```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, thenthe answer (assuming the axiom of choice) is the initial ordinalomega_{alpha + 1} where alpha is the ordinal such that c =aleph_{alpha}. But that's the answer given in the post you werereplying to, so I guess you're looking for some other kind of answer.
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