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 what does 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 most cases of your writings.
Second, since the paths are nothing but another notation of the binary strings, 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 that are 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 every one up to every n.
