If $X$ and $Y$ are topological spaces with associated Borel $\sigma$-algebras $\mathcal{B}_X$ and $\mathcal{B}_Y$, then the product $\sigma$-algebra $\mathcal{B}_X\otimes \mathcal{B}_Y\subset \mathcal{B}_{X\times Y}$, $X\times Y$ with the product topology. If $X$ and $Y$ are separable metric then equality holds. What is an example when the inclusion is strict?
[Math] Product of Borel $\sigma$-algebras vs Borel $\sigma$-algebra of product
general-topologyset-theory
Best Answer
Take $X=Y=\mathbb 2^R$, embedded with the discrete topology. The topology is Hausdorff, so $\{(x,x)\in X \times X\}$ is closed, and hence in $B_{X\times Y}$, And yet it is not in $B_X \otimes B_Y$