Peano set of naturals, the Peano axioms. more difficult can better be called impossible, as I would hope every sufficiently capable junior high school student to be able to see from looking at the Peano axioms for 10 minutes. > Now let me assume the traditional and standard interpretation of > peano's set of finite natural numbers.( that it should not include > any type of numbers other than finite naturals). > According to that their can be a bijection between an infinite > peano's set and a proper infinite subset of it. Accordingly the > whole can equal the part at infinite level. > And so the kind of reasoning that I hold as very strong and common > sense would be likely erronous. so the set of all evens is w and all > odds is w and all naturals is w. If is is supposed to mean has a cardinality of. > Now you should understand me when I say that since bijection made us > reach to a kind of reasoning against my common sense derived one > above then I will reject bijection itself as a standared for > equality between infinite sets except in certain conditions. Bijection is not a standard for equality. It is a standard for ordering sets according to cardinality, a scalar measure of surjectability. > In reality I have been long wondering about that subject ie using > mapping by bijective functions as a standard for equality between > infinite sets. It is not a standard for equality, just for equal cardinality. Cardinality is an abstraction that is based just on the multiplicity of distinguishable members, without differentiating their values except for their distinctiveness. It is a scalar measure, and that is convenient. Like a determinant as a scalar measure of a matrix, it is not the whole truth about the set. But it is an interesting part of it. > So you should understand me now when I say that I will only accept > bijection as a standard of equality between infinite sets iff it > agrees with the kind of logical common sense reasoning I illustrated > above. But bijection is not a standard of equality. > And I will repeat it for the last time I do really have a strong > intuitive sense ( this is not common sense of coarse) that when Peano > says the set of all finite natural numbers N > are infinite ( no last finite natural number) , this saith is > equivalent to saying that > N= 0,1,2,3,............,w-1 , with w-1 as a member of that set. But it is wrong. There is no last member of the set. > So I think Peano's axioms should be modefined to deal with the > infinite in a full comparative manner. But that would make the set of finite even naturals gained by multiplying every member of N with 2 different from the set of finite even naturals gained by removing every odd number from N. But the finite even naturals are the finite even naturals either way. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: infinity <85u0f9s6n6.fsf@lola.goethe.zz> But that would make the set of finite even naturals gained by multiplying every member of N with 2 different from the set of finite even naturals gained by removing every odd number from N. -------------------------------------- Yes , that is exactly what I am trying to say, They Are Different. You said But the finite even naturals are the finite even naturals either way. WRONG You said:But bijection is not a standard of equality. Well it is a standard for similarity. To say that set has bijection to that set is a standard for equal cardinality of both , I mean that the multiplicity of their members is the same. Or am I wrong? Zuhair === 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 that would make the set of finite even naturals gained by > multiplying every member of N with 2 different from the set of finite > even naturals gained by removing every odd number from N. > -------------------------------------- > Yes , that is exactly what I am trying to say, They Are Different. Then you should be able to name elements that are in one set, but not in the other. This is not the case when starting with the Peano axioms. You want other axioms that would change this. But you have not brought forth an argument which would make it desirable to have different sets of even natural numbers. In fact, it is very undesirable. > But the finite even naturals are the finite even naturals either way. > WRONG > You said:But bijection is not a standard of equality. > Well it is a standard for similarity. To say that set has bijection > to that set is a standard for equal cardinality of both , I mean > that the multiplicity of their members is the same. Sure, more or less. That's the idea. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === 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 > Consider the set N of finite natural numbers, beginning at 0. > You mean of all finite natural numbers. > because if you mean of a finite subset of all finite natural numbers > then > it is clear that 0,1,2,3,4,......,n will have a bijection to > 1,2,3,4,........,n+1 and n+1 is > not present in the first set. Sure. > So you meant the set of all finite natural numbers which an infinite > set > this set has w-1 as a ending number in it so it is > 0,1,2,3,.............,w-1 No. The set of all natural numbers has no ending number, since one Peano axioms explicitly states that if n is a natural number, so is n+1. > This is what we mean when we say the set of all finite natural numbers, > it is strange that > this set differes from other sets in that it contains an infinite > number at its end yet it is a member of it. It has no end, and it contains no infinite number as a member. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: infinity Ok , their is bijection since N and N' only contains finites. but comparatively speaking N' is not a subset of N, because of the dynamic definition I illustrated in the Kangaroo example. I will continue discussing that with you tommorow( if I live).I should go to sleep now!!! Zuhair === Subject: Re: infinity > Ok , their is bijection since N and N' only contains finites. > but comparatively speaking N' is not a subset of N, because of the > dynamic definition I illustrated in the Kangaroo example. > I will continue discussing that with you tommorow( if I live).I should > go to sleep now!!! > Zuhair I don't think sleep will help much. === Subject: Re: infinity > Ok , their is bijection since N and N' only contains finites. > but comparatively speaking N' is not a subset of N, If N' is obtained by removing 0 from N, then it is a subset of N. > because of the > dynamic definition I illustrated in the Kangaroo example. Here is the definition of a subset: If the following implication holds x is an element of A => x is an element of B then A is a subset of B. This implication holds for N' and N. N' is a subset of N. Every element of N' is an element of N. That makes it a subset. Sets do not have dynamic definitions. Sets are collections of things meeting a membership test. That is not dynamic. - 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 > This is a mapping from the finite values {0,1,2, ...} to the > finite values {1,2,...}. Is this mapping not a bijection? > Why not? > ------------------------------------ > It is a bijection of coarse, but set 1,2,3,4,....... is not a subset of > 0,1,2,3,.......... Uh, every element of N has a successor in N, and different elements have different successors. Mapping each element of N to its successor means that all the mappings are also different. That makes the operation a surjection into N. And 0 is not the successor of any number in N, so it is onto a subset of N. But 0 is in the original set. See: just 4 of the Peano axioms are sufficient for showing that the mapping of N to N{0} is a bijection, and that no new elements are introduced by the mapping. > because the first set contains a number which is not present in the > second, you can see that if you biject to the end of each set, None of those sets has an end. > if you do not do that of coarse you cannot see that and you will > think that the first set is a subset of the other. Just reread those axioms. You seem to forget them too easily. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: infinity So anyways, I think that as q goes to infinity, that the graph of F_q(1) looks just like f(x)=x but it's everywhere discontinuous. Where the integral of f(x)=x evaluated over [0,1] is 1/2, I think the integral of the everywhere discontinuous function is 1. It's definitely realized that those proper subsets, a set is a subset of itself but not a proper subset, have by many meanings less elements than the set. Also by several meanings they have the same, infinite amount. Then, there are sets where infinitely many proper subsets are proper supersets of the subset, much greater in extent. That's where a set setminus its proper subset is an infinite set, where the complement of a set is infinite, where a set is not cofinite. Then a function between them demands multiplication of a variable. For the natural/unit equivalencyfunction, in the first case c_i is never in the list as the list starts with infinitely many sequences starting with infinitely many zeros in a decimal expansion, Cauchy sequence, or Dedekind cut, and it's not done. So each b_i =1, a_i= (2^i-1)/ 2^i, and S_i contain (2^n-1)/2^n for n from zero to i. Anyways T is not in L, because it's zero and empty. In that sense, I want to address whether it is not possible to map the naturals to the Dedekinf cuts of each real element of any interval, and if so, whether it's a finite, semi-infinite, or infinite interval. I claim that infinite sets are equivalent, with dual representation for the antidiagonal argument, and/or ubiquitous ordinals or naturals, and that a function between the naturals and reals is an implicit composition with EF. Keith presented a statement that he could map a proper subset of the naturals bijectively to the reals. What's the deal with that? That reminds me that the direct sum of infinitely many copies of N is empty, but not, by exceptional definition. I think the ur-element is zero, and sometimes infinity, and one. === 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 > Is there any value of x for which x+1 is not an > element of the original set? > ----------------- > Yes w-1 in set 0,1,2,3,.......... There is no w-1 in the set 0,1,2,3,... > has 1-1 relation with w in set 1,2,3,4..........., There is no w in the set 1,2,3,4,... > and w is not present in set 0,1,2,3,............. w-k (with k in N) is not present in either N or N{0}. There is no last natural number: if k is an element of N, then so is k+1. Reread the Peano axioms. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: infinity <85oe5j0zlo.fsf@lola.goethe.zz> There is no last natural number: if k is an element of N, then so is k+1. Reread the Peano axioms. ---------------------------- There is no last FINITE natural number. w doesn't have a finite natural number as it's immediate predecessor in order to inherit finitude from it, but this makes it inherites induction but not finitude, because it is defined to be after all finite naturals and derived by summation in a similar way to von neuman's definition but in infinite manner so w= { 0 , S(0), S(S(0)),........................} were S means sucessor, according to the above definition w is natural an so inductive but not finite, in reality w can be viewed to inherit the infinitude of succeesion of immediate finite natural predecessors befor it, so it inherits their naturality and infinitude, ie w is an inductive infinite number.And it is the first of them. Zuhair === 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 > There is no last natural number: if k is an element of N, then so is > k+1. Reread the Peano axioms. > ---------------------------- > There is no last FINITE natural number. There are no others. > w doesn't have a finite natural number as it's immediate predecessor > in order to inherit finitude from it, Then it isn't a finite number. Reread the Peano axioms, in particular the fifth. > but this makes it inherites induction but not finitude, because it > is defined to be after all finite naturals and derived by summation > in a similar way to von neuman's definition but in infinite manner > so The Peano axioms don't talk about any w, and they preclude such values being in the set of naturals. > w= { 0 , S(0), S(S(0)),........................} > were S means sucessor, according to the above definition w is natural > an so inductive > but not finite, in reality w can be viewed to inherit the infinitude of > succeesion of immediate finite natural predecessors befor it, so it > inherits their naturality and infinitude, ie There is no such inheritence in the Peano axioms. Induction forbids it. Properties that are passed onto the successor, like finiteness, are properties of every member of the set if they are valid for 0. > w is an inductive infinite number.And it is the first of them. Whatever it is, it is not a natural number. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: infinity <85oe5j0zlo.fsf@lola.goethe.zz> <85fyqu22qy.fsf@lola.goethe.zz> Then it isn't a finite number. Reread the Peano axioms, in particular the fifth. ----------------------- That is exactly what I am saying. w is not a finite natural number, w is an infinite natural number. There is no such inheritence in the Peano axioms. Induction forbids it. Properties that are passed onto the successor, like finiteness, are properties of every member of the set if they are valid for 0. ---------------------------- I know ! I didn't say it is Peano's .This idea is not to be traced to Peano, it is my tpye of inheritance.And I think it is right. w is an inductive infinite number.And it is the first of them. Whatever it is, it is not a natural number. -- wrong this is not Cantors w , it is mine, and it is a natural number but not finite. Zuhair Zuhair === Subject: Re: infinity <85oe5j0zlo.fsf@lola.goethe.zz> <85fyqu22qy.fsf@lola.goethe.zz> Then it isn't a finite number. Reread the Peano axioms, in particular the fifth. ----------------------- That is exactly what I am saying. w is not a finite natural number, w is an infinite natural number. There is no such inheritence in the Peano axioms. Induction forbids it. Properties that are passed onto the successor, like finiteness, are properties of every member of the set if they are valid for 0. ---------------------------- I know ! I didn't say it is Peano's .This idea is not to be traced to Peano, it is my tpye of inheritance.And I think it is right. Zuhair Zuhair === 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 > Then it isn't a finite number. Reread the Peano axioms, in particular > the fifth. > ----------------------- > That is exactly what I am saying. w is not a finite natural number, w > is an infinite natural number. It is not a natural number, period. Whatever it happens to be (and it usually happens to be defined as the cardinality of the set of naturals), it is not a natural number. > There is no such inheritence in the Peano axioms. Induction forbids > it. Properties that are passed onto the successor, like finiteness, > are properties of every member of the set if they are valid for 0. > ---------------------------- > I know ! I didn't say it is Peano's .This idea is not to be traced to > Peano, it is my tpye of inheritance.And I think it is right. But then you have no business talking about the properties of natural numbers, which are defined using the Peano axioms. You can fantasize then about Zuhair numbers, for which you are free to invent your own axioms. Just don't complain if nobody is interested in incoherent and useless ill-defined constructs. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: infinity ... > > First of all you don't argue on my claim of the thread. > > > Second, your above argueing is not clear to me, since both sets are > > > well-ordered. But it's nice, so I give this: > > > N > > > {1},{2},{3}, ... > > > N/{1},N/{2},N/{3}, ... > > > {1,2},{1,3},{2,3},{1,4},... > > > N/{1,2}, ... > > > ... > > Is this supposed to be a list? My reading > > of this is that your list is: > > N, > > {1}, > > {2}, > > {3}, > > ... > > N/{1}, > > Right here we have a problem. What is the element > > before N/{1} in your list? ... > > > Now count in diagonal sequence. You may think of Cantor's first > > > diagonal proof. > > What diagonal? > > Come on, come on! He means the zigzag diagonal, as in the standard > > demonstration that the rationals _are_ countable. This isn't Cantor's > > first diagonal proof, > I see, you are the real checker. You knows it all. You are famous. You > are apodictic. All the authors who speak of the first diagonal proof of > Cantor are wrong. > > but if you're going to argue with cranks you > > must expect them to be pretty muddled about things. > > Anyway, it's obvious that *if* the OP shows a list of lists that > > include all the subsets that is enough. In practice, of course he's > > given the standard crank non-list. > > Brian Chandler > > http://imaginatorium.org > You and many of the other checkers are not able to discuss my starting > argument. You are only able to respond to the usual wrong arguments you > know. And I think, your answers are memorized because you are unable to > think your own thoughts. > AS > Albrecht, much as I appreciate some of your ideas, I am missing this one a > little. If you are trying to create an enumeration, then you must be doing what > Brian suggests, with the zigzag diagonal like for the enumeration of the > rationals. However, you only have one set on the top line and the bottom, which > you don't show, but which would have the null set. There is not really a zigzag > diagonal covering the set listed this way, as far as I can tell, because it's > not really a rectangular list. I highly recommend the binary natural ordering. > -- > Smiles, > Tony I beg your pardon. But it was just a fake. AS === Subject: Re: infinity ... > > First of all you don't argue on my claim of the thread. > > > Second, your above argueing is not clear to me, since both sets are > > > well-ordered. But it's nice, so I give this: > > > > N > > > {1},{2},{3}, ... > > > N/{1},N/{2},N/{3}, ... > > > {1,2},{1,3},{2,3},{1,4},... > > > N/{1,2}, ... > > > ... > > > > Is this supposed to be a list? My reading > > > of this is that your list is: > > > > N, > > > {1}, > > > {2}, > > > {3}, > > > ... > > > N/{1}, > > > > Right here we have a problem. What is the element > > > before N/{1} in your list? ... > > > > Now count in diagonal sequence. You may think of Cantor's first > > > diagonal proof. > > > > What diagonal? > > Come on, come on! He means the zigzag diagonal, as in the standard > > demonstration that the rationals _are_ countable. This isn't Cantor's > > first diagonal proof, > > I see, you are the real checker. You knows it all. You are famous. You > > are apodictic. All the authors who speak of the first diagonal proof of > > Cantor are wrong. > > but if you're going to argue with cranks you > > must expect them to be pretty muddled about things. > > Anyway, it's obvious that *if* the OP shows a list of lists that > > include all the subsets that is enough. In practice, of course he's > > given the standard crank non-list. > > Brian Chandler > > http://imaginatorium.org > > You and many of the other checkers are not able to discuss my starting > > argument. You are only able to respond to the usual wrong arguments you > > know. And I think, your answers are memorized because you are unable to > > think your own thoughts. > > AS > Albrecht, much as I appreciate some of your ideas, I am missing this one a > little. If you are trying to create an enumeration, then you must be doing what > Brian suggests, with the zigzag diagonal like for the enumeration of the > rationals. However, you only have one set on the top line and the bottom, which > you don't show, but which would have the null set. There is not really a zigzag > diagonal covering the set listed this way, as far as I can tell, because it's > not really a rectangular list. I highly recommend the binary natural ordering. > -- > Smiles, > Tony > I beg your pardon. But it was just a fake. > AS You can biject all constructable subsets of P(N) to N, but not the unconstructable. That's an argument against the diagonal proof, since, if a set will be constructable in form of an antidiagonal of a list, it's a constructable set and therefore it is element of the set of the sets which are bijectable with N. --> The diagonal argument can't proof if a list containes all subsets of N since there occures a contradiction. AS === Subject: Re: infinity ... unconstructable. That's an argument against the diagonal proof, since, > if a set will be constructable in form of an antidiagonal of a list, > it's a constructable set and therefore it is element of the set of the > sets which are bijectable with N. Then that constructable set will be in the original list, and the diagonal set will be different from it, and it can't be a set constructed from the diagonal. But that just proves that you can't construct a set from the diagonal of the list of sets while allowing that set to also be in the list. No such set exists in the list. Therefore, the constructed set must be a set that does not exist in the mapping, and therefore the mapping does not map every possible set in the list. === Subject: Re: infinity ... unconstructable. That's an argument against the diagonal proof, since, > if a set will be constructable in form of an antidiagonal of a list, > it's a constructable set and therefore it is element of the set of the > sets which are bijectable with N. > Then that constructable set will be in the original list, and the > diagonal set will be different from it, and it can't be a set > constructed from the diagonal. > But that just proves that you can't construct a set from the > diagonal of the list of sets while allowing that set to also be > in the list. No such set exists in the list. Therefore, the > constructed set must be a set that does not exist in the mapping, > and therefore the mapping does not map every possible set in > the list. And you are shure the nonconstructable exists? And what meaning did they have to you? You like them. Are they nice. What kind of properties they have (elese than be unconstructable)? And what do you do with them then knowing they are? AS === Subject: Re: infinity ... > > You can biject all constructable subsets of P(N) to N, but not the > > unconstructable. That's an argument against the diagonal proof, since, > > if a set will be constructable in form of an antidiagonal of a list, > > it's a constructable set and therefore it is element of the set of the > > sets which are bijectable with N. > Then that constructable set will be in the original list, and the > diagonal set will be different from it, and it can't be a set > constructed from the diagonal. > But that just proves that you can't construct a set from the > diagonal of the list of sets while allowing that set to also be > in the list. No such set exists in the list. Therefore, the > constructed set must be a set that does not exist in the mapping, > and therefore the mapping does not map every possible set in > the list. > And you are shure the nonconstructable exists? WE are sure [sic] that the 'diagonal' is just as constructible as the list itself is. > And what meaning did > they have to you? What unconstructables' is AS referring to? > You like them. Are they nice. What kind of properties > they have (elese than be unconstructable)? And what do you do with them > then knowing they are? As we have no idea what ASA is nattering on about, perhaps he would explicate. > AS === Subject: Re: infinity ... albstorz@gmx.de says... >You can biject all constructable subsets of P(N) to N, but not the >unconstructable. By constructable do you mean definable? Or do you mean computable? Or do you mean finite, or what? Let me list the combinations here: 1. There is no bijection between N and the set of all subsets of N. 2. There is no computable bijection between N and the set of all computable subsets of N. 3. There is no definable bijection between N and the set of all definable subsets of N. Whatever notion of function we want, it is true that there is no function mapping N to the set of all functions from N into {0,1}. However, if you mix your notions of functions, you can get mixed results: 2'. There *is* a definable bijection between N and the set of all computable subsets of N, but that bijection is not computable. 3'. There *is* a bijection between N and the set of all definable (in the language of arithmetic) subsets of N, but that bijection is not definable in the language of arithmetic. 3' can be extended further: 3''. There *is* a bijection between N and the set of all definable (in the language of ZFC) subsets of N, but that bijection is not definable in the language of ZFC. If you stick to any one consistent notion of function and subset, then Cantor's theorem tells you that there is no bijection (according to that notion) between N and the set of subsets (according to that notion) of N. -- Daryl McCullough Ithaca, NY === Subject: Re: infinity ... You can biject all constructable subsets of P(N) to N, but not the >unconstructable. > By constructable do you mean definable? Or do you mean computable? > Or do you mean finite, or what? Let's have a deal: By unconstructable I mean Unconstructable. May be. > Let me list the combinations here: > 1. There is no bijection between N and the set of all subsets of N. > 2. There is no computable bijection between N and > the set of all computable subsets of N. > 3. There is no definable bijection between N and the > set of all definable subsets of N. > Whatever notion of function we want, it is true that there > is no function mapping N to the set of all functions from > N into {0,1}. No. You are just kidding? Nothing of this is truth. > However, if you mix your notions of functions, you can > get mixed results: > 2'. There *is* a definable bijection between N and the > set of all computable subsets of N, but that bijection > is not computable. > 3'. There *is* a bijection between N and the set of all > definable (in the language of arithmetic) subsets of N, > but that bijection is not definable in the language of > arithmetic. > 3' can be extended further: > 3''. There *is* a bijection between N and the set of all > definable (in the language of ZFC) subsets of N, but > that bijection is not definable in the language of ZFC. > If you stick to any one consistent notion of function and > subset, then Cantor's theorem tells you that there is no > bijection (according to that notion) between N and the set of > subsets (according to that notion) of N. > -- > Daryl McCullough > Ithaca, NY Oh, it's just minding, nothing more. Show me one computable infinite subset of N. In totally. Cantor's dream leads to a nondenumerable mass of schwachsinn. Nothing more. AS === Subject: Re: infinity ... > > > Because I can prove it (and it's a very old proof). A powerset of > > > a nonempty set contains more elements that the set. Can you prove > > > otherwise? > > This argument is stupid. Is there any magic in the powerfunction? > > Proofs are not stupid until they can be refuted. The proof that for an > > arbitrary set S, Card(S) < Card(P(S)) has not been refuted by anyone. > Even if you think that the powersets of finite and infinite sets have > both a greater cardinality than their starting sets, you would not > really think it depends on the same cause in both cases. > You must proof it independently for finite and for infinite sets. In > this sense the argument is stupid. > AS > Albrecht, do you accept the axiom of induction? If so, it is easily provable > inductively that the power set of a set of size n has size 2^n, and since this > is an equality property, it holds for the infinite case. The power set of an > infinite set is infinite, but a larger infinity than the set. > -- > Smiles, > Tony No Tony. If you think so, you should also argue, that since the cartesian product of two sets like {1,2}x{3,4}={{1,3},{1,4},{2,3},{2,4}} in the finite case is always of graeter cardinality than the starting sets, it should be also in the infinite case. But it isn't. AS === 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 >> albstorz@gmx.de said: >> > > > > > > Because I can prove it (and it's a very old proof). A powerset of >> > > a nonempty set contains more elements that the set. Can you prove >> > > otherwise? >> > > This argument is stupid. Is there any magic in the powerfunction? >> > > Proofs are not stupid until they can be refuted. The proof that for an >> > arbitrary set S, Card(S) < Card(P(S)) has not been refuted by anyone. >> Even if you think that the powersets of finite and infinite sets have >> both a greater cardinality than their starting sets, you would not >> really think it depends on the same cause in both cases. >> You must proof it independently for finite and for infinite sets. In >> this sense the argument is stupid. >> Albrecht, do you accept the axiom of induction? If so, it is easily >> provable inductively that the power set of a set of size n has size >> 2^n, and since this is an equality property, it holds for the >> infinite case. The power set of an infinite set is infinite, but a >> larger infinity than the set. > No Tony. > If you think so, you should also argue, that since the cartesian > product of two sets like {1,2}x{3,4}={{1,3},{1,4},{2,3},{2,4}} in the > finite case is always of graeter cardinality than the starting sets, it > should be also in the infinite case. But it isn't. You are arguing with the guy who thinks that the cardinality of the set of even naturals is less than that of the naturals. The amusing thing is that although Tony is talking rubbish here, powersets _do_ have a larger cardinality than the original set, since any presumed mapping f:S->P(S) can't cover {x in S| x not in f(x)} since that set is different from all f(x) for all x in S. So while you win this particular argument against Tony (not too hard to do), you are still wrong in the main. The above proof does not depend on powers, calculation, finiteness or infiniteness. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: infinity ... <85y84n42v5.fsf@lola.goethe.zz> albstorz@gmx.de said: >> > > > > > > Because I can prove it (and it's a very old proof). A powerset of >> > > a nonempty set contains more elements that the set. Can you prove >> > > otherwise? >> > > This argument is stupid. Is there any magic in the powerfunction? >> > > Proofs are not stupid until they can be refuted. The proof that for an >> > arbitrary set S, Card(S) < Card(P(S)) has not been refuted by anyone. >> Even if you think that the powersets of finite and infinite sets have >> both a greater cardinality than their starting sets, you would not >> really think it depends on the same cause in both cases. >> You must proof it independently for finite and for infinite sets. In >> this sense the argument is stupid. >> Albrecht, do you accept the axiom of induction? If so, it is easily >> provable inductively that the power set of a set of size n has size >> 2^n, and since this is an equality property, it holds for the >> infinite case. The power set of an infinite set is infinite, but a >> larger infinity than the set. > No Tony. > If you think so, you should also argue, that since the cartesian > product of two sets like {1,2}x{3,4}={{1,3},{1,4},{2,3},{2,4}} in the > finite case is always of graeter cardinality than the starting sets, it > should be also in the infinite case. But it isn't. > You are arguing with the guy who thinks that the cardinality of the > set of even naturals is less than that of the naturals. > The amusing thing is that although Tony is talking rubbish here, > powersets _do_ have a larger cardinality than the original set, since > any presumed mapping f:S->P(S) can't cover {x in S| x not in f(x)} > since that set is different from all f(x) for all x in S. This proof needs the expression x in S| x not in f(x) has to be meaningful. But it isn't in all the proofs about infinite sets. The antidiagonal is an unicorn. That's all. The proof of the existence of unicorns don't bring them in existence. They don't exist in spite of any proof about it. The other way makes a shoe out of this rubbish: If a proof leads to the result, that unicorns exist, you have proofed your proof (respectively your assumptions) to be wrong. Understand this, and you are out of cantorian fog. It will fall like scales from your eyes: all will be clear and easy. > So while you win this particular argument against Tony (not too hard > to do), you are still wrong in the main. The above proof does not > depend on powers, calculation, finiteness or infiniteness. > -- > David Kastrup, Kriemhildstr. 15, 44793 Bochum Powers of integers could be brake down to multiples of integers which could be brake down to sums of integers. Why should power be able to give a result which is unreachable by multiplication or addition? That's magic math you depend on. AS === 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 albstorz@gmx.de said: > >> >> > >> > > Because I can prove it (and it's a very old proof). A powerset of > > a nonempty set contains more elements that the set. Can you prove > > otherwise? >> > This argument is stupid. Is there any magic in the powerfunction? >> > Proofs are not stupid until they can be refuted. The proof > that for an arbitrary set S, Card(S) < Card(P(S)) has not > been refuted by anyone. > Even if you think that the powersets of finite and infinite sets have both a greater cardinality than their starting sets, you would not really think it depends on the same cause in both cases. > You must proof it independently for finite and for infinite sets. In this sense the argument is stupid. Albrecht, do you accept the axiom of induction? If so, it is easily provable inductively that the power set of a set of size n has size 2^n, and since this is an equality property, it holds for the infinite case. The power set of an infinite set is infinite, but a larger infinity than the set. >> No Tony. >> If you think so, you should also argue, that since the cartesian >> product of two sets like {1,2}x{3,4}={{1,3},{1,4},{2,3},{2,4}} in >> the finite case is always of graeter cardinality than the >> starting sets, it should be also in the infinite case. But it >> isn't. >> You are arguing with the guy who thinks that the cardinality of the >> set of even naturals is less than that of the naturals. >> The amusing thing is that although Tony is talking rubbish here, >> powersets _do_ have a larger cardinality than the original set, >> since any presumed mapping f:S->P(S) can't cover {x in S| x not in >> f(x)} since that set is different from all f(x) for all x in S. > This proof needs the expression x in S| x not in f(x) has to be > meaningful. So where is the problem with its meaning for you? S was _assumed_ to be a set. And f(x) was _assumed_ to be defined as a mapping function from S, so f(x) is defined for all x in S. And since f(x) was assumed to be a member of P(S), then f(x) is again a set. So why can't I just take all x in S that have the property of not being an element of f(x)? > But it isn't in all the proofs about infinite sets. Irrelevant. We are talking about the above proof. Of course, for every truth you'll be able to find millions of invalid proofs. But that does not change a truth, if there is just a single valid proof. So what is wrong with the above proof? > The antidiagonal is an unicorn. That's all. The proof of the > existence of unicorns don't bring them in existence. You are babbling nonsense because you are out of arguments. > They don't exist in spite of any proof about it. The other way > makes a shoe out of this rubbish: If a proof leads to the result, > that unicorns exist, you have proofed your proof (respectively your > assumptions) to be wrong. Understand this, and you are out of > cantorian fog. It will fall like scales from your eyes: all will be > clear and easy. Sorry, but I don't have the required amount of alcoholic beverages and mind-altering drugs in the house. >> So while you win this particular argument against Tony (not too >> hard to do), you are still wrong in the main. The above proof does >> not depend on powers, calculation, finiteness or infiniteness. > Powers of integers could be brake down to multiples of integers > which could be brake down to sums of integers. Why should power be > able to give a result which is unreachable by multiplication or > addition? That's magic math you depend on. You are babbling nonsense. I repeat: the above proof does not depend on powers, calculation, finiteness or infiniteness. No magic math is involved, and no powers are involved. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: infinity ... Because I can prove it (and it's a very old proof). A powerset of >> a nonempty set contains more elements that the set. Can you prove >> otherwise? > This argument is stupid. Is there any magic in the powerfunction? A > hidden megabooster for transcendental overflow? What is the very > special aspect of the powerfunction to be so magic? [...] > I'm very sensible about this because this argument is found in very > much books although it's total meaningless. > (Weak minds might be impressed by the big numbers which are easily > produced by powerfunction.) > What in finity holds may not (or do not) hold in infinity. > So what is your proof? The Gedanken in the context is, if finite powersets are always greater than there sarting sets, is there any connectivity to the idea that infinite powersets are greater than there starting sets. There is no connectivity. AS === 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 Because I can prove it (and it's a very old proof). A powerset of a nonempty set contains more elements that the set. Can you prove otherwise? >> This argument is stupid. Is there any magic in the powerfunction? A >> hidden megabooster for transcendental overflow? What is the very >> special aspect of the powerfunction to be so magic? [...] >> I'm very sensible about this because this argument is found in very >> much books although it's total meaningless. >> (Weak minds might be impressed by the big numbers which are easily >> produced by powerfunction.) >> What in finity holds may not (or do not) hold in infinity. >> So what is your proof? > The Gedanken in the context is, if finite powersets are always > greater than there sarting sets, is there any connectivity to the > idea that infinite powersets are greater than there starting sets. > There is no connectivity. You are babbling. The standard proof about the larger cardinality of powersets makes no use of finiteness, infiniteness or, in fact, the power function at all. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: infinity ... <853bmv5ih7.fsf@lola.goethe.zz>> Because I can prove it (and it's a very old proof). A powerset of a nonempty set contains more elements that the set. Can you prove otherwise? >> This argument is stupid. Is there any magic in the powerfunction? A >> hidden megabooster for transcendental overflow? What is the very >> special aspect of the powerfunction to be so magic? [...] >> I'm very sensible about this because this argument is found in very >> much books although it's total meaningless. >> (Weak minds might be impressed by the big numbers which are easily >> produced by powerfunction.) >> What in finity holds may not (or do not) hold in infinity. >> So what is your proof? > The Gedanken in the context is, if finite powersets are always > greater than there sarting sets, is there any connectivity to the > idea that infinite powersets are greater than there starting sets. > There is no connectivity. > You are babbling. The standard proof about the larger cardinality of > powersets makes no use of finiteness, infiniteness or, in fact, the > power function at all. > -- > David Kastrup, Kriemhildstr. 15, 44793 Bochum To someone, who don't know what is the issue, argueing seems to be babbling. The question was: is there any meaning about the fact, that finite powersets are greater than their starting sets concerning the size of infinite sets? This and only just this point was under discussion. AS === Subject: Re: infinity ... > So, if there is an infinite set there is an infinite number. > Do you mean that an infinite set (or natural numbers) must contain an > infinite number as a member (which is false)? Or do you mean that > the size of an infinite set is represented by an infinite number > (which is partially true)? Not only partially. If there is no infinite number there is no infinite set. If sets consist of discret, distinguishable, individual elements, and sets are definde like this, the natural numbers are just representative for the elements and also for the sets. {1,2,3} means a set with element #1, element #2, element #3, this is the ordinal aspect of numbers. The set with cardinality 3 is just this set {1,2,3}, and at the same time it represents all sets with 3 elements. That's the open secret of the numbers. ordinal = cardinal = natural AS === Subject: Re: infinity ... >> So, if there is an infinite set there is an infinite number. >> Do you mean that an infinite set (or natural numbers) must contain an >> infinite number as a member (which is false)? Or do you mean that >> the size of an infinite set is represented by an infinite number >> (which is partially true)? > Not only partially. If there is no infinite number there is no infinite > set. > If sets consist of discret, distinguishable, individual elements, and > sets are definde like this, the natural numbers are just representative > for the elements and also for the sets. > {1,2,3} means a set with element #1, element #2, element #3, this is > the ordinal aspect of numbers. The set with cardinality 3 is just this > set {1,2,3}, and at the same time it represents all sets with 3 > elements. That's the open secret of the numbers. > ordinal = cardinal = natural Only for finite sets or numbers. For infinite sets we must use infinite ordinals and infinite cardinals. But these ordinals and cardinals do not equate to any naturals because there are no infinite natural numbers. What is the number for set N = {0,1,2,3,...}? If it's w, then how can w be a member of N? If it's w+1, then how can w+1 be a member of N? If it's w-1 (whatever that means), then how can w-1 be a member of N? === Subject: Re: infinity ... >> So, if there is an infinite set there is an infinite number. >> Do you mean that an infinite set (or natural numbers) must contain an >> infinite number as a member (which is false)? Or do you mean that >> the size of an infinite set is represented by an infinite number >> (which is partially true)? > Not only partially. If there is no infinite number there is no infinite > set. > If sets consist of discret, distinguishable, individual elements, and > sets are definde like this, the natural numbers are just representative > for the elements and also for the sets. > {1,2,3} means a set with element #1, element #2, element #3, this is > the ordinal aspect of numbers. The set with cardinality 3 is just this > set {1,2,3}, and at the same time it represents all sets with 3 > elements. That's the open secret of the numbers. > ordinal = cardinal = natural > Only for finite sets or numbers. > For infinite sets we must use infinite ordinals and infinite > cardinals. But these ordinals and cardinals do not equate to any > naturals because there are no infinite natural numbers. Yes. And the cause of this aspect is very good: there are no infinite sets in the sense of a size. Infinite means just: sizeless big. Modern math calls this aleph_0. ok. but you can do nothing with this. Else you will earn contradictions. > What is the number for set N = {0,1,2,3,...}? There is no number. > If it's w, then how can w be a member of N? what is w? Nothing more than unicorn. > If it's w+1, then how can w+1 be a member of N? ... > If it's w-1 (whatever that means), then how can w-1 be a member of N? Whatever this means? Nothing. AS === Subject: Re: infinity ... So biject two non-well-orderable sets. > > With or without an axiom of choice? > I think it is more interesting for the most people without AC, since AC > is not widely loved by the mathematics. > But you may show both if you want. > AS > That shows how little AS knows about things, since given the axiom of > choice, there is no such thing as a non-well-orderable set (though > admittedly there are some sets known that have yet to be well-ordered). W.9frg.9fl === Subject: Re: infinity ... >> A natural number is a set, a cardinality is an equivalence class. >> You make me hopefull. One sees a >> correspondence between natural numbers and von Neumann sets after all. >> And natural numbers don't behave in any other way than sets. So, if >> there is an infinite set there is an infinite number. If there is no >> infinite number there is no infinite set. And vic versa. > You have made exactly this mistake before. Yes every number is a set. > I agree that we can assign a correspondence between natural numbers > and certain sets. But which sets? > von Neumann sets are only one such kind of set, where each one has > a successor: > 0 = {} = { } > 1 = {0} = { {} } > 2 = {0,1} = { {}, {{}} } > 3 = {0,1,2} = { {}, {{}, {{}} } } > etc. > But there are other, equally valid sets that can be associated with > natural numbers. E.g.: > 0 = {} > 1 = {a} u {{0}} = { a, {{{}}} } > 2 = {a} u {{1}} = { a, {{{a, {{{}}} }}} } > 3 = {a} u {{2}} = { a, {{{a, {{{a, {{{}}} }}} }}} } > etc. > This is just one example of the many kinds of sets that can be > associated with natural numbers. > http://en.wikipedia.org/wiki/Natural_number#Constructions_based_on_set_theor y > So when Albrecht says that natural numbers are sets, I ask the > question, which sets? Any kind of set in particular? Yes. The natural numbers are sets which build a kind of stamp form or placeholder of every set with the same cardinality. The natural numbers are the equivalent classes of the sets with the same number of elements. Natural numbers counts sets. What else? AS === Subject: Re: infinity ... > > Coincidently natural numbers and cardinalities are undistinguishable in > > finity. > > They are very similar, but they are not quite undistinguishable. > > A natural number is a set, a cardinality is an equivalence class. > You make me hopefull. Some experts make .8a.8a.8ah, h.9amm and > .9f.9f.9fh if I said A natural number is a set. One sees a > correspondence between natural numbers and von Neumann sets after all. > You are free to say A natural number is a set. without .8a.8a.8ah-, > h.9amm- and .9f.9f.9fh- comments. Be lucky. You are right. > And natural numbers don't behave in any other way than sets. So, if > there is an infinite set there is an infinite number. If there is no > infinite number there is no infinite set. And vic versa. > You have made exactly this mistake before. What mistake? > Yes every number > is a set. No, not every set is a number. For example > {peach, apple, plum, fiddle} is a set but not a number. Wrong. It's a number. The number is this aspect of the set which it have in common with the set {diddle, daddle, doddle, duddle}. To say, a set is a number or a set has a number is a slight difference which has no effect in this considerations. The important aspect is, that every set has a number of elements, since a set, totally independent of the kind of the members of the set, behaves like a set of numbers with same cardinality in all math aspects. But it's sysiphosiousely done. > Just because you have a set does not mean you have a number. > So yes, there is an infinite set. But this does not mean > that this set is a number. Indeed, no infinite set is a number. > - William Hughes > P.S Actually it is not true that natural numbers must be sets, but > they can be. As you insist on using a model in which the natural > numbers are sets, I am playing along to be polite. Oh, how polite. Go play with sand. You are not able to grasp the overview to see the inconsistency of math. You hang on your indoctrination. You are unable to think independently. AS === Subject: Re: infinity ... > > > Coincidently natural numbers and cardinalities are undistinguishable in > > > finity. > > They are very similar, but they are not quite undistinguishable. > > A natural number is a set, a cardinality is an equivalence class. > > You make me hopefull. Some experts make .8a.8a.8ah, h.9amm and > > .9f.9f.9fh if I said A natural number is a set. One sees a > > correspondence between natural numbers and von Neumann sets after all. > > You are free to say A natural number is a set. without .8a.8a.8ah-, > > h.9amm- and .9f.9f.9fh- comments. Be lucky. You are right. > > And natural numbers don't behave in any other way than sets. So, if > > there is an infinite set there is an infinite number. If there is no > > infinite number there is no infinite set. And vic versa. > You have made exactly this mistake before. > What mistake? Assuming that since every number is a set that every set must be a number. > Yes every number > is a set. No, not every set is a number. For example > {peach, apple, plum, fiddle} is a set but not a number. > Wrong. It's a number. The number is this aspect of the set which it > have in common with the set {diddle, daddle, doddle, duddle}. >To say, a > set is a number or a set has a number is a slight difference which has > no effect in this considerations. No, it is crucial. Your thesis is not that every set is a number, but that every set has a number of elements. If your thesis were every set is a number, then to state that there is no infinite number is to deny the existence of an infinite set. If your thesis is that every set has a number of elements, then to say there is no infinite number does not mean that an infinite set does not exists, just that an infinite set does not have an infinite number of elments (at least in the way you define number). >The important aspect is, that every > set has a number of elements, since a set, totally independent of the > kind of the members of the set, behaves like a set of numbers with same > cardinality in all math aspects. Yes all sets have a cardinality. Before you say that this is false, look up the definition of cardinality. This may save you from saying somethiing really silly. Notee. the cardinality of an infinite set does not behave in the same way as the cardinality of a finite set. -William Hughes === Subject: Re: infinity ... >Yes all sets have a cardinality. That's your Choice. === Subject: Re: infinity ... >Yes all sets have a cardinality. > That's your Choice. At least if one chooses to have choice. === Subject: Re: infinity ... That's your Choice. Am I assuming AC here? My understanding was that, while AC was needed to put a total ordering on the equivalence classes under bijection, one could assert the existence of these classes even without AC. Or are the cardinalities defined not as all possible equivalence classes but as the equivalence classes on the ordinals? -William Hughes === Subject: Re: infinity ... > >Yes all sets have a cardinality. > That's your Choice. > Am I assuming AC here? > My understanding was that, while AC was needed to > put a total ordering on the equivalence classes > under bijection, one could assert the existence > of these classes even without AC. Or are the > cardinalities defined not as all possible > equivalence classes but as the equivalence classes > on the ordinals? > -William Hughes As I understand it, without AC, one cannot be sure of creating any sort of function at all from one arbitrary set to another, much less guarantee that such a function, even if creatable, be an injection or a surjection. === Subject: Re: infinity ... Yes all sets have a cardinality. > > That's your Choice. > Am I assuming AC here? > My understanding was that, while AC was needed to > put a total ordering on the equivalence classes > under bijection, one could assert the existence > of these classes even without AC. Or are the > cardinalities defined not as all possible > equivalence classes but as the equivalence classes > on the ordinals? > -William Hughes > As I understand it, without AC, one cannot be sure of creating any sort > of function at all from one arbitrary set to another, much less > guarantee that such a function, even if creatable, be an injection or a > surjection. Modern math concept leads to garbage thinking. AS === Subject: Re: infinity ... > > >Yes all sets have a cardinality. > > That's your Choice. > > Am I assuming AC here? > > My understanding was that, while AC was needed to > > put a total ordering on the equivalence classes > > under bijection, one could assert the existence > > of these classes even without AC. Or are the > > cardinalities defined not as all possible > > equivalence classes but as the equivalence classes > > on the ordinals? > > -William Hughes > As I understand it, without AC, one cannot be sure of creating any sort > of function at all from one arbitrary set to another, much less > guarantee that such a function, even if creatable, be an injection or a > surjection. > Modern math concept leads to garbage thinking. > AS As AS does't seem to be able to think at all, it can make no difference to him what kind of thinking is involved. === Subject: Re: infinity ... >> Yes every number is a set. No, not every set is a number. For example >> {peach, apple, plum, fiddle} is a set but not a number. >> Wrong. It's a number. The number is this aspect of the set which it >> have in common with the set {diddle, daddle, doddle, duddle}. >> To say, a >> set is a number or a set has a number is a slight difference which has >> no effect in this considerations. Albrecht, if every number is a set, which set is 3? What number is the set {a, {b}, {{c}}, {{{d}}}, ...}? If sets are numbers, then adding sets should obey the same rules as adding numbers, right? What is {1,2,3} + {0,2,4}? Is it 3+3 = 6? === Subject: Re: infinity ... >> Yes every number is a set. No, not every set is a number. For example >> {peach, apple, plum, fiddle} is a set but not a number. >> Wrong. It's a number. The number is this aspect of the set which it >> have in common with the set {diddle, daddle, doddle, duddle}. >> To say, a >> set is a number or a set has a number is a slight difference which has >> no effect in this considerations. > Albrecht, if every number is a set, which set is 3? > What number is the set {a, {b}, {{c}}, {{{d}}}, ...}? > If sets are numbers, then adding sets should obey the same rules as > adding numbers, right? What is {1,2,3} + {0,2,4}? Is it 3+3 = 6? 3 is every set with 3 members. Shurely the sets are different. And they have different properties. But in the aspect of size of membership they are exact same. Natural numbers are the essence of the sets concerning their size. David,try to be logic and honest to yourself. It's not a play to win or loose. It's a play to understand or to stay weak in mind. You might belief in a one century old tale or in a more than a thousand years old known thruth. AS === Subject: Re: infinity ... > >> Yes every number is a set. No, not every set is a number. For example > >> {peach, apple, plum, fiddle} is a set but not a number. > >> Wrong. It's a number. The number is this aspect of the set which it > >> have in common with the set {diddle, daddle, doddle, duddle}. > >> To say, a > >> set is a number or a set has a number is a slight difference which has > >> no effect in this considerations. > Albrecht, if every number is a set, which set is 3? > What number is the set {a, {b}, {{c}}, {{{d}}}, ...}? > If sets are numbers, then adding sets should obey the same rules as > adding numbers, right? What is {1,2,3} + {0,2,4}? Is it 3+3 = 6? > 3 is every set with 3 members. In what sense? If 3 were every set with 3 members then 3 would not have 3 members. > Shurely the sets are different. And they > have different properties. But in the aspect of size of membership they > are exact same. Natural numbers are the essence of the sets concerning > their size. Not in NBG. > David,try to be logic and honest to yourself. > It's not a play to win or loose. It's a play to understand or to stay > weak in mind. AS seems to have the stay weak in mind part down pat, but needs more practice with understand. > You might belief in a one century old tale or in a more than a thousand > years old known thruth. > AS === Subject: Re: infinity ... > stephen@nomail.com said: > > stevendaryl3016@yahoo.com said: > >> albstorz@gmx.de says... > >> > >>I don't know what you are talking about. The proof for finite sets > >>needs just a complete induction. This will not hold for infinity I > >>think. > >> > >> There is no induction involved in the finite case, and the > >> infinite case is *exactly* the same proof as the finite case. > >> > >> Here it is once again: > >> > >> Let A be any set whatsoever, finite or infinite, it doesn't matter. > >> Let f be any function from A to P(A). > >> Let w = { x in A | x is not an element of f(x) }. > >> Let x = any set in A. > >> Let u = f(x). We prove that u is not equal to w. > >> > >> By definition of w, we have x in w <-> x is not an element of f(x). > >> So x in w <-> x is not an element of u. That means that there are > >> two cases: Case 1: x in w, and x is not in u. In that case, u cannot > >> equal w. Case 2: x is not in w, and x is in u. In that case, u cannot > >> equal w. > >> > >> So what we have proved is that forall x, w is not equal to f(x). So > >> w is not in the image of f. So f is not a bijection between A and P(A). > >> > >> There's no induction. There's no assumption that A is finite. > > But there is an assumption that y is in S. > > Of course there is an assumption that y is in S. The > goal is to find a bijection f from S to P(S). That > means that for every element x in P(S), there must be > an element y in S such that f(y)=x. w is an element of P(S). > In order for f to be a bijection, there must exist an > element y in S such that f(y)=w. > > If y is not in S, then it is irrelevant to the question > of whether or not f is a bijection from S to P(S). > > Do you consider the following a bijection from S={a,b,c} to its > power set? > > f(a) = {} > f(b) = {a} > f(c) = {b} > f(d) = {c} > f(e) = {a,b} > f(g) = {a,c} > f(h) = {b,c} > f(i) = {a,b,c} > > If w= { x : x in S and x not in f(x) } > then w= {a,b,c}. > > Is there a y in S such that f(y) = {a,b,c}? No. > > Is there a y such that f(y) = {a,b,c}? Yes, f(i)={a,b,c}, but > i is not in S, and is not part of a bijection from S to P(S). > > The above function is a bijection from {a,b,c,d,e,g,h,i} > to P({a,b,c}). It is not a bijection from {a,b,c} to P({a,b,c}). > > Stephen > > But, if this set went on forever, then i WOULD be in the set, and for any > subset, you could identify an x such that it mapped to that set. It is true > that no bijection is possible in the finite case. Then by TO-induction, if it is true for all the finite cases, requires that it must be true for the infinite cases also. > So tell me, what rule of bijection did I break, and at what point does the > mapping through the infinite binary strings break down. For lack of answers > to > thses questions, my bijection stands. That lack of answers only occurs in TOmatics. Outside TOmatics, there are a surplus of answers which show no such bijection can exist. === Subject: Re: infinity ... About the leading zeros, maybe it's one way to illustrate that the antidiagonal process as applied to an infinite list that is supposed to be a list of all real numbers is a contradiction in terms, because each element of the set of reals is distinct from each other, because a set is not a multiset. Here we are seeing that an antidiagonal different in representation than each element of an infinite list is not an element of that list, nor of the range of the function. It's kind of interesting that all rational numbers have at least dual representation as simple continued fractions that are finite. As well, it appears that an element of a continued fraction might be an arbitrarily large number. Is it not so that a rational number can be represented with an infinite continued fraction, in a similar way to how a root of a power series might be an integer? Consider other ways to represent each of the set of real numbers as a sequence, besides trivially representing each real number as the sequence (0), representations that give each real number at least one unique identifying sequence. Congratulations. Most of the sequences are infinite and most of the arguments about binary integer modulus representation apply. I model only binary numbers where there is exactly and only one antidiagonal of the matrix of elements thus constructed, all elements of the unit interval of reals as infinite bitstrings with a beginning, and dual representation allows the existence of an antidiagonal, another extension. The Equivalency Function is defined in a way thus that all values of EF(n) except for n=0 are indefinite, and greater than zero, and less than or equal to one. This is where the set of integers is an infinite set. I might suggest a completely different tack on the antidiagonal: not that it shows that there is no mapping between a set and its powerset, but that infinity is dually represented as zero. I think more about the antidiagonal of an infinite list of particularly all possible sequences of binary elements. Thus, either a does not exist or a is dually represented. Why would that not be so? Alternatively, as the reals are uncountable, you might say, or infinite, x_n, is never the last element of the reals to affect the antidiagonal, yet x_n is always itself. I consider the Equivalency Function, and its list of values, and what its binary antidiagonal would be. It would start with infinitely many 1's, then, at some point there might be some values that are not 1's, for around EF^-1(1/2), then, I'm not quite sure what happens. An alternative consideration is that the antidiagonal would be all ones, then perhaps some ones and zeros, and eventually all zeros. Another consideration is that it is ill-defined as all the elements are indefinite, or that the antidiagonal is .111.... So say you want to add that to the list and generate a new antidiagonal. The list elements are sorted in ascending order, so you have to put any list element you consider in the correct location. If it's .111..., that's greater than or equal to any element of the range, and the infinite list has no end, so there's no place to put it. === Subject: Re: infinity ... > William Hughes said: > Because we are discussing standard mathematics here and there > are no infinite natural numbers (only infinite ordinals and cardinals > which are not the same thing). Yes it is true that the semi-mythical > TO naturals have infinite members and that induction sometimes > holds, but that is irrelevent here. > No it's not at all. The bijection I offered is between *N, which has infinite > values as members, and its power set. Except that one must assume that result to obtain it. To starts with the set of all endless binary strings and wants to represents sets of them by similar strings. thus one member sets would be strings with one 1 and the rest all zeros. If this could really be done, one could get by with representing all such infinite strings using only finite strings, the one element set strings, with all the zeros beyond the one 1 omitted. Thus if TO could do what he claims, he would only need a set of finite strings each with a single 1 to represent all his infinite strings. === Subject: Re: infinity ... keep mumbling about numbers. Why on earth should the y be larger > than any element in w? We've proved that y cannot exist at all, but > there is no a priori reason any notion of largeness is involved. If > we were considering the set of all finite simple groups or the set of > all Platonic solids, there would be no larger, but the proof applies > exactly the same. > Did I not give what appears to be a valid bijection between the NUMBERS in *N > and the SETS in P(*N)? If *N is a set, in the sense of normal set theory, then no you didn't. (Am I remembering right? Is *N the set of TOnats, the set of infinitely leftward extending digits laced with colons and other decorations?) > If the bijection is not valid, please state what mistake > I made in constructing it. Well, you didn't state it in a way that anyone could derive for any element of P(*N) what element of *N was supposed to map to it. Saying 'map 0 to {0}, dot dot [defining finite cases] ... carry on into the infinite' leaves the problems to the mumbled dots at the end. > When I say the element y is larger, perhaps what I > should say is its position is greater than the size of S, that is, it is in the > set[*], but not in the segment of the set[**] considered as S. Well, there you are. S is a set. In set theory that means that S is the set S. It contains every element that is in S, and does not contain any that are not in S. Only S is the set containing every element in S and infinite sets too: if T is an infinite set it means we cannot even in principle lists the elements at some constant readable point size on any four-sided piece of paper, however large. Even so, the elements of T are indeed all elements that belong to T. (I hope I've remembered right: we consider a mapping from S to P(S), and y is the putative element of mapping to the 'magic subset' of those elements which map to subsets not containing themselves. Yes? So y, an element of S cannot exist.) But in your version, first you're still stuck on this numbers thing, so you can't describe set membership without irrelevancies like greater than. Then you have four references to sets. Can you please explain which sets they are: > When I say the element y is larger, perhaps what I > should say is its position is greater than the size of S, This S really is the set we are talking about, no? > that is, it is in the set[*], The set?? Which one is that? Not S? but not in the segment of the set[**] Is this again some other (the same other?) set that isn't S? What is the relevance of this? >... considered as S. Oh no, the last the set[**] is considered as S. Well, normal set theory doesn't have a defined relation considered as. Is this the set S, or is it not? > When you ask for an > element that maps to the entire set, that assumes you have a bitstring that > that element should be equal to. How many 1's are in that bit string? Where > does it end? A bijection to the power set requires something to map onto every subset, including the set itself, which is a (non-proper [technical term]) subset. I'm a bit lost, but I expect the answer to Where does it end? is It doesn't. Brian Chandler http://imaginatorium.org === Subject: Re: infinity ... > Virgil said: > > > albstorz@gmx.de said: > > > > > You must proof it independently for finite and for infinite > > sets. In this sense the argument is stupid. > > > > > > > > > > Albrecht, do you accept the axiom of induction? If so, it is > > easily provable inductively that the power set of a set of size n > > has size 2^n, and since this is an equality property, it holds > > for the infinite case. The power set of an infinite set is > > infinite, but a larger infinity than the set. > > That is only true if 'size' means cardinality, since it is based > entirely on injection/surjection/bijection arguments. > No, cardinality only comes to this conclusion through the clumsiest > of methods, like a bear that scratches his back by getting himself > run over by a truck. Then TO must be claiming that not only can sets of the same cardinality have different sizes, but also that sets of the same size may have different cardinalities. I should very much like to see any example of the latter situation. > > And TO correcting albstorz is a perfect illustration of the blind > leading the blind. > Yes, that's how it looks, to the blind. AS TO is one of them, he can speak for them. === Subject: Re: infinity ... > Virgil said: > > Except for the obvious bijection between *N and P(*N). But, hey, what's > > one > > measly counterexample? > > In this case it is quite measly, since it specifically requires that > one assume most of the result in order to prove the result. > Assume what, specifically? That there is a bijection between the set of all binary strings and the set of only those binary strings having a single non-zero digit. > But any such countable well-ordering would prove that the set of endless > sequences of binary digits is TO-finite. > Not I am talking about actual infinitely long bitstrings, such as would > represent an actual infinite value in binary numbers. But the set of them having only a single non-zero digit IS TO-finite. And if a set of binary strings is to be represented by 1's for each member, a one element set can only have one 1. === Subject: Re: infinity ... > Virgil said: > > > Virgil said: > > > > > David R Tribble said: > > > > But that's an incomplete mapping, because there are not enough > > > infinite > > > binary strings in *N to enumerate all of the subsets of *N. Try > > > it, > > > if you don't believe me. > > > > > > > > > Not enough infinite binary strings? > > > > Precisely. > > > > So, you need more than an infinite amount? > > You need more then you have, in the same sense that there are more > reals than rationals even though there is an endless supply of both. > > There are also more rationals than naturals Except that one can easily inject the rationals into the naturals, which assures that the 'size' of the rationals must be the same size as a subset of of the naturals. > The problem with the reals is that you > don't have a natural well-ordering of them. I can give you that, when you're > ready, which will probably never be. A valid explicit well-ordering of the reals would be of some actual mathematical interest. If TO could provide one, he might be worth a tinker's dam after all, but I am not going to bate my breath til he does! === Subject: Re: infinity ... > Virgil said: > > > Did you have an objection to the bijection between *N and P(*N)? What > > did I do wrong there, bijection-wise? > > You assume your result in order to prove it. > > Virgil, you're being a lunkhead. It wasn't a proof, and I didn't assume any > conclusion. TO's inability to see what he is actually doing is one of the reasons TOmatics is so self-destructive. > I described a bijection What TO described was nowhere near to being either a complete description nor a bijection, and reasons why it is not a bijection have been repeateedly provided as well as reasons why it cannot be one. That TO will not acknowledge either does not vitiate their validity. > No one has answered that question yet. Perhaps TO is losing his memory, or what little literacy he has. > Discredit the > counterexample with established rules For whatever S and whatever f:S -> P(S), To wants to propose, TO cannot find any s in S with f(s) = {x in S:x not in f(x)}. Absent such an s, f is not a surjection and thus not a bijection either. or accept it as a refutation of the > proof. You claim to be logical? Obey logic. === Subject: Re: infinity ... > > Since the argument is not about sizes but about lack of any > surjection/bijection from any set X to its power set P(X), size in TO's > sense, is not an issue, but bjiection is. > > > It's clearly not. I just see a bijection between them. > > TO accepts the proof of no bijection then immediately turns around and > claims to see a bijection. > No, I accept the result that the power set is larger, but not that a > bijection > is impossible. You need to focus. Try ginseng. As usual TO is off point. No one has claimed that the sets are of equal size, unless it be TO, so that equality of size has never been the issue. But there has been stated and proved that there is never a surjection, much less an injection, from any set to its power set, and that IS the issue. So far, it is an issue that TO has carefully avoided. === Subject: Re: infinity ... > Virgil said: > > Since 'more' in the context above means that there cannot be any > bijection, TO is now claiming that where there *are* reasons why no > bijections can exist, there also are no reasons why bijections > cannot exist. > I have said repeatedly and consistently that bijection alone does not > indicate equality of size for infiite sets. You know that, > dorkenburger. TO is, as usual, missing the point! The issue is whether where bijection is impossible, equality of size can still occur. Does TO claim that? If so, TO should give us an example of it occuring. === Subject: Re: infinity ... > Virgil said: > For what set is TO assuming that each member of its power set can be > represented by a single bit in an infinite sequence of bits? > I never said that. If each subset of *N is to be represented by an infinite binary sequence of digits with 1 in some position representing the presence of a member *N and 0 representing its absence, then one element sets must be represented by strings with one 1 in them. So that, while he may not have been aware of it, TO was saying precisely that. That TO is often unaware of the menning of what he is saying has long been apparent. > If that set is the set of all Dedekind infinite binary strings, which is > uncountably infinite, or any set bijectable with it, then TO's > assumption is false. Unless we are operating in TOmatics where an > infinite string of bits can contain an uncountable number of bits. > What is the limit on the number of bits? One bit for every real number in the > entire real interval. How's that? To do that in a string, TO must not only well-order the reals, but order them isomorphically to the set of finite naturals. > > > > > I got a little confused, > > but you still haven't proven anything like > > the impossibility of a bijection with the power set. > > We have to those who are not so permanently confused. > > > > In other cases > > bijections are performed without regard to such discrepancies. > > The 'discrepancy' is that when mapping a set to its power set, there > must always be a member of the codomain which is not the image of > anything in the domain. > > When that happens, whatever one does have, it is not a bijection. > So, which element of the power set does not have a natural mapped to it? {x in S:x not in f(x)} === Subject: Re: infinity ... <85oe5liuuk.fsf@lola.goethe.zz> <851x2h9z6h.fsf@lola.goethe.zz> David Kastrup said: >> Wrong. N is a different set from N{1}, but that does not mean that >> the sets have different cardinality. To illustrate this for the >> really dense people with a simpler example: 2 is in {2} and not in >> {1}, and that means that the set {2} is different from the set {1}, >> even though both sets have the same cardinality, namely 1. > What about {1}/{1,2,3}? I think that has a negative number of elements, > don't you David? It has an unidentifiable number of elements. === Subject: Re: infinity ... Tony Orlow: >> What do you want me to try, anyway, and infinite mapping, >> element-by-element? A bijection's a bijection, right? David R Tribble: >> Yes, that would be nice. Please show us your bijection. Tony Orlow: >> f(0) = ...000 = {} >> f(1) = ...001 = {0} >> f(2) = ...010 = {1} >> f(3) = ...011 = {0,1} >> f(4) = ...100 = {2} >> f(5) = ...101 = {0,2} >> f(6) = ...110 = {1,2} >> f(7) = ...111 = {0,1,2} >> etc. Tony Orlow: > Notice above that no natural maps to a subset which contains it, so w is all > of N. Imagining any completed w leads to a contradiction, since the natural > that would map to it is always bigger than every natural in that set. > That's okay though, because for every natural, there's a larger one. If x > exists, 2^x exists. The sets are infinite, so the bijection continues, > [...] You realize, of course, that you've just proved that there is no natural x that can map all the members of N, don't you? You say: Choose a natural x in N that maps f(x) to all the naturals in N. But x must be bigger than all naturals in N, so choose 2^x (which we know exists, because x exists), and call it x'. Oh, but now x' must be larger than all naturals in N, so choose x'' = 2^x'. Oh, but now, x'' must be larger, ... and so on. We conclude: Therefore we cannot find an x in N that maps f(x) to N. Therefore there is no x in N that maps f(x) to N. Therefore x does not exist. Therefore f(x) is not a surjection from N to P(N). Therefore f(x) is not a bijection from N to P(N). === Subject: Re: infinity ... . > You are assuming a finite natural number of bits, but that is simply > not the case with *N, or it would ony contain finite values. Are we > talking about the same set? Only TO has any clue as to what he is talking about. > The conclusion is that there does not exist a surjection from the > members of *N to the members of P(*N), and therefore no bijection > between them is possible. > But that conclusion assumes a last element, whence it draws its > contradictions. What 'last element' is that? The disproof of TO's alleged bijection merely requires the production of a member of the power set not in the image of the function, which does not require any 'last element', no matter how much TO claims otherwise. === Subject: Re: infinity > Mathematics is not founded by logics , these two subjects are parallel > to each other. > many parts of mathematics are not explained by logics at all, example: > what is number? > still their is no final answer to that question. > Of coarse it would be good if we have a logical godfatherhood on math. > but that is not the case. > Zuhair Does Zuhair, in his infinite wisdom, know of any part of mathematics in conflict with logic? Mathematicians do not. === Subject: Re: infinity conflict with logic? Mathematicians do not. ZF is inconsistent. A set containing nothing is trivially regular. Consider the humble predicate: true. Extend Cantor's results. Skolemize, all the way out. Infinite sets are equivalent. interpretation: powerset. Ord: less than nothing. Virgil, you speak for yourself. Ross === Subject: Re: infinity I didn't say conflict, I said are not fully explained by logics Zuhair === Subject: Re: infinity > I didn't say conflict, I said are not fully explained by logics > Zuhair That merely is an admission that logic and mathematics are not the same thing. If they were the same, we would not need two names. === Subject: Re: infinity No. I do really believe that their are parts of mathematics that cannot be explained by formal logic, mathematics is something not only analytic, it is inductive(experiental) also,or what Kant call's it composite . formal logic cannot make one know composite facts that necessitate observation for their validation , mathematics has a kind of such a reality ,so logic alone is not enough. Tell me Virgil what is the meaning of number by logic??? Proove to me in a logical manner what is 1+1=2 some things are just common sense derived , that cannot be explained by logics. This is a long story Virgil you don't need to heart you head with it. Zuhair === Subject: Re: infinity be explained by formal logic, Yes, we call those axioms. But everything else is derived from the axioms by formal logic. > mathematics is something not only > analytic, it is inductive(experiental) also, No, it most certainly is not. > or what Kant call's it > composite . formal logic cannot make one know composite facts that > necessitate observation for their validation , mathematics has a kind > of such a reality ,so logic alone is not enough. Tell me Virgil what is > the meaning of number by logic??? > Proove to me in a logical manner what is 1+1=2 This is provable from the axioms, but the formal proof is actually surprisingly long. > some things are just common sense derived No, in mathematics things are not common sense derived. - Randy === Subject: Re: infinity Rand please anwer the following questions: On what thing axioms are based? My answer is common sense derived experience. Or more clarily what generates axioms? What is the difference between an axiom and a postulate? Is all of what is present in mathematics axiomatic formally derived? What is the impact of Godel's incompleteness theorum on the assumption of logical axiomatic foundation of mathematics ? I agree with you Randy that many mathematical facts can be translated into logical language, which gives the impression that formal logic is more basic than math.which is not the truth. mathematical entities are imperical, but fine imperical.( Platonisim). The formal logics only describe the form of mathematics but not the content of it. The content of mathematical subjects is based in it's depth on common sense , and that common sense cannot always be traced into formal logical aximoatic system as you think. When I say mathematics is the art of common sense reasoning I meant at the base of it. Common sense is itself imperical or experiental. However this is another subject . Zuhair === Subject: Re: infinity On what thing axioms are based? An attempt to give a rigorous self-consistent foundation to a useful or interesting mathematical structure (number theory, probability, geometry, topology, ...) > My answer is common sense derived experience. Your answer is incorrect. The development of axioms is precisely to remove the ambiguities, self-contradictions, and vagueness which common sense might dictate. To put things on a firmer foundation than common sense or experience. > Or more clarily what generates axioms? Mathematicians. > What is the difference between an axiom and a postulate? I think they can be used interchangeably in mathematics. In science, postulate can be used in the same way, to derive results from. However, a science postulate is a guess about Mother Nature's real postulates, and the derived results are predictions which can be tested and falsified. This is not true in mathematics. There is no real postulate which is the truth against which a mathematical axiom is tested. > Is all of what is present in mathematics axiomatic formally derived? Yes, it is formally derived from axioms. Every mathematics course I've ever seen beyond a certain level starts on day 1 with here are our axioms, and then proves theorems from those axioms for the rest of the semester. > What is the impact of Godel's incompleteness theorum on the assumption > of logical axiomatic foundation of mathematics ? I'm less sure of my ground here. I believe Godel's Theorem established that you must always have an axiom, something accepted as true and unprovable within your formal system. > I agree with you Randy that many mathematical facts can be translated > into logical language, I wonder when I said that? However, with some caution I think I will agree that this is true, since it is the basis of automatic proof systems. (I think). > which gives the impression that formal logic is > more basic than math.which is not the truth. It doesn't give that impression to me. > mathematical entities are imperical, No, they are not. > When I say mathematics is the art of common sense reasoning I meant at > the base of it. No, it is not. > Common sense is itself imperical or experiental. An engineer, a physicist, and a mathematician are sitting on a hill in Scotland looking at a flock of sheep. The engineer says the sheep in Scotland are black. The physicist says Correction. There are at least 20 black sheep in Scotland. The mathematician says You're both wrong. There are at least 20 sheep in Scotland which are black on at least one side. The engineer is applying common sense logic, and the conclusion may easily be wrong. - Randy === Subject: Re: infinity > I'm less sure of my ground here. I believe Godel's Theorem > established that you must always have an axiom, something > accepted as true and unprovable within your formal system. In other words, that we can't prove everything, but need some basic assumptions or axioms? This has been well understood since antiquity. No need to invoke Godel. === Subject: Re: infinity established that you must always have an axiom, something > accepted as true and unprovable within your formal system. > In other words, that we can't prove everything, but need some basic > assumptions or axioms? This has been well understood since > antiquity. No need to invoke Godel. Ah, the Liar paradox. This statement is not provable. - Goedel All Cretans are liars. - Epimedes This statement is not true. - Liar For any false statement, the statement is not provably true. It's provably false. The problem with that is where it claims itself to be false. Temporally, that's true. http://www.iep.utm.edu/p/par-liar.htm In the null axiom theory, the opposite of all possible statements is the same set of statements. Ross === Subject: Re: infinity I'm less sure of my ground here. I believe Godel's Theorem > > established that you must always have an axiom, something > > accepted as true and unprovable within your formal system. > In other words, that we can't prove everything, but need some basic > assumptions or axioms? This has been well understood since > antiquity. No need to invoke Godel. > Ah, the Liar paradox. > This statement is not provable. - Goedel > All Cretans are liars. - Epimedes Excuse me, Epimenides. > This statement is not true. - Liar > For any false statement, the statement is not provably true. It's > provably false. The problem with that is where it claims itself to be > false. Temporally, that's true. > http://www.iep.utm.edu/p/par-liar.htm > In the null axiom theory, the opposite of all possible statements is > the same set of statements. Ross === Subject: Re: infinity established that you must always have an axiom, something > accepted as true and unprovable within your formal system. > In other words, that we can't prove everything, but need some basic > assumptions or axioms? This has been well understood since > antiquity. No need to invoke Godel. Then axiomatize, everything. Ross === Subject: Re: infinity ... > Virgil said: > That shows how little AS knows about things, since given the axiom of > choice, there is no such thing as a non-well-orderable set (though > admittedly there are some sets known that have yet to be well-ordered). > > Gee, like what, Virgil? Like one version of the axiom of choice, which merely states that every set is well-orderable, but gives no methodology for doing it. === Subject: Re: infinity ... Virgil said: > Virgil said: > > That shows how little AS knows about things, since given the axiom of > > choice, there is no such thing as a non-well-orderable set (though > > admittedly there are some sets known that have yet to be well-ordered). > > > Gee, like what, Virgil? > Like one version of the axiom of choice, which merely states that every > set is well-orderable, but gives no methodology for doing it. Well, that sounds like rock-solid proof to me! Stupid arbitrary axioms. Make it make sense. -- Smiles, Tony === Subject: Re: infinity ... > Virgil said: > > > Virgil said: > > > That shows how little AS knows about things, since given the > > axiom of choice, there is no such thing as a non-well-orderable > > set (though admittedly there are some sets known that have yet > > to be well-ordered). > > > > Gee, like what, Virgil? > > Like one version of the axiom of choice, which merely states that > every set is well-orderable, but gives no methodology for doing it. > > Well, that sounds like rock-solid proof to me! Stupid arbitrary > axioms. Make it make sense. It is quite impossible to have things make sense while insulated from reality in the dream world of TOmatica. Until TO leaves that world, there are a lot of things that make perfect sense outside it that will not make sense to him. === Subject: Re: infinity ... > Virgil said: > Then lets see one of your alleged bijections from some set X to its > power set P(X). > > The only valid disproof of a proof of non-existence is an example > of what is alleged to be non-existent . > Hmm, so what was the problem with my bijection again? What rule did I > break. You failed to come up with any s in S such that f(s) = {x in S:x not in f(x)} > No one has been able to tell me, no matter how many times I > beg for assistance in understanding this deep and mysterious subject. To has been told, but his illiteracy prevents him from knowing it. > > If anyone counterclaims agains the proof of no surjection from and > set X to its power set P(X), that person owes us an example of such > a mapping. > And if anyone wants to claim their proof stands in the face of an > obvious counterexample, that person owes us an explanation of how > that counterexample does not apply. So, what of it, Virgie? It is TO claiming the counter-example to a standard proof. Let him find the counter-example, an S and a f:S -> P(S), and an s in S with f(s) = (x in S: x not in f(x)}. > What rule > does my bijection between *N and P(*N) break? Take your time.... See above! === Subject: Re: infinity ... Virgil said: > Virgil said: > > Then lets see one of your alleged bijections from some set X to its > > power set P(X). > > > > The only valid disproof of a proof of non-existence is an example > > of what is alleged to be non-existent > Hmm, so what was the problem with my bijection again? What rule did I > break. > You failed to come up with any s in S such that > f(s) = {x in S:x not in f(x)} I failed to name the last natural? Shame on me! > No one has been able to tell me, no matter how many times I > beg for assistance in understanding this deep and mysterious subject. > To has been told, but his illiteracy prevents him from knowing it. What rule of construction did I break in my bijection? What did I do wrong? > > > > If anyone counterclaims agains the proof of no surjection from and > > set X to its power set P(X), that person owes us an example of such > > a mapping. > And if anyone wants to claim their proof stands in the face of an > obvious counterexample, that person owes us an explanation of how > that counterexample does not apply. So, what of it, Virgie? > It is TO claiming the counter-example to a standard proof. > Let him find the counter-example, an S and a f:S -> P(S), > and an s in S with f(s) = (x in S: x not in f(x)}. 2^oo-1 maps to {0,...,oo-1}. > What rule > does my bijection between *N and P(*N) break? Take your time.... > See above! What did I do wrong, besides failing to name the last natural? -- Smiles, Tony === Subject: Re: infinity ... > Virgil said: > You failed to come up with any s in S such that > f(s) = {x in S:x not in f(x)} > I failed to name the last natural? Shame on me! Why TO thinks that misrepresenting the issue of finding an s such that f(s) = {x in S:x not in f(x)} is connected to the issue of whether there is a last member of S will get him out of the fix he is in is just another of the mysteries continually surrounding the foggy land of TOmatica. > > > No one has been able to tell me, no matter how many times I > > beg for assistance in understanding this deep and mysterious subject. > > To has been told, but his illiteracy prevents him from knowing it. > What rule of construction did I break in my bijection? What did I do wrong? It may have started with getting born. > > > > If anyone counterclaims agains the proof of no surjection from and > > set X to its power set P(X), that person owes us an example of such > > a mapping. > > > And if anyone wants to claim their proof stands in the face of an > > obvious counterexample, that person owes us an explanation of how > > that counterexample does not apply. So, what of it, Virgie? > > It is TO claiming the counter-example to a standard proof. > > Let him find the counter-example, an S and a f:S -> P(S), > and an s in S with f(s) = (x in S: x not in f(x)}. > 2^oo-1 maps to {0,...,oo-1}. Without knowing S and f, we cannot determine whether {0,...,oo-1} = {x in S:x not in f(x)} or not, so TO's answer is insufficient. > > > What rule > > does my bijection between *N and P(*N) break? Take your time.... > > See above! > What did I do wrong, besides failing to name the last natural? Suggesting that there is one, for starters! === Subject: Re: infinity ... > What about {1}/{1,2,3}? I think that has a negative number of elements, don't > you David? If TO had any idea of what was really going on he would not make such an ass of himself. What does he think the definition of set relative differences, AB, is. > I knew you'd agree! I bet you can help me with imaginary sets too. > Perhaps we can collaborate? That'll be great! All of TO's set seem to be equally imaginary, as is everything in TOmatics. === Subject: Re: infinity ... Virgil said: > What about {1}/{1,2,3}? I think that has a negative number of elements, don't > you David? > If TO had any idea of what was really going on he would not make such > an ass of himself. What does he think the definition of set relative > differences, AB, is. Alas, the extent of poor Virgil's humor seems to be laughing at the fly whose wings he pulls off. It was a joke Virgil. A joke. Sheesh! > I knew you'd agree! I bet you can help me with imaginary sets too. > Perhaps we can collaborate? That'll be great! > All of TO's set seem to be equally imaginary, as is everything in > TOmatics. And Virgil's little purple friend Cato. -- Smiles, Tony === Subject: Re: infinity ... > Virgil said: > > > > What about {1}/{1,2,3}? I think that has a negative number of > > elements, don't you David? > If TO had any idea of what was really going on he would not make > such an ass of himself. What does he think the definition of set > relative differences, AB, is. > > Alas, the extent of poor Virgil's humor seems to be laughing at the > fly whose wings he pulls off. It was a joke Virgil. A joke. Sheesh! With TO's extraordinarily remote contact with reality, one can never be quite sure that what in others might be intended to be funny is not TO being serious. > > > I knew you'd agree! I bet you can help me with imaginary sets > > too. Perhaps we can collaborate? That'll be great! > > All of TO's set seem to be equally imaginary, as is everything in > TOmatics. > And Virgil's little purple friend Cato. Another thing that only exists in the weird wonderland of TOmatica, where most of us, fortunately, need never go. === Subject: Re: infinity ... > Is it > impossible to biject an infinite ordered set with its power set? Nope Repeated posting of a falsehood does not make it any less false. === Subject: Re: infinity ... > Is it > impossible to biject an infinite ordered set with its power set? Nope Wrong again, TO! At least everywhere except possibly in that wonderland of TOmatics. === Subject: Re: infinity ... Tony Orlow: >> I already showed you the bijection between binary *N and P(*N). >> What didn't you like about it? It is valid. David R Tribble: >> No, you showed a mapping between *N and R, which is equivalent >> to a mapping between *N and P(N). That's easy. Tony Orlow: >> What do you want me to try, anyway, and infinite mapping, >> element-by-element? A bijection's a bijection, right? David R Tribble: >> Yes, that would be nice. Please show us your bijection. Tony Orlow: >> f(0) = ...000 = {} >> f(1) = ...001 = {0} >> f(2) = ...010 = {1} >> f(3) = ...011 = {0,1} >> f(4) = ...100 = {2} >> f(5) = ...101 = {0,2} >> f(6) = ...110 = {1,2} >> f(7) = ...111 = {0,1,2} >> etc. Any questions? David R Tribble: >> Where are the infinite subsets, such as the set of even numbers? Tony Orlow: > Well, infinitely far down the list of course! > f(N/3) = 0:010101.....010101 = {0,2,4,6,8,....} > f(2N/3)= 0:101010.....101010 = {1,3,5,7,9,....} All of the sets you listed, finite and infinite, are composed of only finite naturals. Which means that they are all subsets of N. You're using N (and N/3, 2N/3, etc.), which is a member of *N. So all you're doing is showing a bijection between *N and P(N), which is what I said you were doing in the first place. That's easy, and is equivalent to a bijection between R and P(N). But we're asking for a bijection between *N and P(*N), or between N and P(N). You haven't adequately shown those yet. > Oh, now comes the list of primes (sigh) > Uh, um.... > f(N/pi) = ....1010001010110 = {2,3,5,7,11,13,...} > PS, I'm kidding about the N/pi part (I think). The infinite natural x in *N that maps f(x) to the infinite set of primes is: x = 2^2 + 2^3 + 2^5 + 2^7 + 2^11 + 2^13 + ... But like all the other subsets your f mapping defines, the set of primes contains only finite numbers, which means that it is one of the subsets of N. Do keep trying, though. === Subject: Re: infinity ... David R Tribble said: > Tony Orlow: >> I already showed you the bijection between binary *N and P(*N). >> What didn't you like about it? It is valid. > David R Tribble: >> No, you showed a mapping between *N and R, which is equivalent >> to a mapping between *N and P(N). That's easy. > Tony Orlow: >> What do you want me to try, anyway, and infinite mapping, >> element-by-element? A bijection's a bijection, right? > David R Tribble: >> Yes, that would be nice. Please show us your bijection. > Tony Orlow: >> f(0) = ...000 = {} >> f(1) = ...001 = {0} >> f(2) = ...010 = {1} >> f(3) = ...011 = {0,1} >> f(4) = ...100 = {2} >> f(5) = ...101 = {0,2} >> f(6) = ...110 = {1,2} >> f(7) = ...111 = {0,1,2} >> etc. Any questions? > David R Tribble: >> Where are the infinite subsets, such as the set of even numbers? > Tony Orlow: > Well, infinitely far down the list of course! > . > . > . > f(N/3) = 0:010101.....010101 = {0,2,4,6,8,....} > f(2N/3)= 0:101010.....101010 = {1,3,5,7,9,....} > All of the sets you listed, finite and infinite, are composed of > only finite naturals. Which means that they are all subsets of N. Incorrect. See the digits to the left of the ellipses? Those are in infinite positions, representing infinite evens and odds. > You're using N (and N/3, 2N/3, etc.), which is a member of *N. Yes we are talking aboout *N. > So all you're doing is showing a bijection between *N and P(N), which > is what I said you were doing in the first place. That's easy, > and is equivalent to a bijection between R and P(N). No, it's P(*N). The sets include elements of infinite value. > But we're asking for a bijection between *N and P(*N), or between > N and P(N). You haven't adequately shown those yet. It's done. > Oh, now comes the list of primes (sigh) > Uh, um.... > f(N/pi) = ....1010001010110 = {2,3,5,7,11,13,...} > PS, I'm kidding about the N/pi part (I think). > The infinite natural x in *N that maps f(x) to the infinite set of > primes is: > x = 2^2 + 2^3 + 2^5 + 2^7 + 2^11 + 2^13 + ... > But like all the other subsets your f mapping defines, the set of > primes contains only finite numbers, which means that it is one of > the subsets of N. If there are no infinite primes, then the subset would correspond to a finite value. Not all subsets have to have infinite values. I have only listed finite numbers here, but do take note of the ellipses. Are there infinite primes? If not, then there arent an infinite number of them. > Do keep trying, though. -- Smiles, Tony === Subject: Re: infinity ... > David R Tribble said: > > f(N/3) = 0:010101.....010101 = {0,2,4,6,8,....} f(2N/3)= > > 0:101010.....101010 = {1,3,5,7,9,....} > > All of the sets you listed, finite and infinite, are composed of > only finite naturals. Which means that they are all subsets of N. > Incorrect. See the digits to the left of the ellipses? Those are in > infinite positions, representing infinite evens and odds. So that 0:010101.....101010 and 0:101010.....010101 must both represent sets of half evens and half odds. > No, it's P(*N). The sets include elements of infinite value. If TO is saying that each 1 bit in x:xxxxx.....xxxxxx represents the presents of a member of *N in the set represented by that string, and each 0 represents the absence, then each 1 member set can be represented by a single 1 bit, with all others zero, so that it member can also be represented with a single 1 bit as well, and all the zeros on one side or the other of that 1 digit are extraneous. Then TO is saying that every member of *N can be represented by a single 1 preceded (or followed) by an appropriate string of 0's. s this represents the set whose only element is that number, it equally represents that number! > > But we're asking for a bijection between *N and P(*N), or between N > and P(N). You haven't adequately shown those yet. > It's done. Not outside TOmatica. > If there are no infinite primes, then the subset would correspond to > a finite value. Not all subsets have to have infinite values. I have > only listed finite numbers here, but do take note of the ellipses. > Are there infinite primes? If not, then there arent an infinite > number of them. There are infinitely many finite primes outside of TOmatica, where the sun still shines and infinite still means Dedekind infinite. === Subject: Re: infinity ... David R Tribble said: >> The infinite natural x in *N that maps f(x) to the infinite set of >> primes is: >> x = 2^2 + 2^3 + 2^5 + 2^7 + 2^11 + 2^13 + ... >> But like all the other subsets your f mapping defines, the set of >> primes contains only finite numbers, which means that it is one of >> the subsets of N. > If there are no infinite primes, then the subset would correspond to a > finite value. > Not all subsets have to have infinite values. I have only listed finite > numbers here, but do take note of the ellipses. Are there infinite primes? If > not, then there arent an infinite number of them. Well, there are no infinite primes. There can't be, because any infinite natural must have at least a finite number of prime factors. (If N is infinite, then N/2 must also exist, according to your rules. Which means that no infinite N is prime, right?) So since there are no infinite primes, there must only be a finite number of them, according to you. So therefore there must be a largest prime, right? Oh, but I suppose that it's finite but unidentifiable, right? Euclid must have gotten this one wrong, huh? === Subject: Re: infinity ... David R Tribble said: >> The infinite natural x in *N that maps f(x) to the infinite set of >> primes is: >> x = 2^2 + 2^3 + 2^5 + 2^7 + 2^11 + 2^13 + ... >> But like all the other subsets your f mapping defines, the set of >> primes contains only finite numbers, which means that it is one of >> the subsets of N. > If there are no infinite primes, then the subset would correspond to a > finite value. Really? So x (above) must be a finite number? What about n = 1+1+1+1+..., which is obviously less than x? Is it also finite? === Subject: Re: infinity ... Tony Orlow: >> f(0) = ...000 = {} >> f(1) = ...001 = {0} >> f(2) = ...010 = {1} >> f(3) = ...011 = {0,1} >> etc. Any questions? David R Tribble: >> Where are the infinite subsets, such as the set of even numbers? Tony Orlow: >> Well, infinitely far down the list of course! >> . >> . >> . >> f(N/3) = 0:010101.....010101 = {0,2,4,6,8,....} >> f(2N/3)= 0:101010.....101010 = {1,3,5,7,9,....} David R Tribble: >> All of the sets you listed, finite and infinite, are composed of >> only finite naturals. Which means that they are all subsets of N. Tony Orlow: > Incorrect. See the digits to the left of the ellipses? Those are in > infinite positions, representing infinite evens and odds. See the elements within the braces? They are all finite. All those sets are members of P(N). So you're mapping from *N to P(N), which like I said before, is easy. You'd have to use numbers like your 1:000...001 [*] before you map to any subsets containing infinite naturals. But then you'd be outside the members of *N. [*] I think. Who knows? === 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 > David R Tribble said: >> Tony Orlow: I already showed you the bijection between binary *N and P(*N). What didn't you like about it? It is valid. >> David R Tribble: No, you showed a mapping between *N and R, which is equivalent to a mapping between *N and P(N). That's easy. >> Tony Orlow: What do you want me to try, anyway, and infinite mapping, element-by-element? A bijection's a bijection, right? >> David R Tribble: Yes, that would be nice. Please show us your bijection. >> Tony Orlow: f(0) = ...000 = {} f(1) = ...001 = {0} f(2) = ...010 = {1} f(3) = ...011 = {0,1} f(4) = ...100 = {2} f(5) = ...101 = {0,2} f(6) = ...110 = {1,2} f(7) = ...111 = {0,1,2} etc. Any questions? >> David R Tribble: Where are the infinite subsets, such as the set of even numbers? >> Tony Orlow: >> Well, infinitely far down the list of course! >> . >> . >> . >> f(N/3) = 0:010101.....010101 = {0,2,4,6,8,....} >> f(2N/3)= 0:101010.....101010 = {1,3,5,7,9,....} >> All of the sets you listed, finite and infinite, are composed of >> only finite naturals. Which means that they are all subsets of N. > Incorrect. See the digits to the left of the ellipses? Those are in infinite > positions, representing infinite evens and odds. >> You're using N (and N/3, 2N/3, etc.), which is a member of *N. > Yes we are talking aboout *N. >> So all you're doing is showing a bijection between *N and P(N), which >> is what I said you were doing in the first place. That's easy, >> and is equivalent to a bijection between R and P(N). > No, it's P(*N). The sets include elements of infinite value. Then you should be able to tell us whether e, the index of the set of even numbers E such that f(e)=E, is an element of E. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: infinity ... > Is it > impossible to biject an infinite ordered set with its power set? Nope For any set S, finite or infinite, and any function f:S -> P(S), the set {x in S: x not in f(s)} is not of form f(s) for any s in S. === Subject: Re: infinity ... > stephen@nomail.com said: > > David R Tribble said: > >> > >> > >> What do you want me to try, anyway, and infinite mapping, > >> element-by-element? A bijection's a bijection, right? > >> > >> Yes, that would be nice. Please show us your bijection. > > f(0) = ...000 = {} f(1) = ...001 = {0} f(2) = ...010 = {1} f(3) = > > ...011 = {0,1} f(4) = ...100 = {2} f(5) = ...101 = {0,2} f(6) = > > ...110 = {1,2} f(7) = ...111 = {0,1,2} > > If we define w= { x : x not in f(x) } then we get w = {0, 1, 2, > 3, ...... } > > So for what y does f(y) = { 0, 1, 2, 3, ..... }? > > And is y in f(y)? > > Let me guess. You will claim that F(N) = { 0, 1, 2, 3, ..... }. So > is N in { 0, 1, 2, 3, ..... }? If it is, then N is not in w, and > F(N) does not equal w. If it is not, then N is in w, and F(N) does > not equal w. > But I have constructed a bijection between the two using an > > intermediate binary representation. What is the specific rule I > > have broken concerning the construction of bijections. If I > > haven't broken any such rule, then is it true that a bijection > > between two sets means that have the same size, or even > > cardinality? > > No you have not. > > Stephen > > Notice above that no natural maps to a subset which contains it, so w > is all of N. Imagining any completed w leads to a contradiction, > since the natural that would map to it is always bigger than every > natural in that set. That's okay though, because for every natural, > there's a larger one. If x exists, 2^x exists. The sets are infinite, > so the bijection continues, even though there is a difference between > the values which are in the subsets and the values which denote the > subsets. None of this handwaving and doubletalk designates any member of *N which maps to {x in *N: x not in f(x)}. And without it, TO's bijection is not even a surjection. === Subject: Re: infinity ... Virgil said: > stephen@nomail.com said: > > David R Tribble said: > >> > >> > >> What do you want me to try, anyway, and infinite mapping, > >> element-by-element? A bijection's a bijection, right? > >> > >> Yes, that would be nice. Please show us your bijection. > > f(0) = ...000 = {} f(1) = ...001 = {0} f(2) = ...010 = {1} f(3) = > > ...011 = {0,1} f(4) = ...100 = {2} f(5) = ...101 = {0,2} f(6) = > > ...110 = {1,2} f(7) = ...111 = {0,1,2} > > > > If we define w= { x : x not in f(x) } then we get w = {0, 1, 2, > > 3, ...... } > > > > So for what y does f(y) = { 0, 1, 2, 3, ..... }? > > > > And is y in f(y)? > > > > Let me guess. You will claim that F(N) = { 0, 1, 2, 3, ..... }. So > > is N in { 0, 1, 2, 3, ..... }? If it is, then N is not in w, and > > F(N) does not equal w. If it is not, then N is in w, and F(N) does > > not equal w. > > But I have constructed a bijection between the two using an > > intermediate binary representation. What is the specific rule I > > have broken concerning the construction of bijections. If I > > haven't broken any such rule, then is it true that a bijection > > between two sets means that have the same size, or even > > cardinality? > > > > No you have not. > > > > Stephen > > > Notice above that no natural maps to a subset which contains it, so w > is all of N. Imagining any completed w leads to a contradiction, > since the natural that would map to it is always bigger than every > natural in that set. That's okay though, because for every natural, > there's a larger one. If x exists, 2^x exists. The sets are infinite, > so the bijection continues, even though there is a difference between > the values which are in the subsets and the values which denote the > subsets. > None of this handwaving and doubletalk designates any member of *N which > maps to {x in *N: x not in f(x)}. And without it, TO's bijection is > not even a surjection. So, it is required that we determine the very last element in an infinite bijection? Why must I name the end of the set in order for the bijection to be valid? You cannot name anything except some conceptual end of the unending set as a point where the bijection breaks down. I swear this theory just seems more and more insane to me. How do you tolerate it? Ugh! Please line up by the basketball courts for your head-whacking. Maybe that'll work. At least it'll be fun. -- Smiles, Tony === Subject: Re: infinity ... > Virgil said: > > > stephen@nomail.com said: > > > David R Tribble said: > > >> > > >> > > >> What do you want me to try, anyway, and infinite mapping, > > >> element-by-element? A bijection's a bijection, right? > > >> > > >> Yes, that would be nice. Please show us your bijection. > > > f(0) = ...000 = {} f(1) = ...001 = {0} f(2) = ...010 = {1} f(3) = > > > ...011 = {0,1} f(4) = ...100 = {2} f(5) = ...101 = {0,2} f(6) = > > > ...110 = {1,2} f(7) = ...111 = {0,1,2} > > > > If we define w= { x : x not in f(x) } then we get w = {0, 1, 2, > > 3, ...... } > > > > So for what y does f(y) = { 0, 1, 2, 3, ..... }? > > > > And is y in f(y)? > > > > Let me guess. You will claim that F(N) = { 0, 1, 2, 3, ..... }. So > > is N in { 0, 1, 2, 3, ..... }? If it is, then N is not in w, and > > F(N) does not equal w. If it is not, then N is in w, and F(N) does > > not equal w. > > > But I have constructed a bijection between the two using an > > > intermediate binary representation. What is the specific rule I > > > have broken concerning the construction of bijections. If I > > > haven't broken any such rule, then is it true that a bijection > > > between two sets means that have the same size, or even > > > cardinality? > > > > No you have not. > > > > Stephen > > > > Notice above that no natural maps to a subset which contains it, so w > > is all of N. Imagining any completed w leads to a contradiction, > > since the natural that would map to it is always bigger than every > > natural in that set. That's okay though, because for every natural, > > there's a larger one. If x exists, 2^x exists. The sets are infinite, > > so the bijection continues, even though there is a difference between > > the values which are in the subsets and the values which denote the > > subsets. > > None of this handwaving and doubletalk designates any member of *N which > maps to {x in *N: x not in f(x)}. And without it, TO's bijection is > not even a surjection. > > So, it is required that we determine the very last element in an infinite > bijection? Who says anything about last? Only TO! Order of elements is not relevant in finding an element of *N which f will map to {x in *N: x not in f(x)}. TO's continuing obsession with such irrelevances to the issue as order can only be to support his attempts to avoid facing the issue squarely and admitting his error. Why must I name the end of the set in order for the bijection to > be > valid? You cannot name anything except some conceptual end of the unending > set > as a point where the bijection breaks down. I swear this theory just seems > more > and more insane to me. How do you tolerate it? Ugh! Please line up by the > basketball courts for your head-whacking. Maybe that'll work. At least it'll > be > fun. === Subject: Re: infinity ... None of this handwaving and doubletalk designates any member of *N which > maps to {x in *N: x not in f(x)}. And without it, TO's bijection is > not even a surjection. > So, it is required that we determine the very last element in an infinite > bijection? Where do you see anybody talking about the order, let alone the very last element? No it is not required to talk about the very last element. All that is required to prove a mapping is not a bijection from set A to set B is to identify one element of either set which is not part of the mapping. There is no requirement that it be the last. Nobody but you has used the term last. Nobody but you has talked about the ordering of the mapping. What we are pointing out is that this set IS NOT MAPPED by any element. It need not be the mythical end that this place in the mapping is missing. It's just missing. End of story. - Randy === Subject: Re: infinity ... > Randy Poe said: > > > David R Tribble said: > > Yes, that would be nice. Please show us your bijection. > > f(0) = ...000 = {} > > f(1) = ...001 = {0} > > f(2) = ...010 = {1} > > f(3) = ...011 = {0,1} > > f(4) = ...100 = {2} > > f(5) = ...101 = {0,2} > > f(6) = ...110 = {1,2} > > f(7) = ...111 = {0,1,2} There is no *n in this 'listing', or any other, for which f(*n) = { x in *N: x not in f(x)}. > and > I accept the inductive proof that shows it is 2^n for set size n, finite or > infinite. My point is that a bijection can be formed, without apparently > breaking any rules of bijection. If this bijection is disallowed, I would > like > to know by which rule this occurs. What makes this bijection invalid? The fact that we can specify a member of P(*N) which is not the image of any member of *N. === Subject: Re: infinity ... > David R Tribble said: > >> I already showed you the bijection between binary *N and P(*N). > >> What didn't you like about it? It is valid. > > David R Tribble said: > >> No, you showed a mapping between *N and R, which is equivalent > >> to a mapping between *N and P(N). That's easy. > > >> What do you want me to try, anyway, and infinite mapping, > >> element-by-element? A bijection's a bijection, right? > > David R Tribble said: > >> Yes, that would be nice. Please show us your bijection. > > > f(0) = ...000 = {} > > f(1) = ...001 = {0} > > f(2) = ...010 = {1} > > f(3) = ...011 = {0,1} > > f(4) = ...100 = {2} > > f(5) = ...101 = {0,2} > > f(6) = ...110 = {1,2} > > f(7) = ...111 = {0,1,2} > > etc. Any questions? > > Where are the infinite subsets, such as the set of even numbers? > > > Well, infinitely far down the list of course! And where in that list is {x in *N: x not in f(x)}? === Subject: Re: infinity ... Virgil said: > David R Tribble said: > >> I already showed you the bijection between binary *N and P(*N). > >> What didn't you like about it? It is valid. > > > > David R Tribble said: > >> No, you showed a mapping between *N and R, which is equivalent > >> to a mapping between *N and P(N). That's easy. > > > >> What do you want me to try, anyway, and infinite mapping, > >> element-by-element? A bijection's a bijection, right? > > > > David R Tribble said: > >> Yes, that would be nice. Please show us your bijection. > > > > f(0) = ...000 = {} > > f(1) = ...001 = {0} > > f(2) = ...010 = {1} > > f(3) = ...011 = {0,1} > > f(4) = ...100 = {2} > > f(5) = ...101 = {0,2} > > f(6) = ...110 = {1,2} > > f(7) = ...111 = {0,1,2} > > etc. Any questions? > > > > Where are the infinite subsets, such as the set of even numbers? > > > > > Well, infinitely far down the list of course! > And where in that list is {x in *N: x not in f(x)}? At the bottom. -- Smiles, Tony === Subject: Re: infinity ... > Virgil said: > > > David R Tribble said: > > >> I already showed you the bijection between binary *N and P(*N). > > >> What didn't you like about it? It is valid. > > > > > David R Tribble said: > > >> No, you showed a mapping between *N and R, which is equivalent > > >> to a mapping between *N and P(N). That's easy. > > > > > >> What do you want me to try, anyway, and infinite mapping, > > >> element-by-element? A bijection's a bijection, right? > > > > > David R Tribble said: > > >> Yes, that would be nice. Please show us your bijection. > > > > > f(0) = ...000 = {} > > > f(1) = ...001 = {0} > > > f(2) = ...010 = {1} > > > f(3) = ...011 = {0,1} > > > f(4) = ...100 = {2} > > > f(5) = ...101 = {0,2} > > > f(6) = ...110 = {1,2} > > > f(7) = ...111 = {0,1,2} > > > > etc. Any questions? > > > > Where are the infinite subsets, such as the set of even numbers? > > > > > > Well, infinitely far down the list of course! > > And where in that list is {x in *N: x not in f(x)}? > > At the bottom. If there is *n in*N such that f(*n) = {x in *N: x not in f(x)}, is *n in {x in *N: x not in f(x)} or is *n not in {x in *N: x not in f(x)}? In either case, it can only happen in the looking glass land of TOmatica where things can be sinmultaneoulsy true and false. === Subject: Re: infinity ... > Did I not give what appears to be a valid bijection between the > NUMBERS in *N and the SETS in P(*N)? No! At least it does not have even the appearance of being valid outside TOmatics. > If the bijection is not valid, > please state what mistake I made in constructing it. If the mapping is called f, you did not show that there is some *n in *N such that f(*n) = {x in *N: x not in f(x)}. === Subject: Re: infinity ... > Randy Poe said: > > > David R Tribble said: > > >> I already showed you the bijection between binary *N and > > >> P(*N). What didn't you like about it? It is valid. > > > David R Tribble said: > > >> No, you showed a mapping between *N and R, which is > > >> equivalent to a mapping between *N and P(N). That's easy. > > > >> No, it was specifically a bijection between two sets of > > >> infinite binary strings representing, on the one hand, the > > >> whole numbers in *N starting from 0, both finite and > > >> infinite, in normal binary format, and on the other hand, > > >> the specification of each subset of whole numbers in *N, > > >> where each bit which, in the binary number, represents 2^n > > >> denotes membership of n in the subset. This is a bijection > > >> between the whole numbers in *N and P(*N), using an > > >> intermediate bijection with a common set of infinite binary > > >> strings. > > > David R Tribble said: > > >> But that's an incomplete mapping, because there are not > > >> enough infinite binary strings in *N to enumerate all of the > > >> subsets of *N. Try it, if you don't believe me. > > > > Not enough infinite binary strings? How many in *N and in > > > P(*N)? Are you saying that I cannot construct a bijection > > > between the two on an element-by-element basis which > > > continues infinitely through the set of binary strings? > > That's exactly what I'm saying. > > card(*N) = c, but card(P(*N)) = 2^c, and c < 2^c. > > At what point does the bijection break down? > > What is the bijection? > The bijection I offered between *N and P(*N), using the > intermediate set of infinite binary strings, remember? It's > described, above, in this very post. > The proof that card(S) < card(P(S)) shows that no bijection exists. > That proof has been given a few times in this thread and I see > you're trying to get through it. Whether you believe the proof or > not yet, you do realize that the proposition is Let f(S) be any > mapping from S to P(S). Then f can't be a bijection. Right? > Yes, the proof purports to show no bijection can exist by positing a > complete infinite set and deriving a contradiction from the fact that > the element mapping to it is not in it. What is this allegedly complete infinite set that TO claims the proof requires? In the proof that any f: S -> P(S) is not surjective, a certain set is described, but there is no need for the certain set to be infintie, and under certain circumstances, it can even be the empty set, for example when f(x) = {x} for each x in S. then {x in S:x not in f(x)} = {}. But it is still a subset of S and a member of P(S). Perhaps if TO would learn to read at above kindergarten level, he would make fewer of theses stupid mistakes. > For any set, there exists a number mapping to it, and every number > maps to a unique set. The fact that it has no end is no different > from an y other infinite bijection. So, I ask again, what rule of > bijection did I break, and where does this bijection break down, > since it seems to hold for all finite cases? Does TO claim that there is some finite set (or even, gasp, a Dedekind-finite set) that allows a surjection to its power set? Please show us an example of such an anomalous object! > Did I break some > specific bijection-forming rule? Yes! > If the proof is correct (as it is), then ALL bijections break down. > The proof consists of showing that no matter what mapping you > choose, there's an unmapped element of P(S). > > > It certainly works for all finite cases, does it not? > > Absolutely not. Let S = {1,2,3}. Then P(S) has 8 elements: {}, {1}, > {2}, {3}, {1,2}, {2,3}, {1,3}, {1,2,3} > > There is no bijection between the 3 elements of S and the 8 > elements of P(S). > You snipped what i was referring to as working. I believe it was that > no element is a member of the set to which it maps. This is certainly > the case here and for all finite cases. let S = {1,2,3}, and f(x) = {x}, then EVERY element is a member of the set to which it maps, but there are 5 sets in P({1,2,3}) which are not values of f. > > > What makes you think it falls apart at some point? > > It falls apart for EVERY set. > No, in the infinite case, the bijection exists for every natural and > for every subset. It is a perfect 1-1 correspondence. Please name a > natural or a subset which is left out of the mapping. First give us the mapping, as the subsets left out depend critically on which mapping is used. In general terms on subset left out by any function f:S -> P(S) is {x in S: x not in f(x)}. As S gets larger, there are more sets left out. > > > For which element is there not a corresponding subset? For each element there is a subset, but not cconversely. > > It breaks apart in the other direction: there is (at least one) > subset for which there is no corresponding element. > Name it please. {x in S: x not in f(x)} > > First you have to define your mapping rule. Given any mapping rule, > a subset can be identified for which there is no corresponding > element. > I gave the mapping rule: natural <-> binary string <-> subset, > ordered naturally from 0. The calling this mapping from *N to P(*N) by some name, say f:*N -> P(*N), then {x in *N: x not in f(x)} is not in Image(f). === Subject: Re: infinity ... > William Hughes said: > You have made exactly this mistake before. > For example > {peach, apple, plum, fiddle} is a set but not a number. > Huh! I coulda swore that was 4! To is often foresworn. > Just because you have a set does not mean you have a number. > So, not every set has a size? Not according to TO's definition of size. === Subject: Re: infinity > zuhair said: > And I > should say God help you Tony!! Eve God, being logical, lacks the power to make TO's dreamworld into reality. > that's what it's going to take? Ha ha. Perhaps. === Subject: Re: infinity > zuhair said: > simple correction: > > > In a simialr way the set 2,4,6,............. at virgils speach has w/2 > > members, but at Zuhair's > it has 2w/2= w members and at TO's 1.5 w. > > > --- > sorry I made a mistake : I meant at Virgils speed > > bye > zuhair > > > I think Virgil's speech goes at w/2 also. ;D At least my speech is no faster than my thinking, which cannot properly be said of either TO nor Zuhair. === 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 >> zuhair said: >> simple correction: >> >> >> In a simialr way the set 2,4,6,............. at virgils speach has w/2 >> >> members, but at Zuhair's >> it has 2w/2= w members and at TO's 1.5 w. >> >> --- >> sorry I made a mistake : I meant at Virgils speed >> I think Virgil's speech goes at w/2 also. ;D > At least my speech is no faster than my thinking, which cannot > properly be said of either TO nor Zuhair. You can't hold a candle to the speed at which their thinking jumps to conclusions. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: infinity >> zuhair said: >> simple correction: >> >> >> In a simialr way the set 2,4,6,............. at virgils speach has w/2 >> >> members, but at Zuhair's >> it has 2w/2= w members and at TO's 1.5 w. >> >> --- >> sorry I made a mistake : I meant at Virgils speed >> I think Virgil's speech goes at w/2 also. ;D > At least my speech is no faster than my thinking, which cannot > properly be said of either TO nor Zuhair. > You can't hold a candle to the speed at which their thinking jumps to > conclusions. Particlarly unwarranted ones. === Subject: Re: infinity ... > stephen@nomail.com said: >> stevendaryl3016@yahoo.com said: albstorz@gmx.de says... I don't know what you are talking about. The proof for finite sets needs just a complete induction. This will not hold for infinity I think. There is no induction involved in the finite case, and the infinite case is *exactly* the same proof as the finite case. Here it is once again: Let A be any set whatsoever, finite or infinite, it doesn't matter. Let f be any function from A to P(A). Let w = { x in A | x is not an element of f(x) }. Let x = any set in A. Let u = f(x). We prove that u is not equal to w. By definition of w, we have x in w <-> x is not an element of f(x). So x in w <-> x is not an element of u. That means that there are two cases: Case 1: x in w, and x is not in u. In that case, u cannot equal w. Case 2: x is not in w, and x is in u. In that case, u cannot equal w. So what we have proved is that forall x, w is not equal to f(x). So w is not in the image of f. So f is not a bijection between A and P(A). There's no induction. There's no assumption that A is finite. >> But there is an assumption that y is in S. >> Of course there is an assumption that y is in S. The >> goal is to find a bijection f from S to P(S). That >> means that for every element x in P(S), there must be >> an element y in S such that f(y)=x. w is an element of P(S). >> In order for f to be a bijection, there must exist an >> element y in S such that f(y)=w. >> If y is not in S, then it is irrelevant to the question >> of whether or not f is a bijection from S to P(S). >> Do you consider the following a bijection from S={a,b,c} to its >> power set? >> f(a) = {} >> f(b) = {a} >> f(c) = {b} >> f(d) = {c} >> f(e) = {a,b} >> f(g) = {a,c} >> f(h) = {b,c} >> f(i) = {a,b,c} >> If w= { x : x in S and x not in f(x) } >> then w= {a,b,c}. >> Is there a y in S such that f(y) = {a,b,c}? No. >> Is there a y such that f(y) = {a,b,c}? Yes, f(i)={a,b,c}, but >> i is not in S, and is not part of a bijection from S to P(S). >> The above function is a bijection from {a,b,c,d,e,g,h,i} >> to P({a,b,c}). It is not a bijection from {a,b,c} to P({a,b,c}). >> Stephen > But, if this set went on forever, then i WOULD be in the set, and for any > subset, you could identify an x such that it mapped to that set. No it would not. > that no bijection is possible in the finite case. In the proof, a last element > of the set is assumed when we assume there is some string representing the > entire set. There is no mention of strings in the proof. Can't you read a simple proof? There is no mention of strings or last elements. Just read the proof. Stop making up stuff. The proof really only makes use of four things. A set S, its power set P(S), a function f:S->P(S), and the set w = { x : x not in f(x) }. Given that P(S) is determined by S, and that w is determined by f, the proof really only depends on two things. There are no strings. There is no largest element. > However, to whatever extent we have considered the set, say to N > elements, we can always consider it to N+1, or 2^N elements. If this is the set > of all naturals, for instance, then N in the set implies 2^N in the set. There > is no end to the bijection. The proof assumes one. No. The proof does not assume an end to the bijection. If you think it does, please point out where it makes such an assumption. Remember, the rest of us are talking about the entire set, and the entire power set, not just an extent of it. > So tell me, what rule of bijection did I break, and at what point does the > mapping through the infinite binary strings break down. For lack of answers to > thses questions, my bijection stands. The set of even naturals does not show up in your bijection. You even admitted it. You can only account for the evens that are less than some N, but that is not all the evens. Stephen === Subject: Re: infinity ... stephen@nomail.com said: > stephen@nomail.com said: >> stevendaryl3016@yahoo.com said: albstorz@gmx.de says... I don't know what you are talking about. The proof for finite sets needs just a complete induction. This will not hold for infinity I think. There is no induction involved in the finite case, and the infinite case is *exactly* the same proof as the finite case. Here it is once again: Let A be any set whatsoever, finite or infinite, it doesn't matter. Let f be any function from A to P(A). Let w = { x in A | x is not an element of f(x) }. Let x = any set in A. Let u = f(x). We prove that u is not equal to w. By definition of w, we have x in w <-> x is not an element of f(x). So x in w <-> x is not an element of u. That means that there are two cases: Case 1: x in w, and x is not in u. In that case, u cannot equal w. Case 2: x is not in w, and x is in u. In that case, u cannot equal w. So what we have proved is that forall x, w is not equal to f(x). So w is not in the image of f. So f is not a bijection between A and P(A). There's no induction. There's no assumption that A is finite. >> But there is an assumption that y is in S. >> >> Of course there is an assumption that y is in S. The >> goal is to find a bijection f from S to P(S). That >> means that for every element x in P(S), there must be >> an element y in S such that f(y)=x. w is an element of P(S). >> In order for f to be a bijection, there must exist an >> element y in S such that f(y)=w. >> >> If y is not in S, then it is irrelevant to the question >> of whether or not f is a bijection from S to P(S). >> >> Do you consider the following a bijection from S={a,b,c} to its >> power set? >> >> f(a) = {} >> f(b) = {a} >> f(c) = {b} >> f(d) = {c} >> f(e) = {a,b} >> f(g) = {a,c} >> f(h) = {b,c} >> f(i) = {a,b,c} >> >> If w= { x : x in S and x not in f(x) } >> then w= {a,b,c}. >> >> Is there a y in S such that f(y) = {a,b,c}? No. >> >> Is there a y such that f(y) = {a,b,c}? Yes, f(i)={a,b,c}, but >> i is not in S, and is not part of a bijection from S to P(S). >> >> The above function is a bijection from {a,b,c,d,e,g,h,i} >> to P({a,b,c}). It is not a bijection from {a,b,c} to P({a,b,c}). >> >> Stephen >> > But, if this set went on forever, then i WOULD be in the set, and for any > subset, you could identify an x such that it mapped to that set. > No it would not. Sure it would. > that no bijection is possible in the finite case. In the proof, a last element > of the set is assumed when we assume there is some string representing the > entire set. > There is no mention of strings in the proof. Can't you > read a simple proof? There is no mention of strings or > last elements. Just read the proof. Stop making > up stuff. I am referring also to my bijection, to which no one has raised any serious objection except on pronciple. > The proof really only makes use of four things. > A set S, its power set P(S), a function f:S->P(S), > and the set w = { x : x not in f(x) }. And the question as to what element maps to the completed set. This is a largest natural argument. > Given that P(S) is determined by S, and that w is > determined by f, the proof really only depends on > two things. An ended unending set and a largest natural number. > There are no strings. There is no largest element. > However, to whatever extent we have considered the set, say to N > elements, we can always consider it to N+1, or 2^N elements. If this is the set > of all naturals, for instance, then N in the set implies 2^N in the set. There > is no end to the bijection. The proof assumes one. > No. The proof does not assume an end to the bijection. > If you think it does, please point out where it makes > such an assumption. It asks for the largest natural when it asks for the natural that maps to the entire completed set. It may be masked somewhat, but that's precisely what's going on. The first element maps to the null set, the last to the entire set. Is there a last element in the set? No? Then I guess it cannot map to anything. > Remember, the rest of us are talking about the entire set, > and the entire power set, not just an extent of it. Yes, and you are asking which element maps to the entire set, which is like asking which is the last element. Think about it. > So tell me, what rule of bijection did I break, and at what point does the > mapping through the infinite binary strings break down. For lack of answers to > thses questions, my bijection stands. > The set of even naturals does not show up in your bijection. What are you talking about? I offered a bijection between *N and P(*N), remember? With the binary strings? (sigh) Of course you forget. Everyone wishes it would just go away. All the hand waving only makes the bijection mad. Calm down or you might get stung. > You even admitted it. You can only account for the evens > that are less than some N, but that is not all the evens. How did you get on the subject of the evens? Take a vitamin. > Stephen -- Smiles, Tony === Subject: Re: infinity ... > I am referring also to my bijection, to which no one has raised any > serious objection except on pronciple. I and others have raised serious objections, some of us many times, to which TO has offered no serious counters. > > The proof really only makes use of four things. A set S, its power > set P(S), a function f:S->P(S), and the set w = { x : x not in f(x) > } . > And the question as to what element maps to the completed set. As nothing but completed sets are involved. Outside TOmatica there are only completed sets, anything uncompleted is not a set in the outside world. > This is a largest natural argument. Why does TO keep insisting that any such thing exists, when he knows it does not? > > Given that P(S) is determined by S, and that w is determined by f, > the proof really only depends on two things. > An ended unending set and a largest natural number. Those things only exist in TOmatica. Why TO ever comes out of his ambiguous land of TOmatica to face the harsh winds of reality is a mystery. > No. The proof does not assume an end to the bijection. If you > think it does, please point out where it makes such an assumption. > It asks for the largest natural when it asks for the natural that > maps to the entire completed set. But it does not do that. TO is the only one who insists that it maps to the entire completed set. WE only say what it does NOT map to, namely the set {x in S: x not in f(x)} which can NEVER, NEVER be the completed set. > It may be masked somewhat, but that's precisely what's going on. The > first element maps to the null set There need not be a first, as S need not be ordered at all, much less ordered so as to have any first. S could be the naturally ordered set of ratinals for instance whichdoes not have anything like a first. , the last to the entire set. What is the last rational? Nobody but TO needs S even to be ordered at all, much less that it be ordered so as to have a first and a last. TO is clearly off in his cloud cuckoo land of TOmatica again. > Is there a last element in the set? No? Then I guess it cannot map to > anything. A circle has no last member but can map into the plane. What sort of stuff is TO getting off on today? > > Remember, the rest of us are talking about the entire set, and the > entire power set, not just an extent of it. > Yes, and you are asking which element maps to the entire set False, we are asking which member of the entire set S maps to {x in S: x to in f(x)}. Answer: none! > I offered a bijection between *N and P(*N), remember? TO has said there is such a thing, but has not given anything more subtantial than that claim of existence for us to go on. TO must give a specific rule, f: *N -> P(*N), telling how each member of *N determines a subset of *N in such detail as to allow a disinterested person to determine whether there is any *n in *N for which f(*n) = {x in *N: x not in f(x)}. Until he has done that, TO is just handwaving. > You even admitted it. You can only account for the evens that are > less than some N, but that is not all the evens. > How did you get on the subject of the evens? How does TO avoid it? === Subject: Re: infinity ... > stephen@nomail.com said: > >> stevendaryl3016@yahoo.com said: Tony: > I am referring also to my bijection, to which no one has raised any serious > objection except on pronciple. Indeed. I think it's fair to say that if (counterfactually) mathematics were postmodern stamp-collecting, or art appreciation, there would be bound to be at least one person in the class who said I really like your bijection - it's rather cute. And over the objections of those who pointed out that it isn't a bijection, we could all say Well done, Tony - you've shown some creativity, some determination not to be rulebound, to break the mould, cast the die, mix the metaphor. What is a metaphor, anyway? I've got one, and I'd love to use it. OK, we'll meet next week, and look at Phil's worms in a new light... Brian Chandler http://imaginatorium.org === Subject: Re: infinity ... >There are no strings. There is no largest element. Is there, at least, a spoon? === Subject: Re: infinity Introduction to mathematical philosophy: an infinite set is a set that > can have one to one correspondance with some of it's proper part. > A complete non sense. > To tell me that at infinity the whole can be as big as a proper part of > it is complete non sense. > That is what I am objecting to. Consider the set S: S = { a, ab, abb, abbb, abbbb, ... } In other words, S is the set of 'a' and its successors, where each element 'x' has a successor 'xb'. Obviously S is an infinite set, containing an infinite number of elements. Now consider this subset of S: T = S {a} T = { ab, abb, abbb, abbbb, ... } Obviously, T is also an infinite set. Now consider the mappings f(x) = xb for all x in S g(xb) = x for all xb in T It's obvious that f maps every member of S to every member of T, and g maps every member of T to every member of S (they are inverses of each other), so they define a one-to-one mapping (a bijection) between S and T. So S and its proper subset T, which are both infinite sets, have exactly the same number of members. Since you believe that this is nonsense (i.e., you think that S is a bigger set than T), please tell us which members of S or T are not included in the one-to-one mapping defined by f and g. If S and T really are different sizes, then at least one member in S or T must be left out of the mapping, right? === Subject: Re: infinity about it again and again for months . And I am telling you it is a false priniciple. such a priniciple should never be an axiom to define infinite set by, It is contraversial. it is not only me who says that , it is well known that this is contraversial. These mappings between T and S are suffecient to define the infinite set S as a set which can have a function f:R-->R that is bijective between any FINITE subset of it and another finite subset of it. And in reality this is a better definition of S being infinite than that of reflexion. I will show you lateron were you are wrong. Zuhair === Subject: Re: infinity S = { a, ab, abb, abbb, abbbb, ... } > In other words, S is the set of 'a' and its successors, where each > element 'x' has a successor 'xb'. Obviously S is an infinite set, > containing an infinite number of elements. > Now consider this subset of S: > T = S {a} > T = { ab, abb, abbb, abbbb, ... } > Obviously, T is also an infinite set. > Now consider the mappings > f(x) = xb for all x in S > g(xb) = x for all xb in T > It's obvious that f maps every member of S to every member of T, > and g maps every member of T to every member of S (they are inverses > of each other), so they define a one-to-one mapping (a bijection) > between S and T. > So S and its proper subset T, which are both infinite sets, have > exactly the same number of members. > Since you believe that this is nonsense (i.e., you think that S is > a bigger set than T), please tell us which members of S or T are > not included in the one-to-one mapping defined by f and g. If S > and T really are different sizes, then at least one member in S or > T must be left out of the mapping, right? > These mappings between T and S are suffecient to define the infinite > set S as a set which can have a function f:R-->R that is bijective > between any FINITE subset of it and another finite subset of it. And in > reality this is a better definition of S being infinite than that of > reflexion. So bijecting a finite subset of set A to another finite subset of A makes set A infinite? > I will show you lateron were you are wrong. Oh, please do. === Subject: Re: infinity No, if every finite subset of set A can be bijected along the same direction by a unidirectional function f with another finite subset of set A, then set A is infinite. so set F={ 1,2} is not infinite because if the bijective function f is increamental then number 2 cannot be bijected to heigher number in F, while if f is decreamental then 1 cannot be bijected ot a lower number in F. I think the definition above is good to define infinitude of a set. However if the set is well ordered then a simpler definition can be formed in terms of the ordering relation of that set R. Like saying: set A is infinite with respect to R if for every finite member of set A their exist another finite member of set A that have the relation R to it. These two definitions above serve a good definition of infinitude of sets. So why we should leave these definitions to go and adopt an equivocal one like that of Dedekind. My argument in a set theoratic manner ( although I hate that way ) is set A according to the first definition is the union of all it's finite subsets, Now this whole set A cannot have the same property as it's members otherwise it will be a member of itself.and this is contradictive , so the property every finite subset of an infinite subset has that is it's ability to be bijected with in the way defined above , this property of members of A cannot be possessed by the whole set A , as I said becasue set A will be a member of itself and that it contradictive. Since A cannot be a set and a member at the same time. Zuhair === Subject: Re: infinity let me clarify: A set P is called a subset of set A if every member of set P is also a member of set A, while the converse is not true, ie not every member of set A is also a member of set P. Now set P is a subset of A. set P is a proper subset of set A , if set A defined above actually contained members that are not in set P. Now logically speaking set A cannot be a proper subset of itself. Now we define set A as infinite if every proper subset of set A can be bijected to another proper subset of set A, provided that set A is not a binary set( two membered set). This can be verified for any finite subset of A and for any proper infinite subset of A also. like in saying that the set of all evens in N=1,2,3,4,......... can be bijected with set of all odds in N. Now if set A itself can be bijected to a proper subset of it , then this makes set A a proper subset of itself , which is contradictive. Therefore set A cannot be bijected to a proper subset of it. Proving the above makes us return to common sense logical calculation of the infinite that is if we say that the number natural numbers in N=1,2,3,4,........ is w (better to write that as N=1,2,3,4,......,w ) , then the number of even numbers in N is w/2 similarily the number of odds would be w/2. This common sense reasoning is better than any venture made on vague grounds. And accordingly the priniciple of reflexion will be distroyed, and what is believed to be a counterintuitive fundemental of the infinite would be discarded as a failed venture to understand it. So alot of the inductive properties the above priniciple washed from w will be returned to it. Now as I said in previous posting about Counter sets , many objected to the dynamic examples I've set and said that they are outside the stationary scope of the set theory. well today I will translate that dynamic version of mine to the stationary set theory. Now if we form N+=1,2,3,........... in such a manner that it has 2-1 correspondance with set N=1,2,3,....,w , then N+ has 2w of natural numbers in it and these numbers are 1,2,3,4,..................,w , w+1,w+2,w+3,.....................,2w-3,2w-2,2w-1,2w So the number which present in N+ and not present in N are: w+1,w+2,w+3,.....................,2w-3,2w-2,2w-1,2w As Toney answered for me before. of coarse w is the first natural( summation based ) infinite number. w is infinite but inductive. Zuhair === Subject: Re: infinity N=1,2,3,4,........ is w > (better to write that as N=1,2,3,4,......,w ) , then the number of > even numbers in N is w/2, similarily the number of odds would be w/2. > This common sense reasoning is better than any venture made on vague > grounds. > [...] > Now if we form N+=1,2,3,........... in such a manner that it has 2-1 > correspondance with set N=1,2,3,....,w , then N+ has 2w of natural > numbers in it and these numbers are > 1,2,3,4,..................,w , > w+1,w+2,w+3,.....................,2w-3,2w-2,2w-1,2w > So the number which present in N+ and not present in N are: > w+1,w+2,w+3,.....................,2w-3,2w-2,2w-1,2w But I can form a one-to-one correspondence between all the members of N and all the members of N+: Define function m, which is mapping (injection) from N to N+: m(k) = k/2 if k is even, or (k-1)+w if k is odd, for all k in N. This maps all the members of N to all the members in N+. Define function g, the inverse of m, which is mapping (injection) from N+ to N: g(k) = 2k if k < w, or (k-w)+1 if k >= w, for all k in N+. This maps all the members of N+ to all the members in N. m and g together define a bijection between N and N+, so the two sets are exactly the same size. Isn't this fun? === Subject: Re: infinity > 1,2,3,4,..................,w , > w+1,w+2,w+3,.....................,2w-3,2w-2,2w-1,2w > So the number which present in N+ and not present in N are: > w+1,w+2,w+3,.....................,2w-3,2w-2,2w-1,2w > But I can form a one-to-one correspondence between all the members > of N and all the members of N+: > Define function m, which is mapping (injection) from N to N+: > m(k) = k/2 if k is even, or (k-1)+w if k is odd, > for all k in N. Shouldn't that be (k - 1)/2 + w if k is odd? === Subject: Re: infinity Now we define set A as infinite if every proper subset of set A can > be bijected to another proper subset of set A, provided that set A is > not a binary set( two membered set). Given set B, which is not a two-membered set: B = {1,2,3} We biject every proper subset of B to another proper subset of B: {} <-> {} (?) {1} <-> {2} {2} <-> {3} {3} <-> {1} {1,2} <-> {1,3} {1,3} <-> {2,3} {2,3} <-> {1,2} So by your definition, B must be an infinite set. === Subject: Re: infinity Got me. I admit the definition is wrong. But I think I can save it if I say that their should not be common member between the bijected sets. So Set A is infinite if every proper subsets of set A can be bijected to another proper subset of set A that is disjoint from it. provided that set A is not a binary set. Zuhair === Subject: Re: infinity > Got me. > I admit the definition is wrong. > But I think I can save it if I say that their should not be common > member between the bijected sets. > So Set A is infinite if every proper subsets of set A can be bijected > to another proper subset of set A that is disjoint from it. provided > that set A is not a binary set. > Zuhair Dedkind: A set is infinite if and only if there is an injection from that set to some proper subset. According to The Zuhair definition above, the set of all real numbers is not an infinite set, since its proper subset of non-zero real numbers cannot be bijected to any disjoint subset of the reals. If the set of reals cannot be called infinite, what set can be? === Subject: Re: infinity OK , Virgil Correction: Set A is infinite if every finite proper subset of A can be bijected to another disjoint proper subset of A.Provided that set A is not a binary set. Zuhair === Subject: Re: infinity > OK , Virgil > Correction: > Set A is infinite if every finite proper subset of A can be bijected to > another disjoint proper subset of A.Provided that set A is not a binary > set. > Zuhair What does Zuhair mean by a binary set? And, if that means something other than a set with 2 members, how does one test whether a binary set is finite or infinite? How does one apply this Zuhair definition to, say, the set of rational numbers? How would one go about proving that EVERY finite subset of the rationals can be bijected to some disjoint subset? iy can be done, but it isn't as trivial as for Dedekind's definition. The Dedekind definition is trivially easy to use by comparison. For example f(x) = x/sqrt(x^1 + 1) bijects the rationals onto a proper subset of the rationals, namely onto the open interval of rationals (-1,1), so the rationals form a Dedekind infinite set. Note that the same function works for Dedekind infiniteness of every dense subset of the reals. === 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 > The Dedekind definition is trivially easy to use by comparison. > For example f(x) = x/sqrt(x^1 + 1) bijects the rationals onto a proper > subset of the rationals, namely onto the open interval of rationals > (-1,1), so the rationals form a Dedekind infinite set. Ouch. Ouch, ouch, ouch. Try f(x) = x*|x|/(1+x^2) instead. > Note that the same function works for Dedekind infiniteness of every > dense subset of the reals. Does it? -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: infinity In what way? Do you believe that the set {0,1,2,...} does NOT have a bijection with {1,2,3,...}? Or is it that you believe that the set is not infinite? ----------------------------------- The answer to that depends on the definition of both sets 0,1,2,3,.......... and 1,2,3,..... My main objection is to think that set 1,2,3,4....... is a proper subset of 0,1,2,3,.... and at the same time claim that a bijection exists between them? this is impossible! simply because any set has bijection with itself , so accordingly if 1,2,3,..... is a proper subset of 0,1,2,3,....... then it will have a bijection with 1,2,3,...... in 0,1,2,3,.......... and so number 1 in set 1,2,3,........ will have two arguments in set 0,1,2,3,....... that are numbers 0 and 1, and so the real relation between these two sets so defined is surjection without injection. While if you define set 1,2,3,.......... as f(x)=x+1 for x=0,1,2,3,........ then of coarse their is bijection between these sets , but what many people do not realize is that set 1,2,3,..... here is not a proper subset of set 0,1,2,3,........ , since it has a number that do not belong to set 0,1,2,3,.............., if one thinks in a dynamic manner like I presented in the Kangaroo examples , then it is obvious that at any one-one comparison between these sets their is always an extra-finite at 1,2,3,..... not present at 0,1,2,3,..... Zuhair === Subject: Re: infinity > Here's what I think you need to do: > > For a subset of the reals R, the range is defined as a duple: > > where X e R U {oo} // X is a real, or a formal symbol 'oo' Q e {0, > 1, 2} // Q is 0, 1, or 2 > > Informally (since I can't be bothered, and I've forgotten what the > point of this was supposed to be), X is the difference between the > greatest lower bound and least upper bound of the set if both > exist, 'oo' otherwise; Q is 0 is the set has both min and max, 1 if > it has either but not both, and 2 if the set has neither min nor > max. > > HTH > Sure, that's a good suggestion, although I am not sure what > difference it makes if the set has no min, no max, or neither. We have previously noted your ignorance on the point of these existences. Also your ignorance on the difference between a LUB and a max and the difference between a GLB and a min. > The > range would still be arbitrarily close to what it would be if there > were a max and a min. So, < corresponds to both 1 and 2, and <= > corresponds to 0. But, your way works also. On the contrary, TO's method does not work, but Brians would. So would the use of 'diameters' in place of TO's hopeless 'value ranges'. === Subject: dumb question for sanity check If I need function defined on R, that is rieman integrable and not lebegue integrable, and I have function f defined on [0, +oo) with such property, and I define function g on R by it being equal to f on [0,+oo), and to 0 on (-oo, 0), then g is a function R that is not lebegue but is rieman, right? thx, sorry for wasting your time === Subject: Re: dumb question for sanity check > If I need function defined on R, that is rieman integrable and not > lebegue It's Lebesgue, not Lebegue and it's Riemann, not Rieman! > integrable, and I have function f defined on [0, +oo) with such I assume that when you talk about a Riemann-integrable function _f_ defined in [0,+oo), what you mean is that the function _f_ is Riemann-integrable in each interval [0,M] and that, furthermore, the limit lim_{ M -> +oo } int_0^M f exists. > property, and I define function g on R by it being equal to f on > [0,+oo), and to 0 on (-oo, 0), then g is a function R that is not > lebegue but is rieman, right? Sure. Jose Carlos Santos === Subject: Re: Me and David C. Ullrich > On this forum I have argued the question, Two coins were flipped and > at least one is a head. What are the chances that there are two heads? > Many mathematicians get it confused with The probability for two > heads, given at least one head? > I say that given at least one head, and told, at least one head > mean two different things. > Dr. Ullrich has stated in this forum, that the two mean the same thing. > If for no other reason than that he said so. Also because everyone > assumes them to be the same. (everyone does not, I don't) This question has drawn much debate. Consider the formula: P(A|B)=[P(B|A)P(A)] / P(B), then, Two coins were flipped and we were told 'at least one is a head'. P(HH|toldat least one H) = (1/4) / P(B) B is the event which has happened and P(B) is the prior probability that it happens. We start with four equally likely outcomes. HH, HT, TH, TT. P(B) equals the probability that we were told at least one heads at HH, plus the probability that we were told at least one heads at HT, Plus the probability that we were told at least one heads at TH, plus the probability that we were told at least one heads at TT. The answer to our question definitely depends upon these probabilities. Alter P(B) and get a different answer to our question. P1. Let P(B) = 1/4 + 0 + 0 + 0 = 1/4 and get answer 1. P2. Let P(B) = 1/4 + 1/2*1/4 + 1/2*1/4 + 0 = 1/2 and get answer 1/2. P3. Let P(B) = 1/4 + 1/4 + 1/4 + 0 = 3/4 and get answer 1/3. P1. P2. and P3. define three different coin flip sequences. In each, we hear the phrase, at least one is a head. Professor Ullrich says that when we hear the phrase at least one is a head we can assume P3. As the three sequences get different answers, obviously this is an erroneous assumption. When the numbers in the formula change, the words in the problem statement have to change also. Ullrich, and those like him make the erroneous assumptions, I don't. I get accused of making probability hard. Probability is hard. Making erroneous assumptions doesn't make it easier. Eldon:) === Subject: Re: Me and David C. Ullrich > You confuse posters by reading more into this problem than > is already there. Probability problems are difficult enough. > Let's ask Dave Rusin, an expert in Probability Theory on this > forum, to give us the definitive answer to this problem as > posed by you. Maybe he will help us, maybe not. Dave? > It tends to be considered slightly bad form to call out > or solicit replies from individual posters by name, unless > they've been participating in the discussion, because they > may feel one is imposing an obligation on them to reply > (although Dave has posted in this thread, so maybe it's > OK after all). > I think at last we're zeroing in on the source of > Elmo's misunderstanding of the following statement > of the problem: > Two coins were tossed, and at least one is a head Two coins were tossed. (past tense) We're talking about an event which has already happened. The coins were tossed, and the statement was made about the toss. The coins landed HH, and the statement was made, or the coins landed HT, and the statement was made, or, the coins landed TH and the statement was made. Prior to the statement, there were four equally likely ways for the coins to exist. After the statement, there are only three. The coins were tossed, and a head was discovered. Is it just as likely that the head was discovered at HH, than at HT? (I don't think so) Eldon doesn't seem to be the one with the misunderstanding. Eldon === Subject: Re: Me and David C. Ullrich > > You confuse posters by reading more into this problem than > > is already there. Probability problems are difficult enough. > > Let's ask Dave Rusin, an expert in Probability Theory on this > > forum, to give us the definitive answer to this problem as > > posed by you. Maybe he will help us, maybe not. Dave? > It tends to be considered slightly bad form to call out > or solicit replies from individual posters by name, unless > they've been participating in the discussion, because they > may feel one is imposing an obligation on them to reply > (although Dave has posted in this thread, so maybe it's > OK after all). > I think at last we're zeroing in on the source of > Elmo's misunderstanding of the following statement > of the problem: > Two coins were tossed, and at least one is a head > Two coins were tossed. (past tense) We're talking about an event which > has already happened. > The coins were tossed, and the statement was made about the toss. > The coins landed HH, and the statement was made, or the coins > landed HT, and the statement was made, or, the coins landed TH > and the statement was made. > Prior to the statement, there were four equally likely ways for > the coins to exist. After the statement, there are only three. Two ways, you daft twat! HH, or HT (in either order). The probabilities of HT versus TH versus HH are immaterial once the coins have been tossed, at which time your question is asked - By then the outcome is a done deal. I really no longer have the time or patience to discuss this with you any any further. === Subject: Re: Me and David C. Ullrich > > You confuse posters by reading more into this problem than > > is already there. Probability problems are difficult enough. > > Let's ask Dave Rusin, an expert in Probability Theory on this > > forum, to give us the definitive answer to this problem as > > posed by you. Maybe he will help us, maybe not. Dave? > > It tends to be considered slightly bad form to call out > > or solicit replies from individual posters by name, unless > > they've been participating in the discussion, because they > > may feel one is imposing an obligation on them to reply > > (although Dave has posted in this thread, so maybe it's > > OK after all). > > I think at last we're zeroing in on the source of > > Elmo's misunderstanding of the following statement > > of the problem: > > Two coins were tossed, and at least one is a head > Two coins were tossed. (past tense) We're talking about an event which > has already happened. > The coins were tossed, and the statement was made about the toss. > The coins landed HH, and the statement was made, or the coins > landed HT, and the statement was made, or, the coins landed TH > and the statement was made. > Prior to the statement, there were four equally likely ways for > the coins to exist. After the statement, there are only three. > Two ways, you daft twat! HH, or HT (in either order). > The probabilities of HT versus TH versus HH are immaterial > once the coins have been tossed, at which time your question > is asked - By then the outcome is a done deal. The probabilities are not immaterial. This is a probability question. HT, TH, and HH were equally likely, prior to the statement. After the statement, are they still equally likely? That is not immaterial, it is the key to our answer. Hint: It is more likely that a H was discovered at HH, than at HT. > I really no longer have the time or patience to discuss this > with you any any further. Leave whenever you have to, but try to understand my argument before you go. Eldon === Subject: Re: Me and David C. Ullrich <87ll0ycuz9.fsf@phiwumbda.org> the conditions under which the person would tell you that >at least one is a head. In the case HT/TH, they might >be equally likely to say that one is a tail. > This is a reasonable concern. > But it's equally true that you don't know the conditions under which > they will tell you that two coins were tossed. (Maybe they only do it > when they both come down heads.) > -- Richard If there was a prejudice toward either H or T, it should be part of the problem statement. We know that the coins were tossed, and the statement was made. We know that the coins landed HH and the statement was made, or, the coins landed HT and the statement was made, or, the coins landed TH and the statement was made. We know that the writer of the statement observed Hx, or xH. We have no evidence that the writer knows anything about coin x. Eldon === Subject: Re: riemann vs lebesgue integrals Quote >Take f:[0,+oo[ ---> R defined by >f(x) = 1 if 0 <= x < 1 >f(x) = -1/2 if 1 <= x < 2 >f(x) = 1/3 if 2 <= x < 3 >f(x) = -1/4 if 3 <=x < 4 >and so on. Then f is not Lebesgue-integrable, but its Riemann integral is log(2). >Jose Carlos Santos I see why this is not lebesgue integrable, but why is this rieman integrable ? the domain is not bounded === Subject: Re: riemann vs lebesgue integrals <13084703.1129999373719.JavaMail.jakarta@nitrogen.mathforum.org>, hujin > Quote >Take f:[0,+oo[ ---> R defined by >f(x) = 1 if 0 <= x < 1 >f(x) = -1/2 if 1 <= x < 2 >f(x) = 1/3 if 2 <= x < 3 >f(x) = -1/4 if 3 <=x < 4 >and so on. Then f is not Lebesgue-integrable, but its Riemann integral > is log(2). > I see why this is not lebesgue integrable, but why is this rieman integrable ? the > domain is not bounded Improper Riemann integral === Subject: Re: riemann vs lebesgue integrals Quote >> Quote >>Take f:[0,+oo[ ---> R defined by >>f(x) = 1 if 0 <= x < 1 >>f(x) = -1/2 if 1 <= x < 2 >>f(x) = 1/3 if 2 <= x < 3 >>f(x) = -1/4 if 3 <=x < 4 >>and so on. Then f is not Lebesgue-integrable, but its Riemann integral >> is log(2). >> I see why this is not lebesgue integrable, but why is this rieman integrable ? the >> domain is not bounded >Improper Riemann integral Ok, why does the improper Riemann integal exists and converges? === Subject: Re: riemann vs lebesgue integrals In sci.math, hujin <40802.1130002098751.JavaMail.jakarta@nitrogen.mathforum.org>: > Quote > Quote >Take f:[0,+oo[ ---> R defined by > >f(x) = 1 if 0 <= x < 1 > >f(x) = -1/2 if 1 <= x < 2 > >f(x) = 1/3 if 2 <= x < 3 > >f(x) = -1/4 if 3 <=x < 4 > >and so on. Then f is not Lebesgue-integrable, but its Riemann integral > is log(2). > > > I see why this is not lebesgue integrable, but why is this rieman integrable ? the > domain is not bounded >>Improper Riemann integral > Ok, why does the improper Riemann integal exists and converges? Does it converge to a known value? The series 1 - 1/2 + 1/3 - 1/4 + ... is notorious for requiring special handling; it converges but not absolutely. One could probably prove that lim (n -> +oo) sum(i = 1, +oo) f(i/n) = log(2) where f(r) is as defined above. However, does this mean that lim (d -> 0+) sum(i = 0, +oo) f(i * d) is properly defined? (Can anyone prove this either way?) As for being Lebesgue-integrable -- I for one don't quite see the problem here. In particular, Lebesgue can integrate things like g(x) = 0, x rational g(x) = x, x irrational without difficulty (over [0,1] integral g(x) dx = 1/2, for example). Riemannian integration couldn't hope to handle this properly. -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: riemann vs lebesgue integrals >>Take f:[0,+oo[ ---> R defined by >> f(x) = 1 if 0 <= x < 1 >> f(x) = -1/2 if 1 <= x < 2 >> f(x) = 1/3 if 2 <= x < 3 >> f(x) = -1/4 if 3 <=x < 4 >>and so on. Then f is not Lebesgue-integrable, but its Riemann integral >>is log(2). >why is this non lebesgue integrable? my definition of lebesgue >integrable of a nonnegative function f over measurable E is >int_E f = sup_{h <= f} int_E h That's the definition of the _integral_ of a non-negative function. It's not the definition of _integrable_. The function f is _integrable_, by definition, if f is measurable and the integral of |f| is _finite_. >where h is bounded, measurable and vanishes everywhere outside a set of >finite measure. can't we take h_n to be the step function that agrees >with f up until x=n, at which point it vanishes? then doesnt h_n --> f >and int h_n --> log 2? >what am I missing? ************************ David C. Ullrich === Subject: Is any harmonic distribution a usual infinitly differentiable function? for example $uin D^{prime}(Omega)$ and $triangle u=0$ in $Omega$ that is for all $varphyin C^{infty}_{c}(Omega)$, $u(trianglevarphy)=0$ is it then true that $uin C^{infty}(Omega)$???? === Subject: Statistically worse poker hand It's said that 7,2 off-suit is statistically the worse starting poker hand (in Texas hold-em). Statistics change over time so is it possible for this hand to be replaced by a worse starting hand? If everyone knows that 7,2 off-suit is the worse starting hand, and it gets folded most of the time, wouldn't a new hand take this title? Or is this based on a static set of rules (number of cards, winning hands, pre-flop wins, etc). Is 7,2 off-suit the worse starting hand in all flavors of poker or only in Texas hold-em? === Subject: Re: Statistically worse poker hand Statistically worse poker hand >It's said that 7,2 off-suit is statistically the >worse starting >poker hand (in Texas hold-em). Statistics change over >time so is >it possible for this hand to be replaced by a worse >starting hand? >If everyone knows that 7,2 off-suit is the worse >starting hand, and >it gets folded most of the time, wouldn't a new hand >take this >title? Or is this based on a static set of rules number >of cards,winning hands, pre-flop wins, etc). >Is 7,2 off-suit the worse starting hand in all flavors >of poker or only in Texas hold-em? I would say over all, yes! Speaking as just pure odds. You have to look at all the scenarios, as you know a lot of it is not about odds but I will not get into that here. Probably on small stakes texas hold-um it is just about odds. So speaking about just odds--- Ok, you have [2,7] off suit hole cards prior to the flop,turn and river so a chance for a straight or straight flush push their odds even more against you where now a full house or 4 of a kind are easier to get odds wise then just a plain old simple off suit straight . You would only need three matching cards to get a full house, or 4 of a kind the odds are still the same for each, but better than a straight or even straight flush by holding a [2,7] off suit because now you need 4 cards for a straight or straight flush. I am speaking here about matching full house against a straight and 4 of a kind against a straight flush. In other words it just increases the odds on getting the straight or the straight flush in this case. To clarify a little further a straight flush is always more difficult to get than 4 of a kind but in this case it makes it much more difficult. As far as other pairs or three of a kind for off suit [2,7] they are low cards so its obvious they probably are the worst hole cards to have. On other poker games, there are soooo.. many I could not make a general statement! I fold here! ;-} Dan === Subject: Re: Statistically worse poker hand > It's said that 7,2 off-suit is statistically the worse starting > poker hand (in Texas hold-em). Statistics change over time so is > it possible for this hand to be replaced by a worse starting hand? > If everyone knows that 7,2 off-suit is the worse starting hand, and > it gets folded most of the time, wouldn't a new hand take this > title? Or is this based on a static set of rules (number of cards, > winning hands, pre-flop wins, etc). > Is 7,2 off-suit the worse starting hand in all flavors of poker or > only in Texas hold-em? What do they mean when they say, the worst starting hand? They are talking about holdem. They are talking about a hand's (a priori) potential to win (connect with the flop, turn and river) if there is a showdown. For example, these two cards cannot be used in combination to make straights (unlike 6,2; 5,2; 4,2. 3,2 unsuited) or flushes (unlike two suited cards). If nobody improves, 7-2-off loses (or ties on high-card power) to all other 7's and above. So, it has the least high-card power of all hands that do not make straights. === Subject: Universe is You I'll find you in the sunlight, I'll find you in the night, In morning and in twilight In hearing and in sight In conjured and in real In Venus and in Mars In truth incorporeal Of billions of stars In water and in fire In earthquake and in flood In Mount Kilimanjaro That is the House of God In Amazonian jungle And overhanging clifss In the Siberian tundra And on the coral reefs On surface of the moon and In heart of Milky Way Within volcano plumes and In crashing ocean wave In laughter of a child In tears of the bereaved At home and in the wild In everything that lived In passion and in reason In body and in soul In insight and in wisdom In all that I extol - Nirvana and desire The journey and the goal Phenomenal and maya Combined with noumenal - It all exists within you And in you it takes place The ultimate continuum That in you manifests - In sides of dialectic In substance of the light In prayer and in action In wings of bird in flight, In essence and existence In fractals and in chaos In goodness that consists of The yearning and the blessed, In thought and inspiration In everything that's true I, without hesitation, Say: Universe is you. Ilya Shambat === Subject: Re: Universe is You Wrong again. > I'll find you in the sunlight, > I'll find you in the night, > In morning and in twilight > In hearing and in sight > In conjured and in real > In Venus and in Mars > In truth incorporeal > Of billions of stars > In water and in fire > In earthquake and in flood > In Mount Kilimanjaro > That is the House of God > In Amazonian jungle > And overhanging clifss > In the Siberian tundra > And on the coral reefs > On surface of the moon and > In heart of Milky Way > Within volcano plumes and > In crashing ocean wave > In laughter of a child > In tears of the bereaved > At home and in the wild > In everything that lived > In passion and in reason > In body and in soul > In insight and in wisdom > In all that I extol - > Nirvana and desire > The journey and the goal > Phenomenal and maya > Combined with noumenal - > It all exists within you > And in you it takes place > The ultimate continuum > That in you manifests - > In sides of dialectic > In substance of the light > In prayer and in action > In wings of bird in flight, > In essence and existence > In fractals and in chaos > In goodness that consists of > The yearning and the blessed, > In thought and inspiration > In everything that's true > I, without hesitation, > Say: Universe is you. > Ilya Shambat === Subject: Re: Universe is You what a delightful little play of words! for their sake, have no faith in masta Owen here, or others who claim you are right or wrong. for they know not what they are doing, as we continue to pray dearly for all souls, including theirs. > Wrong again. >> I'll find you in the sunlight, >> I'll find you in the night, >> In morning and in twilight >> In hearing and in sight >> In conjured and in real >> In Venus and in Mars >> In truth incorporeal >> Of billions of stars >> In water and in fire >> In earthquake and in flood >> In Mount Kilimanjaro >> That is the House of God >> In Amazonian jungle >> And overhanging clifss >> In the Siberian tundra >> And on the coral reefs >> On surface of the moon and >> In heart of Milky Way >> Within volcano plumes and >> In crashing ocean wave >> In laughter of a child >> In tears of the bereaved >> At home and in the wild >> In everything that lived >> In passion and in reason >> In body and in soul >> In insight and in wisdom >> In all that I extol - >> Nirvana and desire >> The journey and the goal >> Phenomenal and maya >> Combined with noumenal - >> It all exists within you >> And in you it takes place >> The ultimate continuum >> That in you manifests - >> In sides of dialectic >> In substance of the light >> In prayer and in action >> In wings of bird in flight, >> In essence and existence >> In fractals and in chaos >> In goodness that consists of >> The yearning and the blessed, >> In thought and inspiration >> In everything that's true >> I, without hesitation, >> Say: Universe is you. >> Ilya Shambat === Subject: using base vectors from a span. When finding the base vectors for a span these can be used to express other vectors from the set. But I don't understand why its necessary make a row reduction on a matrix to find these expressions: http://photos3.blogger.com/blogger/3626/1346/1600/combi.jpg === Subject: Every open interval contains a rational? I have recently learned on this group that every open interval of R contains at least one rational number. I tried to prove it to myself starting from the definition of an open interval and rational number, but with no success. Can anyone help, please? === Subject: Re: Every open interval contains a rational? >I have recently learned on this group that every open interval of R >contains at least one rational number. >I tried to prove it to myself starting from the definition of an open >interval and rational number, but with no success. >Can anyone help, please? Say w is the length of your interval and r is a rational less than w. Could the set { nw } where n is an integer miss your interval? --Lynn === Subject: [ANN] Java Applet: Epycicles, FFT and Batman http://www.professores.uff.br/hjbortol/arquivo/2005.2/applets/fft/batman.htm l This applet JAVA illustrates the trigonometric polynomial interpolation using FFT (or how to make planetary motions draw arbitrary shapes on the sky, in our case, the Batman sign). Comments and suggestions are welcome! Humberto Jose Bortolossi === Subject: Conjecture on Partitioning of Lucas Numbers Let L be the sequence of Lucas numbers given by http://www.research.att.com/projects/OEIS?Anum=A000204 1,3,4,7,11,18,29,47 Conjecture: for all in naturals, the n-th Lucas number can be written as L(n) = p(1) + p(2) + p(3) + ... + p(s) where p(1), ..., p(s) are positive integers such that 1. each p(i) divides n 2. there exists exactly one index 0 < i < s + 1 such that p(i) = 1 3. there exists at least one index 0 < i < s + 1 such that p(i) = n Note: A comment given for sequence http://www.research.att.com/projects/OEIS?Anum=A000032 -i.e. the above sequence with a prepended 2, is: For distinct primes p,q, L(p) is congruent 1 mod p, L(2p) is congruent 3 mod p, and L(pq) is congruent 1+q(L(q)-1) mod p. Also, L(m) divides F(2km) and L((2k+1)m), k,m >=0. Since L(p) is congruent 1 mod p for prime p, we get that L(p) = 1 + p + ... + p (p > 1). Therefore, the statement holds for prime p. n = 3: L(3) = 1 + 3 = 4 Note: 1 and 3 divide 3 n = 4: L(4) = 1 + 4 + 2 = 7 Note: 1, 2, 4 divide 4 n = 5: L(5) = 1 + 5 + 5 = 11 Note: 1, 5 divide 5 n = 6: L(6) = 1 + 2 + 3 + 6 + 6 = 18 Note: 1, 2, 3, 6 divide 6 n = 7: L(7) = 1 + 7 + 7 + 7 + 7 = 29 n = 8: L(8) = 1 + 2 + 4 + 8 + 8 + 8 + 8 + 8 = 47 n = 9: L(9) = 1 + 3 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 = 76 n = 10: L(10) = 1 + 2 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 Can anyone prove the above conjecture or point to something similar in the literature? If you are interested in how floretions contributed to the following conjecture (be it trivial or not), you might want to look at my earlier post http://mathforum.org/kb/thread.jspa?threadID=1262398&messageID=4019448#40194 48 Creighton Dement