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?


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 $B_X \otimes B_Y$

  • $\begingroup$ This isn't right. Like you said, the diagonal is closed so it is in the sigma algebra generated by (even) open rectangles. In fact, the answer here (due to Kunen) is that this independent of ZFC. See math.stackexchange.com/questions/455797/… $\endgroup$ – hot_queen Aug 6 '14 at 16:29
  • $\begingroup$ Your example will work if you replace $\mathbb{R}$ by any set of higher cardinality with the diagonal as a witness. $\endgroup$ – hot_queen Aug 6 '14 at 16:42
  • $\begingroup$ My bad. Anyway, ive editted the comment $\endgroup$ – S.B. Aug 6 '14 at 17:11
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    $\begingroup$ I think your answer will be more helpful if you also add an explanation for why the diagonal fails to be in $B_X \otimes B_X$. $\endgroup$ – hot_queen Aug 6 '14 at 17:14

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