Date: Dec 31, 2012 3:34 AM
Author: Butch Malahide
Subject: Re: Uncountable List

On Dec 31, 2:26 am, William Elliot <ma...@panix.com> wrote:
> On Sun, 30 Dec 2012, Butch Malahide wrote:
> > On Dec 30, 9:22 pm, William Elliot <ma...@panix.com> wrote:
> > > On Sun, 30 Dec 2012, David C. Ullrich wrote:
> > > > William Elliot <ma...@panix.com> wrote:
>
> > > > >How long does an uncountable list with no duplicates,
> > > > >of infinite binary sequences (IBS) have to be to force
> > > > >the list to contain all the IBS's?

>
> > > > It's obvious that there is no length long enough to force this. A "list"
> > > > of length c (the cardinality of the set of infinite binary sequences)
> > > > need not contain all the sequences, and a list of length greater than c
> > > > must contain duplicates (and still need not contain all the sequences).

>
> > > Whoops.
>
> > > How long does an uncountable list have
> > > to be before it must contain a duplicate?

>
> > If you're still talking about lists of infinite binary sequences, then
> > the answer (assuming the axiom of choice) is the initial ordinal
> > omega_{alpha + 1} where alpha is the ordinal such that c =
> > aleph_{alpha}. But that's the answer given in the post you were
> > replying to, so I guess you're looking for some other kind of answer.

>
> Nope, I revised the problem upon the advise of David.


Well, then, what is your revised problem? I thought it was, "How long
does an uncountable list of infinite binary sequences have to be
before it must contain a duplicate?"; the answer to that, as I just
said, is the smallest initial ordinal greater than c. If that's not
your question, what is it?