Cantor Confusion
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From: William Hughes on 22 Nov 2006 14:29 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > 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. > > > > You are confusing bijections within the model to bijections > > outside of the model. > > And who told you that you were outside? > > But you misunderstood me. W and N are both in "our" universe. You are noting that it is not possible to contruct a bijection in a nonstandard model and that it is possible to contruct a bijection in the standard model. This is true, however, it is not a contradiction. - William Hughes > > Regards, WM
From: William Hughes on 22 Nov 2006 14:32 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. - William Hughes
From: Virgil on 22 Nov 2006 14:36 In article <1164192364.831131.322070(a)b28g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Randy Poe schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > It is not "hard in practice", but it is impossible. There is no chance. > > > That means "in principle". > > > > No, that is not the meaning of "in principle". > > It is. What you mean is "in the magical realm of > transfinity-worshippers" > > Regards, WM Only someone without principles could misinterpret what "in principle" means so badly.
From: Virgil on 22 Nov 2006 14:39 In article <1164192462.824282.134620(a)m7g2000cwm.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > No. You can develop "potential set theory". Just define > > an element of a potential set to be an element of any one > > of the arbitrary sets that can be produced. Now go through > > set theory and add the word "potential" in front of each > > occurence of the word set. There is no difference between > > saying "a potentially infinite set exists" and saying "an actually > > infinite set exists". > > A potentially infinite set has no cardinal number. It cannot be in > bijection with another infinite set because it does not exist > completely. Not true in any extant axiom system. In every axiom system that has been presented so far, there are finite sets and possibly infinite sets, but nothing in between. Something which is potentially, but not actually, infinite cannot be a set at all in any set theory yet seen here.
From: Virgil on 22 Nov 2006 14:45
In article <1164195797.399821.219210(a)h48g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > One counter example contradicts ZFC: > > > There is not one single element of the diagonal which is not contained > > > in a line. This line contains this and all preceding elements. > > > > While true, it is irrelevant to the issue of whether there is one line > > containing every element of the diagonal, which there is not. > > > > WM conflates "every element of the diagonal is in SOME line", which is > > true, with "every element in the diagonal is in THE SAME line", which > > is false. > > Tha means we need at least two lines for the elements o the diagonal? Not to anyone who understands logic. What it does mean is that we need infinitely many finite lines to get all of the infinitely many members of the diagonal. At least to people of any sense. > Please give an example which requires that at least two different lines > are needed to contain two elements of the diagonal. Please give any line for which every element of the diagonal is in THAT line. |