$2^{\omega_1}$ separable? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T07:31:32Z http://mathoverflow.net/feeds/question/20224 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/20224/2-omega-1-separable $2^{\omega_1}$ separable? David R. MacIver 2010-04-03T10:41:06Z 2010-04-04T08:31:14Z <p>I was rereading an answer to an old question of mine and it included a reference to the fact that $2^{\omega_1}$ was separable. I'm having a hard time finding a reference for this fact, and the proof is not immediately obvious to me. Can anyone provide me with a cite and/or a proof? </p> http://mathoverflow.net/questions/20224/2-omega-1-separable/20231#20231 Answer by David R. MacIver for $2^{\omega_1}$ separable? David R. MacIver 2010-04-03T11:38:52Z 2010-04-03T11:38:52Z <p>Should have searched a bit harder before asking this one. This is an immediate consequence of the Hewitt-Marczewski-Pondiczery theorem:</p> <p>Let $m \geq \aleph_0$. If ${X_s : s \in S}$ are topological spaces with $d(X_s) \leq m$ and $|S| \leq 2^m$ then $d(\prod_s X_s) \leq m$.</p> http://mathoverflow.net/questions/20224/2-omega-1-separable/20295#20295 Answer by Henno Brandsma for $2^{\omega_1}$ separable? Henno Brandsma 2010-04-04T08:31:14Z 2010-04-04T08:31:14Z <p>This is indeed the Hewitt-Marczewski-Pondiczery theorem. My proof, following Engelking, is <a href="http://at.yorku.ca/cgi-bin/bbqa?forum=ask_a_topologist;task=show_msg;msg=0487.0001" rel="nofollow">here</a>. It's in fact not that hard, the fact for a product of copies of 2 point discrete spaces already implies the general theorem pretty quickly.</p>