8
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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 '18 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 '18 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. $\endgroup$ – Nate Eldredge Jun 14 '18 at 17:13
  • 3
    $\begingroup$ @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 '18 at 17:19
  • 1
    $\begingroup$ @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 '18 at 17:45
11
$\begingroup$

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 $\aleph_{\omega^2+138}$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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 '18 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 '18 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 '18 at 20:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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