I am given the following exercise:
Let $X_{\alpha}$ be a measureable space with $\sigma$-algebra $M_{\alpha}$ , mark $X\triangleq{\displaystyle \prod_{\alpha\in A}X_{\alpha}}$ and $\pi_{\alpha}:X\rightarrow X_{\alpha}$ .
define $\otimes_{\alpha\in A}M_{\alpha}$ as the $\sigma$-algebra that is created from sets of form $\pi_{\alpha}^{-1}\left(E_{\alpha}\right)$ where $E_{\alpha}\in M_{\alpha}$
Part 1 of the question asked to prove that if $A$ is countable then $\otimes_{\alpha}M_{\alpha}$ is created by sets of form ${\displaystyle \prod_{\alpha\in A}E_{\alpha}}$ where $E_{\alpha}\in M_{\alpha}$ .
Given a topological space $X$ denote the Borel's $\sigma$ -algebra on $X$ by $\mathcal{B}_{X}$
Consider $X_{1},\dots,X_{n}$ metric spaces and define $X=X_{1}\times\dots \times X_{n}$
Now the question asks:
a) Prove $\otimes_{i=1}^{i=n}\mathcal{B}_{X_{i}}\subseteq\mathcal{B}_{X}$.
b) Assume each $X_{i}$ is separable and prove $\otimes_{i=1}^{i=n}\mathcal{B}_{X_{i}}=\mathcal{B}_{X}$.
I am having some problems with this question, mainly because I don't have a strong memory of what the product topology is exactly (I have a definition near me and so I am trying to work with that).
For $a$ I want to start with some element of the form $\Pi_{\alpha}E_{\alpha}$and prove that it is an element of $\mathcal{B}_{x}$ but to be honest, I am not completely sure how $\mathcal{B}_{X}$ looks like - how does the open set in $X$ look like and why something of the form $\Pi_{\alpha}E_{\alpha}$ (or $\pi^{-1}(E_{\alpha})$) is such a set.
After I understand $a$ I hope I will have an idea for $b$
I would really appreciate the help!