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?
Measure Theory – Is Countable Generated Sigma-Algebra Separable?
measure-theory
Best Answer
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 $\lnot(*)$.