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**C.E** Could someone please help me with the following question?

Show that the Borel sigma algebra of the product of two seperable metric spaces is generated by products of Borel sets.

I am not sure how to start. I think I need to show that any open set in the product space is a countable union of products of Borel sets but I don't know how to do this. Any ideas?

It depends on how you're topologizing the product space--but I'll assume you mean you're giving it the product topology. Then, let $\displaystyle B$ be Borel in $\displaystyle X\times Y$. Then, by definition there exists countably many open sets $\displaystyle \left\{O_n\right\}_{n\in\mathbb{N}}$ such that, say for this case, $\displaystyle \displaystyle B=\bigcap_{n\in\mathbb{N}}O_n$. But, by definition of the product topology each $\displaystyle O_n$ may be written as $\displaystyle \displaystyle \bigcup_{\alpha\in\mathcal{A}}U_{\alpha\in\mathcal {A}}\times V_{\alpha,n}$ where $\displaystyle U_{\alpha,n},V_{\alpha,n}$ are open in $\displaystyle X,Y$ respectively. But, for any second countable topological space (and thus in particular any separable metric space--since second countability and separability are equivalent for metric spaces) every open cover of a subset has a countable subcover (this is Lindelof's theorem--a proof can be found here at problem 26). Thus, we may assume that $\displaystyle \#\left(\mathcal{A}\right)\leqslant\aleph_0$. The result then just falls out since

$\displaystyle \displaystyle \begin{aligned}B &=\bigcap_{n\in\mathbb{N}}O_n\\ &=\bigcap_{n\in\mathbb{N}}\bigcup_{\alpha\in\mathc al{A}}\left(U_{\alpha,n}\times V_{\alpha,n}\right)\\ &=\bigcup_{\alpha\in\mathcal{A}}\bigcap_{n\in\math bb{N}}\left(U_{\alpha,n}\times V_{\alpha,n}\right)\\ &= \bigcup_{\alpha\in\mathcal{A}}\underbrace{\left(\b igcup_{n\in\mathbb{N}}U_{\alpha,n}\right)}_{\text{ Borel}}\times\underbrace{\left(\bigcup_{n\in\mathb b{N}}V_{\alpha,n}\right)}_{\text{Borel}}\end{align ed}$