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Here $\omega_1$ is the first uncountable ordinal, and $\mathcal{P}(\omega_1)$ denotes the power set of $\omega_1$. Separable means countably generated as a $\sigma$-algebra.

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    $\begingroup$ The $\sigma$-algebra generated by a countable set has at most $2^\omega$ elements, while $\mathcal P(\omega_1)$ has $2^{\omega_1}$ elements. Assuming the continuum hypothesis, one has that $2^\omega<2^{\omega_1}$, so the answer to the question is no. However I'm not sure whether $2^\omega=2^{\omega_1}$ under a different set of axioms. $\endgroup$
    – Ruy
    Jun 14, 2018 at 16:41
  • $\begingroup$ It is consistent that $2^{\aleph_1}$ is equal to the continuum, but it doesn't necessarily mean the algebra is separable. $\endgroup$
    – Wojowu
    Jun 14, 2018 at 17:07
  • $\begingroup$ A countably generated measurable space that separates points is isomorphic to a separable metric space with its Borel $\sigma$-algebra. So, is there an uncountable separable metric space with every subset Borel? That seems impossible but I don't immediately see how to prove it. $\endgroup$ Jun 14, 2018 at 17:13
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    $\begingroup$ @NateEldredge: It's not impossible. Martin's Axiom guarantees, for example, that every size-$(<\!\mathfrak{c})$ subset $X$ of $\mathbb R$ has every subset of $X$ Borel in $X$ (in fact, every subset of $X$ is a relative $F_\sigma$). See section 3 of this paper (arxiv.org/pdf/math/0603691.pdf) for more stuff like this. $\endgroup$
    – Will Brian
    Jun 14, 2018 at 17:19
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    $\begingroup$ @WillBrian Assuming MA + not CH and taking a subset of $\mathbb R$ of size $\aleph_1$, doesn't the induced topological space with its Borel algebra provide a positive answer to the question? If so, you might want to turn it into an answer. $\endgroup$
    – Wojowu
    Jun 14, 2018 at 17:45

1 Answer 1

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Whether $\mathcal P(\omega_1)$ is separable is independent of ZFC.

If $2^{\aleph_0} \neq 2^{\aleph_1}$ (which is consistent with ZFC -- it is implied by CH for example), then $\mathcal P(\omega_1)$ (which has size $2^{\aleph_1}$) is larger than any countably generated $\sigma$-algebra (which has size at most $2^{\aleph_0}$).

On the other hand, $MA + \neg CH$ (which is also consistent with ZFC) implies that every size-$(<\!\mathfrak{c})$ subset $X$ of the real line is a "$Q$-set." This means that every subset of $X$ is a relative $G_\delta$ (a countable intersection of open subsets of $X$). Suppose $X$ is such a set with $|X| = \aleph_1$. Then the induced topology on $X$ generates the $\sigma$-algebra $\mathcal P(X)$. That is, the $\sigma$-algebra $\mathcal P(X)$ is generated by a countable collection of subsets of $X$, namely, the basic open sets of $X$ as a subspace of $\mathbb R$. Re-indexing the points of $X$ with $\omega_1$, we see that $\mathcal P(\omega_1)$ must be countably generated too.

The same argument shows that $\mathcal P(\omega_2)$, $\mathcal P(\omega_{42})$, $\mathcal P(\omega_{\omega^2+137})$, and every such set in between can, consistently, be countably generated as well. All you need to do is live in a model of set theory where Martin's Axiom holds and the continuum is at least $\aleph_{\omega^2+138}$.

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  • $\begingroup$ Will, thanks for clarifying that $2^{\aleph_0}=2^{\aleph_1}$ is also independent from ZFC. This made me curious as to what exactly one needs to prove that $2^X=2^Y \Rightarrow X=Y$. $\endgroup$
    – Ruy
    Jun 15, 2018 at 16:56
  • $\begingroup$ @Ruy: The Continuum Hypothesis (CH) suffices to prove that if $|2^X| = 2^{\aleph_0}$ then $|X| = \aleph_0$. But these two things are not equivalent: there are models where CH fails, and yet it is still true that if $|2^X| = 2^{\aleph_0}$ then $|X| = \aleph_0$. Similarly, the Generalized Continuum Hypothesis (GCH) implies that if $|2^X| = |2^Y|$ then $|X| = |Y|$. But again, these two things are not equivalent: it is consistent to have the GCH fail, and yet to have that if $|2^X| = |2^Y|$ then $|X| = |Y|$. So the (G)CH is enough for what you're asking about, but it's not exactly the same. $\endgroup$
    – Will Brian
    Jun 15, 2018 at 17:09
  • $\begingroup$ Thanks, @Will. As you say, (CH) implies (LOG), if you allow me to refer to the property "$2^X=2^Y \Rightarrow X=Y$" by that acronym. Thus (LOG) may be seen as a weaker form of (CH) and it would be interesting to decide whether it is equivalent to any well known axiom in Set Theory. $\endgroup$
    – Ruy
    Jun 15, 2018 at 20:23

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