Obviously, what you're looking for is some kind of counter example. I
think the problem lies in not being able to determine at any point of
the calculation just how many digits of the limit you have found. For
the counterexample what we need is a computable series, which we know
converges, yet we cannot compute the limit. 

Perhaps the more mathematically nimble might try the following example
x_i = \sum_{j=0}^i 1/p_j, where p_j is the jth prime number. I suspect
this is a convergent sequence, yet converges too slowly to compute the
limit.

                                                Cheers

Brent Meeker wrote:
> 
> 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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