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:
- Correspondences between Borel algebras and topological spaces [MSE]
- Is every sigma-algebra the Borel algebra of a topology? [MO]
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.
