Let $(X_1, d_1), \dots, (X_n, d_n)$ be separable metric spaces. Endow $X = \prod\limits_{i=1}^n X_i$ with the metric $d$ given by $d(u, v) = \sqrt{\sum\limits_{i=1}^n d_i(u_i, v_i)^2}.$ For a topological/metric space $A,$ let $\mathcal{B}(A)$ denote the Borel algebra on $A,$ i.e. the smallest sigma algebra containing all open sets of $A.$ Prove that $R = \mathcal{B}(X) = \prod\limits_{i=1}^n \mathcal{B}(X_i) = T$ (letters introduced for the purpose of making the following discussion easier to read).
Attempt: Let $Y_i$ be a countable dense subset of $X_i.$ Then $Y = \prod\limits_{i=1}^n Y_i$ is a countable dense subset of $X.$ I used the density of $Y$ to show that any open set $U \subseteq X$ can be written as a countable union of balls centered at points of $Y.$ Thus, $\mathcal{B}(X) = \sigma(\{B_d(y,\epsilon) : y \in Y\}) \, (*).$ Now suppose $Z = Z_1 \times \dots \times Z_n$ where $Z_i \in \mathcal{B}(X_i).$ We need to show that $Z \subseteq R.$
I don't see how we're supposed to continue. I strongly suspect that among the statements $R \subseteq T, T \subseteq R,$ one is supposed to be trivial and the other is shown using separability, but I don't know which one is trivial and which one to use separability on. I used separability to show $(*),$ but this doesn't seem to lead anywhere.
My original plan was to consider $S = \sigma(U_1 \times \dots \times U_n : U_i \subseteq X_i \text{ open})$ and show that $R = S = T.$ We clearly have $S \subseteq R,$ and if we show $\sigma(T) = T$ then it follows $S \subseteq T,$ leaving only $T \subseteq S, R \subseteq S.$ I thought that maybe $R \subseteq S$ follows from separability although I couldn't figure out exactly how, and had no idea how to show $T \subseteq S.$
The biggest obstacle to working with $T$ is that there is no mechanism for taking the union, complement, or intersection of elements of the product of Borel algebras. What is $(Z_1 \times Z_2) \cup (Z_3 \times Z_4)$ where $Z_i \in \mathcal{B}(X_{2-(i \bmod 2)})$? It's just $\{(a,b) : a \in Z_1, b \in Z_2 \text{ or } a \in Z_3, b \in Z_4\}.$ Not very helpful. What exactly is $\sigma(U_1 \times X_2 : U_1 \subseteq X_1 \text{ open})$? Unfortunately, it's just $\mathcal{B}(X_1) \times \{X_2\}.$ Any hints, approaches, or ideas on how to use separability properly and how to deal with products of Borel algebras?
Note: I was given the hint that you need to imitate the proof that $\mathcal{B}(\mathbb{R}) = \sigma(\{(a,b)\}).$ The analogue of open intervals for metric spaces is open balls, and I did just that to show $(*).$ The issue is that I have no idea how to prove $\sigma(\{B_d(y,\epsilon) : y \in Y\}) = T.$
Update: I have almost solved the problem. Let $\pi_i : X \to X_i$ be the projection mapping onto $X_i$ and let $A_i = \{U \subseteq X_i : \pi_i^{-1}(U) \in R\}.$ Then $A_i$ is a sigma algebra since $R$ is and $$\pi_i^{-1}(U^c) = \pi_i^{-1}(U)^c, \pi_i^{-1}(\bigcup\limits_i U_i) = \bigcup\limits_i \pi_i^{-1}(U_i).$$ Furthermore, $A_i$ contains all open subsets of $X_i$ since $\pi_i^{-1}(U) = X_1 \times \dots \times U \times X_{i+1} \times \dots \times X_n$ is an open subset of $X$ whenever $U$ is open.
It follows $\mathcal{B}(X_i) \subseteq A_i \Rightarrow \pi_i^{-1}(\mathcal{B}(X_i)) \subseteq R \Rightarrow T = \bigcup\limits_{i=1}^n \pi_i^{-1}(\mathcal{B}(X_i)) \subseteq R.$
I've shown that if $U \subseteq X$ is open, then $U$ is a countable union of products of balls. To finish the proof, I need to show $T$ is a sigma algebra, which would imply this countable union is in $T,$ hence $U \subseteq T,$ hence $R = \sigma(\{U\}) \subseteq \sigma(T) = T.$
Last update: $T$ is a sigma algebra by definition, completing the proof.