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I'm working on Exercise 2.28 in Prof. David Marker's notes http://homepages.math.uic.edu/~marker/math512/dst.pdf on refining the topology to make Borel sets clopen.

Question: Suppose $X$ is a Polish space and $ B \subseteq X \times X $ is Borel. Is it always possible to put a new Polish topology on $X$ such that $B$ is clopen in the new product topology on $X \times X$?

Theorem 2.24 in the notes above tells us that we can find a new topology $\tau'$ on $X \times X$ such that $B$ is clopen, then is it possible to make this $\tau'$ a product topology?

What I can see is, maybe we can refine $\tau'$ again with projection maps on each cordinate. Denote it as $\tau''$. But it only shows that $\tau''$ is at least finer than the product topology.

Any help will be appreciated!

Hans
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Nice question! Here's a hint to start you off: is it possible to have an uncountable separable metric space $X$ such that the diagonal $\Delta:=\{(x,x) : x \in X\}$ is an open subset of the product space $X \times X$?

Julian Newman
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    I got it. If the diagonal is open in the product topology, then the original topology must be discrete. Thanks! – Hans Jul 22 '23 at 16:31
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    Also in here. https://math.stackexchange.com/questions/1355886/prove-that-if-diagonal-is-open-in-product-topology-then-the-original-topology-i – Hans Jul 22 '23 at 16:32