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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1$\begingroup$ Related question on MSE: math.stackexchange.com/questions/97413/… $\endgroup$– Martin SleziakCommented Jan 20, 2014 at 12:12
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2 Answers
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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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6$\begingroup$ 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 $\endgroup$ Commented Apr 3, 2010 at 15:47
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2$\begingroup$ 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? $\endgroup$ Commented Apr 4, 2010 at 18:41
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$\begingroup$ @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 $\alpha \neq \beta$. $\endgroup$ Commented Apr 4, 2010 at 19:37 -
$\begingroup$ Is it not the case that $\aleph_1 \leq 2^{\aleph_0}$ without the axiom of the choice? I think the following should work: By constructing with transfinite induction we can find $g_\alpha : [0, \alpha) \to \mathbb{Q}$ an injection (using the standard argument that countable ordinals embed in the rationals), and so $f_\alpha : \alpha \to \omega$ a bijection. We now define `$f : \omega_1 \to \{0, 1\}^\omega$ inductively as a diagonalisation of $f(\beta)$ for $\beta < \alpha$, which we can do without AC because we can explicitly count $\alpha$ $\endgroup$ Commented Apr 22, 2010 at 8:28
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2$\begingroup$ @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? $\endgroup$– bofCommented Jan 24, 2014 at 23:13
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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.
answered Apr 4, 2010 at 8:31
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$\begingroup$ @MartinSleziak Thanks for the tip, I fixed the link. $\endgroup$ Commented Jan 24, 2014 at 18:37
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$\begingroup$ The link seems to be down (or moved?), I'll add at least here in the comment a Wayback Machine link. $\endgroup$ Commented Oct 21, 2020 at 18:24