2
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ 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? $\endgroup$ Jan 30, 2021 at 18:51
  • $\begingroup$ OK, I got it, thank you. $\endgroup$
    – jenda358
    Jan 31, 2021 at 8:40
  • $\begingroup$ 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 $\{(x,x,\ldots)\mid x\in X\}\cong X$. $\endgroup$
    – user632577
    Feb 1, 2021 at 16:09

1 Answer 1

0
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ OK, but why is the new topology completely metrizable? $\endgroup$
    – jenda358
    Jan 30, 2021 at 14:36
  • $\begingroup$ @jenda358 you can probably see the idea in the proof of the quoted theorem. $\endgroup$ Jan 30, 2021 at 14:45
  • $\begingroup$ I see, thank you. $\endgroup$
    – jenda358
    Jan 31, 2021 at 8:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.