Timeline for $2^{\omega_1}$ separable?
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| Jan 24, 2014 at 23:13 | comment | added | bof | @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? | |
| Oct 15, 2010 at 21:11 | comment | added | user5810 | You can't necessarily "define `$f : \omega_1 \to \{0, 1\}^\omega$ inductively as a diagonalisation of $f(\beta)$ for $\beta < \alpha$" without AC because even though you can explicitly count $\alpha$, $\alpha$ isn't explicitly counted. | |
| Apr 22, 2010 at 8:28 | comment | added | David R. MacIver | 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 $\alpha$ | |
| Apr 4, 2010 at 19:37 | comment | added | François G. Dorais |
@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 $\alpha \neq \beta$.
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| Apr 4, 2010 at 18:41 | comment | added | Gerald Edgar | 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? | |
| Apr 3, 2010 at 15:47 | comment | added | Gerald Edgar | 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 | |
| Apr 3, 2010 at 11:38 | history | answered | David R. MacIver | CC BY-SA 2.5 |