mm-470 === Subject: Re: Continuous prime number function>There are an infinite number of them.>How about>g(x) = 2 for x<=1>g(x) = f(x) for x positive integral>g(x) = f(floor(x)) + (x-floor(x))*(f(floor(x+1)-f(floor(x)) for x not>integral>(floor(x) being the integral part of x).>This g(x) goes through all of the points of f(n), and just has straight>lines linearly interpolated between these points. Obviously you couldhave>any continuous function between f(n) and f(n+1) replacing the straight>lines.> The last line of the g definition probably should be: g(x) = f(floor(x)) +(x-floor(x))*(f(floor(x+1))-f(floor(x))) for x not integral.> Function g defined by this way is continuous, but its derivative is not.> With the constrain of continuous derivative how can be expressed g?> Can be constructed function g with polynomial regression of function f?> If so will it be a turing computable?> MarcI almost put in the case with continuous derivatives. Its easy to see howsuch a function could be constructed.Instead of connecting the points that pass through primes with straightlines, use a series of curves. One simple way would be to use curved linesegments that have slope zero at the two endpoints. Any continuous functiony(x) with a continuous derivative that has dy/dx = 0 at x=0 and x=1 could bescaled to run between the points (n,f(n)) and (n+1, f(n+1)). A curve made upof lots of little sections like this would have a continuous derivative.This could easily be expanded to provide a function that has a continuousnth derivative. It should be possible to use a section of a polynomial ofdegree n+1 (say the part from [0,1]) appropriately scaled to run betweeneach (n,f(n)) and (n+1, f(n+1)) to make a curve that has an nth derivative.I can't prove this, but it seems clear by analogy with the straight linesegment case.This gets you over the finite number of continuous derivative case, howabout the infinitely continuous derivative (I don't know the correctterminology - I mean where (d^n)y/(dx)^n is defined for all n).This case cannot be resolved using simply segments of polynomials in x toconnect (n,f(n)) and (n+1,f(n+1)). If such a function existed, then we couldtake a Taylor series expansion around any point (all derivatives aredefined). As all derivatives higher than some n will be zero, this willproduce a fixed polynomial in x. But we know there is no polynomial functionwhich produces only primes.What we need is a continuous function that has f^n(x) defined for all0<=x<=1 and all n, f^n(0)=0 for all n, but isn't just f(x)=0. I recall frommy calculus days of 30 years ago that such functions exist. If we canmanipulate this so it also has f^n(1)=0 for all n, then a series of suchcurves joining each data point would be continuous and infinitelydifferentiable. === Subject: shift a line in 2D by 's units' along the direction of the line - using vectorsI have a line whose start and end points are (x1,y1) and (x2,y2). Iwant to shift this line by 's units' along the direction of the lineusing vectors.Below is the function I have written which does the above. When Imeasure the length of the line before and after the shift there issome difference in length. Not sure what the problem is?Please help.//////////////////C- Functionvoid newpts(float x1,float y1,float x2,float y2,float s,float&newx1,float&newy1,float& newx2,float& newy2){ float ux,uy,vx,vy;ux=x1-x2;uy=y1-y2;vx=ux / sqrt(ux*ux + uy*uy);vy=uy / sqrt(ux*ux + uy*uy);newx1=ux - s*vx + x2;newy1=uy - s*vy + y2;newx2=-s*vx + x2;newy2=-s*vy + y2;} === Subject: Re: shift a line in 2D by 's units' along the direction of the line - using vectors> I have a line whose start and end points are (x1,y1) and (x2,y2). I> want to shift this line by 's units' along the direction of the line> using vectors.> Below is the function I have written which does the above. When I> measure the length of the line before and after the shift there is> some difference in length. Not sure what the problem is?It looks like it would be a rounding error. Is the difference in length that you're getting more than would be expected from that? The algorithm itself appears to work on paper.> Please help.C- Function> void newpts(float x1,float y1,float x2,float y2,float s,float&> newx1,float&> newy1,float& newx2,float& newy2)> { > float ux,uy,vx,vy;> ux=x1-x2;> uy=y1-y2;> vx=ux / sqrt(ux*ux + uy*uy);> vy=uy / sqrt(ux*ux + uy*uy);> newx1=ux - s*vx + x2;> newy1=uy - s*vy + y2;> newx2=-s*vx + x2;> newy2=-s*vy + y2;> }newpts(0,0,3,4,10,x1,y1,x2,y2);gives me x1=6, y1=8, x2=9, y2=12 on paper.Will Twentymanemail: wtwentyman at copper dot net === Subject: Re: Utter ConfusionI have always dislike this proof, because it uses the Fundamental Thereom ofArithmetic - that all numbers have a unique representation as a product ofprimes - without actually identifying that it is being used - just a littlehandwaving at the if p|b^2 then p|b stage.The Fundamental Theorom of Arithmetic is somewhat harder to prove than theirrationality of sqrt(2) (well, a few lines harder), and not intuitivelyobvious (at least to me). While I can see that if 2|b^2 then 2|b, I cannotintuitively see that if 17|b^2 then 17|b. A proper proof of theirrationality of sqrt(p) should either refer directly to the FTA as a provedtheorem of prove FTA in its construction.>What went wrong?>Earlier when you said that>b^2= 3 x^2>implies that 3 divides b^2, therefore 3 divides b, you are using the>rule that says that>a divides b^2 implies a divides b ** if a and b are relatively prime**.> Surely not! He must have used the rule that says a divides b^2> implies a divides b if a is a prime.> Your rule would give that a = +/- 1, since if a and b are relatively> prime, and a|b, then 1 = gcd(a,b) = |a|. === Subject: Re: Utter Confusion> I have always dislike this proof, because it uses the Fundamental Thereom of> Arithmetic - that all numbers have a unique representation as a product of> primes - without actually identifying that it is being used - just a little> handwaving at the if p|b^2 then p|b stage.> The Fundamental Theorom of Arithmetic is somewhat harder to prove than the> irrationality of sqrt(2) (well, a few lines harder), and not intuitively> obvious (at least to me). While I can see that if 2|b^2 then 2|b, I cannot> intuitively see that if 17|b^2 then 17|b. A proper proof of the> irrationality of sqrt(p) should either refer directly to the FTA as a proved> theorem of prove FTA in its construction.Or neither. Seehttp://www.maths.ex.ac.uk/~rjc/courses/irr.dvi . === Subject: Re: Utter Confusion>What went wrong?>Earlier when you said that>b^2= 3 x^2>implies that 3 divides b^2, therefore 3 divides b, you are using the>rule that says that>a divides b^2 implies a divides b ** if a and b are relatively prime**.> Surely not! He must have used the rule that says a divides b^2> implies a divides b if a is a prime.> Your rule would give that a = +/- 1, since if a and b are relatively> prime, and a|b, then 1 = gcd(a,b) = |a|.>I have always dislike this proof, because it uses the Fundamental Thereom of>Arithmetic - that all numbers have a unique representation as a product of>primes - without actually identifying that it is being used - just a little>handwaving at the if p|b^2 then p|b stage.>The Fundamental Theorom of Arithmetic is somewhat harder to prove than the>irrationality of sqrt(2) (well, a few lines harder), and not intuitively>obvious (at least to me). While I can see that if 2|b^2 then 2|b, I cannot>intuitively see that if 17|b^2 then 17|b. A proper proof of the>irrationality of sqrt(p) should either refer directly to the FTA as a proved>theorem of prove FTA in its construction.Or avoid using FTA altogether as in http://www.whim.org/nebula/math/ratalint.html === Subject: linear transformation questionthis is not homework.the problem is this:If L and L' are non-zero linear spaces (with the same scalars (real orcomplex)) then there exists a non-zero linear transformation of L intoL'.do i need to fool around with a basis to make this work or is there asimpler way?i was thinking something like, let y=/=0 and y in L'if x in L then has a unique rep. x = a1x1 + ... + anxn where the xi'sare part of a basis.i would then map T: L -> L' like T(x) = (a1 + ... + an)y it looks like T(x+x')= T(x) + T(x'), and T(ax) = aT(x) to me. the basis elements xi must be non zero, so we pick one...T(xi) = y which is not zero, so T is not the zero lineartransformation.i don't really care for this much, though. i would rather construct afunction without reference to a basis. === Subject: Re: linear transformation question> this is not homework.> the problem is this:> If L and L' are non-zero linear spaces (with the same scalars (real or> complex)) then there exists a non-zero linear transformation of L into> L'.> do i need to fool around with a basis to make this work or is there a> simpler way?Can you do the case when L' is one-dimensional?Can you do the case when L is one-dimensional?Can you combine those two special cases to get a solution to the wholeproblem? http://www.math.ohio-state.edu/~edgar/ === Subject: Re: linear transformation question>this is not homework.>the problem is this:>If L and L' are non-zero linear spaces (with the same scalars (real or>complex)) then there exists a non-zero linear transformation of L into>L'.>do i need to fool around with a basis to make this work or is there a>simpler way?Consider nonzero linear mappings T: V -> L' for linear subspacesV of L, use Zorn's Lemma to get a maximal one, and show that if V is notall of L you can extend T to a slightly larger subspace. === Subject: Re: another boring critisism of Cantor's Theorem> Could you explain why V can't be countable? Certainly you can't> prove that in any consistent first order formalism for which the> Loewenheim Skolem theorem applies.> It's a triviality to prove V can't be countable in ZFC. The> Loewenheim-Skolem theorem says that if ZFC is consistent, it has a> countable model. But that's not V.> Philosophically speaking, if we are discorsing in the first-order> language of set theory and uttering theorems of ZFC, we can always> suppose, without making any of our theorems false, that our discourse> is relative to a countable model. But as I say, that's different to> saying V can be countable.> Somehow, I don't think you explained anything. You merely restated> what you said the first time.> If you can prove V exists, you can prove that it is countable > relative to some meta-language. But you claim otherwise. Why?> I suppose I should admit that I don't know a great deal about> this topic, but you have hardly increased my knowledge of it.If ZFC is consistent, and we are uttering theorems of ZFC in thefirst-order language of set theory, we can always entertain thepossibility, without making any of our theorems false, that ourdiscourse, let's call it the object language, is relative to somecountable model. Then a metalanguage will be available in which whatwe referred to as V in our object language, will now in fact becountable. This possibility *may* hold of our language, I don't thinkit *has* to hold, this may be a point of difference between us.But in any case, you have to decide what language you're talking in.Whether you're talking in the object language or the metalanguage, itwon't be true that V is countable. In both the object language and themetalanguage, all theorems of ZFC are true, and it's a trivial theoremof ZFC that V is not countable. === Subject: Re: another boring critisism of Cantor's Theorem>Could you explain why V can't be countable? Certainly you can't>prove that in any consistent first order formalism for which the>Loewenheim Skolem theorem applies.> It's a triviality to prove V can't be countable in ZFC. The> Loewenheim-Skolem theorem says that if ZFC is consistent, it has a> countable model. But that's not V.> Philosophically speaking, if we are discorsing in the first-order> language of set theory and uttering theorems of ZFC, we can always> suppose, without making any of our theorems false, that our discourse> is relative to a countable model. But as I say, that's different to> saying V can be countable.> Somehow, I don't think you explained anything. You merely restated> what you said the first time.> If you can prove V exists, you can prove that it is countable> relative to some meta-language. But you claim otherwise. Why?> I suppose I should admit that I don't know a great deal about> this topic, but you have hardly increased my knowledge of it.> If ZFC is consistent, and we are uttering theorems of ZFC in the> first-order language of set theory, we can always entertain the> possibility, without making any of our theorems false, that our> discourse, let's call it the object language, is relative to some> countable model. Then a metalanguage will be available in which what> we referred to as V in our object language, will now in fact be> countable. This possibility *may* hold of our language, I don't think> it *has* to hold, this may be a point of difference between us.> But in any case, you have to decide what language you're talking in.> Whether you're talking in the object language or the metalanguage, it> won't be true that V is countable. In both the object language and the> metalanguage, all theorems of ZFC are true, and it's a trivial theorem> of ZFC that V is not countable.But then, just like the original thing only _seemed_ uncountable,but in reality countable, your 'meta world' could also be onlyseemingly uncountable, but be 'meta-meta' countable.Or do i miss something?Herman Jurjus === Subject: Re: another boring critisism of Cantor's Theorem> Could you explain why V can't be countable? Certainly you can't> prove that in any consistent first order formalism for which the> Loewenheim Skolem theorem applies.> It's a triviality to prove V can't be countable in ZFC. The> Loewenheim-Skolem theorem says that if ZFC is consistent, it has a> countable model. But that's not V.> Philosophically speaking, if we are discorsing in the first-order> language of set theory and uttering theorems of ZFC, we can always> suppose, without making any of our theorems false, that our discourse> is relative to a countable model. But as I say, that's different to> saying V can be countable.> Somehow, I don't think you explained anything. You merely restated> what you said the first time.> If you can prove V exists, you can prove that it is countable > relative to some meta-language. No, you can't.> But you claim otherwise. Why?Why do you claim you can?I clarified the correct statement of what you're trying to say here.Why do you persist in stating this confused version of it?> I suppose I should admit that I don't know a great deal about> this topic, but you have hardly increased my knowledge of it. === Subject: Re: differentiability of multivariable function> Define f(0,0) = 0 and f(x, y) = x^3/(x^2+y^2) for (x,y) != (0,0)> Let gamma be a differentiable mapping of R into R^2 with gamma(0) => (0, 0) and |gamma'(0)| > 0. Put g(t) = f(gamma(t)) and prove that g is> in C' (i.e. has a continuous derivative) if gamma is in C'Hint: Check that everything is fine if if gamma is a straight line, gamma(t) = (at,bt). Also check that the partials of f, while not continuous at 0, are bounded near 0. This should help you show that f(gamma(t)) is close enough to f(gamma'(0)*t) for small t to get the answer. === Subject: Re: Path connectedness> Is that true ?> Every Hausdorff path connected topological space is arc connectedYes it is true. It is part of Hahn-Mazurkiewicz Theorem. A proof can befound in Hocking and Young's Topology.> At fist glance, we can suppose with no loss of generality that X is > compact. (Im f is compact)> We can follow the steps of Uryson proof.[deletia]> by induction, we can build> fn such fn(k/2^n) distincts for all k and f_n (k/2^n)=fm(k/2^n) if m>n.> So we can define a one to one function f* from the set all diadic > numbers to X.> I'm not so sure f* is continuousI believe you can use the following observation to prove this. For anyinterval [u,v], you can find another interval [u',v'] such that v'-u'<=v-u and f_k([u,v]) subset f([u',v']). Combine this observationwith a compactness argument.> If yes we can prolonge f* to a map f** :[0,1] to X> But I'm not so sure f** will be one to one.Unfortunately this will not be true. It is easy to construct a pathin the plane having a single self-intersection, which can be arrangedto occur at non-dyadic values. === Subject: Re: Path connectedness===Subject: Re: Path connectedness > Is that true ? > Every Hausdorff path connected topological space is arc > connected >Yes it is true. It is part of Hahn-Mazurkiewicz Theorem.Here is some discussion about HM and corollary which I have been unable tofathom. Perhaps you can make sense of it and give incite how to showcorollary. === Subject: Re: continuous function on a bounded set> The first good thing to know is that a compact, connected metric> space with exactly two non-cut points (i.e. points p where X{p} is> connected) is homeomorphic to the unit interval. This is tricky to> prove.>Indeed, nary a notion how to preceed, yet much intuition about snags.Let a and b be the non-cut points. Define an ordering of X via xX. > Now show that the space in question is arcwise connected by > intersecting chains of open sets with compact closures between two > points and using the result in the previous paragraph (this > bypasses your concerns with convergence).>Yes by the chain theorem about connectedness, I can show there's a>overlaping finite chain of open sets with compact closures between anytwo>points a,b. I can even make the open sets of the chain path-connected and>bounded to within 1/n. Now for each n, I squeeze down on the path from a>to b to obtain a path from a to b with the premise of the 'good thing'.>Seems I'm repeating portions of A Theorem. So now the space is acclaimed>arcwise connected.Yes.> Then, take a continuous map from the Cantor set onto X (an exercise> in itself) and link up images of the endpoints using arc> connectedness. This will give f from [0,1] onto X.>Huh? Not visualizing what's happening.>This all looks a lot worse than the accounting nightmare of A Theorem.Imagine an onto map of the Cantor set g:C->X. For each pair of adjacentendpoints x,y of C, find an arc from g(x) to g(y). This can be made smallenough so that the resulting map of [0,1] is still continuous.> And yes, the corollary is that a path connected Hausdorff space> is arc connected.>How is it a corollary about general spaces has come of a complex theorem>about compact metric spaces?For a path, consider the image of the path and do arc connectedness inthat image.---- === Subject: Re: Path connectednessOriginator: grubb@lola> And yes, the corollary is that a path connected Hausdorff space> is arc connected.>How is it a corollary about general spaces has come of a complex theorem>about compact metric spaces?>For a path, consider the image of the path and do arc connectedness in>that image.OK. We say that X is a Peano space if it is a compact, connected, locallyconnected, metrizable space. There are two results to know:1. Any Peano space is arcwise connected.We've played with this. The proof uses chains of sets between pointsand a characterization of [0,1].2. A Hausdorff space is a continuous image of the unit interval ifand only if it is a Peano space. (This is the HM result).We've discussed this also, although mostly the 'if' part. Find amap of the Cantor ternary set onto X and link up endpoints using arcconnectedness. The 'only if' part is not too bad except for metrizability.This requires some hard work with metrizability results. In particular,a compact, continuous image of a metric space is metrizable.Now suppose that Y is a Hausdorff, path connected space. Let x, ybe in Y. Let p:[0,1]->Y be a path with p(0)=x, p(1)=y. Then, the imagep([0,1]) is a Hausdorff image of [0,1], so is a Peano space (by 2.)Thus, p([0,1]) is arc connected (by 1.). Thus, we can join x and y byan arc in p([0,1]), and hence in Y.--Dan Grubb === Subject: Re: Path connectedness> Unfortunately this will not be true. It is easy to construct a path> in the plane having a single self-intersection, which can be arranged> to occur at non-dyadic values. === Subject: Re: Path connectedness>Unfortunately this will not be true. It is easy to construct a path>in the plane having a single self-intersection, which can be arranged>to occur at non-dyadic values.However I think a slight variation of your argument can be made to work.Namely you can arrange that the images under f_n of the open intervals](k-1)/2^n, k/2^n[ k=1,2,...,2^n are disjoint. === Subject: Re: Path connectedness Every Hausdorff path connected topological space is arc connected >arc connected = if a and b are distinct points in X then we can find >an ONE TO ONE (injectif) path f from [0,1] to X with >f(0)=a and f(1)=b.Arc connected requires more that that. In the case X is Hausdorff, then fis homeomorphism and f([0,1]) is homeomorph [0,1]. However consider [0,1]with the cofinite topology and f the identity. f is a path from 0 to 1and it is an arc by your definition, however it's not an arc sincef([0,1]) isn't a homeomorph of [0,1]. >Let u = Sup{ t | f(t)=a} >and v = Inf{ t | f(t)=b} >Because X is Hausdorf we have f(u)=a and f(v)=bYou can have the humorous situation where v < u.To avoid this define u = sup f^-1({a})as before but v = inf (f^-1({b}) / [0,u])---- === > Subject: Path connectedness>Every Hausdorff path connected topological space is arc connected>arc connected = if a and b are distinct points in X then we can find>an ONE TO ONE (injectif) path f from [0,1] to X with>f(0)=a and f(1)=b.> Arc connected requires more that that. In the case X is Hausdorff, then f> is homeomorphism and f([0,1]) is homeomorph [0,1]. However consider [0,1]> with the cofinite topology and f the identity.This topology is not hausdorff>f is a path from 0 to 1> and it is an arc by your definition, however it's not an arc since> f([0,1]) isn't a homeomorph of [0,1]. === Subject: Rubik's cube operatorsI have a good collection of edge operators for my Rubik's cube, but Idon't know any corner operators. Can anyone give me any of thefollowing:A double-corner swapA meson (twist one corner cubie one way, another corner cubie theother way)A baryon (twist three corner cubies the same way) === Subject: Re: Rubik's cube operators> I have a good collection of edge operators for my Rubik's cube, but I> don't know any corner operators. Can anyone give me any of the> following:You can use the incredible Cube Explorer program to find a short(est)way to achieve anything on a Rubik's cube:http://home.t-online.de/home/kociemba/cube.htmThe sequences it comes up with are not always comfortable to turnhowever.> A double-corner swap> A meson (twist one corner cubie one way, another corner cubie the> other way)The classic way to twist two corners without moving any edges is:FD'F'R'D'R, to twist the FRU corner anti-clockwise,U, U2, or U' to bring the other corner to FRU positionR'DRFDF', to twist this other corner clockwise (inverse of first part)U, U2, or U' to bring the top layer back to where it was (inverse ofsecond part).> A baryon (twist three corner cubies the same way)RUR'URU2R'U2, but this also does an edge 3-cycle.To get really fast and efficient ways of solving the cube you can dolittle better than starting at www.speedcubing.com.JaapJaap's Puzzle Page:http://www.geocities.com/jaapsch/puzzles/ === Subject: Re: Rubik's cube operators> You can use the incredible Cube Explorer program to find a short(est)> way to achieve anything on a Rubik's cube:> http://home.t-online.de/home/kociemba/cube.htm> The sequences it comes up with are not always comfortable to turn> however.When you have found an optimal solution you have the option tocontinue the search to find all other optimal (and suboptimal)solutions.H. Kociemba === Subject: Re: Rubik's cube operatorsDo you Know what is de group accting in the cube? why? === Subject: Re: Rubik's cube operatorsrupertmccallum@yahoo.com says...> I have a good collection of edge operators for my Rubik's cube, but I> don't know any corner operators. Can anyone give me any of the> following:> A meson (twist one corner cubie one way, another corner cubie the> other way)Rupert,This type of move (or operator) you can easily create yourselfthrough the use of the group-theoretic concept of _commutation_.The idea is very simple. Say you have a move sequence P thattwists one corner cubelet, while scrambling large portionsof the cube, _except_ for the one of the face layers thecubelet lies in. For example, the sequence R'DRFDF' twiststhe FRU cubelet CW and leaves the U layer untouched.If you now perform the move sequence P' (P backwards) the cube will be restored. However, consider what happens if youperform the move sequence Q consisting of the single move Ubefore you do P'. You'll (1) move the twisted cubelet at FRUto FLU. You'll twist the new cubelet at FRU ccw, leavingthe top layer otherwise intact and you'll unscramble thebottom two layers into their original state. Finish offwith Q' to restore the top layer back into position andyou'll have performed your meson move!The sequence PQP'Q' (= R'DRFDF' U FD'F'R'D'R U') which youjust performed is called a commutator of P and Q.> A baryon (twist three corner cubies the same way)Can you see how you can create a move sequence for thisusing a commutation of the sequence for the meson?Here's a good page talking about the relevance of grouptheory to puzzles like the Rubik's Cube:http://www.geocities.com/jaapsch/puzzles/theory.htmEnjoy! Christer EricsonSony Computer Entertainment, Santa Monica === Subject: Re: Rubik's cube operators>I have a good collection of edge operators for my Rubik's cube, but I>don't know any corner operators. Can anyone give me any of the>following:>A double-corner swapHold the cube so the upper face has the four corners in question. Themoves shown below will swap the front two corners with the back twocorners, leaving the bottom two layers undisturbed (assuming nopictures on the faces giving additional orientation, and maybe eventhen, I don't know for sure).Perform R'U'F'UFR, where R, U, F represent right, upper, and frontfaces rotated clockwise and R', U', F' rotate them counterclockwise. === Subject: martingale inequality, indep. random variablesAlright, I'm obviously missing something easy here, but suppose X_n isa martingale with sup E[(X_n)^2]. Write X_n = X_0 + Y_1 + ... + Y_n,with X_0 and the Y_i being independent random variables.Define K_n = sup(m>n) |X_m - X_n|. Show that E[(K_m)^2] < = 4 SUM(n+1,oo) E[(Y_i)^2]. (*)Now, I know thatSUM(n+1,oo) E[(Y_i)^2] = E [( SUM(n+1,oo) Y_i )^2].If I could somehow take the sup outside of the expectation in (*),then I would have the inequality but in fact I wouldn't need the 4,and I guess that's what's puzzling me...any hints? === Subject: Re: martingale inequality, indep. random variables>Alright, I'm obviously missing something easy here, but >[1] suppose X_n is>a martingale with sup E[(X_n)^2]. Write >[2] X_n = X_0 + Y_1 + ... + Y_n,>with X_0 and the Y_i being independent random variables.It's not clear to me whether [2] is supposed to be anotherhypothesis, or whether you're under the impression that[2] is possible because of [1]. Just in case, someone shouldpoint out that [2] definitely does _not_ follow from [1];sums of independent random variables give a martingalebut not conversely, the notion of a martingale is moregeneral.>Define K_n = sup(m>n) |X_m - X_n|. Show that >E[(K_m)^2] < = 4 SUM(n+1,oo) E[(Y_i)^2]. (*)>Now, I know that>SUM(n+1,oo) E[(Y_i)^2] = E [( SUM(n+1,oo) Y_i )^2].>If I could somehow take the sup outside of the expectation in (*),>then I would have the inequality but in fact I wouldn't need the 4,>and I guess that's what's puzzling me...any hints? === Subject: Re: martingale inequality, indep. random variables> Alright, I'm obviously missing something easy here, but suppose X_n is> a martingale with sup E[(X_n)^2]. Write X_n = X_0 + Y_1 + ... + Y_n,> with X_0 and the Y_i being independent random variables.> Define K_n = sup(m>n) |X_m - X_n|. Show that > E[(K_m)^2] < = 4 SUM(n+1,oo) E[(Y_i)^2]. (*)> Now, I know that> SUM(n+1,oo) E[(Y_i)^2] = E [( SUM(n+1,oo) Y_i )^2].> If I could somehow take the sup outside of the expectation in (*),> then I would have the inequality but in fact I wouldn't need the 4,> and I guess that's what's puzzling me...any hints?For general martingales one could obtain an inequality like this from Doob's inequality, although it seems to me that I would get 16 instead of 4. For sums of independent random variables there is Levy's inequality, which I think really would give you the constant 4. Neither of these inequalities are obvious (in the sense that one is likely to figure them out by oneself). Look in the book Some random series of functions by Kahane - Levy's inequality is in Chapter 2 or 3, I think. === Subject: Re: martingale inequality, indep. random variables> Alright, I'm obviously missing something easy here, but suppose X_n is> a martingale with sup E[(X_n)^2]. Write X_n = X_0 + Y_1 + ... + Y_n,> with X_0 and the Y_i being independent random variables.> Define K_n = sup(m>n) |X_m - X_n|. Show that> E[(K_m)^2] < = 4 SUM(n+1,oo) E[(Y_i)^2]. (*)> Now, I know that> SUM(n+1,oo) E[(Y_i)^2] = E [( SUM(n+1,oo) Y_i )^2].> If I could somehow take the sup outside of the expectation in (*),> then I would have the inequality but in fact I wouldn't need the 4,> and I guess that's what's puzzling me...any hints?> For general martingales one could obtain an inequality like this from > Doob's inequality, although it seems to me that I would get 16 instead > of 4. For sums of independent random variables there is Levy's > inequality, which I think really would give you the constant 4. Neither > of these inequalities are obvious (in the sense that one is likely to > figure them out by oneself). Look in the book Some random series of > functions by Kahane - Levy's inequality is in Chapter 2 or 3, I think.Also, I recall seeing a rather simpler argument for this, which is part of a proof of the law of large numbers. I think it is due to Kolmogorov. In any case, the result is not so obvious, and you need to find it in a book somewhere. === Subject: Re: martingale inequality, indep. random variables> Alright, I'm obviously missing something easy here, but suppose X_n is> a martingale with sup E[(X_n)^2]. Write X_n = X_0 + Y_1 + ... + Y_n,> with X_0 and the Y_i being independent random variables.> Define K_n = sup(m>n) |X_m - X_n|. Show that> E[(K_m)^2] < = 4 SUM(n+1,oo) E[(Y_i)^2]. (*)> Now, I know that> SUM(n+1,oo) E[(Y_i)^2] = E [( SUM(n+1,oo) Y_i )^2].> If I could somehow take the sup outside of the expectation in (*),> then I would have the inequality but in fact I wouldn't need the 4,> and I guess that's what's puzzling me...any hints?> For general martingales one could obtain an inequality like this from > Doob's inequality, although it seems to me that I would get 16 instead > of 4. For sums of independent random variables there is Levy's > inequality, which I think really would give you the constant 4. Neither > of these inequalities are obvious (in the sense that one is likely to > figure them out by oneself). Look in the book Some random series of > functions by Kahane - Levy's inequality is in Chapter 2 or 3, I think.> Also, I recall seeing a rather simpler argument for this, which is part of a > proof of the law of large numbers. I think it is due to Kolmogorov. In any > case, the result is not so obvious, and you need to find it in a book somewhere.OK, I found Levy's inequality:Suppose X_i are i.r.v.s such thatP {|X_i + ... + X_n| >= L/2} <= dfor 1 <= i <=n.ThenP {sup |X_1 + ... + X_n| > = L} <= d/(1-d).But I still don't see how this applies...it gives results about supsof the sum, but I want something on the expectation of the square ofthat sup.... === Subject: Ques. re curves in 3-space - with clarification.I am posting this again with a clarification.I would really appreciate your insight into the following which seems to be true but I don't see an easy proof: Let C be a smooth curve in R^3 and p, q, r be 3 points that occur in that order on C. Then there is a pair of tangents to C at two points between p and r such that the angle between them is the same as that between pq and qr. *** Of course, the tangents at 2 points of C need not be coplanar.Simply treat the tangents as vectors (direct C arbitrarily) and findthe angle between them, i.e. cos^-1 u.v/|u||v|, or same as moving bothto start at the origin and measure the angle between them.***This is trivial if C is in R^2: there is a point s between p and q so that the tangent at s is parallel to pq, there is a point t between q and r so that the tangent at t is parallel to qr, ... But, what is the 3D analogue of Lagrange's Th.? Clearly, the tangent need not be parallel to pq at any point in between p an q in the 3D case. But it does seem likely that the tangent must turn as much as the chords. Please cc a rsponse to guha@ait.ac.th. Sumanta Guha === Subject: Re: Ques. re curves in 3-space - with clarification.>Let C be a smooth curve in R^3 and p, q, r be 3 points that occur in >that order >on C. Then there is a pair of tangents to C at two points between p >and r such that the angle between them is the same as that between pq >and qr. >*** Of course, the tangents at 2 points of C need not be coplanar.>Simply treat the tangents as vectors (direct C arbitrarily) and find>the angle between them, i.e. cos^-1 u.v/|u||v|, or same as moving both>to start at the origin and measure the angle between them.***So you're measuring the angle as a scalar... that should be easy to prove.Consider a roller-coaster car which moves along the curve at a constantspeed. The car's direction defines its pitch and yaw (roll is irrelevant).If you combine the changing of pitch and yaw into a single scalar we'll callcurvature, the angle between the pq and qr vectors can never exceed theintegral of curvature over the distance pr. It can be a lot less, though. Consider a spring shape. Curvature magnitudecould be constant while p, q, and r are colinear. But that's OK, you canalways move your tangent points arbitrarily close together. --Keith Lewis klewis {at} mitre.orgThe above may not (yet) represent the opinions of my employer. === Subject: Nick Cook's The Hunt For Zero PointTo appear in my book Super Cosmos:Again Nick's book is an exciting read on the conventional technology and the Nazi history, but he stumbles seriously where knowledge of physics is really needed.1. On the Aurora and the large triangular craft reported by NIDS near Area 41, Palmdale CA etc.The Townsend-Brown effect is not anti-gravity.charges on the two plates obeys Newton's 3rd law and there is no spontaneous acceleration of the capacitor's center of mass in a vacuum. This is in contrast to the ZPE vacuum propeller with /zpf > 0 at the stern and equal and opposite /zpf < 0 in the bow. Here, as shown by Forward), there will be an acceleration of the center of mass. This is not a violation of Newton's third law, but is a consequence of it! Furthermore, one can metric engineer the exotic vacuum zero point field /zpf in the fuselage of the saucer to fit Alcubierre's warp drive solution with zero g-force and effective faster-than-light travel even though the occupants inside the saucer are weightless like the astronauts in free float orbiting Earth with rockets off and no rotation about center of mass of the shuttle. This is real anti-gravity acceleration field (Paul Hill) G-Engine (George Trimble) breakthrough propulsion.Now some kind of electrostatic charging may be used in Aurora for some purpose that has an advantage in flight through air. Perhaps some kind of MHD effect, but it is not zero g-force anti-gravity and it would not work in a vacuum.Now to un-garble a key quote by Nick that appears at least twice.I ... found five possible pathways to antigravity: manipulating an object's mass and/or inertia; exploitation of the zero point energy field; perturbations of the space-time continuum; faster-than-light travel; and gravity shielding. p. 117Let's take them one at a time:1. manipulating an object's mass and/or inertiaThis idea has been propounded by Puthoff & Co.The idea is fundamentally confused as far as anti-gravity propulsion is concerned.When one looks at Hal Puthoff's PV papers it becomes obvious that Hal waffles on some serious battle-tested ideas in mainstream physics.1-1 Hal waffles on the principle of tensor covariance, i.e. that the map is not the territory - no intrinsic meaning to coordinates. Hal says tensors are not needed in his dielectric engineering model in which he with a variable speed of light c' = c/K. He is not able to solve the problem of a rotating source with this scheme. That alone is enough to junk the model. However, he does write a SSS metric with goo = K^-1 and gxx = gyy = gzz = K = e^2GM/c^2r. Whether his g's obey a tensor law is not clear.1-2 Hal is confused about the meaning of Einstein's equivalence principle. His model only works for asymptotically flat space times and he thinks the equivalence principle is nonlocal relating measurements at r --> infinity to those at finite r. Hal never considers the locality of the equivalence principle which connects instantaneously coincident inertial timelike geodesic LIF observers with non-inertial timelike non-geodesic LNIF observers BOTH at the same local event P, i.e. within a small neighborhood of P, small compared to the local radii of curvature of space-time.Mathematically this is the local tetrad map at P where LIF Alice's metric is that of special relativity nuv to a good approximation. Alice lives in the quasi-flat tangent space-time even though the 4th rank curvature tensor may not vanish in the LIF. LNIF Bob sees the curved metric guv. Bob's LNIF cousin Ray will see gu'v' wheregu'v' = Xu'^uXv'^vguvYou will find none of this battle-tested physics in any of Hal Puthoff's PV papers on metric engineering for interstellar flight. None of the NASA BPP managers seem to have enough basic physics understanding to have noticed this point either.It's because Hal is confused on these basic ideas that he entertained the irrelevant idea of changing the inertia of the ship that Nick Cook alludes to above.The idea is irrelevant because it shows a fundamental misunderstanding of Einstein's equivalence principle that the inertia completely cancels out of the problem in timelike geodesic motion with zero g-force and small tidal curvature force gradients. Indeed, even Galileo knew this in rudimentary form.In terms of high school Newtonian physics:Weight = W = Inertia x g-field of gravity = mgW = mgAny real anti-gravity propulsion leaves m alone, it merely changes g!In fact it makes g = 0 even though the self-controlled timelike geodesic path of the saucer can make U turns at normally might look like 1000 g(surface of Earth) to the outside observer.m will obey special relativity to the outside observer. Right now by m I mean what is felt on board the saucer. Think of m as the mass of an alien conscious AI Gray inside the saucer piloting it with its artificial brain micro-waves.W = 0 for all objects inside the saucer when Nick Cook's anti-gravity switch is ON.m does not change at all!Even if you could change m you do not want to do it because you then change e/m, and h/mc and you destroy the delicate balance of matter as explained by Sir Martin Rees in Just Six Numbers, by Paul Davies in Accidental Universe and by John Barrow and Frank Tipler in The Anthropic Cosmological Principle.Michael Ibison tried to wipe some of the egg off Hal's face on this by writing to me that making m smaller would be an advantage in conventional rocket lifting. Ibison however took no account of what I just mentioned.So even if you could change m, you do not want to do so because it would destroy the delicate metastability of all real matter! Also, you do not need to do at all for real anti-gravity propulsion in the sense of George Trimble's G-Engine (1956) that Nick Cook is keen on and rightly so.So trash (1) as a false lead and a very bad ill-conceived idea!2. exploitation of the zero point energy fieldYes, that is the only idea that works. That is what my theory of the /zpf field is all about.The basic definition of the zero point energy field is my original equation not found in any previous physics paper or book by anyone from this Planet and this time:/zpf = (Quantum of Area)^-1[(Quantum of Volume)|Vacuum Coherence|^2 - 1]The Quantum of Area and the Quantum of Volume come from Loop Quantum Gravity based on Roger Penrose's spin networks forWheeler's IT FROM BIT.'The local zero point dark energy/matter stress tensor for exotic vacuum istuv(ZPE) = (Sakharov's Metric Elasticity)^-1/zpfguvMetric Elasticity = (String Tension)^-1The ordinary weightless vacuum has /zpf = 0 at the specified scale.Experimentallytoo(ZPE) ~ 10^-7 ergs/cce.g. Measuring and Understanding the Universe Rev. Mod. Phys. 75, Oct This is an experimental measurement - a fact!The scale here is that of the FLRW metric 10^2 - 10^4 megaparsecsi.e. dark energy density Omega = 0.73 +- 0.04Table I p. 1435There are no measurements of the Planck scale. But if you make the guess that G(Newton) holds at all scales you gettoo(ZPE) ~ 10^115 ergs/cc when Vacuum Coherence = 0It is the vacuum coherence that solves the Cosmological Constant Paradox, i.e. the 122 Powers of Ten discrepancy between these two numbers of 10^-7 ergs/cc and 10^115 ergs/cc.3. perturbations of the space-time continuumThis is a mirage. I mean it is already included in 2. It is not a linearly independent basic idea. In other words it is redundant. Nick, in his ignorance of the physics, has made a false distinction. It is of the very essence of metric engineering to shape the space-time continuum to our will like Q in Star Trek! So it is with what M. Kaku calls The Masters of Hyperspace visiting us in their saucers like Gulliver visited the Lilliputans. ;-)4. faster-than-light travelThis is a consequence of 2 as explained by Alcubierre. Aliens who get here in their flying saucers do it globally faster than light even though locally they are, in effect, standing still.5. gravity shieldingLike 3 this is a redundant idea already included in 2.Therefore, of Nick's 5 ideas above. The first is simply a bad idea that shows a deep misunderstanding of what Einstein's great idea is all about.3 and 5 are false distinctions already included in 2, which is the correct idea. 4 is a consequence of 2. Therefore, of Nick's 5 ideas, only one is needed to understand George Trimble's G-Engine. === Subject: Re: Nick Cook's The Hunt For Zero PointI don't pretend to understand ALL the math you list, but I'm trying to graspsome of your finer points. Had a few questions.1) Hunt for Zero Point is on my list To Read Soon, you seem to trash it,is it worth the read?2) Your references to Zero Point seem to indicate it as a necessarycomponent to Anti-Gravity. My superficial exposure to the subject seemsto suggest Zero Point is nothing more than an energy source that can begrafted from Vacuum. This would make it a separate issue from a devicecapable of generating an Anti-Gravity field. Once such a device is inventedusing conventional power sources, Zero Point could be used to eliminate theneed to carry large amounts of traditional fuel. Am I way off base here?3) Sort of a side issue, you mention AI Gray. Is this somethingmentioned in Cook's book? Not clear if you think the Gray's are some sortof animated meat robots, or simply capable of thought projection.FIF> To appear in my book Super Cosmos:> Again Nick's book is an exciting read on the conventional technology and> the Nazi history, but he stumbles seriously where knowledge of physics> is really needed.> 1. On the Aurora and the large triangular craft reported by NIDS near> Area 41, Palmdale CA etc.> The Townsend-Brown effect is not anti-gravity.> charges on the two plates obeys Newton's 3rd law and there is no> spontaneous acceleration of the capacitor's center of mass in a vacuum.> This is in contrast to the ZPE vacuum propeller with /zpf > 0 at the> stern and equal and opposite /zpf < 0 in the bow. Here, as shown by> Forward), there will be an acceleration of the center of mass. This is> not a violation of Newton's third law, but is a consequence of it!> Furthermore, one can metric engineer the exotic vacuum zero point field> /zpf in the fuselage of the saucer to fit Alcubierre's warp drive> solution with zero g-force and effective faster-than-light travel even> though the occupants inside the saucer are weightless like the> astronauts in free float orbiting Earth with rockets off and no rotation> about center of mass of the shuttle. This is real anti-gravity> acceleration field (Paul Hill) G-Engine (George Trimble)> breakthrough propulsion.> Now some kind of electrostatic charging may be used in Aurora for some> purpose that has an advantage in flight through air. Perhaps some kind> of MHD effect, but it is not zero g-force anti-gravity and it would not> work in a vacuum.> Now to un-garble a key quote by Nick that appears at least twice.> I ... found five possible pathways to antigravity: manipulating an> object's mass and/or inertia; exploitation of the zero point energy> field; perturbations of the space-time continuum; faster-than-light> travel; and gravity shielding. p. 117> Let's take them one at a time:> 1. manipulating an object's mass and/or inertia> This idea has been propounded by Puthoff & Co.> The idea is fundamentally confused as far as anti-gravity propulsion is> concerned.> When one looks at Hal Puthoff's PV papers it becomes obvious that Hal> waffles on some serious battle-tested ideas in mainstream physics.> 1-1 Hal waffles on the principle of tensor covariance, i.e. that the map> is not the territory - no intrinsic meaning to coordinates. Hal says> tensors are not needed in his dielectric engineering model in which he> with a variable speed of light c' = c/K. He is not able to solve the> problem of a rotating source with this scheme. That alone is enough to> junk the model. However, he does write a SSS metric with goo = K^-1 and> gxx = gyy = gzz = K = e^2GM/c^2r. Whether his g's obey a tensor law is> not clear.> 1-2 Hal is confused about the meaning of Einstein's equivalence> principle. His model only works for asymptotically flat space times and> he thinks the equivalence principle is nonlocal relating measurements at> r --> infinity to those at finite r. Hal never considers the locality of> the equivalence principle which connects instantaneously coincident> inertial timelike geodesic LIF observers with non-inertial timelike> non-geodesic LNIF observers BOTH at the same local event P, i.e. within> a small neighborhood of P, small compared to the local radii of> curvature of space-time.> Mathematically this is the local tetrad map at P where LIF Alice's> metric is that of special relativity nuv to a good approximation. Alice> lives in the quasi-flat tangent space-time even though the 4th rank> curvature tensor may not vanish in the LIF. LNIF Bob sees the curved> metric guv. Bob's LNIF cousin Ray will see gu'v' where> gu'v' = Xu'^uXv'^vguv> You will find none of this battle-tested physics in any of Hal Puthoff's> PV papers on metric engineering for interstellar flight. None of the> NASA BPP managers seem to have enough basic physics understanding to> have noticed this point either.> It's because Hal is confused on these basic ideas that he entertained> the irrelevant idea of changing the inertia of the ship that Nick Cook> alludes to above.> The idea is irrelevant because it shows a fundamental misunderstanding> of Einstein's equivalence principle that the inertia completely cancels> out of the problem in timelike geodesic motion with zero g-force and> small tidal curvature force gradients. Indeed, even Galileo knew this in> rudimentary form.> In terms of high school Newtonian physics:> Weight = W = Inertia x g-field of gravity = mg> W = mg> Any real anti-gravity propulsion leaves m alone, it merely changes g!> In fact it makes g = 0 even though the self-controlled timelike geodesic> path of the saucer can make U turns at normally might look like 1000> g(surface of Earth) to the outside observer.> m will obey special relativity to the outside observer. Right now by m> I mean what is felt on board the saucer. Think of m as the mass of an> alien conscious AI Gray inside the saucer piloting it with its> artificial brain micro-waves.> W = 0 for all objects inside the saucer when Nick Cook's anti-gravity> switch is ON.> m does not change at all!> Even if you could change m you do not want to do it because you then> change e/m, and h/mc and you destroy the delicate balance of matter as> explained by Sir Martin Rees in Just Six Numbers, by Paul Davies in> Accidental Universe and by John Barrow and Frank Tipler in The> Anthropic Cosmological Principle.> Michael Ibison tried to wipe some of the egg off Hal's face on this by> writing to me that making m smaller would be an advantage in> conventional rocket lifting. Ibison however took no account of what I> just mentioned.> So even if you could change m, you do not want to do so because it would> destroy the delicate metastability of all real matter! Also, you do not> need to do at all for real anti-gravity propulsion in the sense of> George Trimble's G-Engine (1956) that Nick Cook is keen on and rightlyso.> So trash (1) as a false lead and a very bad ill-conceived idea!> 2. exploitation of the zero point energy field> Yes, that is the only idea that works. That is what my theory of the> /zpf field is all about.> The basic definition of the zero point energy field is my original> equation not found in any previous physics paper or book by anyone from> this Planet and this time:> /zpf = (Quantum of Area)^-1[(Quantum of Volume)|Vacuum Coherence|^2 - 1]> The Quantum of Area and the Quantum of Volume come from Loop Quantum> Gravity based on Roger Penrose's spin networks for> Wheeler's IT FROM BIT.'> The local zero point dark energy/matter stress tensor for exotic vacuum is> tuv(ZPE) = (Sakharov's Metric Elasticity)^-1/zpfguv> Metric Elasticity = (String Tension)^-1> The ordinary weightless vacuum has /zpf = 0 at the specified scale.> Experimentally> too(ZPE) ~ 10^-7 ergs/cc> e.g. Measuring and Understanding the Universe Rev. Mod. Phys. 75, Oct> This is an experimental measurement - a fact!> The scale here is that of the FLRW metric 10^2 - 10^4 megaparsecs> i.e. dark energy density Omega = 0.73 +- 0.04> Table I p. 1435> There are no measurements of the Planck scale. But if you make the guess> that G(Newton) holds at all scales you get> too(ZPE) ~ 10^115 ergs/cc when Vacuum Coherence = 0> It is the vacuum coherence that solves the Cosmological Constant> Paradox, i.e. the 122 Powers of Ten discrepancy between these two> numbers of 10^-7 ergs/cc and 10^115 ergs/cc.> 3. perturbations of the space-time continuum> This is a mirage. I mean it is already included in 2. It is not a> linearly independent basic idea. In other words it is redundant. Nick,> in his ignorance of the physics, has made a false distinction. It is of> the very essence of metric engineering to shape the space-time> continuum to our will like Q in Star Trek! So it is with what M. Kaku> calls The Masters of Hyperspace visiting us in their saucers like> Gulliver visited the Lilliputans. ;-)> 4. faster-than-light travel> This is a consequence of 2 as explained by Alcubierre. Aliens who get> here in their flying saucers do it globally faster than light even> though locally they are, in effect, standing still.> 5. gravity shielding> Like 3 this is a redundant idea already included in 2.> Therefore, of Nick's 5 ideas above. The first is simply a bad idea that> shows a deep misunderstanding of what Einstein's great idea is all about.> 3 and 5 are false distinctions already included in 2, which is the> correct idea. 4 is a consequence of 2. Therefore, of Nick's 5 ideas,> only one is needed to understand George Trimble's G-Engine. === Subject: Re: Frequentist probability confusion <405EC787.2070703@univie.ac.at As to the original question about natural numbers. If one requires that> probability measures be countably additive, then there is on probability> measure defined on the integers that I would want to label uniform.> Countably additive means that if {A_n} is a countable collection of> disjoint (measurable) sets, then the probability of the union of the A_n> is equal to the sum of the probabilities of each of the A_n. I certainly> want my probability measures to be countably additive, but some authors> argue that it is enough to be finitely additive. It seems to me that no countable probability space for which eachelement has probability zero can have a countably additive meausure. By a countably additive measure we mean that we have the usefultheorem that a countable union of sets of measure zero has measurezero. In the hypothesized uniform measure over the integers eachsingleton set has measure zero. The integers are a countable union ofsingleton sets. So a countably additive measure must have a measureof zero for the entire space and it fails to be a probability space.> If one allows finitely> additive probability measures, then one can defined a probability measure> over the natural numbers that some might want to label uniform. However,> the construction of this measure uses the the axiom of choice (or at least> some large portion of the axiom of choice).> Daniel Waggoner === Subject: Commutative rings aren't physical spaces (This Week's Finds 205)>We get a little >dictionary for translating between geometry and algebra, like this:> GEOMETRY ALGEBRA> spaces commutative rings> maps homomorphismsThis may be well and good for mathematical space on sci.math, but it'snot how physical space works on sci.physics.* (or shouldn't be).The difference between a direct product and a tensor product is thatthe multiplicands of a direct product don't talk to each: they can'tnoncommunicating parallel worlds each oblivious to the other.Intercomponent communication requires tensor product.At the level of individual points of a physical geometry, it takesnoncommutativity of product to create intercomponent communication.If you believe that the algebra of physical space is commutative thenyour thinking is still in its pre-linear-logic period. A key insightof linear logic (not the only one) is that the additive operations ofdirect product and direct sum (which coincide for vector spaces but notfor algebras and most other mathematical objects) juxtaposenoninteracting universes, whereas tensor product creates universesalong with communication channels between them. (You don't have toknow linear logic to know this -- it's already a basic differencebetween tensor product and direct sum of vector spaces -- but if youknow that then you're in good shape to make a start on linear logic.)A simple example is the quaternions, which depend on noncommutativity(namely ij = -ji) to allow rotation about an arbitrarily orientedaxis. As Hamilton finally realized in the mid-19th century, acommutative geometry has serious problems trying to swing one axisaround to another in any physically meaningful way, regardless ofwhether the angle of swing is circular or hyperbolic (I don't know ifHamilton considered the latter).This is why no Clifford algebra whose underlying vector space is ofdimension greater than two is commutative. If two nonreal axes of anassociative algebra (more precisely their unit vectors) commute, itmeans that they can't communicate with each other. In particular aThe Clifford algebras of dimension at most two are commutative becausethey don't *have* two nonreal axes and may as well be treated asscalars, as they usually are, but shouldn't be when more dimensions arebrought in. The real axis is ironically named since it is the oneaxis that isn't a physical axis, consisting rather of the observablevalues as genuine scalars. Complex numbers aren't real, they are tworeal numbers delivered by two concurrent observers separated by quarterof a wavelength, though a sixth of a wavelength works better for someapplications. (Sorry, starting to ramble there... :)Vaughan PrattDon't contact me at pratt@boole.stanford.edu, substitute cs for boole instead. === Subject: Re: Commutative rings aren't physical spaces (This Week's Finds 205) We get a little >dictionary for translating between geometry and algebra, like this:> GEOMETRY ALGEBRA> spaces commutative rings> maps homomorphisms> This may be well and good for mathematical space on sci.math, but it's> not how physical space works on sci.physics.* (or shouldn't be).> The difference between a direct product and a tensor product is that> the multiplicands of a direct product don't talk to each: they can't> noncommunicating parallel worlds each oblivious to the other.> Intercomponent communication requires tensor product.> At the level of individual points of a physical geometry, it takes> noncommutativity of product to create intercomponent communication.> If you believe that the algebra of physical space is commutative then> your thinking is still in its pre-linear-logic period. A key insight> of linear logic (not the only one) is that the additive operations of> direct product and direct sum (which coincide for vector spaces but not> for algebras and most other mathematical objects) juxtapose> noninteracting universes, whereas tensor product creates universes> along with communication channels between them. (You don't have to> know linear logic to know this -- it's already a basic difference> between tensor product and direct sum of vector spaces -- but if you> know that then you're in good shape to make a start on linear logic.)> A simple example is the quaternions, which depend on noncommutativity> (namely ij = -ji) to allow rotation about an arbitrarily oriented> axis. As Hamilton finally realized in the mid-19th century, a> commutative geometry has serious problems trying to swing one axis> around to another in any physically meaningful way, regardless of> whether the angle of swing is circular or hyperbolic (I don't know if> Hamilton considered the latter).> This is why no Clifford algebra whose underlying vector space is of> dimension greater than two is commutative. If two nonreal axes of an> associative algebra (more precisely their unit vectors) commute, it> means that they can't communicate with each other. In particular a> The Clifford algebras of dimension at most two are commutative because> they don't *have* two nonreal axes and may as well be treated as> scalars, as they usually are, but shouldn't be when more dimensions are> brought in. The real axis is ironically named since it is the one> axis that isn't a physical axis, consisting rather of the observable> values as genuine scalars. Complex numbers aren't real, they are two> real numbers delivered by two concurrent observers separated by quarter> of a wavelength, though a sixth of a wavelength works better for some> applications. (Sorry, starting to ramble there... :)> Vaughan PrattThis is a rather presumptuous post! Mr. Pratt presumes to know howthings should be in physics. Everyone knows that the mathematicsused in physics is just a model for describing the real world. Anyonewho says that the real world is or is not some mathematicalstructure doesn't really get it. Take for example supposedly curvedspace as in general relativity. Whitehead showed you could dorelativity in flat space. It's all a device for thinking about thingsand shouldn't be confused with reality. === Subject: a common problemSuppose that there is an Object. For this object, there are N statesS_i(i=1,2...N). For each state, there are events stimulating the state ofthe object to the other state. A simple example is shown below. I need toknow :for a very long time duration, the time the object will stay in each stateS_i(i=1,2...N).What's the related mathematical knowledge? semi-markovian? event e12 event e23 event e34S1 ------------> S2 --------------> S3 -------------> S4| ^|----------------------------------------| event e13ZHANG Yanhttp://www.ntu.edu.sg/home5/pg01308021 === Subject: number of indep. soloutions to diffyqs?I've taken some classes in solving differential equations, and I understandall the different techniques, and everything, but am confused as to how wecan be assured that our techniques yield *all* possible solutions.Especially when we do things like, if x1 is a solution, put x2(x) =x1(x)*f(x) into the differential equation and obtain another solution. Isthere a general way to tell how many linearly independent solutions adifferential equation has, (especially regardless of homogeneity andlinearity)? I understand why these methods work, of course, but can't seeexactly how we can be sure we are obtaining all the solutions. I can seeintuitively a little why it should be the case (especially when usingeigenvalues of a matrix in the matrix form of a diffyq x'=Ax, but would likesomething more rigorous).Jeremy === Subject: Re: number of indep. soloutions to diffyqs?> I've taken some classes in solving differential equations, and I understand> all the different techniques, and everything, but am confused as to how we> can be assured that our techniques yield *all* possible solutions.> Especially when we do things like, if x1 is a solution, put x2(x) => x1(x)*f(x) into the differential equation and obtain another solution. Is> there a general way to tell how many linearly independent solutions a> differential equation has, (especially regardless of homogeneity and> linearity)? I understand why these methods work, of course, but can't see> exactly how we can be sure we are obtaining all the solutions. I can see> intuitively a little why it should be the case (especially when using> eigenvalues of a matrix in the matrix form of a diffyq x'=Ax, but would like> something more rigorous).Check out the Wronskian. Nonlinear differential equations are harder to deal with and may havea variety of solutions. -- Christopher Heckman === Subject: Re: number of indep. soloutions to diffyqs?> I've taken some classes in solving differential equations, and I understand> all the different techniques, and everything, but am confused as to how we> can be assured that our techniques yield *all* possible solutions.> Especially when we do things like, if x1 is a solution, put x2(x) => x1(x)*f(x) into the differential equation and obtain another solution. Is> there a general way to tell how many linearly independent solutions a> differential equation has, (especially regardless of homogeneity and> linearity)? I understand why these methods work, of course, but can't see> exactly how we can be sure we are obtaining all the solutions. I can see> intuitively a little why it should be the case (especially when using> eigenvalues of a matrix in the matrix form of a diffyq x'=Ax, but would like> something more rigorous).Good textbooks (the ones with proofs) include existence anduniqueness theorems. That is what you want here. It is somethingseparate from techniques for finding solutions in closed form.For example...Coddington, An Introduction to Ordinary Differential Equations.Dover reprint, around $10. http://www.math.ohio-state.edu/~edgar/ === Subject: Re: functions that haltIn sci.logic, |-|erc:> oo> ____|mn> / /_/ / _ - Herc, The Unrecognised Truman> / K-9/ /_/ - Join www.chatty.net -> /____/_____ - Nanotechnology is gonna be HUGE... (RMF)> --------------> Like all diagonal arguments, this one ASSUMES the existence of something (in> this case, some computable total F that enumerates all and only the computable> total functions on N->N), and then shows that that leads to a contradiction,> thus demonstrating that the assumption was false (in this case, demonstrating> that there can be no such F).> WRONG!> I'll call elements of F constructable functions. Now tell us where has anyone> defined constructable functions to allow calls to other functions?> G = lambda x (.... F x ....)> Constructable functions do not allow all function calls. Doing so would only> require an additional small set of primitive functions, eg. Succ(), Twice() to make> constructable> functions equivalent to all possible TMs and therefore amenible NOT to halt.> F halts, G calls F, by inspection if F halts then G halts.> THEREFORE G belongs to the set of halting TMs,> NOT necessarily to the set of constructible functions.> Constructible functions are a proper subset of halting functions,> and G is not constructable.> Herc> I for one don't see a problem here. If f() is a primitive recursive> function (I'm assuming this is what you mean by constructable;> I'm also assuming that a primitive recursive function is a function> that is either totally primitive (e.g., f(x) = x + 1) or a> function that is implemented using fixed-limit for loops> (e.g., f(x) = x! can be implemented using the code> int p = 1;> for(i = 1; i <= x; i++) p = p * i;> return p;> )> and g() is a primitive recursive function, then so is their> sum, difference, product, quotient, and composition.> The only thing not allowed is the general mu-loop:> while(!done) { ... }> and function composition doesn't need that.> while (x) {> abc> }> is equivalent to tail recursion :> fun w (x) {> if (x) {> abc> w (x)> }> }> Correct.> So...> The only thing not allowed is the general mu-loop:> while(!done) { ... }> and function composition doesn't need that.> function naming is not allowed either, because it is equivalent> to allowing 'while'.I fail to see how.> Actually while (x) is allowed if it can be 'calculated' that the> loop terminates.> Examine :> i = 1> while (i < 10) {> i++> }Yes, that's a variant (in C, anyway) of for(i=1;i<10;i++) { ... }.Makes life interesting for C programmers on occasion, as for() canalso be used to write mu-loops:for(init(); test(); incr()){}is perfectly valid in C, but gives theoreticians major headaches.> Calculate the loop invariant, this is something that> does not change throughout> the execution of the loop. Here it is i>0.There are several loop invariants, of which 'i>0' is but one.> Then solve simultaneous equations given the ending condition for the loop and> the loop invariant :> i >= 10> i > 0> This has a solution when i = 10, which demonstrates the loop can terminate.> Constructable functions can be defined to remove> general function calls. In practice they might be> tested if they were safe to call, but in theory named> functions are not required for computability at all..> You haven't coded until you do nested lamda definitions on the fly.> Guess I haven't coded then. I'm not familiar enough with LISP> to do that, and in any event my speciality is mostly C++.> In practice humans like to name functions and keep everything modular,> in theory TMs and pure functions are just a single function body.It doesn't matter. One can formalize the problem in variousinteresting ways; the method you're using works fine.What is a name, anyway? It's an association of a string ofcharacters with some object; that object may very well bea lambda-closure. A call to that function would be the sameas calling the lambda-closure.Some languages allow *changing* the object to which a name points.> You might not see a problem but the diagonalisation proves squat,> you assumed the class of halting functions under consideration> perform function calls.> You are correct; the diagonalization of a computable list, as it> turns out, proves absolutely nothing. The reason is that a> computable list is a list of FINITE algorithms [*]. The diagonal> is not a finite algorithm.> clarify finite algorithm?A function/procedure that can be implemented in a Turing machine(with a finite number of states).Obviously, if L is an infinite list of finite algorithms, thediagonal function G cannot be a finite algorithm.> When you rewrite G in terms of evaluating_indexed_functions (F)> and extrapolating those function bodies from their index,> guess what? G won't halt. Eventually G will come> to its own (prospective) index number and representation,> construct its own body and evaluate that. Either G is accepted> because it recognises the infinite recursion and outputs null for> its own diagonal digit, or G will be recognised by the specifications> of constructable functions (part of F) as failing to halt and is not> in the set of constructables.> The diagonal is not computable by constructable functions.> You have yet to prove what limitation the absense of self> referencing paradoxical non functions actually is.> There is no limitation. G simply won't halt.> It will go into an infinite loop 5,4,5,4,5,4... and one can> wait until doomsday, but it won't halt. Eventually a> real machine would crash as it runs out of stack space.> Mea culpa. However, I don't see how this proves that the> set of real numbers is finite and/or denumerable.> Does it disprove Barbs diagnonalistion attack on a theory of> guaranteed halting functions being equivalent in power to the> class of TMs?Yes, as the diagonalization function cannot be placed in a TM.(Sorry Barb, if you're still reading this. I only just realizedthis a day or so ago myself. :-) )> Herc#191, ewill3@earthlink.netIt's still legal to go .sigless. === Subject: Re: functions that halt> In sci.logic, |-|erc> Does it disprove Barbs diagnonalistion attack on a theory of> guaranteed halting functions being equivalent in power to the> class of TMs?> Yes, as the diagonalization function cannot be placed in a TM.Oh, but it can.The point of the guaranteed halting functions (GHF) is that you *can*enumerate them, i.e. you can write a function that gives you the k'thGHF. If you grant a mechanism that allows to to actually call such afunction, or, what amounts to the same thing, emulate it (i.e. parseit and see what it does), you can write the diagonalisation function(DF).The only problem is that you can't guarantee the DF to halt if it is aGHF, because it then calls itself (or a simile of itself) in an infiniterecursion.In other words, there is no contradiction if the diagonalisationfunction is not a GHF.MichaelFeel the stare of my burning hamster and stop smoking! === Subject: Re: functions that halt oo ____|mn / /_/ / _ - Herc, The Unrecognised Truman / K-9/ /_/ - Join www.chatty.net -/____/_____ - Nanotechnology is gonna be HUGE... (RMF)--------------> Does it disprove Barbs diagnonalistion attack on a theory of> guaranteed halting functions being equivalent in power to the> class of TMs?> Yes, as the diagonalization function cannot be placed in a TM.> Oh, but it can.> The point of the guaranteed halting functions (GHF) is that you *can*> enumerate them, i.e. you can write a function that gives you the k'th> GHF.> If you grant a mechanism that allows to to actually call such a> function, or, what amounts to the same thing, emulate it (i.e. parse> it and see what it does), you can write the diagonalisation function> (DF).> The only problem is that you can't guarantee the DF to halt if it is a> GHF, because it then calls itself (or a simile of itself) in an infinite> recursion.> In other words, there is no contradiction if the diagonalisation> function is not a GHF.Right! If a function is proven to halt does that make it a member of GHF?Guarantee can be interpreted either way, guarantees have to be 'given'.PROOF that guarantees have to be given.Assume any program that is proven to halt is in GHF. (1)Assume the standard F(i,j) being the jth output of the ith GHF.Let G(j) = F(j,j) mod 2 + 1Since F halts by inspection G halts (2)(1),(2) -> G is in GHF.But G is not any F(i), CONTRADICTION.Therefore not all programs that halt are in GHF.A systematic Guarantee could include not granting any mechanism tocall a function. This would ensure any encapsulation of G into a formof Godel numbering would be impossible.Herc === Subject: Re: functions that halt> In sci.logic, |-|erc> Does it disprove Barbs diagnonalistion attack on a theory of> guaranteed halting functions being equivalent in power to the> class of TMs?> Yes, as the diagonalization function cannot be placed in a TM.>Oh, but it can.>The point of the guaranteed halting functions (GHF) is that you *can*>enumerate them, i.e. you can write a function that gives you the k'th>GHF.No you can't. That was the whole point of the diagonal proof I posted originally:IF there were some total F such that F(i) = an encoding of the i'th total function on N->N (the i'th GHF in your terms), then any diagonal DF constructed so that DF(j) =/= F(j)(j) is clearly not in the F list -- for each F(i), DF is different from it (in particular, it is guaranteed to be different for input i).>If you grant a mechanism that allows to to actually call such a>function, or, what amounts to the same thing, emulate it (i.e. parse>it and see what it does), you can write the diagonalisation function>(DF).Exactly right. And of course you *can* emulate it (e.g. via a Universal Turing Machine).>The only problem is that you can't guarantee the DF to halt if it is a>GHF,?? By *definition*, a total function (a GHF) is guaranteed to halt on all inputs. So if DF is a total function then it clearly must halt, even when the input is an encoding of itself.>because it then calls itself (or a simile of itself) in an infinite>recursion.No it doesn't.>In other words, there is no contradiction if the diagonalisation>function is not a GHF.That would be true, but BY CONSTRUCTION of DF it *is* a total function, since it's simply a slightly modified version of F, which is ASSUMED (incorrectly it turns out) to be a total function.But you're making progress (unlike, e.g., Herc, who has ideological reasons for refusing to understand diagonal arguments in general).>Michael---------------------------| BBB b Barbara at LivingHistory stop co stop uk| B B aa rrr b || BBB a a r bbb | | B B a a r b b | | BBB aa a r bbb | ----------------------------- === Subject: Re: functions that halt Does it disprove Barbs diagnonalistion attack on a theory of> guaranteed halting functions being equivalent in power to the> class of TMs?> Yes, as the diagonalization function cannot be placed in a TM.Barb misses things, I think she's a 60 year old spinster been lecturing thesame text book for so long anything that doesn't fit she can only gloss over.> But you're making progress (unlike, e.g., Herc, who has ideological reasons> for refusing to understand diagonal arguments in general).and most contributors to the thread all see this ideological objection to your blindapplication of diagonalisation as evident.Herc === Subject: Re: functions that haltIn sci.logic, |-|erc:> oo> ____|mn> / /_/ / _ - Herc, The Unrecognised Truman> / K-9/ /_/ - Join www.chatty.net -> /____/_____ - Nanotechnology is gonna be HUGE... (RMF)> --------------> Does it disprove Barbs diagnonalistion attack on a theory of> guaranteed halting functions being equivalent in power to the> class of TMs?> Yes, as the diagonalization function cannot be placed in a TM.> Barb misses things, I think she's a 60 year old spinster been> lecturing the same text book for so long anything that doesn't> fit she can only gloss over.I'm not sure I care if Barb is a 60+-year-old spinster with liverspots or a hot 18 year old Asian model with huge -- erm, well,in any event, how does her physical attributes affect this argument?(Besides, 60-year-olds are more intelligent than teenagers -- experienceand all that. Me, I'm closer to 60 than 18 myself, although I've morethan two decades before I have to worry about retirement.)> But you're making progress (unlike, e.g., Herc, who has ideological> reasons for refusing to understand diagonal arguments in general).> and most contributors to the thread all see this ideological> objection to your blind application of diagonalisation as evident.I'm still trying to figure out your proof regarding the countabilityof real numbers. I think we've shown that your proof regardingthe countability of computable functions cannot be attacked bythe standard diagonalization argument. (Besides, the set of allusable programs is a proper subset of all integers anyway, by anyreasonable mapping.)> Herc#191, ewill3@earthlink.netIt's still legal to go .sigless. === Subject: Re: functions that haltThe Ghost In The Machine Does it disprove Barbs diagnonalistion attack on a theory of> guaranteed halting functions being equivalent in power to the> class of TMs?> Yes, as the diagonalization function cannot be placed in a TM.> and most contributors to the thread all see this ideological> objection to your blind application of diagonalisation as evident.> I'm still trying to figure out your proof regarding the countability> of real numbers. I think we've shown that your proof regarding> the countability of computable functions cannot be attacked by> the standard diagonalization argument. (Besides, the set of all> usable programs is a proper subset of all integers anyway, by any> reasonable mapping.)There were 3 arguments against countable reals.One was every time I suggested a list of computable numbers peopleused Barb's diagonal attack on countability of *functions* before I couldget started. Now that is dismissed we may have an *actual* list of all computablereals on our hands!Two was that sci.logic insists unlimited sequences are shorter than infinite sequences.I showed powersets of countable sequences map to all reals, whether thosepowersets where uncountable or not. The infinite countable set still *contains*the address of every real, its just not_accessible. Find some word to replace*contains* here, you know that sentence has a truthful meaning, you won'tlet me express it.Therefore computable numbers contain every sequence of digits. Thereforethe diagonal attack on that statement is ALSO wrong. That's 2 out of 3 diagonalattacks down, one to go.Herc === Subject: Re: functions that halt[...]> Two was that sci.logic insists unlimited sequences are shorter than infinite sequences.> I showed powersets of countable sequences map to all reals, whether those> powersets where uncountable or not. The infinite countable set still *contains*> the address of every real, its just not_accessible. Find some word to replace> *contains* here, you know that sentence has a truthful meaning, you won't> let me express it.[...]contains <==> expresses <==> defines It's possible to define sqrt(2) unambiguously as follows: For n>=1, let a_n be the largest positive integer such that [ a_n/(10^n) ]^2 < 2. (a_1 = 14, a_2 = 141, ... ) and sqrt(2) := lim_{n->oo} ( a_n/(10^n) ).The same can be done for Pi: Pi := 4*( 1 - 1/3 + 1/5 - 1/7 + 1/9 - ... + .... ).It's even possible to define some uncomputable numbers, such asChaitin's Omega.Any one of the definitions above could be written down usinga finite number of words and mathematical symbols.Can every real number be defined finitely?Cantor would say no.David Bernier === Subject: Re: functions that halt> The Ghost In The Machine Does it disprove Barbs diagnonalistion attack on a theory of> guaranteed halting functions being equivalent in power to the> class of TMs?> Yes, as the diagonalization function cannot be placed in a TM.> and most contributors to the thread all see this ideological> objection to your blind application of diagonalisation as evident.> I'm still trying to figure out your proof regarding the countability> of real numbers. I think we've shown that your proof regarding> the countability of computable functions cannot be attacked by> the standard diagonalization argument. (Besides, the set of all> usable programs is a proper subset of all integers anyway, by any> reasonable mapping.)> There were 3 arguments against countable reals.> One was every time I suggested a list of computable numbers people> used Barb's diagonal attack on countability of *functions* before I could> get started. Now that is dismissed we may have an *actual* list of all computable> reals on our hands!So what is the next countable real greater than Pi?> Two was that sci.logic insists unlimited sequences are shorter than infinite sequences.> I showed powersets of countable sequences map to all reals, whether those> powersets where uncountable or not. The infinite countable set still *contains*> the address of every real, its just not_accessible. Find some word to replace> *contains* here, you know that sentence has a truthful meaning, you won't> let me express it.The infinite countable set does *not* contain the address of everyreal because there is no way to map them in a 1-1 way. Tell me whatthe 'next' real is in any set of reals. You can't.> Therefore computable numbers contain every sequence of digits. Therefore> the diagonal attack on that statement is ALSO wrong. That's 2 out of 3 diagonal> attacks down, one to go.Keep dreaming. === Subject: Re: functions that haltOriginator: tchow@lagrange.mit.edu.mit.edu (Timothy Chow)>Tell me what the 'next' real is in any set of reals. You can't.I don't understand your argument. Surely it is not:1. In the natural ordering of the reals, there is no next real afterany given real. Therefore the reals are uncountable.That's certainly fallacious, as may be seen by replacing real byrational. So maybe you mean:2. The reals are uncountable, and therefore can't be well ordered(i.e., there's no way to totally order the reals so that after anygiven real there's a next real).That's also fallacious, since the existence of a well-ordered uncountableset can be proved, even without using the axiom of choice.Tim Chow tchow-at-alum-dot-mit-dot-eduThe range of our projectiles---even ... the artillery---however great, willnever exceed four of those miles of which as many thousand separate us fromthe center of the earth. ---Galileo, Dialogues Concerning Two New Sciences === Subject: Z-moduleIf G is a subgroup of Z x Z (Z = set of integers) generated by (12, 0) and(3, 1), what is S = (Z x Z)/G? How can I find the free rank and elementarydivisors of S? I tried to express G as G1 x G2 but I really have no clue.Can anyone help? === Subject: Re: Z-module> If G is a subgroup of Z x Z (Z = set of integers) generated by (12, 0) and> (3, 1), what is S = (Z x Z)/G? How can I find the free rank and elementary> divisors of S? I tried to express G as G1 x G2 but I really have no clue.> Can anyone help?You have an abelian group generated by two elements a = (1,0) and b =(0,1) subject to the relations 3a + b = 0 and 12a = 0. The way todeal with this is to put these relations into a matrix [3,1;12,0]. Multiplying by an invertible matrix on the left is equivalent tochoosing a new set of relations while multiplying by an invertiblematrix on the right is equivalent to choosing a new set of generators(I leave it to the reader to convince him/herself of these facts). Asa consquence of the fact that Z is a PID, the left multiplication isequivalent to carrying out a series of invertible elementary rowoperations and the right multiplication to column operations. However, only invertible operations so the only scalar multiplicationallowed is multiplication by 1 or -1.Now it is obvious in this case, but can be shown fairly easily ingeneral that the matrix can be reduced by a series of elementary rowand column operations to a diagonal matrix, in this case [1,0;0,12]. This means that there is a new generating set, say c and d such that c= 0 and 12d = 0. So the group is just Z/12Z. Of course, this worksfor any finitely generated abelian group. === Subject: Re: Z-module> If G is a subgroup of Z x Z (Z = set of integers) generated by (12, 0) and> (3, 1), what is S = (Z x Z)/G? How can I find the free rank and elementary> divisors of S? I tried to express G as G1 x G2 but I really have no clue.> Can anyone help?Hint: each (x,y) in Z x Z can be written as a.(1,0) + b.(3,1) in asingle way (with a and b in Z).Jose Carlos Santos === Subject: Re: Z-moduleSo each (x, y) has the form (12a + 3b, b)but still there is b in both x and y. Then they are not independent. How canI still write it as G1 x G2?What I have is:y (mod 4) x (mod 12)0 01 32 63 9But how do I proceed? === Subject: Re: Z-module> So each (x, y) has the form (12a + 3b, b)> but still there is b in both x and y. Then they are not independent. How can> I still write it as G1 x G2?Each (x,y) has the form a.(1,0) + b.(3,1). So, Z x Z is the direct sumZ.(1,0) + Z.(3,1) and so (Z x Z)/G is isomorphic to Z/(12 Z) + Z/Z,which is isomorphic to Z_{12}.Jose Carlos Santos === Subject: Re: Z-module Adjunct Assistant Professor at the University of Montana.>So each (x, y) has the form (12a + 3b, b)>but still there is b in both x and y. Then they are not independent. How can>I still write it as G1 x G2?>What I have is:>y (mod 4) x (mod 12)>0 0>1 3>2 6>3 9>But how do I proceed?Each element is expressible UNIQUELY as a*(12,0) + b*(3,1). Forgetabout the fact that they are written as pairs of integers(12a+3b,b). Isn't it enough to know what a and b are in order to saywhat the element is? If so, then isn't this the same as just thepair (a,b)? And conversely, if you know that a given element (x,y) isin the subgroup, then can you not figure out what a and b must be? ============Subject: Re: Z-moduleI doubt why EVERY integer pair can be expressed as a*(12,0) + b*(3,1)....If I have (x,y) = (3,2), then b = 212*a + 3*b = 12*a + 3*2 = 3, then a = -1/4 which is not an integer...Thus (3,2) is not an element in the subgroup. === Subject: Re: Z-module>I doubt why EVERY integer pair can be expressed as a*(12,0) + b*(3,1)....I did not say that every pair of integers can. What I said was thatthe subgroup they generate is ->isomorphic<- to Z x Z.>If I have (x,y) = (3,2), then b = 2>12*a + 3*b = 12*a + 3*2 = 3, then a = -1/4 which is not an integer...>Thus (3,2) is not an element in the subgroup.I did not say so. But if you already KNOW that (x,y) is an element ofthe subgroup, then you ->know<- that b = y and that a = (1/12)(x-3y),which will necessarily be an integer (see below).But, since you put the question. What are the elements which are in<(12,0),(3,1)>?What follows is very long, and most of the time, very unnecessary. Butperhaps instructive.The multiples of (3,1) are exactly the elements (x,y) that satisfyx-3y = 0.Since any element is a multiple of (3,1) plus a multiple of (12,0),then you can readily identify the elements in the subgroup by notingthat x-3y must be a multiple of 12. That is: it consists exactly ofall elements (x,y) such that x-3y = 0 (mod 12).If you first mod out by (12,0), then you have the group (Z/12Z) xZ. To mod out by the subgroup in question, you will now have to modthis subgroup by the element (3,1). Now, in (Z/12Z) x Z, one mutipleof (3,1) is (0,4). So you can actually first mod out by (0,4) to get(Z/12Z)x(Z/4Z). And now you are trying to mod out by the element (3,1)of ->this<- group. This element generates the subgroup whose elementsare (0,0), (3,1), (6,2), and (9,3). What remains after you mod out?How many elements does it have? The group had 48 elements, you aremoding out by a subgroup with 4 elements, so there are 12 elements inthe quotient. Two elements (a,b), (a,c) will be congruent modulo thesubgroup if and only if b = c. An element of the form (a,b), with0<=a<3 will be in the subgroup if and only if a=b=0. So one set ofrepresentatives is given by all the pairs(a,b): 0<=a<3, 0<=b<4.That gives the desired 12 elements. Now the only question is whetherit is cyclic of order 12, or a cyclic group of order 2 times a cyclicgroup of order 6. Is the group of exponent 6? Well, 6*(1,1) = (6,6),which is not in the subgroup, so it is not exponent six. Since everyelement of a cyclic group of order 12 is either a generator or ofexponent 6, it follows that the quotient is generated by (1,1), and iscyclic of order 12. ============Subject: Re: Z-moduleSo this is my original question:If G is a subgroup of Z x Z (Z = set of integers) generated by (12, 0) and(3, 1), what is S = (Z x Z)/G? How can I find the free rank and elementarydivisors of S?Now I have S = Z/12Z. So the free rank is 0 and the elementary divisor is 12= 2^2 * 3?Shouldn't an elementary divisor be a power of a prime? === Subject: Re: Z-module> So this is my original question:> If G is a subgroup of Z x Z (Z = set of integers) generated by (12, 0) and> (3, 1), what is S = (Z x Z)/G? How can I find the free rank and elementary> divisors of S?The real answer to this and similar problems in higher dimensions isfind the Smith normal form. === Subject: Re: Z-module Adjunct Assistant Professor at the University of Montana.>So this is my original question:>If G is a subgroup of Z x Z (Z = set of integers) generated by (12, 0) and>(3, 1), what is S = (Z x Z)/G? How can I find the free rank and elementary>divisors of S?>Now I have S = Z/12Z. It is isomorphic to that, yes.>So the free rank is 0 and the elementary divisor is 12>= 2^2 * 3?>Shouldn't an elementary divisor be a power of a prime?The elementary divisors are the orders of the primary cyclic summandsof the group. If you write Z/12Z as a direct sum of cyclic groups ofPRIME POWER ORDER, then you getZ/12Z = (Z/4Z) + (Z/3Z)so the elementary divisors of S are (2^2,3). ============Subject: Re: Computer based proofs> Perhaps because it never took place... Sorry for that, it may have> been the four color theorem.> IMHO, the belief that the proof of the 4-colour theorem> _depends_ on the computer is wrong.> Computers were certainly used in establishing the proof,> mainly in showing that a certain method was very likely to work.> But the actual proof was expressed in the usual way,> with a large number - several hundred - cases,> each of which could be readily checked by hand.> (In fact I assume they were, as a couple of minor errors were found.)> It doesn't seem to me that the proof of the 4-colour theorem> differs in this from many other results,> eg the classification of finite simple groups,The vast majority of CFSG does not depend on computer calculations,but thereare parts that do. For example, I think I am correct in saying thatthe proof of the existence of some of the sporadic simple groups,including the Lyons group, but not including the Monster and itssubgroups, still depends on computer constructions.Derek Holt.> or results on the packing of spheres in high dimensions> using the Leech lattice. > I'm not sure if there are any results> that have been proved by computer,> where the proof cannot be stated and understood in the usual way.> I suppose there could be a result that had been proved> say for n > 10^10,> and in which the result had been verified for n <= 10^10 by computer,> which it would be impractical to check.> But are there any such results? === Subject: Re: Computer based proofs <0p2r70l98rhonka2ja8b55gadlbmj6294p@no.spam OK. Show that a closed subset of a compact set is compact. Use of a> programmable calculator is allowed. Exactly what kind of answer does> _your_ programmable calculator give here?> Is this a joke?No joke, the student has some proof checking or some prove generatingsoftware. For example the one equation for groups that was found andproven by computer.Cc: tim@birdsnest.maths.tcd.ie === Subject: Re: Computer based proofs> I don't agree.> In the original cyclostyled proof the critical diagrams> were all drawn by hand.> The document was about 100 pages long, as I recall,> and there were about 10 diagrams per page.proof suffices, their 1977 paper contained 50 pages of text anddiagrams, 85 pages filled with almost 2500 additional diagrams, and400 microfiche pages that contained further diagrams and thousands ofindividual verifications of claims made in the 24 lemmas in the mainsection of the text. === Subject: Re: Computer based proofs> I don't agree.> In the original cyclostyled proof the critical diagrams> were all drawn by hand.> The document was about 100 pages long, as I recall,> and there were about 10 diagrams per page.> proof suffices, their 1977 paper contained 50 pages of text and> diagrams, 85 pages filled with almost 2500 additional diagrams, and> 400 microfiche pages that contained further diagrams and thousands of> individual verifications of claims made in the 24 lemmas in the main> section of the text.If there were such microfiche pages I never saw them,and what I saw - which could well have been 135 duplicated pages -claimed to be (and presumably was, modulo a small number of omissions)a complete proof of the 4 colour theorem,If so, the proof could easily have been checked,and therefore falls within the classical notion of a mathematical proof.e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ietel: +353-86-2336090, +353-1-2842366s-mail: School of Mathematics, Trinity College, Dublin 2, IrelandCc: tim@birdsnest.maths.tcd.ie === Subject: Re: Computer based proofs> If there were such microfiche pages I never saw them, Right, but Appel and Haken, as I said, claim that their publishedproof contained all that stuff. === Subject: Re: Computer based proofs> If there were such microfiche pages I never saw them,> Right, but Appel and Haken, as I said, claim that their published> proof contained all that stuff.I didn't understand the remark you quoted to mean that these microficheswere a necessary part of the proof.To repeat myself (for the last time) the 2 volumes of duplicated pages(135 pages in all, according to someone on this group)purported to be, and appeared to be, a complete proof.But surely someone else saw this document, and can confirm or contradict what I say?e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ietel: +353-86-2336090, +353-1-2842366s-mail: School of Mathematics, Trinity College, Dublin 2, IrelandCc: tim@birdsnest.maths.tcd.ie === Subject: Re: Computer based proofs> I didn't understand the remark you quoted to mean that these microfiches> were a necessary part of the proof. It seems a bit odd that they should include in the published versionof the proof stuff that wasn't necessary. === Subject: Re: Computer based proofs> I didn't understand the remark you quoted to mean that these microfiches> were a necessary part of the proof.> It seems a bit odd that they should include in the published version> of the proof stuff that wasn't necessary.I've got the pre-prints (which is what in effect they were)somewhere in my attic, and will see if I can locate them.[This may involve breaking the law, as last time I was up thereI found there were bats - a protected species here - in occupation.]I will give an honest report of what I find.I'm not quite sure what is meant by a published version.Maybe that was a later production?e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ietel: +353-86-2336090, +353-1-2842366s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: Re: Computer based proofs> I didn't understand the remark you quoted to mean that these microfiches> were a necessary part of the proof.> It seems a bit odd that they should include in the published version> of the proof stuff that wasn't necessary.>I've got the pre-prints (which is what in effect they were)>somewhere in my attic, and will see if I can locate them.>[This may involve breaking the law, as last time I was up there>I found there were bats - a protected species here - in occupation.]Goodness. I hope you're not also storing off-prints of Anharmonic polynomial generalizations of the numbers of Bernoulli and Eulerand copies of _Mathematics: Queen and Servant of Science_ up there.I like to think that the Republic, if not indeed the EU, insists that bats be kept in a Bell-free zone. === Subject: Re: Computer based proofs> I don't agree.> In the original cyclostyled proof the critical diagrams> were all drawn by hand.> The document was about 100 pages long, as I recall,> and there were about 10 diagrams per page.> proof suffices, their 1977 paper contained 50 pages of text and> diagrams, 85 pages filled with almost 2500 additional diagrams, and> 400 microfiche pages that contained further diagrams and thousands of> individual verifications of claims made in the 24 lemmas in the main> section of the text.I would then guess that the 400 microfiche pages were computer generated. 400 microfiche pages would be about 500,000 normal pages (at least the last time I saw a microfiche). === Subject: Re: Computer based proofs> I don't agree.> In the original cyclostyled proof the critical diagrams> were all drawn by hand.> The document was about 100 pages long, as I recall,> and there were about 10 diagrams per page.> proof suffices, their 1977 paper contained 50 pages of text and> diagrams, 85 pages filled with almost 2500 additional diagrams, and> 400 microfiche pages that contained further diagrams and thousands of> individual verifications of claims made in the 24 lemmas in the main> section of the text.> I would then guess that the 400 microfiche pages were computer> generated.> 400 microfiche pages would be about 500,000 normal pages (at least the> last time I saw a microfiche).I think you misunderstand.The document I saw consisted of ordinary duplicated material,and the diagrams were all drawn by hand at normal size,and were certainly not computer generated -as is shown by the fact that there were a number of errors or omissions.The quotation above reminded me that there were actually 2 volumes(ie sheaves of stapled pages).My recollection - which could easily be wrong -is that only the first volume was actually required for the proof.In any case, the essential point is that it was well within human capabilityto go through all the cases listed.IMHO, the proof lies well within the classical idea of a mathematical proof,and does not introduce any new principle.I can imagine proofs that do depend intrinsically on computers,but I don't believe the proof of the 4 colour theorem is one of them. e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ietel: +353-86-2336090, +353-1-2842366s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: Re: Computer based proofs> I don't agree.> In the original cyclostyled proof the critical diagrams> were all drawn by hand.> The document was about 100 pages long, as I recall,> and there were about 10 diagrams per page.> proof suffices, their 1977 paper contained 50 pages of text and> diagrams, 85 pages filled with almost 2500 additional diagrams, and> 400 microfiche pages that contained further diagrams and thousands of> individual verifications of claims made in the 24 lemmas in the main> section of the text.> I would then guess that the 400 microfiche pages were computer> generated.> 400 microfiche pages would be about 500,000 normal pages (at least the *******That should have been 50,000 pages. > last time I saw a microfiche).> I think you misunderstand.> The document I saw consisted of ordinary duplicated material,> and the diagrams were all drawn by hand at normal size,> and were certainly not computer generated -> as is shown by the fact that there were a number of errors or omissions.What are 400 microfiche pages? Four hundred pages of ordinary paper, shrunk and copied onto maybe four microfiches, or four hundred microfiches, containing several tenthousand pages? === Subject: Hoelder function x^aHow can you show that the function x -> x^a, defined on [0,1], isa-hoelderian, with 0 How can you show that the function x -> x^a, defined on [0,1], is> a-hoelderian, with 0 0 and consider f(x) = (x+h)^a - x^a on [0,oo). Then f' < 0 on (0,oo). Therefore f(x) = (x+h)^a - x^a <= f(0) = h^a for all x >= 0. === Subject: Re: Hoelder function x^a>How can you show that the function x -> x^a, defined on [0,1], is>a-hoelderian, with 0The case a=1/2 is simple,also a=1, but in general is possible to give a>proof?Suppose that 0 <= x < y <= 1. The fact that the derivative of x^a is_decreasing_ shows that y^a - x^a <= (y-x)^a - 0^a = (y-x)^a.(Write y^a - x^a as the integral of the derivative.) === Subject: Re: Hoelder function x^a> How can you show that the function x -> x^a, defined on [0,1], is> a-hoelderian, with 0 The case a=1/2 is simple,also a=1, but in general is possible to give a> proof?compute the derivative, and apply the mean value theorem === Subject: Re: Hoelder function x^a> compute the derivative, and apply the mean value theoremSo I'll have: x^a-y^a = (a z^(a-1)) (x-y), where y=0. === Subject: Re: y=tan(x)-x => x=?> y=tan(x)-x => x=? (in terms of y ...)> if you have any idea x3j11@softhome.netI deviced something interesting for you just now, but I am not sure if such a function is in existence already, and if so, goes by what name.(googling showed nothing similar). Obviously analogy is withcontinued fractions for irrationals, since a closed form is notpossible here.Temporarily, I call this RMF or RecursiveMultipleFunction. If infinite recursion is admitted, then answer is :x[t_]=ArcTan[t+ArcTan[t+ArcTan[t+ArcTan[t+ .. ]]]] ;Upto 4 parantheses depth, the plot looks OK.x[t_]=ArcTan[t+ArcTan[t+ArcTan[t+ArcTan[t]]]] ;Plot[x[y],{y,10,-10}] ;A DO loop can be written for accuracy and convergence for adequatenumerical accuracy with paranthesis depth.A wide range of hitherto implicit inverse functions can be handled in this way, incorporating two, three or more functional operators, especially with a linear additive term.Playing with it would be educative,hopefully fun and exciting too.Not sure if there would be all around agreement on this.But HTH.. :) === Subject: Re: y=tan(x)-x => x=?> y=tan(x)-x => x=? (in terms of y ...)> if you have any idea x3j11@softhome.net>I deviced something interesting for you just now, but I am not sure >if such a function is in existence already, and if so, goes by >what name.(googling showed nothing similar). Obviously analogy is with>continued fractions for irrationals, since a closed form is not>possible here.>Temporarily, I call this RMF or RecursiveMultipleFunction. >If infinite recursion is admitted, then answer is :>x[t_]=ArcTan[t+ArcTan[t+ArcTan[t+ArcTan[t+ .. ]]]] ;>Upto 4 parantheses depth, the plot looks OK.>x[t_]=ArcTan[t+ArcTan[t+ArcTan[t+ArcTan[t]]]] ;>Plot[x[y],{y,10,-10}] ;>A DO loop can be written for accuracy and convergence for adequate>numerical accuracy with paranthesis depth.>A wide range of hitherto implicit inverse functions can be >handled in this way, incorporating two, three or more >functional operators, especially with a linear additive term.>Playing with it would be educative,hopefully fun and exciting too.>Not sure if there would be all around agreement on this.>But HTH.. :)Actually, that is nothing new. You're just iterating a functioncontaining a parameter to get a fixed point. That's just the implicitfunction theorem or Banach's fixed point theorem in disguise:y = tan(x)-x yields x = arctan(x+y). So for every y you want to find afixed point of the function f: x -> arctan(y+x). Under certainconditions (|f'| <1 at the fixed point x(y)) the iteration convergesto x(y), if the initial value x0 is close enough to x(y).In your example you picked x0 = 0: 0 -> arctan(y+0) -arctan(y+arctan(y)) -> arctan(y+arctan(y+arctan(y)))... -> x(y).Since |f'(x)| < 1 for all x, it is clear, that this sequenceconverges. === Subject: Re: y=tan(x)-x => x=?>y=tan(x)-x => x=? (in terms of y ...)> Not an elementary function.Of course there are series solutions such as > x = v - (2/15) v^3 + (3/175) v^5 - (2/1575) v^7 - (16/202125) v^9 + ...> where v = (3 y)^(1/3). You can also use Newton's Method and the like.BTW, to get all solutions in y = ArcTan[x] (simpler case if other x isabsent), we add 2 k Pi, any integer k. For this case, due to x-shiftand other branches, when not added that way, is there a way to getall solutions? Newton's method may capture inverse function only from === Subject: Re: y=tan(x)-x => x=?>y=tan(x)-x => x=? (in terms of y ...)> Not an elementary function.Of course there are series solutions such> as x = v - (2/15) v^3 + (3/175) v^5 - (2/1575) v^7 - (16/202125) v^9 +> ... where v = (3 y)^(1/3). You can also use Newton's Method and the> like.> BTW, to get all solutions in y = ArcTan[x] (simpler case if other x is> absent), we add 2 k Pi, any integer k.Rather, we add just k*pi.> For this case, due to x-shift and other branches, when not added that> way, is there a way to get all solutions?Yes. All solutions are given by x = k*pi + v - 2/15*v^3 + 3/175*v^5 - 2/1575*v^7 - 16/202125*v^9 + ...where v = (3*(y + k*pi))^(1/3).The trouble with this method is that the series does not seem toconverge rapidly unless v is small.David Cantrell === Subject: Re: y=tan(x)-x => x=?>y=tan(x)-x => x=? (in terms of y ...)> Not an elementary function.Of course there are series solutions such> as x = v - (2/15) v^3 + (3/175) v^5 - (2/1575) v^7 - (16/202125) v^9 +> ... where v = (3 y)^(1/3). You can also use Newton's Method and the> like.> For this case, due to x-shift and other branches, when not added that> way, is there a way to get all solutions?>Yes. All solutions are given by> x = k*pi + v - 2/15*v^3 + 3/175*v^5 - 2/1575*v^7 - 16/202125*v^9 + ...>where v = (3*(y + k*pi))^(1/3). >The trouble with this method is that the series does not seem to>converge rapidly unless v is small.I wouldn't expect it to converge at all unless |v| < (3 pi)^(1/3).Note that if f(x) = tan(x)-x, f'(x) = tan^2(x) = 0 when x = Pi,and f(Pi) = -Pi. So an inverse for f should have a singularityat y = -Pi. You can't have a power series converging past a singularity. === Subject: Re: y=tan(x)-x => x=?>y=tan(x)-x => x=? (in terms of y ...)> Not an elementary function.Of course there are series solutions> such as x = v - (2/15) v^3 + (3/175) v^5 - (2/1575) v^7 -> (16/202125) v^9 + ... where v = (3 y)^(1/3). You can also use> Newton's Method and the like.> For this case, due to x-shift and other branches, when not added that> way, is there a way to get all solutions?>Yes. All solutions are given by> x = k*pi + v - 2/15*v^3 + 3/175*v^5 - 2/1575*v^7 - 16/202125*v^9 + ...>where v = (3*(y + k*pi))^(1/3).As RI has pointed out below, the series above has a small radius ofconvergence. Nonetheless, the answer to GLN's question is still Yes.I should have said:Let F(x) = tan(x) - x, -pi/2 < x < pi/2, and let G be its inverse function.Then all the solutions of y = tan(x) - x are given by x = k*pi + G(y + k*pi), k integer.>The trouble with this method is that the series does not seem to>converge rapidly unless v is small.> I wouldn't expect it to converge at all unless |v| < (3 pi)^(1/3).So my comment above was, at best, a gross understatement of the trouble.> Note that if f(x) = tan(x)-x, f'(x) = tan^2(x) = 0 when x = Pi,> and f(Pi) = -Pi.Hmm. I had thought, in getting the series, that you had been thinking of afunction with domain (-pi/2, pi/2), in other words, my F(x) above. Thenx = pi would not be pertinent.> So an inverse for f should have a singularity at y = -Pi.??? G(y) has no singularity at y = -pi.Anyway, let me now mention a nice way of approximating G:Begin with h(x) = ( x + (pi^2 - 12)/(3*pi^2)*x^3 )/( 1 - (2/pi*x)^2 ),which is a rational approximation for tan(x) on (-pi/2, pi/2). We may thensolve h(x) - x = y algebraically (using cube roots, etc.) However, I thinkthe solution looks somewhat nicer when presented as x = 4*y/pi^2 * ( 2*cos(1/3*Arccos( 3*pi^6/(128*y^2) - 1 )) - 1 )The right side of the above approximates G(y) so that, at its worst,|relative error| is about 0.0014 (occurring when |y| is about 1.2). Andrelative error approaches 0 both as y approaches 0 and as |y| increaseswithout bound.WithapproxG(y) = 4*y/pi^2 * ( 2*cos(1/3*Arccos( 3*pi^6/(128*y^2) - 1 )) - 1 ),we may then approximate all solutions of tan(x) - x = y by x = k*pi + approxG(y + k*pi), k integer.Perhaps it might be nice to conclude with an example:Approximate the solution of tan(x) - x = 20 which liesbetween 5*pi/2 and 7*pi/2.The desired solution is then approximately x = 3*pi + approxG(20 + 3*pi) = 10.96286 .For comparison, the precise solution is 10.963289...David Cantrell === Subject: Re: y=tan(x)-x => x=?your equation involves one transcendental function.ie see http://www.omatrix.com/manual/transcendental.htmyou cant solve x in terms of y in closed form.there are other things you can do, such as series solutions,or some approximations.> y=tan(x)-x => x=? (in terms of y ...)> if you have any idea x3j11@softhome.net === Subject: Simple combinatorial principle (quadrants)I had to use the following lemma yesterday. It must surelybe well known; I wonder whether anyone can tell me more.(Does it have a name?)Suppose we have a country divided into four quadrants | NW | NE | --------------+-------------- | SW | SE |and we know: - the North is inhabited - the South is inhabited - the East is inhabited - the West is inhabitedthen either the NW and SE are both inhabited, or the NE and SW are both inhabited.(Up to symmetry there are only six possible patternsof habitation, so it's easy to verify this claim.)More precisely I used the following corollarySuppose I have two sets X and Y, and an equivalence relation ~on each set. I also have a set S of pairs (x, y); i.e. S is asubset of the cartesian product X x Y.The following are equivalent:1. There are (x, y) and (x', y') in S such that not (x ~ x') and not (y ~ y').2. There are (x, y) and (x', y') in S such that not (x ~ x'); & there are (v, w) and (v', w') in S such that not (w ~ w').Of course (1) => (2) immediately; to prove the converse youcan use the quadrant principle above.RobinTo email me, change spam to c s dot man === Subject: Re: Simple combinatorial principle (quadrants)> I had to use the following lemma yesterday. It must surely> be well known; I wonder whether anyone can tell me more.> (Does it have a name?)> Suppose we have a country divided into four quadrants> |> NW | NE> |> --------------+--------------> |> SW | SE> |> and we know:> - the North is inhabited> - the South is inhabited> - the East is inhabited> - the West is inhabited> then either the NW and SE are both inhabited,> or the NE and SW are both inhabited.Here's a proof:Suppose the premiss. If all four quadrants are inhabited thenthe conclusion follows. If not, there is an uninhabited quadrant:say the NW. Since the North is inhabited, the NE must be, andsince the West is inhabited the SW must be.Has nobody ever come across this before? Seems unlikely.Maybe *because* it's so obvious it's never been remarked on.> Suppose I have two sets X and Y, and an equivalence relation ~> on each set. I also have a set S of pairs (x, y); i.e. S is a> subset of the cartesian product X x Y.> The following are equivalent:> 1. There are (x, y) and (x', y') in S such that not (x ~ x')> and not (y ~ y').> 2. There are (x, y) and (x', y') in S such that not (x ~ x');> & there are (v, w) and (v', w') in S such that not (w ~ w').> Of course (1) => (2) immediately; to prove the converse you> can use the quadrant principle above.Suppose we have (x, y), (x', y'), (v, w), (v', w') as in (2).What you do is to partition S into four quadrants; the element(m, n) is placed according to the following: | x ~ m | not(x ~ m) w ~ n | w ~ n --------------+-------------- x ~ m | not(x ~ m) not(w ~ n) | not(w ~ n) |We know the North is inhabited by (v, w), the South by (v', w'),the West by (x, y) and the East by (x', y'). So there must betwo diagonally-opposite pairs, say (a, b) and (a', b'). Sincethey are in different columns we know not(a ~ a'), and sincethey are in different rows we know not(b ~ b').If X and Y are the reals, this tells us:If we have a set of points on the plane such that the linecontaining any two such points is either horizontal or vertical,then all of the points lie on the *same* horizontal or verticalline.RobinEmail: change spam to c s dot man === Subject: Re: Simple combinatorial principle (quadrants)> Suppose we have a country divided into four quadrants> |> NW | NE> --------------+--------------> SW | SE> |> and we know:> - the North is inhabited> - the South is inhabited> - the East is inhabited> - the West is inhabited(ne + nw)(se + sw)(se + ne)(sw + nw)= (se + sw ne) (nw + ne sw)= ne sw + se nw> then either the NW and SE are both inhabited,> or the NE and SW are both inhabited.> (Up to symmetry there are only six possible patterns> of habitation, so it's easy to verify this claim.)7 patternsne sw; se nwnw sw sene sw sene nw sene nw swne nw sw se> More precisely I used the following corollary> Suppose I have two sets X and Y, and an equivalence relation ~> on each set. I also have a set S of pairs (x, y); i.e. S is a> subset of the cartesian product X x Y.> The following are equivalent:> 1. There are (x, y) and (x', y') in S such that not (x ~ x')> and not (y ~ y').> 2. There are (x, y) and (x', y') in S such that not (x ~ x');> & there are (v, w) and (v', w') in S such that not (w ~ w').> Of course (1) => (2) immediately; to prove the converse you> can use the quadrant principle above. === Subject: Yikes! I've Broken Incommensurability!While Working with the other stuff that I'veposted recently, I Totally-Busted Incommens-urability.And it's Easy to see.Given a Circle, having =any= radius, constructa right-Triangle,1. having the Circle's radius as it's hypotenuse,call it C,2. one side as a semi-chord, with one end shar-ing a point with the circle-end of the hypotenuse,call it A,3. and the other side drawn from the center ofthe Circle, to meet the semi-chord, forming aright-angle, and sharing a point with the 'dangl-ing'-end of the semi-chord, call it B.4. The Pythagorean Theorem applies, andA^2 + B^2 - C^2 = 0,5. and is =ALWAYS= Equals Zero.6. As one varies the angle that the hypotenusemake with the X axis, A and B adjust theirlengths continuously, andA^2 + B^2 - C^2 = 0, always,7. even when A or B becomes Equal toZero,8. at which point, either the semi-chord, A,or the center-pinned line, B, becomes Equalto the hypotenuse, C.9. This seems, to me, to be quite astounding,because Incommensurability is nowhere tobe seen.10. Since the radius can be of =any= length,this Circle operation scales from zero toInfinity, which covers =any= plane, and doesso, Continuously, without a hint of Incom-mensurability.11. It does this be-cause when A and B go-Irrational, they balence each other =Exactly=!12. Which means that Irrational Numbersare no longer a Problem - just construcea Circle with the necessary radious, and, voil.88,there's your Conjugated-pair of Irrationals,ready to do anything that you want to do withthem.13. Of course, you cannot measure theseNumbers, but you can address them, Exactly,through the Circle-Geometry, given above. Youknow, pick your favorite Greek symbol, andits, Bye-bye, 'Irrationals'.Q. E. D.K. P. Collins === Subject: Constucting a better set of generators for this subgrop of Z^2 by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i3FCeOk31874;I am really lost about this question.Does anyone know how to find the free rank and and elementary divisorsof G?H is a subgroup of Z^2 generated by (3 5) and (6 14).G = Z^2/ Hthere is a hint given: try to construct a better set of generators for H. === Subject: Re: Constucting a better set of generators for this subgrop of Z^2 Adjunct Assistant Professor at the University of Montana.>I am really lost about this question.>Does anyone know how to find the free rank and and elementary divisors>of G?>H is a subgroup of Z^2 generated by (3 5) and (6 14).>G = Z^2/ H>there is a hint given: try to construct a better set of generators for H.Indeed; think about it: assume you have the case of Z x Z, and His a subgroup of the form H = < (a,0), (x,y) >, with 0<= x < a. Inthis case, can you solve the problem?Now take H = < (3,5), (6,14) >. Instead of (3,5) and (6,14), you coulduse (3,5) and (6,14)+k(3,5) for any integer k, and still generate thesame subgroup. So pick a k that makes things easier: e.g., set k=-2;then we have that H is also generated by(3,5) and (0,4).Now, instead of (3,5) and (0,4), you could replace them with a set ofthe form (3,5)+k(0,4), (3,5), for some integer k. Pick a good one, forexample, k=-1.So now we have that H is also generated by (3,1) and (0,4).Is it now easier to see the answers for Z^2/H? ============Subject: Re: Analysis Texts suitable as Preparation for Royden's Real Analysis by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i3FCeL831750;>Next year I plan on taking a year-long course that uses Royden's Real Analysis.> My problem is that I don't have much of at background in analysis. Related to>that, I don't know what will be expected of me in terms of prerequisite>knowledge for that textbook. The upside is that I learn pretty well on my>own. I plan on studying from now until Fall quarter starts.>So I was wondering what texts would suitably prepare me for that type of class.> Ideally there would be something in the way of solutions. I've looked at>Rudin's Principles of mathematical analysis but didn't really care for his>style.>I've heard that Spivak's Calculus is more like an intro to mathematical>analysis but I'm not sure whether or not that would be good enough as an>introduction. I know that there is a solution's manual, so for studying on my>own, I like the idea of having that. >My library also has a book called Analysis with an Introduction to Proof by>Steven Lay. Here are the table of contents: >http://www.amazon.com/exec/obidos/tg/detail/ -/0130898791/ref=pm_dp_ln_b_2/103-3178448-9129447?v=glance &s=books&vi=contents>I've seen other books at the library but I'm not really sure WHAT I need to>RandyTry Bruckner,Bruckner,ThomsonThey have a few books but look for the one published after1999.Also try anything that says introduction to analysis === Subject: Re: PATTERNS IN FINGERPRINTS, CACTI PREDICTED <407C3391.BE1EA37E@hate.spam.netIn message <407C3391.BE1EA37E@hate.spam.net>, Uncle Al >1.1 billion (with a b) East Indians, 1 million (with an m) flush>toilets.Never let the facts get in the way of a good rant...http://www.censusindia.netRichard Herring === Subject: Re: PATTERNS IN FINGERPRINTS, CACTI PREDICTED <407C3391.BE1EA37E@hate.spam.net> In message <407C3391.BE1EA37E@hate.spam.net>, Uncle Al>1.1 billion (with a b) East Indians, 1 million (with an m) flush>toilets.> Never let the facts get in the way of a good rant...> http://www.censusindia.nethttp://www.censusindia.net/ 2001housing/S00-017.htmlNo latrine:122,078,136 households 63.6% of all householdsClosed drainage23,925,761 12.5% of all households1.1 billion East Indians nominal, therefore at least 87.5% or962,500,000 East Indians directly discharge their feces into thestreets (when there are streets) each day. Conservatively assigningone pound of feces/person-day (cholera and rampant diarrhea willspiral that upwards) gives a total annual storm of 1.6x10^11kilograms of human covering India each year. Add cows tingin the streets and pray for a bountiful monsoon to flush it all....well, not away but perhaps downstream.BTW, 400 million Eat Indian cows and buffloes drop meadow patties eachday. One sixth of the world's cows and one half of the world'sbuffaloes live in India. Say two kilos/bovine-day conservatively. Now the total storm is 4.5x10^11 kilograms of raw covering India each year. How much is that? Look up the*habitable* area of India and divide. On the average, there isn't anyplace you can scrape your shoe.Hindus have 300 million gods (30 crores of gods). Each god is thenapportioned 1.5 metric tonnes of /year. If religion were worthanything at all, one would expect at least one of those omnisicent,omnipresent, omnibenevolent deities would have divine shut of it. Uncle Alhttp://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals)Quis custodiet ipsos custodes? The Net! === Subject: Convergence in distribution and convergence of expectationsIf X_n -> X in distribution, then for all bounded and continious functions fholdsE f(X_n) -> E f(X)I have a sequence of random variables X_n that converge to the Poissondistribution (by Poisson's Limit Theorem). Are there conditions that allowto conclude E X_n -> E Poiss(a) ? All expectations are finite.Boris Hollas === Subject: Graph Theory Question againI need to rephrase my earlier question because I didn't state it correctly.Anyone have any ideas?Say I have a set of n vertices in the plane, (let me call them 1,2,3,...,n),and I connect the vertices like a polygon (i.e. each vertex is connected toonly 2 vertices by edges and there are no two edges that cross each other.).You could have a crazy assortment of vertices and edges that makes the shapelook very weird, but that's ok, we don't need any regularity here. (I hopeI have not said anything that can cause a mix-up here. I am just assumingwe have n points in the plane where the edges form an n-gon that need not beregular, in fact may look very strange, with each edge length beingdifferent and all angles being different)Now suppose that I place one vertex outside this polygon. Can I connectevery single vertex of the polygon to this vertex with an edge such thatnone of these edges cross each other? These edges do NOT have to bestraight lines, they can be curves (that don't overlap each other and alsomust be in the plane). This seems to be true, but how wouldI prove it? Also, what if the vertex was inside the polygon?Moshe === Subject: Re: Graph Theory Question again Now suppose that I place one vertex outside this polygon. Can I connect> every single vertex of the polygon to this vertex with an edge such that> none of these edges cross each other? These edges do NOT have to be> straight lines, they can be curves (that don't overlap each other and also> must be in the plane). This seems to be true, but how would> I prove it? Also, what if the vertex was inside the polygon?You can prove it with Euler's rule. === Subject: Re: Graph Theory Question again> Say I have a set of n vertices in the plane, (let me call them 1,2,3,...,n),> and I connect the vertices like a polygon.> Now suppose that I place one vertex outside this polygon. Can I connect> every single vertex of the polygon to this vertex with an edge such that> none of these edges cross each other? > This seems to be true, but how would I prove it? Also, what if the> vertex was inside the polygon?I don't know nuthin' 'bout graph theory, but here's a proof that what youdescribe is possible. *Before* you add any edges, it's clear that you canconnect any part of the boundary of the polygon to the outlier P (orinlier as the case may be). This is because the only transition that isblocked is between the inside and outside of the polygon. So any twopoints on the outside; any two points on the inside; or any point on theboundary plus any other point can *always* be connected by a new edge,without edges crossing or leaving the plane.Now, you have to show that you can add new edges, from the vertices v_i toP, such that no other v_i will ever blocked from being able to beconnected to P.For a new edge to stop you being able to connect some other v_i to P, itwould have to divide the plane into two (like the boundary of the polygondoes: inside and outside), with some v_i on one side, and P on the otherside. But adding lines that end at P will *never* put P on the wrongside of anything else, because P is an endpoint of the new edge, so noton any side of the new edge at all.Jeremy === Subject: Re: Graph Theory Question again> straight lines, they can be curves (that don't overlap each other and also> must be in the plane). This seems to be true, but how would> I prove it? Also, what if the vertex was inside the polygon?Take he outlier and the two points whose one and only connection is closest to the outlier. Replace this connection with a zig to the outlier and a zag back to the second point. It can cross any other existing side since the side you replaced by zig and zag was the closest.Bob Kolker === Subject: Re: Graph Theory Question again> straight lines, they can be curves (that don't overlap each other andalso> must be in the plane). This seems to be true, but how would> I prove it? Also, what if the vertex was inside the polygon?> Take he outlier and the two points whose one and only connection is> closest to the outlier. Replace this connection with a zig to the> outlier and a zag back to the second point. It can cross any other> existing side since the side you replaced by zig and zag was the closest.> Bob KolkerI'm not understanding your response. I must have totally screwed up myquestion.The polygon I create keeps its edges and vertices. You're not allowed tobreak the edgesof the polygon (that might not be regular). You only create new edges fromthe vertices of the polygon to the outlier point.The edges you newly create can be curves, but must be in the plane. What'sa zig and what's a zag? We must also connect every vertex ofthe polygon to this outlier point so that none of these connectionsintersect. === Subject: Re: Mass versus WeightIf your posts are intended to assist everyone else in becoming as smart asyou are, why not establish a website and invite anyone who is interestedto visit it? That way sci.math won't suffer from the endless off-topicclutter you post.There are two things you must never attempt to prove: the unprovable --and the obvious.Democracy: The triumph of popularity over principle.http://www.crbond.com === Subject: Re: Mass versus Weight> - Mass versus Weight -> UP we stand; DOWN we fall(;^) 4/4/'04> Mass and weight are easy to confuse, 1) Dumb Donny Head STILL does not know the differences amonginertial, gravitational, active, and passive masses. 2) Dumb Donny Head is a spewing idiot who cannot be educated inhigh school physics. 3) Hey Dumb Donny Head, F=ma and F=GmM/r^2.Uncle Alhttp://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals)Quis custodiet ipsos custodes? The Net! === Subject: Re: Mass versus WeightCut F=ma and F=GmM/r^2 are both wrong. Don't you know anything about>parentheses?Learn something about the order of arithmatic operations before youassume that parentheses are required. In the above examples, they arenot.SquishuasqSuPiAsMhua@hotmail.com(remove the capital letters from my e-mail address to contact me) === Subject: Re: Mass versus Weight> F=ma and F=GmM/r^2 are both wrong. Don't you know anything about>parentheses?>Maybe somebody from sci.math will have to explain it to youappear to be. People do slip of every once in a while, forgettingthis basic fact which must guide any attempts at discussion with you.Gene Nygaardhttp://ourworld.compuserve.com/homepages/Gene_Nygaard/ Gentlemen of the jury, Chicolini here may look like an idiot, and sound like an idiot, but don't let that fool you: He really is an idiot. Groucho Marx === Subject: Re: Mass versus Weight> Mass and weight are easy to confuse... [says Shead]Sigh! Inertia http://scienceworld.wolfram.com/physics/Inertia.htmlWeight http://scienceworld.wolfram.com/physics/Weight.htmlMass http://scienceworld.wolfram.com/physics/Mass.html The time has come, the Walrus said, To talk of many things: Of shoes, and ships, and sealing wax, of cabbages and kings. And why the sea is boiling hot and whether pigs have wings. Makes about as much sense as Shead eternal struggle with inertia, weight and mass. === Subject: Question about Totar orderCorrection and new questions: 2.- Talking about TOTAL order sets, There existe a TOTAL order set A withan injection F:R--->A of the reals to A, preserving orders and sucht thatF(R) is dense in A (with the topology induced by de order) but suchthat |R| < |A| but not = ?3. If the set in 2. existe... can we construct a non finit secuencesof sets s.t. |R| < |A1| < ... and always F(A1) dense in Ai+1 ? dose ithave a maximum?and what about if |R| < |A1| < ... and F(R) dense in Ai for all i? === Subject: Re: Question about Totar order> Correction and new questions: > 2.- Talking about TOTAL order sets, There existe a TOTAL order set A with> an injection F:R--->A of the reals to A, preserving orders and sucht that> F(R) is dense in A (with the topology induced by de order) but such> that |R| < |A| but not = ?The surreal numbers 3. If the set in 2. existe... can we construct a non finit secuences> of sets s.t. |R| < |A1| < ... and always F(A1) dense in Ai+1 ? dose it> have a maximum?> and what about if |R| < |A1| < ... and F(R) dense in Ai for all i?Dense in what topology? My understanding is that every ordered field can berealized as a subfield of the surreals. Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. Correction and new questions:> 2.- Talking about TOTAL order sets, There existe a TOTAL order set A> with an injection F:R--->A of the reals to A, preserving orders and> sucht that F(R) is dense in A (with the topology induced by de order)> but such that |R| < |A| but not = ?No: identify R with F(R), let x in A, not in R, then x induce a (improper)cut in R of the form (-oo, a), [a,+oo) , or of the form (-oo, a], (a,+oo).In the first case, say, x is < a and greater than all reals 3. If the set in 2. existe... can we construct a non finit secuences> of sets s.t. |R| < |A1| < ... and always F(A1) dense in Ai+1 ? dose> it have a maximum?> and what about if |R| < |A1| < ... and F(R) dense in Ai for all i? === Subject: Re: Antidiagonal, Infinity: The Equivalency Function maps to a range of [0,1], you can simply: multiply it by a positive real constant a to get a range [0,a]. Yet,: in order to get an infinite interval for the range, multiplying by a: scalar infinity would lead to the function reducing to f(x)=x: defined on the naturals.: EF(n)= n/oo, range [0,1]: aEF(n)= an/oo, range [0,a]: ooEF(n)=n, range Nn/oo ?For what integer value of n does n/oo = .5 ? How manydigits does the binary representation of the integer n have?What is the smallest value of n such that EF(n) >=.1 ?Stephen === Subject: Re: Antidiagonal, Infinity> Consider what happens if you apply the unmodified diagonal argument to> the following binary list (all numbers are base 2):> 0.0000000...> 0.1111111...> 0.0111111...> 0.0011111...> 0.0001111...> .> .> .> If you simply go down the diagonal and change each 1 to a 0 and each 0 to> a 1, the number you get is 0.100000..., and sure enough, it differs by at> least one digit from each number in the original list.> But that's not sufficient, because 0.1000000... is the same real number> as 0.01111111..., which is the third number in the list. > As has been explained, this problem is easily fixed. So this is the standard you forgot to list the irrational numbersflaw? I don't see any numbers with an infinite number of 0's and 1'slike 0.010101010101010101..., 0.101010101..., or any of the numberswith increasingly many zeros between ones, like0.101001000100001000001.... and so on, or other variations of the samenumbers. Partial list of the rationals of course is countable, so weshouldn't find the failure of the argument suprising, and of coursetaking the time to explain it to me. :) === Subject: Re: Antidiagonal, Infinity> Consider what happens if you apply the unmodified diagonal argument to> the following binary list (all numbers are base 2):> 0.0000000...> 0.1111111...> 0.0111111...> 0.0011111...> 0.0001111...> .> .> .> If you simply go down the diagonal and change each 1 to a 0 and each 0 to> a 1, the number you get is 0.100000..., and sure enough, it differs by at> least one digit from each number in the original list.> But that's not sufficient, because 0.1000000... is the same real number> as 0.01111111..., which is the third number in the list. > As has been explained, this problem is easily fixed. > So this is the standard you forgot to list the irrational numbers> flaw? I don't see any numbers with an infinite number of 0's and 1's> like 0.010101010101010101..., 0.101010101..., or any of the numbers> with increasingly many zeros between ones, like> 0.101001000100001000001.... and so on, or other variations of the same> numbers. I don't see what the presence or absence of irrational numbers in thelist has to do with anything. It's easy to give a similar counterexamplethat contains irrationals. The whole point is that the number producedby the diagonal argument may be a number already in the list, regardlessof which other numbers may be missing from the list.>Partial list of the rationals of course is countable, so we> shouldn't find the failure of the argument suprising, and of course> taking the time to explain it to me. :)The failure of what argument? The diagonal argument applied to decimalnumbers doesn't fail, even though the rationals are still countable. Howdoes countability of the rationals make anything surprising orunsurprising in either case?I am not convinced by your statement that the binary case is stillsufficient regardless of the base, given that I have demonstrated that amodification of the argument is necessary in the binary case. Do youunderstand what sort of modification is required, and why?Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. I don't see what the presence or absence of irrational numbers in the> list has to do with anything. It's easy to give a similar counterexample> that contains irrationals. The whole point is that the number produced> by the diagonal argument may be a number already in the list, regardless> of which other numbers may be missing from the list.Well, consider the number 0.101010101.... for example, it's a rationalnumber, but isn't listed. How can I say it isn't listed? Well, theorder the elements are put in require a number to follow it, with aspecific digit in the next position. There exists a number whichrequires each intesection of that type, namely the number with all 0sbut only the k-th digit as a 1. Now each position on the diagonal isfull, but what about every other number? The nature of this orderingexcludes all the irrational number, and excludes all the rationals ofrepeating decimals! The process of constructing an ordering of thattype misses alot of numbers, namely the number which contain both aninfinite 0 digits and 1 digits. If you're going to argue the realnumbers are uncountable, you have to include all the real numbers.> The failure of what argument? The diagonal argument applied to decimal> numbers doesn't fail, even though the rationals are still countable. How> does countability of the rationals make anything surprising or> unsurprising in either case?The failure of his ordering. By constraining himself to a subset ofthe reals, he was about to show how the anti-digonal could produce anumber already on the list, I'm going further and saying that isimpossible as well because such a construction only produces a subsetof the real numbers, but does not produce all of them. This is aconsequence of imposing order by some means on the list.Note that when we're ordering the list we're doing so according to afunction on the natural numbers. We take a first element on the list,then recursively define which number comes next based on a rule. Thisgives us a first element, and a rule for constructing the next number.Every ordering can be expressed in this manner. Sound familiar?This of course builds by induction a countable subset of the realnumbers by whatever rule the person in question is proposing. This isa very common error when you come across critcism of Cantor'sarguements, so I'm seeing the same theme illustrated in binary. Suchorderings almost always end up with a subset of the rationals,although I've seen the non-repeating rationals used the most. It looksto me that instead they're taking the alternate binary representationthese non-repeating rationals instead, and then making the sameblunder. In general these numbers don't have to be rationals yourright, yet the general idea of imposing a specific order on the listby nature only includes a subset of the real numbers.> I am not convinced by your statement that the binary case is still> sufficient regardless of the base, given that I have demonstrated that a> modification of the argument is necessary in the binary case. Do you> understand what sort of modification is required, and why?No I don't understand why any modification to the arguement isneccesary, as I see it contructing an arguement which requires you toorder the list will always miss uncountably many real numbers.Arguements which aim to demonstrate that the contructed anti-diagonalis simply the alternate representation of a number already listedalways impose an ordering, and therefore selects a countable subset ofthe reals. Am I still missing something here about binary? === Subject: Re: Antidiagonal, Infinity> I don't see what the presence or absence of irrational numbers in the> list has to do with anything. It's easy to give a similar counterexample> that contains irrationals. The whole point is that the number produced> by the diagonal argument may be a number already in the list, regardless> of which other numbers may be missing from the list.> Well, consider the number 0.101010101.... for example, it's a rational> number, but isn't listed. How can I say it isn't listed? It's perfectly obvious that this number isn't listed, but what is yourpoint? I didn't claim that my list contained all the reals. If you're going to argue the real> numbers are uncountable, you have to include all the real numbers.That's not true. Perhaps you missed ZDK's proof that the irrationals areuncountable. Since the irrationals are a subset of the reals, we have aproof that the reals are uncountable that doesn't include all the reals.> The failure of what argument? The diagonal argument applied to decimal> numbers doesn't fail, even though the rationals are still countable. How> does countability of the rationals make anything surprising or> unsurprising in either case?> The failure of his ordering. Huh? That was my ordering. And it didn't fail. It succeeded. You seemto be confused about what the ordering was intended to accomplish. Icertainly was not trying to list all of the real numbers. That is afool's errand, because the reals are uncountable. I thought everyone inthis thread except RAF accepted that.My purpose in presenting that particular ordering was to provide acounterexample to the claim that if you list the numbers in binary, andthen change all the diagonal digits by swapping 0's and 1's, then theresulting number is a new number not in the original list. False. Myexample demonstrates the flaw in that argument. Notice I am not claimingthat the conclusion is wrong; the reals are uncountable, but you need abetter argument than the one given here to establish it.>By constraining himself to a subset of> the reals, he was about to show how the anti-digonal could produce a> number already on the list, I'm going further and saying that is> impossible as well What is impossible as well? Be specific. Do you mean it is impossiblefor the binary diagonal argument to produce a number that is already onthe list? Then how do you explain the fact that I produced a specificexample in which that is precisely what happens?> because such a construction only produces a subset> of the real numbers, but does not produce all of them. This is a> consequence of imposing order by some means on the list.You are not making sense here. None of this has anything to do withwhether my construction is a counterexample to the binary diagonalargument.You seem to be confusing me with RAF, who claims that because of theexistence of a single flawed argument, all other arguments for theuncountability of the reals must be equally flawed. No one else isclaiming that. Of course the argument is easily fixed, and of course thereals are uncountable, but none of this changes the fact that the simplebinary diagonal argument has a flaw. This of course builds by induction a countable subset of the real> numbers by whatever rule the person in question is proposing. This is> a very common error when you come across critcism of Cantor's> arguements, The binary diagonal argument that I criticized is not one of Cantor'sarguments. He knew better than to make a mistake like that. I am not convinced by your statement that the binary case is still> sufficient regardless of the base, given that I have demonstrated that a> modification of the argument is necessary in the binary case. Do you> understand what sort of modification is required, and why?> No I don't understand why any modification to the arguement is> neccesary, as I see it contructing an arguement which requires you to> order the list will always miss uncountably many real numbers.The point of the diagonal argument is to show that, if f: N -> R is anymapping from the naturals to the reals, then f is not a surjection. Ifcarefully formulated, the argument accomplishes this by producing a realnumber x (depending on f according to a specific rule) such that x is notin the range of f.If the argument is to accomplish its stated goal, then it must workproperly for every f. I produced a particular f such that the number xproduced by the diagonal process is already in the range of f. This doesnot, of course, prove that f is a surjection. What it proves is that thesimple diagonal argument has failed at its intended purpose, which was toshow that for every f there is an x (constructible from f by a particularprocess) that demonstrates f is not a surjection. In other words, youpicked the wrong rule. Be more careful in your choice of a rule, and theproof will work.This is why Cantor used decimal numbers instead of binary, and this iswhy he formulated his rule so that the generated number would not end inall 9's or all 0's. He saw the difficulty, even if you don't.> Arguements which aim to demonstrate that the contructed anti-diagonal> is simply the alternate representation of a number already listed> always impose an ordering, and therefore selects a countable subset of> the reals. Am I still missing something here about binary?What you are missing has to do with the nature of a proof and notspecifically with binary.Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. Well, consider the number 0.101010101.... for example, it's a rational> number, but isn't listed. How can I say it isn't listed? > It's perfectly obvious that this number isn't listed, but what is your> point? I didn't claim that my list contained all the reals.I asked about the modifying the binary case of Cantor's arguments, andwhy it was neccesary. You gave me an example of a fallacious reasoningwhich I do not attribute to you personally and I found enough evidenceto convince myself that it's wrong, and the binary case doesn'trequire any modification. I asked why the binary case was special forCantor's argument, if the list isn't assumed to contain all the reals,then it seem to me a bit off topic. Obviously it doesn't contain allreals.> If you're going to argue the real> numbers are uncountable, you have to include all the real numbers.> That's not true. Perhaps you missed ZDK's proof that the irrationals are> uncountable. Since the irrationals are a subset of the reals, we have a> proof that the reals are uncountable that doesn't include all the reals.Just because other uncountable sets exist if you're going to show methat the reals are uncountable I want you to show me the reals, notthe irrationals or any other subset of the reals. If you are going toprove the reals are uncountable, you must include all the reals! Ifyou can inherit that property from a subset of the reals, like thetrasendentals or the irrationals as a whole that's fine, but you mustdemonstrate that as part of the proof leading to the real numbers. Anyproof that stopped at the uncountability of the irrationals would beincomplete. > Huh? That was my ordering. And it didn't fail. It succeeded. You seem> to be confused about what the ordering was intended to accomplish. I> certainly was not trying to list all of the real numbers. That is a> fool's errand, because the reals are uncountable. I thought everyone in> this thread except RAF accepted that.And you're correct, I asked you to give me such a list so I couldfigure out exactly where RAF was going wrong, not because I thought hewas right. Anyone attempting to list the reals by imposing and orderon the list fails, and you gave me a general example of how thatarguement has been demonstrated in the past. I don't consider youpersonally responsibly for that arguement, and I agree that such aline of reasoning designed to disprove the validity of Cantor'sarguement are flawed. I think somehow you got the impression I agreedwith RAF, which of course I don't. > My purpose in presenting that particular ordering was to provide a> counterexample to the claim that if you list the numbers in binary, and> then change all the diagonal digits by swapping 0's and 1's, then the> resulting number is a new number not in the original list. False. My> example demonstrates the flaw in that argument. Notice I am not claiming> that the conclusion is wrong; the reals are uncountable, but you need a> better argument than the one given here to establish it.I guess i'm strengthening Cantor's result by suggesting for ananti-diagonal of the reals, it's impossible to produce a duelrepresentation in any base. Any arguement which demonstrates that aduel representation can be produced tacitly imposes an ordering on thelist. If we could swap members on the list around, it would ruin thearguement. Note that the list you provided by switching elementsaround you would produce a different number, you had to make anassumption about what number comes next each time you proceed down thediagonal list to ensure that you're able to build a duelrepresentation number.So that being said, we just start by SELECTING an initial number,we'll call it a and a must have a specific digit in the first spot. Wethen SELECT a number that gives us the anti-diag we want to go next.we'll call it b. Now, again after b, the next number requires acertain digit to give us the anti-diag we want. So again we SELECT anumber to go next. This process goes on indefinitely, each timeSELECTING a number that follows. Thinking about the definition of theNatural numbers, I recall a set of axioms that went something likethis.1)1 is in the set. 2)Every successor of 1 is in the set.and of course this gives us all the Natural numbers as we think ofthem. Now the contruction of this duel representation anti-diagonalnumber follows the same premise. SELECT a number and map it to 1. Thensince we're SELECTING a single number to succeed a, b, and you canquickly see how this list is countable. So, from the set of realswe've selected a countable subset. You and I both know that taking acountable subset of the reals leaves uncountably many real numbers tochoose from. When someone presents a list like this, it's alwayssufficient to simply choose any of the uncountably many reals thathave not been selected, no matter what the representation, and showthey've been left off the list. Any arguement that requires you toorder the list either explicitly or tacitly will fail for thisreason.If the more ambitious task of talking about all such lists of all suchnumber of all such patterns then I suspect that set of anti-diagonalnumbers would cover the reals, because all the lists could be seens asconverging sequences of real numbers and of course, that's a prettystandard definition if I remember correctly. I mean it's a subtleerror and list does seem to have a notion of order, the key isCantor never assume or proposed any order. His arguement doesn't failor need to be modified because this ordering is a limitation hedismissed.I distinctly recall the first time his diagonal arguement wasdemonstrated to me how my teacher continued to reinforce the idea thatI could choose any number, in any position between 0 and 1. I askedhim about rearranging the order and he said it would produce justanother number. One can note at that point that every possible orderproduces a real, not neccesarily unique, in fact, infinitelyrepetative. I think what we see at this point, is again the notiong ofthe power set being contructed by the anti-diag so to speak. Ireally don't have enough knowledge to explain that any more rigorouslyunfortunately.>By constraining himself to a subset of> the reals, he was about to show how the anti-digonal could produce a> number already on the list, I'm going further and saying that is> impossible as well > What is impossible as well? Be specific. Do you mean it is impossible> for the binary diagonal argument to produce a number that is already on> the list? Then how do you explain the fact that I produced a specific> example in which that is precisely what happens?Ok, it's impossible to construct a list of the reals that produces anyspecific real already on the list. I explain the fact that you did sobecause any such contruction requires an ordering, and that orderingleaves out uncountably many real numbers. It's impossible for for suchan ordering to exist for the real numbers ONLY. Of course this subsetyou've created by this ordering can produce a number already on thelist, you just did so. This would never affect Cantor's arguementthough, because the reals cannot be ordered in this fashion. Withoutmissing uncountably many reals.> You seem to be confusing me with RAF, who claims that because of the> existence of a single flawed argument, all other arguments for the> uncountability of the reals must be equally flawed. No one else is> claiming that. Of course the argument is easily fixed, and of course the> reals are uncountable, but none of this changes the fact that the simple> binary diagonal argument has a flaw.Sorry if you felt this way, I never had you confused at all. I'mtreating this as an explaination of what RAF was saying in terms ofthe binary arguement, since I've heard it before. I don't attributethis in any way to your reasoning, I've got it straight that you'rejust explaining someone else reasoning. None the less I refer to thefallacious arguements you put forth as your arguements, because you'rethe person I'm conversing with, and it gives me a way to speak aboutthat list specifically as opposed to some other list. RAF neverproduced that list, it's not fair to attribute it to him.How could I hold you in poor regard for taking the time to explain tome exactly what i asked to have explained? Your presentation made itvery clear to me where the flaw in that sort of reasoning lies, and Isimply explained why I agreed, and how I see the flaw. So, unless ithasn't been made clear by now - thank you! I really apprechiate thatyou took the time to explain it to me! :)> The point of the diagonal argument is to show that, if f: N -> R is any> mapping from the naturals to the reals, then f is not a surjection. If> carefully formulated, the argument accomplishes this by producing a real> number x (depending on f according to a specific rule) such that x is not> in the range of f.> If the argument is to accomplish its stated goal, then it must work> properly for every f. I produced a particular f such that the number x> produced by the diagonal process is already in the range of f. This does> not, of course, prove that f is a surjection. What it proves is that the> simple diagonal argument has failed at its intended purpose, which was to> show that for every f there is an x (constructible from f by a particular> process) that demonstrates f is not a surjection. In other words, you> picked the wrong rule. Be more careful in your choice of a rule, and the> proof will work.Ok exactly, I think we're both on the same page here. I'm saying thatwhen you construct a duel representation anti-diagonal number that youin fact, only include a countable number reals in the domain of f,therefore you only get a countable number out. The tacit selectionprocess in any such ordering only includes a countable set, thereforenot all reals are in the domain of f. Does that make it a bit moreclear what I'm trying to say? > This is why Cantor used decimal numbers instead of binary, and this is> why he formulated his rule so that the generated number would not end in> all 9's or all 0's. He saw the difficulty, even if you don't.distinction was only to make sure we didn't count each number twiceor something like that. It was more to ensure we were representing thereals uniquely, not to prevent instability in the construction of thediagonal.> Arguements which aim to demonstrate that the contructed anti-diagonal> is simply the alternate representation of a number already listed> always impose an ordering, and therefore selects a countable subset of> the reals. Am I still missing something here about binary?> What you are missing has to do with the nature of a proof and not> specifically with binary.I hope by this point I've made it clear why I have made thisstatement. I think I've got a handle on the arguement at this point. === Subject: Re: Antidiagonal, Infinity> That's not true. Perhaps you missed ZDK's proof that the irrationals are> uncountable. Since the irrationals are a subset of the reals, we have a> proof that the reals are uncountable that doesn't include all the reals.> Just because other uncountable sets exist if you're going to show me> that the reals are uncountable I want you to show me the reals, not> the irrationals or any other subset of the reals. If you are going to> prove the reals are uncountable, you must include all the reals! Why? I can prove a set with at least two members is not empty by proving that it has a proper subset that is not empty.In a similar way I can prove any set to be uncountable by proving some proper subset of it is uncountable === Subject: Re: Antidiagonal, Infinity> Well, consider the number 0.101010101.... for example, it's a rational> number, but isn't listed. How can I say it isn't listed? > It's perfectly obvious that this number isn't listed, but what is your> point? I didn't claim that my list contained all the reals.> I asked about the modifying the binary case of Cantor's arguments, and> why it was neccesary. And that is exactly what I explained, but you are apparently incapable ofunderstanding the explanation.>You gave me an example of a fallacious reasoning> which I do not attribute to you personally and I found enough evidence> to convince myself that it's wrong, and the binary case doesn't> require any modification. What evidence? You have not pointed out an error in the argument.The diagonal proof depends on the assertion that for each f: N -> R thereexists x in R (depending on f) such that x is not in the range of f.Do you agree?I showed that this assertion is violated in the case of the simple binaryversion of the diagonal argument.Do you agree?If not, then be specific. Which of those two statements do you disagreewith, and why?The only fallacious reasoning that is involved here is the assertionthat the simple binary diagonal argument always produces an x that is notin the range of f. It does not. That is what I showed.>I asked why the binary case was special for> Cantor's argument, if the list isn't assumed to contain all the reals,> then it seem to me a bit off topic. Obviously it doesn't contain all> reals.The Cantor argument never assumes that the list contains all the reals.Other people have modified Cantor's argument to include that assumption,but it is not necessary.> If you're going to argue the real> numbers are uncountable, you have to include all the real numbers.> That's not true. Perhaps you missed ZDK's proof that the irrationals are> uncountable. Since the irrationals are a subset of the reals, we have a> proof that the reals are uncountable that doesn't include all the reals.And besides that, I am wondering what made you think that I was trying toargue the real numbers are uncountable in the first place. It happensthat they are, but that's not what the discussion is about. As youexplained above, you began by asking what is wrong with the simple binarydiagonal argument, and that is the only point that I have beenaddressing. Huh? That was my ordering. And it didn't fail. It succeeded. You seem> to be confused about what the ordering was intended to accomplish. I> certainly was not trying to list all of the real numbers. That is a> fool's errand, because the reals are uncountable. I thought everyone in> this thread except RAF accepted that.> And you're correct, I asked you to give me such a list so I could> figure out exactly where RAF was going wrong, not because I thought heYou didn't ask where RAF was going wrong, but it so happens that I alsoexplained that. He went wrong by assuming that since one argument forthe uncountability of the reals turns out to be fallacious, they must allbe fallacious.> was right. Anyone attempting to list the reals by imposing and order> on the list fails, If we assume the axiom of choice, then every set (including the reals)can be well ordered. The issue we were discussing was countability, notwhether the reals can be ordered.>and you gave me a general example of how that> arguement has been demonstrated in the past. Why do you call it a general example? I don't see how I could possiblyhave made the example any more specific than it already is. I gave you aspecific list of real numbers with the property that the diagonal numberaccording to the simple binary argument turns out to be a number alreadyin the list.It's not in any way a general argument, because the conclusion does nothold for general lists. It holds for that specific list that I gave you.It's called a counterexample, and the existence of a singlecounterexample is sufficient to invalidate a proof.>I don't consider you> personally responsibly for that arguement, It's true that I was not the first to make up such an example, butneither was RAF. The existence of examples similar to this is commonknowledge. Certainly I was familiar with such examples long before RAFheard of them.>and I agree that such a> line of reasoning designed to disprove the validity of Cantor's> arguement are flawed. Wrong on two counts. The example is not flawed (you have not pointed outa flaw), and the example does not attempt to disprove the validity ofCantor's argument. Instead, it disproves the validity of the simplebinary diagonal argument, which is not Cantor's argument at all.>I think somehow you got the impression I agreed> with RAF, which of course I don't.I have never thought any such thing. I don't see how you possibly couldhave gotten that impression.On the other hand, I was convinced that somehow *you* got the impressionthat *I* agreed with RAF, which of course I don't. If you don't thinkthat, then why do you accuse me of attempting to refute Cantor, which isRAF's tactic and not mine?> My purpose in presenting that particular ordering was to provide a> counterexample to the claim that if you list the numbers in binary, and> then change all the diagonal digits by swapping 0's and 1's, then the> resulting number is a new number not in the original list. False. My> example demonstrates the flaw in that argument. Notice I am not claiming> that the conclusion is wrong; the reals are uncountable, but you need a> better argument than the one given here to establish it.> I guess i'm strengthening Cantor's result I don't see how you can strengthen Cantor's result by arguing about a>by suggesting for an> anti-diagonal of the reals, it's impossible to produce a duel> representation in any base. Any arguement which demonstrates that a> duel representation can be produced tacitly imposes an ordering on the> list. Oh, so you're one of those cranks who doesn't think 0.999... = 1. If Ihad known that, I probably wouldn't have wasted time starting on thisdiscussion.>If we could swap members on the list around, it would ruin the> arguement. Note that the list you provided by switching elements> around you would produce a different number, you had to make an> assumption about what number comes next each time you proceed down the> diagonal list to ensure that you're able to build a duel> representation number.The list would produce a different answer if I gave you a different list,but I had a reason for giving you exactly the list that I did. It'scalled constructing a counterexample.See, if someone says to you that every even natural number n is a sum oftwo primes, you can disprove the statement by setting n = 2. That's acounterexample. It does no good for the other person to claim that yourexample wouldn't work so well if you had chosen 4 or 6 or 8. You had areason for choosing 2, because that's the case that disproves the claim.So it is with the diagonal argument. If you claim you have a way tochoose an x in R for every f: N -> R such that x is not in the range off, then I get to choose the particular f that I want to consider if Iwant to disprove your claim. It does you no good to protest that yourmethod works fine for some different f. You have to make it work forevery f, and a single counterexample breaks your claim.By constraining himself to a subset of> the reals, he was about to show how the anti-digonal could produce a> number already on the list, I'm going further and saying that is> impossible as well > What is impossible as well? Be specific. Do you mean it is impossible> for the binary diagonal argument to produce a number that is already on> the list? Then how do you explain the fact that I produced a specific> example in which that is precisely what happens?> Ok, it's impossible to construct a list of the reals that produces any> specific real already on the list. I have presented you with an example that does precisely that.>I explain the fact that you did so> because any such contruction requires an ordering, and that orderingYou are talking nonsense. Uncountability of the reals means that if youare given any f: N -> R, then f is not a surjection.Notice that the mapping f: N -> R is part of the hypothesis, and eachsuch f induces an ordering on ran(f), a subset of the reals. That meansan ordering is inescapable. I didn't just make up an ordering and graftit onto the problem; it's right there in the problem statement to beginwith, in the definition of countability.> leaves out uncountably many real numbers. It's impossible for for such> an ordering to exist for the real numbers ONLY. Of course this subset> you've created by this ordering can produce a number already on the> list, you just did so. This would never affect Cantor's arguement> though, because the reals cannot be ordered in this fashion. Without> missing uncountably many reals.Neither Cantor nor I has claimed anything to the contrary. Look again atwhat is to be proved. Hypothesis: f: N -> R is a mapping. Conclusion: f is not a surjection.Not only do we *not* assume f is a surjection, we in fact prove theopposite. So why, exactly, do you keep mindlessly repeating theconclusion, which is that f is not a surjection? Do you think there issomeone here other than RAF who doubts that conclusion? What point areyou trying to make?> You seem to be confusing me with RAF, who claims that because of the> existence of a single flawed argument, all other arguments for the> uncountability of the reals must be equally flawed. No one else is> claiming that. Of course the argument is easily fixed, and of course the> reals are uncountable, but none of this changes the fact that the simple> binary diagonal argument has a flaw.> Sorry if you felt this way, I never had you confused at all. But you just repeated *again* that f: N -> R is not a surjection, as ifyou thought I disagreed. If that's not confusing me with RAF, then Idon't know what is.> How could I hold you in poor regard for taking the time to explain to> me exactly what i asked to have explained? Your presentation made it> very clear to me where the flaw in that sort of reasoning lies, and I> simply explained why I agreed, and how I see the flaw. So, unless it> hasn't been made clear by now - thank you! I really apprechiate that> you took the time to explain it to me! :)It's quite apparent from your subsequent postings that you have notunderstood a single word of my explanation.> The point of the diagonal argument is to show that, if f: N -> R is any> mapping from the naturals to the reals, then f is not a surjection. If> carefully formulated, the argument accomplishes this by producing a real> number x (depending on f according to a specific rule) such that x is not> in the range of f.> If the argument is to accomplish its stated goal, then it must work> properly for every f. I produced a particular f such that the number x> produced by the diagonal process is already in the range of f. This does> not, of course, prove that f is a surjection. What it proves is that the> simple diagonal argument has failed at its intended purpose, which was to> show that for every f there is an x (constructible from f by a particular> process) that demonstrates f is not a surjection. In other words, you> picked the wrong rule. Be more careful in your choice of a rule, and the> proof will work.> Ok exactly, I think we're both on the same page here. I'm saying that> when you construct a duel representation anti-diagonal number that you> in fact, only include a countable number reals in the domain of f, Hypothesis: f: N -> R is a mapping.That has nothing to do with whether there is a dual representationproblem or not. It is part of the hypothesis that ran(f) is countable.I don't see what that trivial observation has to do with explaining theflaw in the simple binary argument.> therefore you only get a countable number out. The tacit selection> process in any such ordering only includes a countable set, therefore> not all reals are in the domain of f. Does that make it a bit more> clear what I'm trying to say?It makes it clear that you keep repeating trivialities without an ounceof understanding. Do you understand why the example I gave is acounterexample to the simple binary diagonal argument?> This is why Cantor used decimal numbers instead of binary, and this is> why he formulated his rule so that the generated number would not end in> all 9's or all 0's. He saw the difficulty, even if you don't.> distinction was only to make sure we didn't count each number twice> or something like that. It was more to ensure we were representing the> reals uniquely, not to prevent instability in the construction of the> diagonal.Counting each element twice in an infinite set has no effect on thecardinality. There is nothing wrong with counting each number twice.The hypothesis says f: N -> R is a mapping; it does not say that f is aninjection. In fact, that is another way to handle the flaw in the simpleargument. Given f: N -> R, we can modify the mapping to purposely counteach dual-representation number twice, so that the number produced by thediagonal argument necessarily differs from all of them.> Arguements which aim to demonstrate that the contructed anti-diagonal> is simply the alternate representation of a number already listed> always impose an ordering, and therefore selects a countable subset of> the reals. Am I still missing something here about binary?> What you are missing has to do with the nature of a proof and not> specifically with binary.> I hope by this point I've made it clear why I have made this> statement. I think I've got a handle on the arguement at this point.I don't. In binary, you only get one rule, and it doesn't work.In binary, you get as many rules as you imagination and creativity are able to create. Mine can create coultably many, and some of them work.Ross has this dingbat delusion that an anti-diagonal algorithm must be as bad as possible, rather than as good as possible.There is no integer base from 2 on up that cannot be made to show the uncountability of the reals. But there is a lot of unaccountability in Ross' so far futile attempts at serious thought. === Subject: Re: Antidiagonal, InfinityWhy could a rule that I make necessarily give a rational number for anantidiagonal of a list of rational numbers? In a base greater than 2there are multiple options, b-1 many, for the anti-element of thelisted expansion, use that flexibility to construct a rational number. Is it thus always the case that rationals may never be completelylisted? No, it is not. By saying that you can construct a rulegenerating a number on a list different than any number on that list,contradictions ensue.The antidiagonal of a list of irrational numbers might be rational. If you claim to be able to make it irrational then the same conceptwould apply to rationals, a set with somewhat differentcharacteristics. I'm interested in this regular continued fractionrepresentation of Kovarik, perhaps the anthyphairetic ratio becausehe is so fond of measured Banach spaces.http://mathworld.wolfram.com/ContinuedFraction.htmlI think it's interesting that Khinchin's constant is the n'th root ofthe product of n elements of the continued fraction for _almost all_ numbers, in the presumably measure-theoretic sense. I wonder ifthey're normal, and see mention of Gosper, a combinatorialist.It's kind of interesting that all rational numbers have at least dualrepresentation as simple continued fractions that are finite. Aswell, it appears that an element of a continued fraction might be anarbitrarily large number. Is it not so that a rational number can berepresented with an infinite continued fraction, in a similar way tohow a root of a power series might be an integer?Consider other ways to represent each of the set of real numbers as asequence, besides trivially representing each real number as thesequence (0), representations that give each real number at least oneunique identifying sequence. Congratulations. Most of the sequencesare infinite and most of the arguments about binary integer modulusrepresentation apply.Like I said, I have some issues with the various decimalnon-diagonals. I introduce things like leading zeros and say thenon-diagonal is not an element of the range, but that's moreindirection that conclusion. How do you address the leading zerosconcern?I model only binary numbers where there is exactly and only oneantidiagonal of the matrix of elements thus constructed, all elementsof the unit interval of reals as infinite bitstrings with a beginning,and dual representation allows the existence of an antidiagonal,another extension.About the Equivalency Function, its antidiagonal is either duallyrepresented or not an element of the range. As well, in light of thenested interval result, it does not generate the contradictoryconditions of the result. In terms of a diagonal as a continuedfraction representation: [0,0,0,...], which number's continuedfraction is [1,1, 1, ...]? Is that exactly equal to one as is theantidiagonal? I want you to present further arguments against mappingbetween the natural and real numbers so I can continue to explain whyor why not they apply.The Equivalency Function is defined in a way thus that all values ofEF(n) except for n=0 are indefinite, and greater than zero, and lessthan or equal to one. This is where the set of integers is aninfinite set.That lets you say dx is 1/oo, sometimes, where there are exactly oomany points between zero and one, and 2*oo between zero and two, thereare some issues with continuity, oo * 1/oo = 1.I might suggest a completely different tack on the antidiagonal: notthat it shows that there is no mapping between a set and its powerset,but that infinity is dually represented as zero. === Subject: Re: Antidiagonal, Infinity> Why could a rule that I make necessarily give a rational number for an> antidiagonal of a list of rational numbers?A Rossian rule might do anything, but a reasonable rule need not. In a base greater than 2> there are multiple options, b-1 many, for the anti-element of the> listed expansion, use that flexibility to construct a rational number.It is not be possible in any base with any fixed rule, as some list will screew it up.> Is it thus always the case that rationals may never be completely> listed? No, it is not. By saying that you can construct a rule> generating a number on a list different than any number on that list,> contradictions ensue.No one, except possibly Ross, is trying to construct a number ON the list, everyone else is trying to construct a number which is NOT on the list. That is the whole point of it all.> The antidiagonal of a list of irrational numbers might be rational.So? > If you claim to be able to make it irrational then the same concept> would apply to rationals, a set with somewhat different> characteristics.Once an anti-diagonal algorithm is defined, the rationality or irrationality of constructed number depends entirely on the list that the algorithm is applied to, though the odds are greatly in favor of an irrational from randomly arranged lists. The rest of the junk deleted. === Subject: Re: Schauder basisskip... > If X has a Schauder basis then for every x in X there exists> a sequence (x_n) of vector______'s such that x_n -> x. Actually, there's a sequence {a_n} of scalars, belonging to the field over whick X is constructed and uniquely determined for each x, and a fixed sequence {e_n} of vectors of X, such that x = Sum (n=1, infinity) (a_n * e_n). That is, every x is an infinite linear combination of the vetors e_1, e_2..... Right? > The first topological basis______ you think of probably won't consist of something> countable, but then if you think about it some more you see how> to modify ______ so there are only countably many basic vectors______'s and> your done.> But first, what's the consequence______ that follows directly from> the definition?>I'm really confused, but an immediate consequence of the definition is>that, for every x, the infinite series Sum (n=1, infinity) (a_n * e_n)>is convergent. x.> True. And the fact that that _sum_ converges to x shows that a> certain _sequence_ converges to x. What sequence?The sequence of partial sums converge to x. This means that, for everyx in X and for every eps>0, if n is sufficiently large, then ||(Sum(k=1,n) ex_k * x_k) - x||0, the open ballcentered at x and of radius eps intersects D. Therefore, D is dense inX. It remains to show (if it's true, of course) that D is countable. Well, if X is countable, this is immediate (in fact, if X is countablewe have nothing to prove), but I'm still wondering how I can prove Dis countable or prove it contains a countable subset that's also densein X.Amanda === Subject: Re: Schauder basis> The sequence of partial sums converge to x. This means that, for every> x in X and for every eps>0, if n is sufficiently large, then ||(Sum> (k=1,n) ex_k * x_k) - x|| x_k) is in the open ball of radius eps and center x. The letter x in> ex is to emphasize that the scalars of the infinite linear combination> of the x_k's that leads to x depends on x.> If we define D = {ex_1 * x_1 ...+ ex_k * x_k | k in N (the naturals),> x in X}, then, for every x in X and every eps>0, the open ball> centered at x and of radius eps intersects D. Therefore, D is dense in> X. It remains to show (if it's true, of course) that D is countable. > Well, if X is countable, this is immediate (in fact, if X is countable> we have nothing to prove), but I'm still wondering how I can prove D> is countable or prove it contains a countable subset that's also dense> in X.D is a countable union of finite-dimensional subspaces. So if you can show each of these subspaces is separable, it would be a good thing. === Subject: Re: Schauder basis>skip... > If X has a Schauder basis then for every x in X there exists> a sequence (x_n) of vector______'s such that x_n -> x. Actually, there's a sequence {a_n} of scalars, belonging to the field over whick X is constructed and uniquely determined for each x, and a fixed sequence {e_n} of vectors of X, such that x = Sum (n=1, infinity) (a_n * e_n). That is, every x is an infinite linear combination of the vetors e_1, e_2..... Right? > The first topological basis______ you think of probably won't consist of something> countable, but then if you think about it some more you see how> to modify ______ so there are only countably many basic vectors______'s and> your done.> But first, what's the consequence______ that follows directly from> the definition?>I'm really confused, but an immediate consequence of the definition is>that, for every x, the infinite series Sum (n=1, infinity) (a_n * e_n)>is convergent. x.> True. And the fact that that _sum_ converges to x shows that a> certain _sequence_ converges to x. What sequence?>The sequence of partial sums converge to x. Right. And the partial sums are linear combinations of the e_n.So if S is the set of all linear combinations of the e_n thenS is dense in X. We'd be done except that of course S isnot countable. But S has a certain countable subset whichis _also_ dense in X...>This means that, for every>x in X and for every eps>0, if n is sufficiently large, then ||(Sum>(k=1,n) ex_k * x_k) - x||x_k) is in the open ball of radius eps and center x. The letter x in>ex is to emphasize that the scalars of the infinite linear combination>of the x_k's that leads to x depends on x.>If we define D = {ex_1 * x_1 ...+ ex_k * x_k | k in N (the naturals),>x in X}, then, for every x in X and every eps>0, the open ball>centered at x and of radius eps intersects D. Therefore, D is dense in>X. It remains to show (if it's true, of course) that D is countable. >Well, if X is countable, this is immediate (in fact, if X is countable>we have nothing to prove), but I'm still wondering how I can prove D>is countable or prove it contains a countable subset that's also dense>in X.D is not countable. That's because the ex_j can be any real numbers,and the reals are uncountable. But...>Amanda === Subject: Re: Schauder basis> skip... > If X has a Schauder basis then for every x in X there exists> a sequence (x_n) of vector______'s such that x_n -> x. Actually, there's a sequence {a_n} of scalars, belonging to the field over whick X is constructed and uniquely determined for each x, and a fixed sequence {e_n} of vectors of X, such that x = Sum (n=1, infinity) (a_n * e_n). That is, every x is an infinite linear combination of the vetors e_1, e_2..... Right? > The first topological basis______ you think of probably won't consist of something> countable, but then if you think about it some more you see how> to modify ______ so there are only countably many basic vectors______'s and> your done.> But first, what's the consequence______ that follows directly from> the definition?>I'm really confused, but an immediate consequence of the definition is>that, for every x, the infinite series Sum (n=1, infinity) (a_n * e_n)>is convergent. x.> True. And the fact that that _sum_ converges to x shows that a> certain _sequence_ converges to x. What sequence?>The sequence of partial sums converge to x. > Right. And the partial sums are linear combinations of the e_n.> So if S is the set of all linear combinations of the e_n then> S is dense in X. We'd be done except that of course S is> not countable. But S has a certain countable subset which> is _also_ dense in X...>This means that, for every>x in X and for every eps>0, if n is sufficiently large, then ||(Sum>(k=1,n) ex_k * x_k) - x||x_k) is in the open ball of radius eps and center x. The letter x in>ex is to emphasize that the scalars of the infinite linear combination>of the x_k's that leads to x depends on x.>If we define D = {ex_1 * x_1 ...+ ex_k * x_k | k in N (the naturals),>x in X}, then, for every x in X and every eps>0, the open ball>centered at x and of radius eps intersects D. Therefore, D is dense in>X. It remains to show (if it's true, of course) that D is countable. >Well, if X is countable, this is immediate (in fact, if X is countable>we have nothing to prove), but I'm still wondering how I can prove D>is countable or prove it contains a countable subset that's also dense>in X.> D is not countable. That's because the ex_j can be any real numbers,> and the reals are uncountable. But...But we can take the set of all rational linear combinations of the x_n(that is, the set of all linear combinations with rationalcoefficients). Since the rationals are countable and are dense in R,we get a countbale set dense in X and we are done. Right?If X is a vector space over the complexes, we can take the complexeswith rational real and rational imaginary parts.The thing gets more complicated if X is vector space over an arbitraryfield F...But, we didn't use the fact that X is Banach. The completeness of Xwas not considered at allAmanda >Amanda> ************************> === Subject: Re: Schauder basis> skip... > If X has a Schauder basis then for every x in X there exists> a sequence (x_n) of vector______'s such that x_n -> x. Actually, there's a sequence {a_n} of scalars, belonging to the field over whick X is constructed and uniquely determined for each x, and a fixed sequence {e_n} of vectors of X, such that x = Sum (n=1, infinity) (a_n * e_n). That is, every x is an infinite linear combination of the vetors e_1, e_2..... Right? > The first topological basis______ you think of probably won't consist of something> countable, but then if you think about it some more you see how> to modify ______ so there are only countably many basic vectors______'s and> your done.> But first, what's the consequence______ that follows directly from> the definition?>I'm really confused, but an immediate consequence of the definition is>that, for every x, the infinite series Sum (n=1, infinity) (a_n * e_n)>is convergent. x.> True. And the fact that that _sum_ converges to x shows that a> certain _sequence_ converges to x. What sequence?>The sequence of partial sums converge to x. > Right. And the partial sums are linear combinations of the e_n.> So if S is the set of all linear combinations of the e_n then> S is dense in X. We'd be done except that of course S is> not countable. But S has a certain countable subset which> is _also_ dense in X...>This means that, for every>x in X and for every eps>0, if n is sufficiently large, then ||(Sum>(k=1,n) ex_k * x_k) - x||x_k) is in the open ball of radius eps and center x. The letter x in>ex is to emphasize that the scalars of the infinite linear combination>of the x_k's that leads to x depends on x.>If we define D = {ex_1 * x_1 ...+ ex_k * x_k | k in N (the naturals),>x in X}, then, for every x in X and every eps>0, the open ball>centered at x and of radius eps intersects D. Therefore, D is dense in>X. It remains to show (if it's true, of course) that D is countable. >Well, if X is countable, this is immediate (in fact, if X is countable>we have nothing to prove), but I'm still wondering how I can prove D>is countable or prove it contains a countable subset that's also dense>in X.> D is not countable. That's because the ex_j can be any real numbers,> and the reals are uncountable. But...>But we can take the set of all rational linear combinations of the x_n>(that is, the set of all linear combinations with rational>coefficients). Since the rationals are countable and are dense in R,>we get a countbale set dense in X and we are done. Right?Yes!>If X is a vector space over the complexes, we can take the complexes>with rational real and rational imaginary parts.Yes.>The thing gets more complicated if X is vector space over an arbitrary>field F...If we're talking about a Banach space then we're talking aboutreal or complex scalars.>But, we didn't use the fact that X is Banach. The completeness of X>was not considered at allAmazing. We've proved more than was required: If X is a normedvector space and there exist e_1, ... such that every x in X canbe written x = a_1 e_1 + ... then X is separable.>Amanda >Amanda> ************************> === Subject: Re: Schauder basis>But, we didn't use the fact that X is Banach. The completeness of X>was not considered at allNor was it mentioned in your original question: I would appreciate a help to show that if a normed vector space has a Schauder basis, then it's separable.Here's a question for you: if a normed vector space is separable, isits (norm-)completion separable? Here's another: if a normed vector spaceis separable, can it have a vector subspace which is non-separable(with the induced norm)? === Subject: Re: Schauder basis>But, we didn't use the fact that X is Banach. The completeness of X>was not considered at all>Nor was it mentioned in your original question:> I would appreciate a help to show that if a normed vector space has a> Schauder basis, then it's separable.>Here's a question for you: if a normed vector space is separable, is>its (norm-)completion separable? Here's another: if a normed vector space>is separable, can it have a vector subspace which is non-separable>(with the induced norm)?Golly, we just figured out one question and now we have two morequestions? Some people...Who was it who said the best time to try to solve a problem isright after you've solved a different problem?>Lee Rudolph === Subject: An interesting function (hyperbola?)Hello all,Haven't posted in a while but I thought some might be interested in aunusal function that has cropped up in my work.I am testing cross-linking rates in polyurathanes and the data seemsto form a line that starts at zero and approaches the line y=23without ever reaching it. The height seems to be determined by onepart of my mix and the rate of approach by the other. None of my statsoftware has suggested an equation that fits well at all but it surelooks like it is should follow a relativly simple relationship. Italways starts at zero and always approaches the same line, reguardlessof how much or little catayst is added.At this point I am tring to take the center of a hyperbola and move itup and left while rotating it until the asymptotes are at parellel tothe x and y axis, but while drawing it is not that difficult, and themovement easy enought to factor into the normal hyperbola, therotation has been diffcult to fit into the equation.Any ideas or suggestions greatly appreciated.Ghostwriter === Subject: Re: An interesting function (hyperbola?)> The height seems to be determined by one part of my mix and the rate of approach by the other. There are many possibilities depending on your curves. To add to thelist of hyperbola and 1-e^(-k t) types of variation, I suggest thetanh hyperbolic function where two factors influence maximumasymptotic level ymax and rate of reaction build-up rate , which canbe changed in the Mathematica program :rate = {.5,.75,1,1.5,3,5 }; ymax=23 ;PUXlink = Map [ymax* Tanh[ #1 * t ] &, rate ];Plot[PUXlink // Evaluate, {t, 0,2},Frame-> True]; === Subject: Re: An interesting function (hyperbola?)>Haven't posted in a while but I thought some might be interested in a>unusal function that has cropped up in my work.>I am testing cross-linking rates in polyurathanes and the data seems>to form a line that starts at zero and approaches the line y=23>without ever reaching it. The height seems to be determined by one>part of my mix and the rate of approach by the other. None of my stat>software has suggested an equation that fits well at all but it sure>looks like it is should follow a relativly simple relationship. It>always starts at zero and always approaches the same line, reguardless>of how much or little catayst is added.>At this point I am tring to take the center of a hyperbola and move it>up and left while rotating it until the asymptotes are at parellel to>the x and y axis, but while drawing it is not that difficult, and the>movement easy enought to factor into the normal hyperbola, the>rotation has been diffcult to fit into the equation.>Any ideas or suggestions greatly appreciated.>GhostwriterHave you considered an exponential approach to the asymptote? Forexample if your horizontal asymptote is y = c, an equation of theform:y = c (1 - e^(-kx))will start at 0 and approach y = c at a rate depending on the size ofk. Choose k to fit your data in some sense. === Subject: Re: An interesting function (hyperbola?)>I am testing cross-linking rates in polyurathanes and the data seems>to form a line that starts at zero and approaches the line y=23>without ever reaching it. The height seems to be determined by one>part of my mix and the rate of approach by the other. None of my stat>software has suggested an equation that fits well at all but it sure>looks like it is should follow a relativly simple relationship. It>always starts at zero and always approaches the same line, reguardless>of how much or little catayst is added.>At this point I am tring to take the center of a hyperbola and move it>up and left while rotating it until the asymptotes are at parellel to>the x and y axis, but while drawing it is not that difficult, and the>movement easy enought to factor into the normal hyperbola, the>rotation has been diffcult to fit into the equation.>Any ideas or suggestions greatly appreciated.Hyperbolas with asymtotes parallel to the axes are the easiest kind todescribe by equation! Start with y = 1/x and transform. In this case,y = 23 - A / (x + B)and since it goes through the origin, you know that 0 = 23 - A / (0 + B)23 * B = Ay = 23 - 23 * B / (x + B)y = 23 * x / (x + B)--Keith Lewis klewis {at} mitre.orgThe above may not (yet) represent the opinions of my employer. === Subject: Little algebra questionS(*) is a commutative regular monoid. Consider the following relation~,defined as follows:a~b iff b=ca (with c in S, c invertible in S).If x in S is invertible, then [x]_~ = S (this is obvious).So in an abelian group the relation ~ is total (each element is in relationwith all others). Why??? === Subject: Re: Little algebra question> S(*) is a commutative regular monoid. Consider the following relation> ~,defined as follows:> a~b iff b=ca (with c in S, c invertible in S).> If x in S is invertible, then [x]_~ = U(S) (this is obvious).> So in an abelian group the relation ~ is total (each element is in relation> with all others). Why???Why not?Let G=U(S) be the group of invertible elements of S.Then G acts on S vie left multiplication g*s := gs (g in G, s in S)and the above equivalence classes are the orbits of this action.In fact, you do not need commutativity or regularity for this.In the special case when S is a group (G=S) this gives the usualaction on G on itself via left multiplication.Marc === Subject: Re: Little algebra questionfake ha scritto nel messaggio> S(*) is a commutative regular monoid. Consider the following relation> ~,defined as follows:> a~b iff b=ca (with c in S, c invertible in S).> If x in S is invertible, then [x]_~ = S (this is obvious).> So in an abelian group the relation ~ is total (each element is inrelation> with all others). Why??? === Subject: Re: Little algebra questionfake ha scritto nel messaggio> S(*) is a commutative regular monoid. Consider the following relation> ~,defined as follows:> a~b iff b=ca (with c in S, c invertible in S).> If x in S is invertible, then [x]_~ = S (this is obvious).***Pardon, here is:If x in S is invertible, then [x]_~ = U(S) (that is the set of invertibleelements of S). ^^^^^ === Subject: List of geometric figures and their volume, area and circumferenceDo any of you know where I can find such a thing? A list, with an example of each figure and the formula to calculate circumference, area and volume if it is an three-dimensional figure.O - Circle Area: pi*r^2 Circumference: 2*pi*r === Subject: Re: Graph theory question>Now suppose that I place one vertex outside this polygon. Can I connect>every single vertex of the polygon to this vertex with an edge such that>none of these edges cross each other? This seems to be true, but how would>I prove it? Also, what if the vertex was inside the polygon?I suspected from the subject line that the connections could be curved, andyou confirmed it in a later post.Call the new node P.The most difficult connection to draw would be the last one, so assume youhave already drawn connections from P to each of 1..N-1. N is eitherexposed to the exterior of your graph so far or contained inside theas-yet-empty 4-sided figure (1,N,N-1,P). Either way, P and N are bothexposed to that region, so you can draw a connection. It's pretty much the same inside the polygon, except the relevance of beingexposed to the exterior is gone. And it's easier to prove in forward order.The (P,1) connection doesn't divide the interior. (P,2) then divides itinto (1,2,P) and (P,2,3...N,1). Since P and 3 are both exposed to theinterior of that, you can draw a connection, dividing the larger regioninto (2,3,P) and (P,3,4...N,1). --Keith Lewis klewis {at} mitre.orgThe above may not (yet) represent the opinions of my employer. === Subject: Re: Graph theory question>Now suppose that I place one vertex outside this polygon. Can I connect>every single vertex of the polygon to this vertex with an edge such that>none of these edges cross each other? This seems to be true, but howwould>I prove it? Also, what if the vertex was inside the polygon?> I suspected from the subject line that the connections could be curved,and> you confirmed it in a later post.> Call the new node P.> The most difficult connection to draw would be the last one, so assume you> have already drawn connections from P to each of 1..N-1. N is either> exposed to the exterior of your graph so far or contained inside the> as-yet-empty 4-sided figure (1,N,N-1,P). Either way, P and N are both> exposed to that region, so you can draw a connection.I don't understand. How can N be contained inside the as-yet-empty 4-sidedfigure (1,N,N-1,P). It isa vertex of that figure, so how can it be inside? Also, what do you mean byN is exposed to the exterior of your graph?> It's pretty much the same inside the polygon, except the relevance ofbeing> exposed to the exterior is gone. And it's easier to prove in forwardorder.> The (P,1) connection doesn't divide the interior. (P,2) then divides it> into (1,2,P) and (P,2,3...N,1). Since P and 3 are both exposed to the> interior of that, you can draw a connection, dividing the larger region> into (2,3,P) and (P,3,4...N,1).> --Keith Lewis klewis {at} mitre.org> The above may not (yet) represent the opinions of my employer. === Subject: Re: Graph theory question The most difficult connection to draw would be the last one, so assume you> have already drawn connections from P to each of 1..N-1. N is either> exposed to the exterior of your graph so far or contained inside the> as-yet-empty 4-sided figure (1,N,N-1,P). Either way, P and N are both> exposed to that region, so you can draw a connection.>I don't understand. How can N be contained inside the as-yet-empty 4-sided>figure (1,N,N-1,P). It is>a vertex of that figure, so how can it be inside? Also, what do you mean by>N is exposed to the exterior of your graph?Contained was an incorrect word on my part. I meant exposed to, whichmeans the same thing as touching. --Keith Lewis klewis {at} mitre.orgThe above may not (yet) represent the opinions of my employer. === Subject: Re: Graph theory question> Excuse my terminology, because I don't know any terminology when it comes to> graph theory. (and I also don't know the slightest bit about graph theory!)> Say I have a set of n vertices in the plane, (let me call them 1,2,3,...,n),> and I connect the vertices like a polygon (i.e. each vertex is connected to> only 2 vertices by edges and there are no two edges that cross each other.).> You could have a crazy assortment of vertices and edges that makes the shape> look very weird, but that's ok, we don't need any regularity here. (I hope> I have not said anything that can cause a mix-up here. I am just assuming> we have n points in the plane where the edges form an n-gon that need not be> regular, in fact may look very strange, with each edge length being> different and all angles being different)> Now suppose that I place one vertex outside this polygon. Can I connect> every single vertex of the polygon to this vertex with an edge such that> none of these edges cross each other? This seems to be true, but how would> I prove it? Also, what if the vertex was inside the polygon?If I understand correctly, no. Take the three verticees(0,0),(1,0),(0,1), conected so as to pake a triangle. When you connectall these with the point (-1,0), you shall have the following twoedges: {(x,0):-1 Excuse my terminology, because I don't know any terminology when itcomes to> graph theory. (and I also don't know the slightest bit about graphtheory!)> Say I have a set of n vertices in the plane, (let me call them1,2,3,...,n),> and I connect the vertices like a polygon (i.e. each vertex is connectedto> only 2 vertices by edges and there are no two edges that cross eachother.).> You could have a crazy assortment of vertices and edges that makes theshape> look very weird, but that's ok, we don't need any regularity here. (Ihope> I have not said anything that can cause a mix-up here. I am justassuming> we have n points in the plane where the edges form an n-gon that neednot be> regular, in fact may look very strange, with each edge length being> different and all angles being different)> Now suppose that I place one vertex outside this polygon. Can I connect> every single vertex of the polygon to this vertex with an edge such that> none of these edges cross each other? This seems to be true, but howwould> I prove it? Also, what if the vertex was inside the polygon?> If I understand correctly, no. Take the three verticees> (0,0),(1,0),(0,1), conected so as to pake a triangle. When you connect> all these with the point (-1,0), you shall have the following two> edges: {(x,0):-1 intersection.> MosheI'm sorry! I should rephrase the question. The edges don't have to bestraight lines they can be curves! (but all in the plane!) === Subject: Wide-ranging functionsAn easy question: is there a real function f such that for any non-emptyopen interval (a, b) and any real number y, there is a number x in (a, b)such that f(x) = y? In other words is there a real function f which, ifrestricted to any non-trivial interval, still has a range of the entire realline?Using the Axiom of Choice, one can easily construct such a real function.Now for the tougher question: is there a measurable function with thisproperty? === Subject: Re: Wide-ranging functions> An easy question: is there a real function f such that for any non-empty> open interval (a, b) and any real number y, there is a number x in (a, b)> such that f(x) = y? In other words is there a real function f which, if> restricted to any non-trivial interval, still has a range of the entire real> line?> Using the Axiom of Choice, one can easily construct such a real function.> Now for the tougher question: is there a measurable function with this> property?Let U1, U2, ... be the open intervals with rational endpoints. Choose a Cantor set K1, of measure 0, contained in U1. U2 K1 has nonempty interior, so there is a Cantor set K2 of measure 0, disjoint from K1, inside U2. Continuing, there exist pairwise disjoint Cantor sets Kn, each of measure 0, with Kn contained in Un. Now each of these Cantor sets has the cardinality of R, so for each n there is a map fn of Kn onto R. Define f = fn on Kn, f = 0 on R (U Kn). Because U Kn has measure 0, f is measurable (in fact Borel measurable). It's easy to check that f has the desired property. === Subject: Re: Wide-ranging functions> An easy question: is there a real function f such that for any non-empty> open interval (a, b) and any real number y, there is a number x in (a, b)> such that f(x) = y? In other words is there a real function f which, if> restricted to any non-trivial interval, still has a range of the entire real> line?> Using the Axiom of Choice, one can easily construct such a real function.> Now for the tougher question: is there a measurable function with this> property?> Let U1, U2, ... be the open intervals with rational endpoints. Choose a> Cantor set K1, of measure 0, contained in U1. U2 K1 has nonempty> interior, so there is a Cantor set K2 of measure 0, disjoint from K1,> inside U2. Continuing, there exist pairwise disjoint Cantor sets Kn, each> of measure 0, with Kn contained in Un. Now each of these Cantor sets has> the cardinality of R, so for each n there is a map fn of Kn onto R.It would be better to use a *continuous* map fn of Kn onto [-n, n] ...> Define f = fn on Kn, f = 0 on R (U Kn). Because U Kn has measure> 0, f is measurable (in fact Borel measurable). It's easy to check> that f has the desired property.... if you want f to be *Borel* measurable. === Subject: Re: Wide-ranging functions> Let U1, U2, ... be the open intervals with rational endpoints. Choose a> Cantor set K1, of measure 0, contained in U1. U2 K1 has nonempty> interior, so there is a Cantor set K2 of measure 0, disjoint from K1,> inside U2. Continuing, there exist pairwise disjoint Cantor sets Kn, each> of measure 0, with Kn contained in Un. Now each of these Cantor sets has> the cardinality of R, so for each n there is a map fn of Kn onto R.> It would be better to use a *continuous* map fn of Kn onto [-n, n] ...> Define f = fn on Kn, f = 0 on R (U Kn). Because U Kn has measure> 0, f is measurable (in fact Borel measurable). It's easy to check> that f has the desired property.> ... if you want f to be *Borel* measurable.Right, the fn's were kind of arbitrary so I only know f is Lebesgue measurable. Do what Fred said if you want Borel measurable. === Subject: Re: Wide-ranging functionsBravo! Well done. === Subject: Re: Wide-ranging functions> An easy question: is there a real function f such that for any> non-empty open interval (a, b) and any real number y, there is a> number x in (a, b) such that f(x) = y? In other words is there a> real function f which, if restricted to any non-trivial interval,> still has a range of the entire real line?> Using the Axiom of Choice, one can easily construct such a real> function. Now for the tougher question: is there a measurable> function with this property?You don't need the axiom of choice for that. You can construct anexample which is as nice as can be for a nowhere continuous function:Borel measurable, Baire class 2, constant almost everywhere. === Subject: Re: Just curious: countable subset> Your last statement is the correct one. One cannot prove in in ZF that > any infinite set (defined as not in a bijection with a finite ordinal) > has a countable subset; one needs at least the axiom of countable choice.> However, one can prove in ZF that a set is Dedekind inifinite (defined > as in a bijection with a proper subset) if and only if it has a > countable subset.Yes, I use the definition of Dedekind infinite.(by the way, is a set infinite in ZFC iff it is dedekind infinite?)Anyway, how do you prove a set is dedekind infinite iff it has acountable subset without using the axiom of choice or the axiom ofcountable choice? === Subject: Re: Just curious: countable subset>(by the way, is a set infinite in ZFC iff it is dedekind infinite?)In ZF, Dedekind infinite implies infinite: Show by induction that a finite set cannot be Dedekind infinite.When we add choice, any set can be well-ordered; with this ordering, the set is isomorphic to an infinite ordinal. Use the standard trick of mapping n to its successor for each n in omega.>Anyway, how do you prove a set is dedekind infinite iff it has a>countable subset without using the axiom of choice or the axiom of>countable choice?To show that the existence of a countable subset implies Dedekind inifinite; use the same trick as in my second paragraph above.Dave Seaman has shown indicated how to show the other direction. (Indeed, most of my remarks here are the same ideas as Dave's.)How much set theory do you know? Have you ever read Halmos, Naive Set Theory? This is really a classic; I highly recommend it. Also, the first chapter of Munkres' Topology text covers most of these issues.As in another thread Greg started, I also recommend to exercises (by T. Huuskonen) on inifinity without choice. These cover the above topics explicitly. Currently, I am pondering question 15 there.explicitly the existence of such sets is consistent with ZF. I assume this is the case. Confirmation, Masters?Finally, I once posed the following exercise to myself to review the matter. In the following, x <= y notates that there exists an injection from x to y. (That is, <= represents the curly inequality). w denotes omega, the least infinite ordinal.Show that the following statements about (a set) x are equivalent in ZFC. Whichimplications can be proven without choice?a) For all n in w, not (x <= n).b) w <= x.c) For all n in w, n <= xd) There exists a proper subset y of x such that x <= y.e) There exists a surjection from x to w.Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Just curious: countable subsetlittle axiomatic set theory. Most set theory I know/use is naive settheory. That's why I'm wondering where you need the axiom of choce andwhen you don't.I haven't read Halmos or Mukres' topology yet. (I'm a first year gradstudent, but pretty busy with other subjects sofar.)Certainly you need the axiom of choice to prove that the cardinalitiesof any two sets are comparable (i.e. <= is a total order oncardinalities). You use this by setting up a well-ordering of the twosets. [By the way, perhaps it's just that no one has found how toprove the same thing in ZF? Or it's proven to be impossible?]But my most important question at this time about set theory iswhether the axiom of choice is needed when I am making countableconstructions. For example,Suppose A1 was already well-ordered.Let a1 denote the smallest element in A1. Define A2 := A1 - {a1}. Let a2 be the smallest elemnt in A2. And so on.For any finite number k, I have unambiguously defined a1... a_k. Butis it therefore enough to make statements about the SEQUENCE a1 ...a_k? I am trying to think of this rigorously and axiomatically -- thesequence is an object by itself. When I make propositions about thesequence, is it enough to make an infinite schema of arguments thatgo, give me a k, and i'll prove it for that k? This isn'tfirst-order logic, right? And I can only use first-order logic inaxiomatic set theory?If the above happens by some chance not to require the axiom ofchoice, how about the same thing but this time, A1 was notwell-ordered and a_k now denotes just some element in A_k. If thisdoesn't require the axiom of choice either, then it's a proof thatevery infinite set has a countable subset.In short, the question isWhen is it ok not to use the axiom of choice when I make some usefulinfinite constructions? If I define sequences or other countablethings, when do I actually make use of the axiom of choice and when doI not need it? === Subject: Re: Just curious: countable subset>little axiomatic set theory. Most set theory I know/use is naive set>theory. That's why I'm wondering where you need the axiom of choce and>when you don't.>I haven't read Halmos or Mukres' topology yet. (I'm a first year grad>student, but pretty busy with other subjects sofar.)Then by all means, read Halmos as soon as possible. It should be required reading of all undergraduate majors, doubly so graduate students.>Certainly you need the axiom of choice to prove that the cardinalities>of any two sets are comparable (i.e. <= is a total order on>cardinalities). You use this by setting up a well-ordering of the two>sets. [By the way, perhaps it's just that no one has found how to>prove the same thing in ZF? Or it's proven to be impossible?]It is consistent with ZF that there exist sets that are incomparable.>But my most important question at this time about set theory is>whether the axiom of choice is needed when I am making countable>constructions. For example,>Suppose A1 was already well-ordered.>Let a1 denote the smallest element in A1. >Define A2 := A1 - {a1}. >Let a2 be the smallest elemnt in A2. And so on.>For any finite number k, I have unambiguously defined a1... a_k. But>is it therefore enough to make statements about the SEQUENCE a1 ...>a_k? I am trying to think of this rigorously and axiomatically -- the>sequence is an object by itself. When I make propositions about the>sequence, is it enough to make an infinite schema of arguments that>go, give me a k, and i'll prove it for that k? This isn't>first-order logic, right? And I can only use first-order logic in>axiomatic set theory?In the above situation, you have not used the axiom of choice. Since A1 is well-ordered, a_i is well-defined (without choice) for each i. The set {a_i : i > 0} exists by the axiom scheme of replacement. The function i |-> a_i also exists by induction.>If the above happens by some chance not to require the axiom of>choice, how about the same thing but this time, A1 was not>well-ordered and a_k now denotes just some element in A_k. If this>doesn't require the axiom of choice either, then it's a proof that>every infinite set has a countable subset.Here, you have used choice to choose the a_i. As long as A1 is infinite, ZF implies the existence of each finite set {a1, a2,..., a_n}. Without some version of choice, you cannot conclude the existence of {a_i : i.>0}>In short, the question is>When is it ok not to use the axiom of choice when I make some useful>infinite constructions? If I define sequences or other countable>things, when do I actually make use of the axiom of choice and when do>I not need it?In most of mathematics, it is always ok to use the axiom of choice. Sometimes, though, one wants to be cognizant of whether one is using the axiom or not. In such cases, it is often of interest whether choice is required.Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Just curious: countable subset> Your last statement is the correct one. One cannot prove in in ZF that > any infinite set (defined as not in a bijection with a finite ordinal) > has a countable subset; one needs at least the axiom of countable choice.> However, one can prove in ZF that a set is Dedekind inifinite (defined > as in a bijection with a proper subset) if and only if it has a > countable subset.> Yes, I use the definition of Dedekind infinite.> (by the way, is a set infinite in ZFC iff it is dedekind infinite?)Yes, because every cardinal in ZFC is also an ordinal. If an ordinal Asatisfies A > n for every natural number n, then we must have A >= sup n= w.> Anyway, how do you prove a set is dedekind infinite iff it has a> countable subset without using the axiom of choice or the axiom of> countable choice?If we have a bijection f: A -> B such that B is a proper subset of A,then choose x in AB and consider the sequence x, f(x), f(f(x)), ....The other implication follows from the fact that every superset of aDedekind infinite set is Dedekind infinite.Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.Any infinite set has a countable subset.It seems obvious that any set has subsets of ALL lesser cardinalities.But the Axiom of Choice is also obvious to me, so maybe I'm biased.Given any set of mutually exclusive nonempty sets, there exists at leastone set that contains exactly one element in common with each of thenonempty sets.Is there an argument against that?--Keith Lewis klewis {at} mitre.orgThe above may not (yet) represent the opinions of my employer. === Subject: Re: Just curious: countable subset>Any infinite set has a countable subset.> It seems obvious that any set has subsets of ALL lesser cardinalities.> But the Axiom of Choice is also obvious to me, so maybe I'm biased.The idea of a set of axioms is that you are stating exactly what youwill then base everything else on, so it behooves one to be extermelycareful.> Given any set of mutually exclusive nonempty sets, there exists at least> one set that contains exactly one element in common with each of the> nonempty sets.> Is there an argument against that?> --Keith Lewis klewis {at} mitre.org> The above may not (yet) represent the opinions of my employer.Try Banach-Tarski sphere paradox. One can't really argue against axioms, but it is possible to show thatthey lead to some strange consequences. === Subject: Re: Just curious: countable subset>Any infinite set has a countable subset.> It seems obvious that any set has subsets of ALL lesser cardinalities.Some consider finite sets to be countable, using the terms countablyinfinite or denumerable to mean a set with the cardinality aleph_0.Evidently this is not what the OP meant, since the question wouldotherwise be trivial (the empty set is countable and is a subset of everyset).> But the Axiom of Choice is also obvious to me, so maybe I'm biased.If infinite is defined to mean having a cardinality that is greaterthan or equal to aleph_0, then proving that every infinite (Dedekindinfinite) set has a denumerable subset is a trivial consequence of thedefinition and does not require AC.If, however, you define infinite to mean not having the cardinality ofany natural number, which is the standard definition in ZF, then you doneed AC, and it's not because a set can fail to have subsets of everysmaller cardinality. The problem is that there can be sets whosecardinality is not comparable with aleph_0 (not greater than, not equalto, and not less than). In other words, there may be a set A such thatneither of the sets A or N can be injected into the other.> Given any set of mutually exclusive nonempty sets, there exists at least> one set that contains exactly one element in common with each of the> nonempty sets.> Is there an argument against that?Just the fact that it can't be proved from the axioms of ZF (if ZF isconsistent), and it leads to results that some consider surprising.Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. Any infinite set has a countable subset.> It seems obvious that any set has subsets of ALL lesser cardinalities.>Some consider finite sets to be countable, using the terms countably>infinite or denumerable to mean a set with the cardinality aleph_0.Right. Of course it's trivially true that all sets have a subset of thesame cardinality as itself.I was under the impression that aleph_0 is the smallest infinitecardinality. >If, however, you define infinite to mean not having the cardinality of>any natural number, which is the standard definition in ZF, then you do>need AC, and it's not because a set can fail to have subsets of every>smaller cardinality. The problem is that there can be sets whose>cardinality is not comparable with aleph_0 (not greater than, not equal>to, and not less than). In other words, there may be a set A such that>neither of the sets A or N can be injected into the other.That's interesting, is there an example A?> Given any set of mutually exclusive nonempty sets, there exists at least> one set that contains exactly one element in common with each of the> nonempty sets.> Is there an argument against that?>Just the fact that it can't be proved from the axioms of ZF (if ZF is>consistent), and it leads to results that some consider surprising.Surprising? I would be interested in an example of that too. Feel free to point me to the appropriate references (especially on the web). My formal education in set theory comes from a single semester of DiscreteMath.--Keith Lewis klewis {at} mitre.orgThe above may not (yet) represent the opinions of my employer. === Subject: Re: Just curious: countable subset>If, however, you define infinite to mean not having the cardinality of>any natural number, which is the standard definition in ZF, then you do>need AC, and it's not because a set can fail to have subsets of every>smaller cardinality. The problem is that there can be sets whose>cardinality is not comparable with aleph_0 (not greater than, not equal>to, and not less than). In other words, there may be a set A such that>neither of the sets A or N can be injected into the other.>That's interesting, is there an example A?As I just pointed out in another post in this thread, see an example, though, with your preparation, you may not be able to understand why S there is such an example.>Given any set of mutually exclusive nonempty sets, there exists at least>one set that contains exactly one element in common with each of the>nonempty sets.>Is there an argument against that?>Just the fact that it can't be proved from the axioms of ZF (if ZF is>consistent), and it leads to results that some consider surprising.>Surprising? I would be interested in an example of that too.Some examples:- Two sets are not necessarily comparable. I.e., there are models of ZF where there are two sets and no injection from either one to the other.- Let C be a nonempty collection of disjoint nonempty sets. Without choice, it is not necessarily the case that there is an injection from C to its union. That is, one cannot show that the number of sets in C is smaller than the number of elements in all these sets.On the other hand, one can say that choice has some surprising consequences, too. For example, every set (e.g., the reals) can be well-ordered.>Feel free to point me to the appropriate references (especially on the web). >My formal education in set theory comes from a single semester of Discrete Math.First, I (again) strongly recommend reading Halmos, Naive Set Theory for a wonderful exposition of ZFC. The independence of Choice from ZF is a well-studied fact, but the development is quite advanced (usually covered in a graduate Set Theory course).Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Just curious: countable subset>Any infinite set has a countable subset.> It seems obvious that any set has subsets of ALL lesser cardinalities.>Some consider finite sets to be countable, using the terms countably>infinite or denumerable to mean a set with the cardinality aleph_0.> Right. Of course it's trivially true that all sets have a subset of the> same cardinality as itself.That's not sufficient to answer the question in ZF.> I was under the impression that aleph_0 is the smallest infinite> cardinality. That's true in ZFC. The question may not make sense in ZF.>If, however, you define infinite to mean not having the cardinality of>any natural number, which is the standard definition in ZF, then you do>need AC, and it's not because a set can fail to have subsets of every>smaller cardinality. The problem is that there can be sets whose>cardinality is not comparable with aleph_0 (not greater than, not equal>to, and not less than). In other words, there may be a set A such that>neither of the sets A or N can be injected into the other.> That's interesting, is there an example A?A Google search for Dedekind finite will turn up lots of hits.> Given any set of mutually exclusive nonempty sets, there exists at least> one set that contains exactly one element in common with each of the> nonempty sets.> Is there an argument against that?>Just the fact that it can't be proved from the axioms of ZF (if ZF is>consistent), and it leads to results that some consider surprising.> Surprising? I would be interested in an example of that too. Feel free to point me to the appropriate references (especially on the web). > My formal education in set theory comes from a single semester of Discrete> Math.Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. By the Nullstellensatz I + J = R, so there are f and g in I and J> resp. with f + g = 1. Then f(X) = {0} and f(Y) = {1}.>I don't quite get why f + g = 1. Would you explain in more detail?>The Nullstellensatz establishes a bijection between certain kinds of>ideals of F[x_1,...,x_n] and certain kinds of subsets of A^n, affine>space over F, when F is algebraically closed. Explicitly, to every>subset S of A^n we associate the ideal of all polynomials that vanish>on S,> I(S) = {f in F[x_1,...,x_n] : f(s) = 0 for all s in S}>and to each ideal I we associate the null set of I,> N(I) = {(x_1,...,x_n) in A^n : f(x_1,...,x_n)=0 for all f in I}.>The Zariski-closed sets and the radical ideals are in one-to-one>inclusion-reversing correspondence; the ideal of the union is the>intersection of the ideals; and the ideal of the intersection is the>sum of ideals.I think this last part is incorrect as stated; it should say theideal of the intersection is the radical of the sum of the ideals.That is the key observation; that if X and Y are Zariski-closed, then I(X intersect Y) = rad( I(X) + I(Y) ).So let me see if I can prove this explicitly, since it seems to be thehard part. Let Z= X intersect Y.By the inclusion reversing correspondence, I(Z) is the smallestradical ideal that contains both I(X) and I(Y) (since Z is the largestZariski-closed subset contained in both X and Y). Since rad(I(X)+I(Y)) is a radical ideal that contains both I(X) and I(Y), itfollows that I(Z) is a subset of rad( I(X) + I(Y) ).Conversely, since Z is contained in both X and Y, it follows that I(X)and I(Y) are both contained in I(Z), so I(X)+I(Y) is contained inI(Z). Therefore,rad( I(X)+I(Y) ) is contained in rad(I(Z)).By the Nullstellensatz, rad(I(Z))=I(Z). So this gives thatrad( I(X)+I(Y) ) is contained in I(Z), giving equality.This is all happening over an algebraically closed field F, of course.>Now, since they are disjoint, X intersect Y is empty. So the ideal>corresponding to X intersect Y must be all of F[x_1,...,x_n]. On the>other hand, the Nullstellensatz says that this is also equal to I(X) +>I(Y), that is, the>I + J = { f+g : f in I, g in J}.>Therefore, I+J = F[x_1,...,x_n].The reason we can do this in this case is that in any commutativering, if rad(I)=(1) then I=(1). So even though we technically we getrad( I(X)+I(Y) ) = F[x_1,...,x_n], we may then conclude immediatelythat I(X)+I(Y) = (1), as desired.I hope I did not make too many errors. Please correct any that are there. ============Subject: Re: Question about first-order logic>It is not the power-set, but the successor set: 5 = {4, {4}}.>Sigh; we seem to be endlessly quibbling over slight mis-statements.Right. There seems to be poor readers in this newsgroup. :-)>I do not think you mean to say 5 = {4,{4}}; you mean to say >5 = 4 union {4} = {0,1,2,3,4}.Right. Sorry for the typo. S(y) := y union {y}; n = {0, ..., n-1}. Hans Aberg === Subject: Re: Question about first-order logic> It is not the power-set, but the successor set: 5 = {4, {4}}.> Where is this peculiar definition to be found?Look for the infinity axiom of, for example, ZF (Zermelo Fraenkel)axiomatic set theory. It should be S(y) := y union {y}; sorry for thetypo. Hans Aberg === Subject: Re: Question about first-order logic> It is not the power-set, but the successor set: 5 = {4, {4}}.> But you suggested that I should look into books on logic to find>the successor of n defined as the power set of n.Nope. Hans Aberg === Subject: Re: Question about first-order logic> Nope. Then I don't understand your comment. Why did you respond, inresponse to my question where we find the oddity of each successiveinteger being defined as the power set of the previous one, thatI should look into books on mathematical logic? === Subject: Re: Question about first-order logic> Nope.> Then I don't understand your comment. Why did you respond, in>response to my question where we find the oddity of each successive>integer being defined as the power set of the previous one, that>I should look into books on mathematical logic?Do you want to have help with the facts, or merely waste time on a uselesspolemics? Forget the powersets; it is wrong.Also look up the theory of ordinal numbers. Applying the successors to theempty set will gnerate the firtt infinte ordinal omega; the ZF infinityaxiom says that it exists. Hans Aberg === Subject: Re: Question about first-order logic> Do you want to have help with the facts, or merely waste time on a useless> polemics? Forget the powersets; it is wrong. So in fact you haven't found it in the literature either? === Subject: Re: Question about first-order logic> Do you want to have help with the facts, or merely waste time on a useless> polemics? Forget the powersets; it is wrong.> So in fact you haven't found it in the literature either?Perhaps a gracious wink would help Hans to understand that you arejerking his chain a bit? I wouldn't want to think you are prolongingthis petty disagreement with any serious intent. *That* would be awaste of time, even if literally speaking you are on the winning side.(As Hans himself can verify by wasting enough more time to reread theentire thread carefully.) === Subject: Re: Question about first-order logic > I wouldn't want to think you are prolonging > this petty disagreement with any serious intent. But there doesn't seem to be any disagreement after all.