# Finer Polish topology which turns countably many given Borel sets to clopen sets

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?

• Let $\tau_n$ be a Polish topology refining $\tau$ and making $B_n$ clopen. Let $\tau_\infty$ be the topology generated by $\bigcup_n\tau_n$. Can you show that this topology is as desired? Jan 30, 2021 at 18:51
• OK, I got it, thank you. Jan 31, 2021 at 8:40
• Further remark: The topology generated by $\bigcup_n \tau_n$ as in the comment above is also the Polish topology $\left(X^\omega,\prod_n \tau_n\right)$ restricted to the diagonal $\{(x,x,\ldots)\mid x\in X\}\cong X$.
– user632577
Feb 1, 2021 at 16:09

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.