On 25-Jun-01, Russell Standish wrote:
> 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).

Thanks for the education vis a vis definition of a continuum, but now I
don't see why the computable numbers don't form a continuum.  Suppose
a0,a1,a2,...ai,... is a convergent sequence of computable numbers with
limit A.  Then it seems that A must be a computable number since given
any number of decimal places (or bits) n there is a value of i=m such
that am is equal to A for the first n places and the digits in those
places will not change for all i>m.  Isn't this the definition of a
computable number - one whose representation can be computed to a given
accuracy in a finite number of steps?  So the computability of the
sequence ai entails computability of the sequences limit.

thnx, Brent Meeker
  I am very interested in the Universe - I am specializing in the    
Universe and all that surrounds it.                                 
      --- Peter Cook


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