Cantor Confusion
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From: Virgil on 22 Nov 2006 14:47 In article <1164195919.471077.233250(a)f16g2000cwb.googlegroups.com>, 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. > > Regards, WM Sequences are functions, a very special sort of set, those with domain N. Since not all sets are sequences, not all sets have the properties that sequences must have in order to be sequences.
From: Virgil on 22 Nov 2006 14:57 In article <1164196369.064190.253870(a)h48g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1164126395.211430.7520(a)h54g2000cwb.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Tell me which point is not accepted: > > > 1) Every line which contains an index of the diagonal, contains all > > > preceding indexes of the diagonal too. > > > 2) Every index of the diagonal is in a line. > > > 3) In order to show that there is no line containing all indexes of the > > > diagonal, there must be found at least one index, which is in the > > > diagonal but not in any line. > > > > This one is flat out false, unless one assumes, a priori, a last line > > and a last member of the diagonal. > > We assume not a last line, but we assume that eery line has finitely > many indexes. And this is true. What you allege in (3) does not follow from this. It is true that given any line there will be a diagonal element not in that line. It is false that given a diagonal element there is no ine which contains it. WM again dyslexes his quantifiers. > > > > It is certainly false in ZF or NBG, where such an assumption is also > > false. > > For finite indexes it is correct. Infinite things are not constrained to behave in all respects like finite ones. That is because "infinite" means "not finite". > > > > It is quite enough to show that for every index in the diagonal, > > except 1, there is some line not containing that index. Then no line > > can contain every index. > > Name two finite indexes which cannot be in one line. That is not at all relevant to what I said. If WM cannot read English, he should get someone to translate it for him. I know I cannot read German well enough to be sure of meanings so I do not pretend otherwise.
From: Virgil on 22 Nov 2006 15:05 In article <1164197100.642614.53260(a)e3g2000cwe.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > MoeBlee schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > 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. > > > > WRONG. It is a theorem that from the set of finite sequences of members > > of a finite alphabet there is an injection into N. > > Nevertheless it is impossible to construct such an injection (as it s > impossible to construct a well order of the reals). Actually, it is quite trivial, Give each character in the alphabet of n characters a different integer value between 0 and n-1 inclusive, and represent the members of N in base n, and the bijection associates each word with the numeral, base n. By induction, at least in ZF or NBG or any system allowing induction, the construction is complete! > > > > > Why should it be > > > possible to construct a bijection between N and N? > > > > Because ANY x is 1-1 with x, by the identity function on x. > > It is an unfounded assumption to believe that all x of the set appear > in this bijection, only by writing x = x. Ridiculous! WM is often ridiculous, but never more so than in denying x = x. > > PS: What about the binary tree in ZFC? Any results? Any idea why it > can't be done? I have no idea why WM cannot do a binary tree in ZFC, as it is easy enough to do.
From: Virgil on 22 Nov 2006 15:08 In article <1164197273.049425.266360(a)b28g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > > > > > > If there is a bijection with all finite line ends, then there is a > > > bijection with one line. > > > > No. > > The diagonal contains only finite line ends. Every element > > d_nn, of the diagonal, is the end of the line which ends > > at the natural number n. > > > > The bijection with all finite line ends is: The line which ends > > in n > > maps to the element d_nn. > > > > There is no bijection with one line. Such a bijection would > > imply that there is a largest d_nn. > > No. It implies only that all d_nn are finite. If you disagree, please > give an example for two elements of the diagonal which cannot be found > in one single line. If you agree, please give an example of 2 = 1. makes as much sense as WM's request. For every line, there is a finite length, say n, so that the diagonal elements beying n, and there are infinitely many of them, are all not in that line. WM's blind spot seems to be growing.
From: William Hughes on 22 Nov 2006 15:29
mueckenh(a)rz.fh-augsburg.de wrote: > If there is no bijection with one line, then there must be an element > of the diagonal outside of every line. No. Not unless the one line contains every element from every line. > (Because what can be done with > several *finite* lines in their linear order, can be done with one > line). What can be done with all lines in their linear order cannot be done with one line. No, you cannot use the fact that a bijection with one line must exist to prove that the set of all lines doesn't exist, then turn around and use the fact the the set of all lines doesn't exist to prove that a bijection with one line must exist - William Hughes |