In article <firstname.lastname@example.org>, WM <email@example.com> wrote:
> On 4 Mai, 22:45, Virgil <vir...@ligriv.com> wrote: > > In article > > <06206984-5944-49e6-9ba6-9f795516c...@p14g2000vbn.googlegroups.com>, > > > > > > > > WM <mueck...@rz.fh-augsburg.de> wrote: > > > On 4 Mai, 18:30, Dan <dan.ms.ch...@gmail.com> wrote: > > > > > > Either they're both relevant to Cantor's argument, or they're both > > > > > > irrelevant . > > > > > > > Cantor's argument > > IS valid! > > > > > > Why should it be considered important under the definition of > > > > equality? If they have at least one different digit , THEY'RE > > > > DIFFERENT. > > > > > But there always only finitely many with at least one different digit > > > whereas there are always infinitely many with none different digit. > > > > Name one! > > A very good question.
So good that WM has no answer to it. >
> > > Thus (reals countable) <==> (reals listable). > > > > But Cantor has shown that there is no complete list of real possible as > > every listing of reals necessarily omits some reals. > > And counterfactually he has assumed that there is a complete list of > natural numbers.
If there isn't then the set of naturals is not countable either. > > > > > My list is constructed such that all possible squences of digits 1 are > > > already in the list. No longer sequence is possible
Except for the concatentation of all those listed finite sequences, which is clearly longer than any of them.
> > Unless your list claims to list all reals, it has no effect on the > > validity of the Cantor diagonal argument, which remains valid in spite > > of it. > > It is true: If the naturals are complete somewhere, the reals cannot > be complete there. Alas this theorem is void since the naturals cannot > be complete anywhere.
Then the set of naturals is also not a countable set.
At least not in Wolkenmuekenheim.
But outside Wolkenmuekenheim, the set of naturals IS countable and the set of reals is NOT. --