# Re: Countable vs Continuous

No - the set of computable numbers does not form a
continuum. Continuity is related to the concept of limits: {x_i} is a
convergent sequence if
\forall \epsilon>0, \exist N: |x_i-x_N|<\epsilon.
A continuous space is one for which every convergent sequence
converges to a limit, ie 

\exists x: \forall\epsilon>0\exists N: |x_i-x|<\epsilon \forall i>N.

we commonly denote x by lim_{i->\infty}x_i.

There are many convergent sequences x_i whose limits cannot be
computed (uncountably many, in fact).

However, my point has always been that the set of computable numbers
is not a discrete set, since between any two computable numbers, a
third can be found. This property is true of the rationals as well.

What should be chucked in the rubbish bin is the concept that only
discrete or continuous universes can exist - obviously other
mathematical structures exist, and I believe we happen to be living in
one.

Cheers

Brent Meeker wrote:
>
> On 22-Jun-01, [EMAIL PROTECTED] wrote:
> >> or continous. Don't the computable numbers form a continuum; hence
> >> even restricting the universe to one we can describe would still
> >> allow it to be continuous?
> >>
> >> Brent Meeker
> >
> > No, the computable numbers do not form a continuum - there are not
> > more than countably many of them. Any real number computable in the
> > limit (such as Pi) has a finite nonhalting program; the set of all
> > such programs cannot have higher cardinality than the integers.
> >
> > Juergen Schmidhuber
> >
> > http://www.idsia.ch/~juergen/
> > http://www.idsia.ch/~juergen/everything/html.html
> > http://www.idsia.ch/~juergen/toesv2/
> >
> Thanks for the reply, Juergen.  I guess I didn't phrase my question
> right.  I know that the cardinality of the computable numbers is the
> same as the integers.  What I was asking was whether the computable
> numbers form a continuum in the topological sense (I'm pretty sure they
> do) - AND - is this a sufficient continuum to provide a model of
> continuous space-time?  Again, I think it is - but I don't know of a
> proof one way or the other.
>
> Brent Meeker
>

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