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Topic: to contruct reals as infinite decimals, do you need axiom of choice
Replies: 5   Last Post: Aug 18, 2013 5:46 PM

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gaya.patel@gmail.com

Posts: 160
Registered: 11/29/05
Re: to contruct reals as infinite decimals, do you need axiom of choice
Posted: Aug 18, 2013 3:36 PM
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On Sunday, August 18, 2013 10:09:26 AM UTC-4, dull...@sprynet.com wrote:
> On Sun, 18 Aug 2013 00:33:40 -0700 (PDT), G Patel
>
> <gaya.patel@gmail.com> wrote:
>
>
>

> >Hi friends of sci.math. I have a question:
>
> >
>
> >To construct real numbers as infinite decimals, do we need axiom of choice?
>
>
>
> No. At least I don't think so - it's easy to use AC without realizing
>
> it.
>
>
>

> >Say for example we are working in base B, then it seems we need to choose one of B natural numbers for an infinite number of positions at the same time.
>
>
>
> Why do we "need" to do this? What are you trying to do, that
>
> requires this?
>
>
>
> Supposing we've defined a real to be an infinite decimal
>
> (mmodulo the appropriate equivalence relation).
>
> We'd "need" to choose infinitely many digits if,
>
> say, we were trying to show that there _exists_
>
> a real number. But we don't need AC for that,
>
> we can just say "let every digit be 0".
>
>
>
> What _are_ you trying to do, where this "need"
>
> ariises?
>


I'm just trying to construct them without reading a professional construction. I failed.

So we just let S = {be set of all equivalence classes of infinite decimals} , where two infinite decimals are equivalent in such and such way.

So no AC needed because in the { } I do not use infinite choice.



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