Cantor Confusion
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From: Virgil on 21 Nov 2006 19:38 In article <MPG.1fcd54d17c2f382e9899a4(a)news.rcn.com>, David Marcus <DavidMarcus(a)alumdotmit.edu> wrote: > mueckenh(a)rz.fh-augsburg.de wrote: > > Ralf Bader schrieb: > > > > > and Mueckenheim took that example for the main issue, ascribing > > > some other weird opinions to Cantor on the way. Although, surprisingly, > > > Mueckenheim's simplified proof of a restricted assertion is in principle > > > correct, it is also telling that he seems to consider it worth to be > > > published what actually is a mildly interesting exercise in basic point > > > set > > > topology. > > > > Unfortunately you missed the clue. The simplified version has only been > > developed, because it can easily be extended to uncountable sets. > > > > I showed that the real plane even stays connected if an *uncountable* > > set of points is left out. So Cantor's theorem is useless. It doesn't > > show a difference between countable and uncountable, as Cantor might > > have tried to suggest. > > So, it turns out that you were proving a different theorem than Ralph > thought. That's reassuring. For a moment there, we thought you might > have given a correct proof of something. Care to show us your proof that > the plane is connected even if an uncountable number of points is > deleted? I will accept that it can be connected if the "right" uncountable set is removed, but I can think of uncountable sets of points whose removal at least appears to totally disconnect the set of those remaining. For example if one removes all the points in the Cartesian plane which have either coordinate non-integral, what is left is about as disconnected as one can imagine.
From: William Hughes on 21 Nov 2006 19:42 mueckenh(a)rz.fh-augsburg.de wrote: > Randy Poe schrieb: > > > > > > > > You never heard of countable uncountable models? > > > > I've never heard of something which is both countable > > (can be bijected to N) and uncountable (can not be bijected > > to N). That would seem to require both P and (not P) to > > be true for some proposition P. > > Indeed, that is my impression too. But ZFC cannot tolerate P and notP. > Therefore people have dispelled it. If ZFC is correct, then, according > to a theorem of Skolem, it must have a countable model. But the most > important theorem of ZFC proves the existence of uncountable sets. > Therefore the worshippers of transfinity proclaim the existence of a > set which is countable (as Skolem requires) "from outside" but does not > contain the bijection with N internally. Therefore "inside the model", > countability cannot be proved - and all is fine. More piffle. Skolem's "paradox" does not concern us. Yes there is a countable model of ZFC. Yes, within this model there is a "set" of natural numbers N', and a "set" of real numbers R' such that within the model there is no "bijection" between N' and R'. But what you claimed was that there was no bijection between N' and N'. - William Hughes
From: Virgil on 21 Nov 2006 19:49 In article <MPG.1fcd6545fad1bf2f9899a5(a)news.rcn.com>, David Marcus <DavidMarcus(a)alumdotmit.edu> wrote: > Virgil wrote: > > In article <1164143109.673986.202100(a)b28g2000cwb.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > William Hughes schrieb: > > > > > > > > Note however that in order to attach a definite cardinal number to > > > > > every set, well-ordering must be possible for every set. That is why > > > > > Cantor insisted on well-ordering of all sets. > > > > > > > > Piffle. Well-ordering has nothing to do with the existence or not of > > > > at least one bijection. > > > > > > > You are wrong. If a set cannot be well-ordered, it cannot be assigned a > > > definite cardinal number. > > > > The existence of a cardinal which bijects to a set S is not a necessary > > condition for the identity function on S to exist. Every set, whether > > well orderable or not, has an identity function, at least in such > > relatively sane systems as ZF and NBG. > > > > If WM wishes to propose a system in which sets can lack identity > > functions, he will have to explicitely present the axiom system for such > > a system before anyone will accept its existence, much less its > > consistency. > > And, much less its sanity. That too.
From: MoeBlee on 21 Nov 2006 19:54 mueckenh(a)rz.fh-augsburg.de wrote: If ZFC is correct, then, according > to a theorem of Skolem, it must have a countable model. If ZFC is consistent, then it has a countable model. If one finds that to be an unappealing feature of ZFC, then fine, one can choose not to use ZFC or to seek some other theory. But ZFC is not inconsistent simply for Skolem's paradox nor is ZFC even rendered prohibitively counterintutitive by Skolem's paradox. There exists a bijection from the universe of the countable model onto N; but that bijection is not itself a member of the universe of the countable model. And why should we assume that the universe of such a countable model must have such a function as a member? MoeBlee
From: MoeBlee on 21 Nov 2006 19:57
My previous post formatted the quote incorrectly. The whole quote: "If ZFC is correct, then, according to a theorem of Skolem, it must have a countable model." is WM's. |