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.