Let $(X,\tau)$ be a Polish space and $(B_n)_{n \in \omega}$ a sequence of Borel sets in $(X,\tau)$. I would like to know if this implies that there is a Polish topology ${\tau}_{\omega}$ on $X$ such that $\tau \subseteq {\tau}_{\omega}$, the Borel $\sigma$-algebra generated by ${\tau}_{\omega}$ is equal to the Borel $\sigma$-algebra generated by $\tau$ and $B_n$ is clopen in $(X,{\tau}_{\omega})$ for every $n \in \omega$.
I know the following theorem:
If $(X,\tau)$ is a Polish space and $B \subseteq X$ is a Borel set in $(X,\tau)$, then there exists a Polish topology $\tau'$ on $X$ such that $\tau \subseteq \tau'$, the Borel $\sigma$-algebra generated by $\tau'$ is equal to the Borel $\sigma$-algebra generated by $\tau$ and $B$ is clopen in $(X,\tau')$.
Using this theorem I would be able to solve the above problem if the sequence of Borel sets was finite (I would just use the theorem finitely many times). But what to do in the case of an infinite sequence?
Best Answer
Let $\tau \cup \{B_n, B_n^\complement: n \in \omega\}$ be the subbase for a new topology on $X$.
It's clearly finer, and still second countable as $\tau$ is. Because we only add $\tau$-Borel sets, the new Borel sets are the same. By definition the $B_n$ are clopen.