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).

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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
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Universe and all that surrounds it.
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