TITLE: Product of Borel sigma algebras
QUESTION [36 upvotes]: If $X$ and $Y$ are separable metric spaces, then the Borel $\sigma$-algebra $B(X \times  Y)$ of the product is the $\sigma$-algebra generated by $B(X)\times B(Y)$.  I am embarrassed to admit that I don't know the answers to: 
Question 1. What is a counterexample when $X$ and $Y$ are non-separable?
Question 2. If $X$ is an uncountable discrete metric space, does
$B(X) \times B(X)$ generate the Borel $\sigma$-algebra on $X \times X$?
Question  3. If $X$ and $Y$ are metric spaces, with $X$ separable, does
$B(X) \times B(Y)$ generate the Borel $\sigma$-algebra on $X \times Y$?

REPLY [9 votes]: This is should probably rather be a comment to Michael Greinecker's answer, but I do not have the necessary privileges.
Michael Greinecker's answer leaves open what happens with a continuum-sized discrete space when one does not assume the continuum hypothesis.
Arnold W. Miller showed in section 4 of On the length of Borel hierarchies that it is consistent relative ZFC that no universal analytic set $U \subset [0,1] \times [0,1]$ belongs to the product $\sigma$-algebra $\mathcal{P}[0,1] \otimes \mathcal{P}[0,1]$. Combined with Rao's result mentioned by Michael Greinecker, this shows that $2^{\mathfrak{c \times c}} = 2^{\mathfrak{c}} \otimes 2^\mathfrak{c}$ is independent of ZFC.
See my answer to Universally measurable sets of $\mathbb{R}^2$ on math.stackexchange.com for related results and more details and references.