```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 longdoes an uncountable list of infinite binary sequences have to bebefore it must contain a duplicate?"; the answer to that, as I justsaid, is the smallest initial ordinal greater than c. If that's notyour question, what is it?
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