Re: [FWP] japhy had a silly idea...

["Daniel S. Wilkerson" <dsw@digital-integrity.com>]

Andrew Pimlott wrote:

> On Tue, May 29, 2001 at 05:48:00AM +0100, Pense, Joachim wrote:
> > Daniel S.Wilkerson wrote:
> > >Therefore, R1 and S have the same cardinality.  By the
> > Schroeder-Cantor-Bernstein
> > >theorem
> > >(http://www.math.lsa.umich.edu/~mathsch/courses/Infinity/Cardinality/Lesson
> > 4.shtml),
> > >there is a 1-1 onto map between R1 and S.  (Perhaps I could be accused of
> >
> > The objective of the sub-discussion is not the existence of an 1-1 mapping
> > between the two sets, but the presentation of an "obvious" one.
>
> People have different ideas of obvious :-)
>
> But his point is that the "obvious" "almost mapping" really is good
> enough, in the sense that the corner cases are negligible--which, to
> a mathematician, is obvious.
>
> Really--this mapping would be immediately accepted by any
> mathematician who wasn't being intentionally pedantic.
>
> Andrew

Recall that a question that lead to this was what is the cardinality of all
countable sequences of elements from a finite alphabet.  This is clearly strictly
greater than countable by Cantor's Diagonalization Argument previously given.
The obviousness of such a mapping seems not relevant to the previous question,
and additionally is a matter of taste.  One might wish for a constructable
demonstration of such a mapping (rather than invoking the Theorem I did as a
smart bomb).  This is doable.

Daniel


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