Is every $\sigma$-algebra a Borel $\sigma$-algebra? [duplicate]

Let $$\mathscr B$$ be a $$\sigma$$-algebra on a set $$X$$. Does there always exist a topology $$\tau$$ on $$X$$ such that $$\mathscr B$$ is the Borel $$\sigma$$-algebra with respect to $$\tau$$, that is, $$\mathscr B$$ is the smallest $$\sigma$$-algebra containing $$\tau$$?

For example, let $$X$$ be an uncountable set and let $$\mathscr B$$ be the family of all sets $$B\subseteq X$$ such that either $$B$$ or $$X\setminus B$$ is countable. Then $$\mathscr B$$ is generated by a topology $$\tau=\{\emptyset\}\cup\{B\subseteq X\!:\left|X\setminus B\right|\le\omega\}$$.

Added: This question is actually a duplicate of the following questions:

This answer to the MO question provides references to two papers, where the problem is solved negatively. For the convenience of a reader, I try to present here a simple argument from the paper of R. Lang.

Given sets $$X,Y$$, let $${}^XY$$ denote the family of all functions from $$X$$ to $$Y$$. Let $$X={}^{\mathbb R}\{0,1\}$$. For $$a\in\mathbb R$$, let $$W_a=\{\mathbf 1_A\!:A\subseteq\mathbb R,\ a\in A\}$$, where $$\mathbf 1_A\in X$$ denotes the characteristic function of a set $$A$$. Let $$\mathscr B$$ be the $$\sigma$$-algebra on $$X$$ generated by $$\{W_a\!:a\in\mathbb R\}$$. We prove that $$\mathscr B$$ cannot be generated by a topology on $$X$$.

For every $$V\in\mathscr B$$ there exists a countable set $$C\subseteq\mathbb R$$ and a set $$Z\subseteq{}^C\{0,1\}$$ such that $$V=\{v\in X\!:v\upharpoonright C\in Z\}$$. It follows that $$\left|\mathscr B\right|\le 2^{\aleph_0}$$. Since the family $$\{W_a\!:a\in\mathbb R\}$$ separates points of $$X$$ (that is, for every distinct $$u,v\in X$$ there exists $$a\in\mathbb R$$ such that $$W_a$$ contains exactly one of $$u,v$$), any family of sets that generates $$\mathscr B$$ must separate points of $$X$$. Hence, every topology $$\tau$$ that generates $$\mathscr B$$ must be $$T_0$$. But this would imply $$\overline{\{u\}}\neq\overline{\{v\}}$$ for all distinct $$u,v\in X$$, hence $$\left|\tau\right|\ge\left|X\right|=2^{2^{\aleph_0}}$$. That is impossible.

• If $X$ is finite, then $\mathcal B$ is itself a topology, so there is a smallest topology generating $\mathcal B$. – daw Feb 6 '19 at 9:04
• math.stackexchange.com/questions/51222/… – daw Feb 6 '19 at 9:07
• Great question! Have a look here on Math.Overflow. – drhab Feb 6 '19 at 9:09
• Also have a look here. – drhab Feb 6 '19 at 9:17
• @drhab Thanks for the links. I expected that the answer is well known, if not trivial. – Peter Elias Feb 6 '19 at 9:23