3
$\begingroup$

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?

$\endgroup$
2
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

5
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.