Cantor Confusion
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From: mueckenh on 23 Nov 2006 06:49 MoeBlee schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > That is not the question any longer. The question is: Can a set > > theorist admit that she is in error? It seems impossible. They all are > > too well trained in defending ZFC. > > In error as to what? Set theorists admit mistakes. It is not uncommon > for books to have errata sheets attached. Would you see a contradiction in these two statements? 1) "The cardinality of omega is |omega| not omega." 2) "The cardinality of omega, also written as |omega|, is omega". Regards, WM
From: mueckenh on 23 Nov 2006 06:59 MoeBlee schrieb: > > If some other theory is consistent, then it has a countable model. > > If a FIRST ORDER theory of predicate logic HAS AN INFINITE model, then > it has a countable model. Of course, I implied these properties. One cannot repeat always everything. But "HAS AN INFINITE" is superfluous, at least the capitals, because if the theory has not an infinite model, then it has to have a countable model too. (Every finite model is per definition countable. The meaning of "countable set" covers "finite set" as well as "denumerable infinite set". Didn't you study set theory and its definitions?) > > > > 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? > > Yes, Skolem was critical of set theory. What do you think, why? Was he too stupid to recognize this big advantage of mathematics? > language is. > > > Why do you think that our > > arithmetic (which is not a model of ZFC) does contain such a bijection? > > I don't know what bijection you are referring to. Exactly what > bijection in exactly what set do you claim existence? No bijection in any infinite set. But why do you think that there is a bijection N <--> Q? Regards, WM
From: mueckenh on 23 Nov 2006 07:10 MoeBlee schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > Countable means: There exists a bijection with N. After having > > recognized that this simple definition leads to a contradiction, > > We've been over this and over this and over this, and you still persist > incorrectly. You've not shown a statement P in the language of a Z set > theory such that both P and ~P are theorems of the theory. > > > people > > defined countable in and outside differently. > > I don't know who does that. I don't. It is necessary to avoid Skolems antinomy. If you have the same definition of countability inside and outside the countable model, then you will have the same result. But that would be a desaster. > > > And if they will > > recognize that there is really a contradiction in ZFC, then they will > > define contradiction differently. I am sure, they will not fail. > > If there is shown a contradiction in ZFC, then I won't change any > definitions; I'll just recognize that ZFC is inconsistent. But you've > not shown a contradiction in ZFC. Yes, the same track should have been followed by set theorists after Russell had shown his antinomy. But I am sure (and in fact I know) that you would not agree that a contradiction is a contradiction. And you are right. That's the simplest way to avoid learning that many of your efforts have been wasted in vain. Regards, WM
From: mueckenh on 23 Nov 2006 07:16 William Hughes schrieb: > > > > If both, W and N, are countable, then renaming the elements of one of > > > > them leads to an identity map. > > > > > > You are confusing bijections within the model to bijections > > > outside of the model. > > > > And who told you that you were outside? And who told you that you were outside? > You are noting that it is not possible to construct a bijection > in a nonstandard model and that it is > possible to construct a bijection in the standard model. This > is true, however, it is not a contradiction. In particular because there is neither a stadard model nor a non-standard model of ZFC. And in particular because the expression "contradiction" is not pat of ZFC. Don't you see that the whole aim of Newspeak is to narrow the range of thought? In the end we shall make thoughtcrime literally impossible, because there will be no words in which to express it. (George Orwell in "1984") Regards, WM
From: mueckenh on 23 Nov 2006 07:20
William Hughes schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > William Hughes schrieb: > > > > > > > > > > The two statments > > > > > > i: P(n) is true for every n an element of N. > > > ii: P(N) is true > > > > > > are not the same (trivial example P(x) is true iff x is an element of > > > N). > > > Induction can be used to prove statements of the form i. > > > (eg all elements of N are finite). Induction cannot be used > > > to prove statements of the form ii (e.g. N is finite). > > > > Here again your mathelogy comes to the surface. N is nothing but the > > collection of all natural numbers. They count themselves. If all are > > finite, then all are finite, > > Yes > > > i.e., then N is finite. > > No see above. The fact that all elements of N are finite > does not mean that N is finite. It does. (The EIT proves it.) The reason is the linear order and the constant distance of the elements. Regards, WM |