
Re: Uncountable List
Posted:
Dec 31, 2012 3:34 AM


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 long does an uncountable list of infinite binary sequences have to be before it must contain a duplicate?"; the answer to that, as I just said, is the smallest initial ordinal greater than c. If that's not your question, what is it?

