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Re: to contruct reals as infinite decimals, do you need axiom of choice
Posted:
Aug 18, 2013 3:36 PM


On Sunday, August 18, 2013 10:09:26 AM UTC4, 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.



