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?

  • $\begingroup$ What if those $C$ are not themselves open or closed sets? $\endgroup$
    – FShrike
    Apr 14, 2023 at 8:30
  • $\begingroup$ @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? $\endgroup$
    – user1123845
    Apr 14, 2023 at 8:36
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    $\begingroup$ 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 $\endgroup$
    – FShrike
    Apr 14, 2023 at 8:51
  • $\begingroup$ 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? $\endgroup$
    – Jochen
    Apr 14, 2023 at 14:47
  • $\begingroup$ 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/… $\endgroup$
    – PhoemueX
    Apr 15, 2023 at 6:20

1 Answer 1


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 $\lnot(*)$.

  • $\begingroup$ Thanks for the answer! Can you explain (or link) what $MA_{\aleph_1}$ is? $\endgroup$
    – PhoemueX
    Apr 24, 2023 at 17:06
  • $\begingroup$ $\textrm{MA}_{\aleph_1}$ 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 $S$. $\endgroup$
    – Edward H
    Apr 24, 2023 at 20:47

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