# Borel $\sigma$-algebra in $\beta \mathbb N \times \beta \mathbb N$

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?

• 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 $\mathfrak c$. Oct 21, 2018 at 10:37
• 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. Oct 22, 2018 at 12:22

## 1 Answer

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 $$\beta(\mathbb{N}) \times \beta(\mathbb{N})$$.