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From: Virgil on 20 Oct 2005 19:02 In article <MPG.1dc1b877f4ab567098a511(a)newsstand.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > Is it > impossible to biject an infinite ordered set with its power set? Nope Repeated posting of a falsehood does not make it any less false.
From: Virgil on 20 Oct 2005 19:12 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)} > No one has been able to tell me, no matter how many times I > beg for assistance in understanding this deep and mysterious subject. To has been told, but his illiteracy prevents him from knowing it. > > > > If anyone counterclaims agains the proof of no surjection from and > > set X to its power set P(X), that person owes us an example of such > > a mapping. > And if anyone wants to claim their proof stands in the face of an > obvious counterexample, that person owes us an explanation of how > that counterexample does not apply. So, what of it, Virgie? It is TO claiming the counter-example to a standard proof. Let him find the counter-example, an S and a f:S -> P(S), and an s in S with f(s) = (x in S: x not in f(x)}. > What rule > does my bijection between *N and P(*N) break? Take your time.... See above!
From: Virgil on 20 Oct 2005 19:14 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.
From: Virgil on 20 Oct 2005 19:25 In article <MPG.1dc1bf82bef7f93f98a517(a)newsstand.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: .. > You are assuming a finite natural number of bits, but that is simply > not the case with *N, or it would ony contain finite values. Are we > talking about the same set? Only TO has any clue as to what he is talking about. > > The conclusion is that there does not exist a surjection from the > > members of *N to the members of P(*N), and therefore no bijection > > between them is possible. > But that conclusion assumes a last element, whence it draws its > contradictions. What 'last element' is that? The disproof of TO's alleged bijection merely requires the production of a member of the power set not in the image of the function, which does not require any 'last element', no matter how much TO claims otherwise.
From: David R Tribble on 20 Oct 2005 19:26
Tony Orlow: >> What do you want me to try, anyway, and infinite mapping, >> element-by-element? A bijection's a bijection, right? > David R Tribble: >> Yes, that would be nice. Please show us your bijection. > Tony Orlow: >> f(0) = ...000 = {} >> f(1) = ...001 = {0} >> f(2) = ...010 = {1} >> f(3) = ...011 = {0,1} >> f(4) = ...100 = {2} >> f(5) = ...101 = {0,2} >> f(6) = ...110 = {1,2} >> f(7) = ...111 = {0,1,2} >> etc. Tony Orlow: > Notice above that no natural maps to a subset which contains it, so w is all > of N. Imagining any completed w leads to a contradiction, since the natural > that would map to it is always bigger than every natural in that set. > That's okay though, because for every natural, there's a larger one. If x > exists, 2^x exists. The sets are infinite, so the bijection continues, > [...] You realize, of course, that you've just proved that there is no natural x that can map all the members of N, don't you? You say: Choose a natural x in N that maps f(x) to all the naturals in N. But x must be bigger than all naturals in N, so choose 2^x (which we know exists, because x exists), and call it x'. Oh, but now x' must be larger than all naturals in N, so choose x'' = 2^x'. Oh, but now, x'' must be larger, ... and so on. We conclude: Therefore we cannot find an x in N that maps f(x) to N. Therefore there is no x in N that maps f(x) to N. Therefore x does not exist. Therefore f(x) is not a surjection from N to P(N). Therefore f(x) is not a bijection from N to P(N). |