```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.
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