Date: Dec 31, 2012 1:55 AM
Author: Butch Malahide
Subject: Re: Uncountable List
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.