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

$$X = \beta N,$$

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

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?

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 silly counterexamples to this statement can be produced for uncountable discrete spaces, etc. However, I was wondering what happens in this particular example of

$X = \beta N$,

the Stone-Cech compactification of the integers? Does this formula still hold?

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

$$X = \beta N,$$

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

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 silly counterexamples to this statement can be produced for uncountable discrete spaces, etc. However, I was wondering what happens in this particular example of

$X = \beta N$,

the Stone-Cech compactification of the integers? Does this formula still hold?

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 silly counterexamples to this statement can be produced for uncountable discrete spaces, etc. However, I was wondering what happens in this particular example of

$X = \beta N$,

the Stone-Cech compactification of the integers? Does this formula still hold?