Cantor Confusion
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From: mueckenh on 24 Nov 2006 08:04 Franziska Neugebauer schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > > 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". > > As 2) may be considered as an "errata sheet" correcting the wording and > rectifiying 1) there is no contradiction. Why did it take so long to switch from "I don't need any advice from you" and from "slightly misleading" to "erratum"? Regards, WM
From: mueckenh on 24 Nov 2006 08:06 William Hughes schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > 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 standard model nor a > > non-standard model of ZFC. > > In which case it is nonsensical to try to talk about > Skolem's "paradox". Yes. It is nonsensical to talk about objects which don't exist. But as you talk about well-ordering of the real numbers and about transfinite numbers and such stuff, I thought it would entertain you, to talk about the structure of non-existing models of non-existing systems. Regards, WM
From: mueckenh on 24 Nov 2006 08:08 William Hughes schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > 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.) > > No it doesn't. Your claim is that assuming > > There exists an infinite set all of whose elements > are finite > > leads to a contradiction. But you have yet > to show a contradiction that doesn't require assuming > > There does not exist an infinite set all of whose elements > are finite Please learn: I use the finiteness of *every* line to change the quantifiers, because for linear and finite sets, this is unrefutably true. And unrefuted every line has a finite number of elements. The usual escape that every number is finite but that there are infinitely many numbers (this paradigm of nonsensical thinking) does not help here. In a line there are not infinitely many elements! Regards, WM
From: mueckenh on 24 Nov 2006 08:10 William Hughes schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > William Hughes schrieb: > > > > > 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. > > > > This holds for every finite line. Every finite line contains every > > element from every preceding line. There are only finite lines. > > > I claim > > There does not exists a line L_1 such that L_1 contains every > element from every line. > > You counter with > > For every line L_1, L_1 contains every element from every line > preceding L_1. > > However, these two statments are not contradictory. > > "every element from every line" > > is not the same thing as > > "every element from every line preceding L_1" > > > Yes it is true that > > For any element of the diagonal, d_nn, there exists a line > L_2, such that L_2 contains d_nn. > > However, the statement > > There exists a line L_1, such that > L_1 contains every element from every line > preceding L_1. > > is not the same as the statement > > There exists a line L_1, such that > L_1 contains every element from every line > preceding L_2. > > So we cannot use > > There exists a line L_1, such that > L_1 contains every element from every line > preceding L_1. > > to show that > > There exists a line L_1, such that L_1 > contains every element from every line For the lines, each of with has a finite number of elements, this *is the same*. If you have a linearly ordered set with a finite number of elements, then there is a maximum. Every line has a finite number of elements. Further the elements are linearly ordered. Therefore the above requirement is satisfied. There is a line which contains all the elements off all lines - unless there were an infinite number of elements. But that would not represent a natural number. Therefore you cannot counter with the usual argument that the elements of any line are all finite but that there are infinitely many of them. This paradigm of uncritical belief does *not* work here. Regards, WM
From: William Hughes on 24 Nov 2006 10:10
mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > 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.) > > > > No it doesn't. Your claim is that assuming > > > > There exists an infinite set all of whose elements > > are finite > > > > leads to a contradiction. But you have yet > > to show a contradiction that doesn't require assuming > > > > There does not exist an infinite set all of whose elements > > are finite > > Please learn: I use the finiteness of *every* line to change the > quantifiers, The fact that *every* line is finite does not mean there are only a finite number of lines unless you assume There does not exist an infinite set all of whose elements are finite > because for linear and finite sets, this is unrefutably > true. And unrefuted every line has a finite number of elements. The > usual escape that every number is finite but that there are infinitely > many numbers (this paradigm of nonsensical thinking) does not help > here. In a line there are not infinitely many elements! The facts that *every* line is finite and that every line L contains every element of every line preceding L does not mean -there are not an infinite number of lines -there is one line that contains every element from every line unless you assume There does not exist an infinite set all of whose elements are finite .. - William Hughes |