Re: Well Ordering the Reals
- From: Tony Orlow <aeo6@xxxxxxxxxxx>
- Date: Wed, 9 Nov 2005 16:31:22 -0500
Daryl McCullough said:
> Tony Orlow says...
> >
> >Daryl McCullough said:
>
> >> >Listen, a countable number of bits, such that each is finite, is
> >> >supposedly sufficient to produce an uncountable number of bit
> >> >strings, ala 2^aleph_0, no?
> >>
> >> Yes, that's right. There are uncountably many infinite bitstrings.
> >Then my well ordering
>
> You don't have a well ordering of the reals.
>
> >only requires countably many bits, each in a finite
> >position, and therefore affording no infinite descending
> >sequence within the set of bitstrings. There is really
> >very little legitiamte doubt that this set is well ordered.
>
> If you are allowing infinite bitstrings, then you don't
> have a well ordering. There are infinite descending chains.
If there are countably many bits, then none is in an infinite location, is it?
> If you are only allowing finite bitstrings, then you do
> have a well ordering, but it doesn't include all the reals,
> only a dense subset.
What makes you say that? Cantor's putuative list of reals has countably many
bits, such that none is in a finite positions.
>
> Once again: Let B = the set of infinite bitstrings (indexed
> by finite natural numbers). Let FB = the set of finite bitstrings.
> Are you claiming to have a bijection between B and R, or are
> you claiming to have a bijection between FB and R. If you mean
> FB, then your bijection doesn't cover all the reals. If you mean
> B, then your set is not well-ordered. In either case, you are
> wrong to claim to have a well-ordering of the reals.
>From what you say, it sounds like any well ordering of the reals is impossible.
Is there something you know that Hilbert didn't?
>
> --
> Daryl McCullough
> Ithaca, NY
>
>
--
Smiles,
Tony
http://www.people.cornell.edu/pages/aeo6/WellOrder/
.
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