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.



Is the measurable space $(\omega_1,\mathcal{P}(\omega_1))$ separable?
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 $\aleph\_{\omega^2+138}$.


