5
$\begingroup$

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?

$\endgroup$
1

2 Answers 2

4
$\begingroup$

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

$\endgroup$
6
  • 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 $\endgroup$ Apr 3, 2010 at 15:47
  • 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? $\endgroup$ Apr 4, 2010 at 18:41
  • $\begingroup$ @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 $\alpha \neq \beta$. $\endgroup$ 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 $\alpha$ $\endgroup$ Apr 22, 2010 at 8:28
  • 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? $\endgroup$
    – bof
    Jan 24, 2014 at 23:13
3
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ @MartinSleziak Thanks for the tip, I fixed the link. $\endgroup$ Jan 24, 2014 at 18:37
  • $\begingroup$ The link seems to be down (or moved?), I'll add at least here in the comment a Wayback Machine link. $\endgroup$ Oct 21, 2020 at 18:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.