# Product of Borel $\sigma$-algebras vs Borel $\sigma$-algebra of product

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?

• See my 5 May 2002 sci.math.research post. I'm not sure if your specific question is answered there (I don't have time right now to check), but even if it doesn't, it should give you enough leads to find an answer. – Dave L. Renfro Aug 6 '14 at 15:38
• – hot_queen Aug 6 '14 at 16:39
• Great refs all. – InTransit Aug 6 '14 at 17:05

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 $B_X \otimes B_Y$
• Your example will work if you replace $\mathbb{R}$ by any set of higher cardinality with the diagonal as a witness. – hot_queen Aug 6 '14 at 16:42
• I think your answer will be more helpful if you also add an explanation for why the diagonal fails to be in $B_X \otimes B_X$. – hot_queen Aug 6 '14 at 17:14