when is it true that $ \mathcal B( \prod_{i \in I} \mathcal X_i )= \bigotimes_{i \in I} \mathcal B( \mathcal X_i ) $

Question

I am following a lecture about stochastic processes, and the professor "randomly" said that :

$$ \mathcal B( \prod_{i \in I} \mathcal X_i )= \bigotimes_{i \in I} \mathcal B( \mathcal X_i ) $$

the conditions are that $I \longleftrightarrow \{ 1, ... , n \}, n > 0$, and that $\forall i \in I, \mathcal X_i$ is a separable metric space.

1. What's important here to prove ? Is the proof by double inclusion? If anyone knows where I can find a proof, that would be great.
2. Can $I \longleftrightarrow \mathbb N$? $I >> N$ (bigger than countable)?
3. Does that still hold for generic topological spaces ? Why is beeing separable important, I believe here one will want to construct a basis (perhaps 3. is already answered by the answer to 1.).

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I precise that I denote by $\mathcal B( \prod_{i \in I} \mathcal X_i )$ the Borel $\sigma$-algebra of the product space endowed with the product metric.

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Thank you. The question is quite vague and I am aware of that. I appreciate any help. I have no additional information in my lecture notes, and I can't imagine searching in 10 different random books about topology to get an answer... I believe it is way more efficient if you help me to orientate my researches.

Answer 1

LHS always contains RHS. For the reverse inclusion we need some way of expressing an open sets in the product space in terms of open sets in the individual spaces using only countable unions and intersections. This is not always possible but it is possible if the spaces are second countable. In this case any open set in the product is a countable union of sets of the form $U_1 \times U_2 \times...\times U_n$ with $U_n$'s open. A metric space is second countable iff it is separable. The result fails for uncountable products even when the spaces are second countable.