Date: Dec 31, 2012 3:26 AM
Author: William Elliot
Subject: Re: Uncountable List

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.