Re: Well Ordering the Reals

Jesse F. Hughes said:
> stevendaryl3016@xxxxxxxxx (Daryl McCullough) writes:
>
> > Jesse F. Hughes says...
> >>
> >>stevendaryl3016@xxxxxxxxx (Daryl McCullough) writes:
> >>
> >>> Tony Orlow says...
> >>>
> >>>>How would one prove that it contained all reals?
> >>>
> >>Daryl, with all due respect, I found your response odd.
> >>
> >>> To prove that a set contains all real numbers, you need to
> >> ^^ One way to...[1]
> >
> > Right. One way.
> >
> >>> show that it contains all rational numbers, and then show
> >>> that your set is complete. To be complete means that for
> >>> every Cauchy sequence of reals in your set converges to a
> >> ^^^^^ rationals
> >
> > No, to say that the space is complete is to say that
> > *every* Cauchy sequence converges, not just Cauchy sequences
> > of rationals. Of course, it amounts to the same thing, because
> > for every Cauchy sequence of reals there is a Cauchy sequence
> > of rationals converging to the same thing.
>
> Well, with your acceptance regarding "one way" above, I don't have a
> real argument here. But your requirement is still apparently
> overstated. My version is better, in my humble opinion.
>
> >
> >>> real in your set.
> >> ^^^ element (which is necessarily real)
> >
> > No, to prove that a set contains all the reals doesn't
> > require proving that it contains *only* reals. It might
> > have other objects that are not reals.
>
> Right, but Tony has a set. It includes Q. He wants to know that it
> includes R. It is sufficient to prove that every Cauchy sequence of
> rationals converges to an element in Tony's happy set.
>
> One can *then* conclude that the element to which such sequences
> converge is a real.
>
> The way you phrased it sounded presumptuous to me. Makes it sound
> like Tony has to prove two things: each Cauchy sequence converges and
> it converges to a real.
>
> Anyway, minor quibbles, I think. You're right whenever it comes to
> things on which we agree, and wrong on the others. Same as everyone
> else.
>
>
I do like your James Harris quotes.

Anyway, are you sure the set includes all rationals? I think this probably
needs to be proven first, but am I being too hard on myself? It may be
difficult to prove with this set. Is it not enough to prove that one can
represent any n to within any finite accuracy epsilon with a finite
representation? Why are the rationals involved, exactly?
--
Smiles,

Tony
http://www.people.cornell.edu/pages/aeo6/WellOrder/
.

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