Virgil
Posts:
4,479
Registered:
1/6/11
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Re: Matheology � 203
Posted:
Jan 30, 2013 5:15 PM
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In article
<d9df6d19-b0a2-4c52-83a9-456b88acffde@f6g2000yqm.googlegroups.com>,
WM <mueckenh@rz.fh-augsburg.de> wrote:
> On 30 Jan., 10:31, William Hughes <wpihug...@gmail.com> wrote:
>
> > For a potentially infinite list L, the
> > antidiagonal of L is not a line of L.
>
> Of course. Every subset L_1 to L_n can be proved to not contain the
> anti-diagonal
> >
> > Does this imply
> >
> > There is no potentially infinite list
> > of 0/1 sequences, L, with the property that
> > any 0/1 sequence, s, is one of the lines
> > of L.
>
> Do you mean potentially infinite sequences?
> Look, everything Cantor does, concerns only finite initial segments.
> You could cut off the sequences behind the digonal digit.
But the anti-diagonal does not nave any merely "diagonal" digit, so must
be endless.
>
> The only thing not terminating, then could be the diagonal itself. But
> then you would claim that the diagonal differs from every entry,
> because it has more digits. In the original argument, the diagonal
> differs at the same places that also exist in the entries. Therefore
> the argument with the diagonal "being longer" is wrong.
Any anti-diagonal differs from each listed entry of the list from which
it is derived in AT LEAST one digit position, so is not a listed entry
of that list. Thus those. like WM, who claim existence of a complete
lists of functions from |N to any set of more than one member are wrong.
>
> So in fact, Cantor shows that the countable set of all terminating
> decimals is uncountable.
WM claims to be able to prove it but his "proof" is only invalid in his
wild weird world of in Wolkenmuekenheim.
What Cantor actually showed was that given a list of functions each
having domain |N and a codomain of at least two members there is no
surjection from |N to that set of functions. Thus any such set satisfies
the standard definition of uncountability.
That WM does not like it is his problem, not ours.
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