```Date: Feb 25, 2013 7:05 AM
Author: mueckenh@rz.fh-augsburg.de
Subject: Re: Matheology § 222 Back to the roots

On 24 Feb., 22:22, Virgil <vir...@ligriv.com> wrote:> There is an equally forceful necessity that, for every  and any member> of your list, unless your list has a fixed last member, the diagonal is> necessarily longer than any given member.And the list is longer then every FIS of lines l_1 to l_n. But whatdoes the remaining part consist of if not of lines?> So just what "isomorphism", other than mere bijection, is WM> imagining in his tiny mind?First let us record that your original assertion is wrong - as in mostcases of your writings.Second, since the paths are nothing but another notation of the binarystrings, we have identity. There remains nothing to prove.>> Simplest logic. Try to find a set that contains its number if it does>> not contain its number. Isn't that simple?> How does that apply to, say, the set of von Neumann natural numbers> in ZF?It applies to Hessenberg's "proof". The set of all subsets of |N thatare not containing that natural number which is mapped upon them,simply does not exist. It is not predicably defined.> > In mathematics we have variables which can assume values.> But variables do not take arguments like A taking A_1.That depends on what you let them take.> What we think is that all your lines are in d, which is, in a sense,> only the union of all your infinitely many lines.Therefore it cannot be more than all lines. And each one is finite,wih no regard how many there are.>> And certainly you don't claim that you can>> find more than one line that would be required to contain what one>> line contains?> No one of your lines contains its own successor so no one of your lines> can contain the line d which contains all successors.There are not *all* successors in potential infinity but only everyone up to every n.Regards, WM
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