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From: Brent Meeker <meekerdb.domain.name.hidden>

Date: Mon, 25 Jun 2001 18:31:55 -0700

On 25-Jun-01, Russell Standish wrote:

*> No - the set of computable numbers does not form a
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*> continuum. Continuity is related to the concept of limits: {x_i} is a
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*> convergent sequence if
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*> \forall \epsilon>0, \exist N: |x_i-x_N|<\epsilon.
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*> A continuous space is one for which every convergent sequence
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*> converges to a limit, ie
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*>
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*> \exists x: \forall\epsilon>0\exists N: |x_i-x|<\epsilon \forall i>N.
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*>
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*> we commonly denote x by lim_{i->\infty}x_i.
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*>
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*> There are many convergent sequences x_i whose limits cannot be
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*> 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

I am very interested in the Universe - I am specializing in the

Universe and all that surrounds it.

--- Peter Cook

Received on Mon Jun 25 2001 - 19:38:28 PDT

Date: Mon, 25 Jun 2001 18:31:55 -0700

On 25-Jun-01, Russell Standish wrote:

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

Received on Mon Jun 25 2001 - 19:38:28 PDT

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