Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
NCTM or The Math Forum.


Virgil
Posts:
8,833
Registered:
1/6/11


Re: to contruct reals as infinite decimals, do you need axiom of choice
Posted:
Aug 18, 2013 5:16 PM


In article <65876c86434742cd9558b5a1c5289c72@googlegroups.com>, Zeit Geist <tucsondrew@me.com> wrote:
> On Sunday, August 18, 2013 12:36:14 PM UTC7, G Patel wrote: > > 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 > > > 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. > > > > What way? It's important. > > First of all, you need to construct the Natural Numbers. > Then, the Integers. > Next, the Rational Numbers. > Finally you construct the Real Numbers.
Or one can do 1. Naturals 2. Positive Rationals (or nonnegative Rationals if 0 is taken as a natural) 3. all Rationals 5. finally the Reals > > Have done the first steps yet? > Do you have Definition for each of those Sets? > > And no, the Axiom of Choice is not need for any of the steps. > > > > > So no AC needed because in the { } I do not use infinite choice. > > Hang on. > Although { } = 0, { } is NOT a real number. > It is a Natural Number, a Finite Ordinal. > Now, 0.000... is a representation of a Real Number. > > Let me know you thoughts. > We can walk through the process. > > ZG 



