# $2^{\omega_1}$ separable?

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?

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Related question on MSE: math.stackexchange.com/questions/97413/… –  Martin Sleziak Jan 20 '14 at 12:12

Should have searched a bit harder before asking this one. This is an immediate consequence of the Hewitt-Marczewski-Pondiczery theorem:

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 $d(\prod_s X_s) \leq m$.

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That's overkill, I think. $2^{[0;1]}$ separable is much easier to prove than the Hewitt-Marczewski-Pondiczery Theorem. In fact, the countable system of Walsh functions is dense in $\{-1;1\}^{[0,1)}$ ... mathworld.wolfram.com/WalshFunction.html –  Gerald Edgar Apr 3 '10 at 15:47
Interesting. Thanks. –  David R. MacIver Apr 3 '10 at 16:03
Of course one also needs $\aleph_1 \le 2^{\aleph_0}$ for this simple proof to work, which is non-constructive (requires Axiom of Choice). Can one prove without choice that $2^{\omega_1}$ is separable? –  Gerald Edgar Apr 4 '10 at 18:41
@Gerald: Good question! The statement appears to imply $\aleph_1 \leq 2^{\aleph_0}$. If $\{d_n\}_{n<\omega}$ enumerates a dense subset then the sets $D_\alpha = \{n<\omega: d_n(\alpha) = 1\}$ must be distinct since $D_\alpha\setminus D_\beta$ cannot be empty when $\alpha \neq \beta$. –  François G. Dorais Apr 4 '10 at 19:37
@GeraldEdgar It would be interesting if one could prove the converse, that separability of $2^{\omega_1}$ implies $\aleph_1\le2^{\aleph_0}$. I don't see how to do that, but it seems to me that separability of $2^{\omega_1}$ does imply the weaker inequality $2^{\aleph_1}\le2^{2^{\aleph_0}}$. Have I got that right? That is weaker, isn't it? –  bof Jan 24 '14 at 23:13

This is indeed the Hewitt-Marczewski-Pondiczery theorem. My proof, following Engelking, is here. 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.

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@MartinSleziak Thanks for the tip, I fixed the link. –  Henno Brandsma Jan 24 '14 at 18:37