Cantor Confusion
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From: mueckenh on 19 Dec 2006 11:23 Franziska Neugebauer schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > > Franziska Neugebauer schrieb: > >> mueckenh(a)rz.fh-augsburg.de wrote: > >> > William Hughes schrieb: > >> [...] > >> >> Assume that there exists an L_D which contains all the numbers > >> >> contained in the diagonal. L_D is bounded, > >> > > >> > If an unbounded diagonal exists, then obviously an unbounded line > >> > must exist. [(*)] > 6. You shall refocus on a proof of (*). Every element of the diagonal is n some line. Do you agree? Every element of the diagonal is a finite natural number, by definition. Up to the finite natural number n: Every element =< n of the diagonal is contained in line n. Induction shows this is valid for any finite natural number. There are only finite numbers in the diagonal. QED. If you cannot understand this proof, then simply try harder. If you don't succeed but nevertheless are convinced of the contrary, remember: N is a well ordered set. So there must be a least number n of the subset for which my statement is not true. Try to find it. Regards, WM
From: mueckenh on 19 Dec 2006 11:25 Franziska Neugebauer schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > > William Hughes schrieb: > [...] > >> L_D is a line and is therefore bounded. > >> > >> Contradiction. > >> Hence, the _assumed_ L_D does not exist. > >> > > Assumed is the diagonal. Premise: "If the diagonal is unbounded." > > L_D is a line and is therefore bounded. > > Yes. And because L_D is bounded it cannot "contain" every element of the > diagonal. Exactly this constitutes the contradiction to your claim P1. Wrong conclusion. Because every line (call it L_D or as you like) is bounded the diagonal cannot be actually infinite. Regards, WM
From: William Hughes on 19 Dec 2006 11:33 mueck...(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > William Hughes schrieb: > > > > > > > Albrecht wrote: > > > > > > > > but in the same time |N isn't complete > > > > > since |N is bijectable to the diagonal and the diagonal is never > > > > > complete since there is no complete line (number) which covers the > > > > > diagonal. > > > > > > > > No, the claim is that there is no line that covers the diagonal. > > > > > > Which results in the conclusion that there is no diagonal. > > > > No. Only properties of the diagonal that you > > have accepted are used. > > The conclusion is that there is no line. > > Every element of the diagonal is n some line. > Every element of the diagonal is a finite natural number, by > definition. > > Up to the finite natural number n: Every element =< n of the diagonal > is contained in line n. Induction shows this is valid for any finite > natural number. > > There are only finite numbers in the diagonal. QED. Given that the diagonal is the potentially infinite set of natural numbers, the fact that there are only fiinite numbers in the diagonal is immediate. The only property of the diagonal that is used is the fact that it is not bounded. In particular -it is assumed that the diagonal contains only finite numbers -it is not assumed that all the elments of the diagonal exist The contradiction holds, using only properties of the potentially infinite set of natural numbers that you have stated. - William Hughes
From: William Hughes on 19 Dec 2006 11:48 mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > William Hughes schrieb: > > [...] > > >> L_D is a line and is therefore bounded. > > >> > > >> Contradiction. > > >> Hence, the _assumed_ L_D does not exist. > > >> > > > Assumed is the diagonal. Premise: "If the diagonal is unbounded." > > > L_D is a line and is therefore bounded. > > > > Yes. And because L_D is bounded it cannot "contain" every element of the > > diagonal. Exactly this constitutes the contradiction to your claim P1. > > Wrong conclusion. Because every line (call it L_D or as you like) is > bounded the diagonal cannot be actually infinite. Moreover, the diagonal cannot be bounded. This is the contradiction. (It is not necessary to assume that the diagonal is "actually infinite" to show that it is unbounded). - William Hughes
From: Franziska Neugebauer on 19 Dec 2006 11:53
mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: >> mueckenh(a)rz.fh-augsburg.de wrote: >> > William Hughes schrieb: >> [...] >> >> L_D is a line and is therefore bounded. >> >> >> >> Contradiction. >> >> Hence, the _assumed_ L_D does not exist. >> >> >> > Assumed is the diagonal. Premise: "If the diagonal is unbounded." >> > L_D is a line and is therefore bounded. >> >> Yes. And because L_D is bounded it cannot "contain" every element of >> the diagonal. Exactly this constitutes the contradiction to your >> claim P1. > > Wrong conclusion. Because every line (call it L_D or as you like) > is bounded the diagonal cannot be actually infinite. Non sequitur. Modus tollens requires the validity of the implication. It is your job to _prove_ it. F. N. -- xyz |