Cantor Confusion
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From: Franziska Neugebauer on 20 Dec 2006 08:44 mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: As usual you cut the relevant parts: ,----[ <4588194e$0$97248$892e7fe2(a)authen.yellow.readfreenews.net> ] | >> >> > 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? [(1)] | > Every element of the diagonal is a finite natural number, by | > definition. [(2)] | > | > 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. [(3)] `---- >> That does not prove (*) since the whole diagonal is _not_ bounded by >> any such n. > > If you doubt my proof, Which proof? You simply stated: (1) forAll elem in diag Exists line (m is in line) (2) forAll elem in diag (elem is finite natural number) (3) forAll natural n (forAll elem in diag <= n is in line n) So please show us _step-by-step_ how (*), which states (*) notExists natural n (forAll m in diag (m <= n)) -> Exists natural n' (notExists natural n(forAll m in line n' (m <=n)), follows from (1), (2) and (3). F. N. -- xyz
From: Franziska Neugebauer on 20 Dec 2006 08:55 Albrecht wrote: > There is not set N. Adieu! F. N. -- xyz
From: Newberry on 20 Dec 2006 09:21 mueckenh(a)rz.fh-augsburg.de wrote: > Newberry schrieb: > > > > >The edges have cardinal number aleph_0. > > > > Certainly. > > > > > A > > > subset of the paths can be shown to be in bijection with the real > > > numbers of the interval [0,1]. > > > > Which subset? Path as a set of intervals or path as a set of eges? How > > does the above follow from this? > > The paths are binary representations of all real numbers of the > interval [0,1]. Some rational reals have double representations, namely > such ending by 000... and such ending by 111... like 1.000... and > 0.111... . So there are not less real numbers in the interval [0, 1] > than paths in the binary tree. We have > > aleph_0 = |{edges}| >= |{paths}| >= |{reals of [0,1]}| = 2^aleph_0 > > while according to set theory > > aleph_0 < 2^aleph_0 > > Regards, WM Well, a Cantorist would say that this part does not hold |{edges}| >= |{paths}| because you considered only the finite paths but not the infinite paths.
From: mueckenh on 20 Dec 2006 12:31 Virgil schrieb: > In article <1166545363.371163.143890(a)i12g2000cwa.googlegroups.com>, > mueckenh(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. > > WM again loses track of the point at issue. > > No one wants anything but finite numbers IN the diagonal, but we all > want infinitely many of them. But there are not actually infinitely many. > > And since WM has used infinite binary trees as justifying his arguments, > he must allow that a complete path in such a tree is like a diagonal of > WM's EIT with its edges being the last members of the lines of that EIT. > > Which gives every line being finite but the diagonal being infinite. > Every edge is in some line but no line contains all edges. The tree proves that, given aleph_0 exists, aleph_0 is not less than 2^aleph_0. The EIT proves that aleph_0 does not exist. Regards, WM
From: mueckenh on 20 Dec 2006 12:31
Virgil schrieb: > In article <1166529653.157491.94810(a)48g2000cwx.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > > So you consider it nonsense that mathematics should consist of clear > > > definitions and proofs of theorems. > > > > No. But I need no proof in order to see that > > 1 + 1/2 + 1/4 + ...+ 1/2^n + ... >= 2 - 1/2^n > 3/2 for n > 2. > > > > Absent some formal definition, "1 + 1/2 + 1/4 + ...+ 1/2^n + ..." is > without any mathematical meaning whatsoever. No. It is less than 10. It was so before your kind of formal definition existed and it will remain so after your kind of formal definiton will long have ceased to be remembered by mankind. Regards, WM |