Cantor Confusion
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From: Eckard Blumschein on 22 Nov 2006 10:17 On 11/21/2006 10:24 PM, Virgil wrote: > In article <456304B0.70705(a)et.uni-magdeburg.de>, > Eckard Blumschein <blumschein(a)et.uni-magdeburg.de> wrote: > >> On 11/12/2006 8:28 PM, William Hughes wrote: >> > mueckenh(a)rz.fh-augsburg.de wrote: >> >> > As you stated "Goes on forever" is a property of the natural >> > numbers. The set {1,2,3,...} is just the natural numbers. So >> > this set must go on forever. >> >> Being admittedly not very familiar with set theory, I nonetheless wonder >> if sets are considered like something going on forever. > > Sequences do, sets do not. > > When one represents the members of a set as members of a sequence, as > {1,2,3,...} does, that little ambiguity should not really mislead anyone. That little ambiguity? Doesn't it make a categorical difference if a set has been a priori set for good?
From: mueckenh on 22 Nov 2006 10:22 William Hughes schrieb: > 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. It does. It is not only a paradox but an antinomy. > 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'. If someone would have claimed, before 1920, that between a countable set N' and a countable set R' no bijection can exist, you would have answered piffle or so. Now you are in the same situation. Regards, WM
From: Eckard Blumschein on 22 Nov 2006 10:25 On 11/22/2006 12:45 PM, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > >> In article <456304B0.70705(a)et.uni-magdeburg.de>, >> Eckard Blumschein <blumschein(a)et.uni-magdeburg.de> wrote: > >> > Being admittedly not very familiar with set theory, I nonetheless wonder >> > if sets are considered like something going on forever. >> >> Sequences do, sets do not. >> > Sequences are sets. Perhaps you correctly learned set theory. So I guess, William Hughes was correct and Virgil this time wrong. I conclude, all elements of an infinite sets are a priori set for good while simultaneously the process of setting is going on forever. Hm, pretty strange. Regards, Eckard
From: mueckenh on 22 Nov 2006 10:31 MoeBlee schrieb: > 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. If some other theory is consistent, then it has a countable model. > But ZFC is not > inconsistent simply for Skolem's paradox nor is ZFC even rendered > prohibitively counterintutitive by Skolem's paradox. That's why Skolem liked ZFC soo much? > 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? Because the bijection from the model (N') onto N *is* a countable set. And if the model contains countable sets (N'), why then does it not contain this set, which bijects N' and N? Why do you think that our arithmetic (which is not a model of ZFC) does contain such a bijection? Regards, WM
From: mueckenh on 22 Nov 2006 10:37
William Hughes schrieb: > > The countability of the set W of all the words of a finite alphabet is > > proved by other means than the usual countability proof. The latter > > consists in constructing a bijection with N, i.e., in constructing a > > list. But it is impossible to construct a list of all words. It is > > impossible to construct a bijection between W and N. Why should it be > > possible to construct a bijection between N and N? > > Because a bijection between W and N cannot be the identity map > but a bijection between N and N can be the identity map. If both, W and N, are countable, then renaming the elements of one of them leads to an identity map. Regards, WM |