# countable generated sigma-algebra is separable?

Let $$X$$ be a metric space and $$\mathcal{B}(X)$$ the Borel-$$\sigma$$-algebra. Assume that $$\mathcal{B}(X)$$ is countably generated, i.e. there exists $$\mathcal{C} \subset \mathcal{B}(X)$$ countable such that $$\sigma(\mathcal{C})=\mathcal{B}(X)$$. Does this imply that $$X$$ is second countable and therfore separable?

• What if those $C$ are not themselves open or closed sets? Apr 14 at 8:30
• @FShrike Then we can not argue that $X$ is second countable. But can this case happen? And if yes, does it directly imply that $X$ can't be secondcountable?
– user1123845
Apr 14 at 8:36
• I don't know the answer to either question. What is the context to this problem? Did an author state it in a paper? Did it come up as a necessary lemma for your solution to an exercise? ... because perhaps with the extra details in that paper / exercise, the question will become more tractable Apr 14 at 8:51
• Could this be just a matter of size? I would guess that the cardinality of countably generated $\sigma$-algebras is that of the continuum. Is the cardinality of the Borel-$\sigma$-algebra of a non-separable metric space always strictly bigger? Apr 14 at 14:47
• This should be helpful: math.stackexchange.com/questions/2854087/… Ok, the argument I had in mind seems to only work if one accepts CH: math.stackexchange.com/questions/2432913/… Apr 15 at 6:20

Let $$(*)$$ denote the proposition of question, namely if $$\mathbf{B}(X)$$ the Borel $$\sigma$$-algebra is countably generated for metrizable $$X$$ then $$X$$ is second countable. We show that $$(*)$$ is independent of ZFC.

For the consistency of $$(*)$$, assume $$2^{\aleph_0}<2^{\aleph_1}$$.

Towards a contradiction let $$(X,d)$$ be a metric space that is not second countable. Since second countability in metric spaces is equivalent with separability, we fix $$\epsilon>0$$ and a sequence $$\langle x_i\in X:i<\omega_1\rangle$$ such that for all $$i\ne j<\omega_1$$, $$d(x_i,x_j)>\epsilon$$. Then for any $$S\subseteq \omega_1$$, there is an open set $$U\subseteq X$$ such that $$x_i\in U\Leftrightarrow i\in S$$. In particular $$\mathopen|\mathbf{B}(X)\mathclose|\ge 2^{\aleph_1}$$. On the other hand if $$A$$ is any countably generated $$\sigma$$-algebra then $$\mathopen|A\mathclose|\le 2^{\aleph_0}$$, so necessarily $$A\ne\mathbf{B}(X)$$ and $$\mathbf{B}(X)$$ is not countably generated. This proves $$(*)$$.

For the consistency of $$\lnot(*)$$, assume $$\textrm{MA}_{\aleph_1}$$.

The space $$X$$ will be the discrete topological space on the underlying set $$\omega_1$$. It is clear that $$X$$ is metrizable, not second countable, and $$\mathbf{B}(X)=\mathscr{P}(\omega_1)$$. It suffices to show that $$\mathscr{P}(\omega_1)$$ is a countably generated $$\sigma$$-algebra.

Fix a family $$A=\{x_i\in \mathscr{P}(\omega):i<\omega_1\}$$ of $$\aleph_1$$ almost disjoint subsets of $$\omega$$, so for all $$i\ne j<\omega_1$$, both $$x_i,x_j\subseteq\omega$$ are infinite but $$x_i\cap x_j$$ is finite. Endowing $$A$$ with the subspace topology identifying $$\mathscr{P}(\omega)$$ with the Cantor space $$2^\omega$$, we now fix a countable base $$B\subseteq\mathscr{P}(A)$$ generating this subspace topology on $$A$$. Since $$\mathopen|A\mathclose|=\aleph_1$$, it only remains to show that every subset $$S\subseteq A$$ is Borel in $$A$$, so that any bijection $$f:A\to\omega_1$$ maps $$B$$ to a countable family generating the $$\sigma$$-algebra $$\mathscr{P}(\omega_1)$$.

Fix $$S\subseteq A$$. By the almost disjoint forcing (see Jech, Set Theory, Theorem 16.20), there is a set $$y\subseteq \omega$$ such that $$S$$ is the set of all $$x\in A$$ where $$x\cap y\subseteq\omega$$ is infinite. It then holds that $$S=\bigcap_{n<\omega}\bigcup_{\substack{n< k<\omega\\k\in y}}\left\{x\in A:k\in x\right\}$$ So $$S\subseteq A$$ is $$G_\delta$$ and hence Borel, and this finishes the proof of $$\lnot(*)$$.

• Thanks for the answer! Can you explain (or link) what $MA_{\aleph_1}$ is? Apr 24 at 17:06
• $\textrm{MA}_{\aleph_1}$ is Martin's axiom for families of $\aleph_1$ dense sets. Here it is used to show that the set $y$ always exists for any choice of $S$. Apr 24 at 20:47