For a second-countable space $X$, we have $${\rm Bor}(X\times X) = {\rm Bor}\, X \otimes {\rm Bor}\, X,$$ that is the Borel $\sigma$-algebra of the product is the product $\sigma$-algebra. Some counterexamples to this statement can be produced for uncountable discrete spaces, etc. However, I was wondering what happens in this particular example of

$$X = \beta \mathbb N,$$

the Stone–Čech compactification of the integers? Does this formula still hold?

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    $\begingroup$ I think the adjective "silly" should be removed. Moreover that it may happen that those "silly" counterexamples will help to answer to original question as $\beta\mathbb N$ does contain a discrete (and hence Borel) subspace of cardinality $\mathfrak c$. $\endgroup$ Oct 21, 2018 at 10:37
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    $\begingroup$ Indeed, since for a discrete space $X$ with cardinal ${} >c$ the diagonal is closed but not in the product sigma-algbra $\mathrm{Bor}(X) \times \mathrm{Bor}(X)$, we easily deduce that the same is true for any Hausdorff space with cardinal ${} > c$, since the Borel sigma-algebra is smaller than for the discrete topology. $\endgroup$ Oct 22, 2018 at 12:22

1 Answer 1


The answer is no.

Jiří Nedoma proved that if $(X,\Sigma)$ is a measurable space $|X| > 2^{\aleph_0}$, then the diagonal is not a measurable subset of $(X\times X, \Sigma \otimes \Sigma)$. (The article is called Note on Generalized Random Variables, the result is Lemma 2, a proof can also be found in Schechter's Handbook of Analysis and Its Foundations section 21.8, which I found out about from David McIver on this very website).

Now, $|\beta(\mathbb{N})| = 2^{2^{\aleph_0}} > 2^{\aleph_0}$, so the diagonal is not an element of $\mathrm{Bor}(\beta(\mathbb{N})) \otimes \mathrm{Bor}(\beta(\mathbb{N}))$. But, as $\beta(\mathbb{N})$ is Hausdorff, the diagonal is closed, and therefore Borel, in $\beta(\mathbb{N}) \times \beta(\mathbb{N})$.


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