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From: Tony Orlow on 21 Oct 2005 14:05 Virgil said: > In article <MPG.1dc1b69de12df94698a510(a)newsstand.cit.cornell.edu>, > Tony Orlow <aeo6(a)cornell.edu> wrote: > > > stephen(a)nomail.com said: > > > Tony Orlow <aeo6(a)cornell.edu> wrote: > > > > David R Tribble said: > > > >> Tony Orlow wrote: > > > >> > > > >> > > > >> > What do you want me to try, anyway, and infinite mapping, > > > >> > element-by-element? A bijection's a bijection, right? > > > >> > > > >> Yes, that would be nice. Please show us your bijection. > > > > f(0) = ...000 = {} f(1) = ...001 = {0} f(2) = ...010 = {1} f(3) = > > > > ...011 = {0,1} f(4) = ...100 = {2} f(5) = ...101 = {0,2} f(6) = > > > > ...110 = {1,2} f(7) = ...111 = {0,1,2} > > > > > > If we define w= { x : x not in f(x) } then we get w = {0, 1, 2, > > > 3, ...... } > > > > > > So for what y does f(y) = { 0, 1, 2, 3, ..... }? > > > > > > And is y in f(y)? > > > > > > Let me guess. You will claim that F(N) = { 0, 1, 2, 3, ..... }. So > > > is N in { 0, 1, 2, 3, ..... }? If it is, then N is not in w, and > > > F(N) does not equal w. If it is not, then N is in w, and F(N) does > > > not equal w. <snip> > > > > > > > But I have constructed a bijection between the two using an > > > > intermediate binary representation. What is the specific rule I > > > > have broken concerning the construction of bijections. If I > > > > haven't broken any such rule, then is it true that a bijection > > > > between two sets means that have the same size, or even > > > > cardinality? > > > > > > No you have not. > > > > > > Stephen > > > > > Notice above that no natural maps to a subset which contains it, so w > > is all of N. Imagining any completed w leads to a contradiction, > > since the natural that would map to it is always bigger than every > > natural in that set. That's okay though, because for every natural, > > there's a larger one. If x exists, 2^x exists. The sets are infinite, > > so the bijection continues, even though there is a difference between > > the values which are in the subsets and the values which denote the > > subsets. > > None of this handwaving and doubletalk designates any member of *N which > maps to {x in *N: x not in f(x)}. And without it, TO's "bijection " is > not even a surjection. > So, it is required that we determine the very last element in an infinite bijection? Why must I name the end of the set in order for the bijection to be valid? You cannot name anything except some conceptual end of the unending set as a point where the bijection breaks down. I swear this theory just seems more and more insane to me. How do you tolerate it? Ugh! Please line up by the basketball courts for your head-whacking. Maybe that'll work. At least it'll be fun. -- Smiles, Tony
From: Randy Poe on 21 Oct 2005 14:09 Tony Orlow wrote: > Virgil said: > > None of this handwaving and doubletalk designates any member of *N which > > maps to {x in *N: x not in f(x)}. And without it, TO's "bijection " is > > not even a surjection. > > > So, it is required that we determine the very last element in an infinite > bijection? Where do you see anybody talking about the order, let alone the "very last element"? No it is not required to talk about the very last element. All that is required to prove a mapping is not a bijection from set A to set B is to identify one element of either set which is not part of the mapping. There is no requirement that it be the "last". Nobody but you has used the term "last". Nobody but you has talked about the ordering of the mapping. What we are pointing out is that this set IS NOT MAPPED by any element. It need not be the mythical "end" that this place in the mapping is missing. It's just missing. End of story. - Randy
From: Tony Orlow on 21 Oct 2005 14:12 David R Tribble said: > Tony Orlow: > >> I already showed you the bijection between binary *N and P(*N). > >> What didn't you like about it? It is valid. > > > > David R Tribble: > >> No, you showed a mapping between *N and R, which is equivalent > >> to a mapping between *N and P(N). That's easy. > > > > Tony Orlow: > >> What do you want me to try, anyway, and infinite mapping, > >> element-by-element? A bijection's a bijection, right? > > > > David R Tribble: > >> Yes, that would be nice. Please show us your bijection. > > > > Tony Orlow: > >> f(0) = ...000 = {} > >> f(1) = ...001 = {0} > >> f(2) = ...010 = {1} > >> f(3) = ...011 = {0,1} > >> f(4) = ...100 = {2} > >> f(5) = ...101 = {0,2} > >> f(6) = ...110 = {1,2} > >> f(7) = ...111 = {0,1,2} > >> etc. Any questions? > > David R Tribble: > >> Where are the infinite subsets, such as the set of even numbers? > > > > Tony Orlow: > > Well, infinitely far down the list of course! > > . > > . > > . > > f(N/3) = 0:010101.....010101 = {0,2,4,6,8,....} > > f(2N/3)= 0:101010.....101010 = {1,3,5,7,9,....} > > All of the sets you listed, finite and infinite, are composed of > only finite naturals. Which means that they are all subsets of N. Incorrect. See the digits to the left of the ellipses? Those are in infinite positions, representing infinite evens and odds. > > You're using N (and N/3, 2N/3, etc.), which is a member of *N. Yes we are talking aboout *N. > > So all you're doing is showing a bijection between *N and P(N), which > is what I said you were doing in the first place. That's easy, > and is equivalent to a bijection between R and P(N). No, it's P(*N). The sets include elements of infinite value. > > But we're asking for a bijection between *N and P(*N), or between > N and P(N). You haven't adequately shown those yet. It's done. > > > > Oh, now comes the list of primes (sigh) > > Uh, um.... > > > > f(N/pi) = ....1010001010110 = {2,3,5,7,11,13,...} > > > > PS, I'm kidding about the N/pi part (I think). > > The infinite natural x in *N that maps f(x) to the infinite set of > primes is: > x = 2^2 + 2^3 + 2^5 + 2^7 + 2^11 + 2^13 + ... > > But like all the other subsets your f mapping defines, the set of > primes contains only finite numbers, which means that it is one of > the subsets of N. If there are no infinite primes, then the subset would correspond to a finite value. Not all subsets have to have infinite values. I have only listed finite numbers here, but do take note of the ellipses. Are there infinite primes? If not, then there arent an infinite number of them. > > Do keep trying, though. > > -- Smiles, Tony
From: Tony Orlow on 21 Oct 2005 14:16 Virgil said: > In article <MPG.1dc1ba2d6d98debc98a513(a)newsstand.cit.cornell.edu>, > Tony Orlow <aeo6(a)cornell.edu> wrote: > > > Virgil said: > > > > Then lets see one of your alleged bijections from some set X to its > > > power set P(X). > > > > > > The only valid disproof of a proof of non-existence is an example > > > of what is alleged to be non-existent > . > > Hmm, so what was the problem with my bijection again? What rule did I > > break. > > You failed to come up with any s in S such that > f(s) = {x in S:x not in f(x)} I failed to name the last natural? Shame on me! > > > No one has been able to tell me, no matter how many times I > > beg for assistance in understanding this deep and mysterious subject. > > To has been told, but his illiteracy prevents him from knowing it. What rule of construction did I break in my bijection? What did I do wrong? > > > > > > If anyone counterclaims agains the proof of no surjection from and > > > set X to its power set P(X), that person owes us an example of such > > > a mapping. > > > And if anyone wants to claim their proof stands in the face of an > > obvious counterexample, that person owes us an explanation of how > > that counterexample does not apply. So, what of it, Virgie? > > It is TO claiming the counter-example to a standard proof. > > Let him find the counter-example, an S and a f:S -> P(S), > and an s in S with f(s) = (x in S: x not in f(x)}. 2^oo-1 maps to {0,...,oo-1}. > > > What rule > > does my bijection between *N and P(*N) break? Take your time.... > > See above! What did I do wrong, besides failing to name the last natural? > -- Smiles, Tony
From: Tony Orlow on 21 Oct 2005 14:18
Virgil said: > In article <MPG.1dc1ba6ecb69b51a98a514(a)newsstand.cit.cornell.edu>, > Tony Orlow <aeo6(a)cornell.edu> wrote: > > > Virgil said: > > > In article <1129721500.027651.21470(a)f14g2000cwb.googlegroups.com>, > > > > That shows how little AS knows about things, since given the axiom of > > > choice, there is no such thing as a non-well-orderable set (though > > > admittedly there are some sets known that have yet to be well-ordered). > > > > > Gee, like what, Virgil? > > Like one version of the axiom of choice, which merely states that every > set is well-orderable, but gives no methodology for doing it. > Well, that sounds like rock-solid proof to me! Stupid arbitrary axioms. Make it make sense. -- Smiles, Tony |