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][1], *etc*. However, I was wondering what happens in this particular example of 

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

the [Stone–Čech compactification][2] of the integers? Does this formula still hold?


  [1]: https://mathoverflow.net/questions/39882/product-of-borel-sigma-algebras
  [2]: https://en.wikipedia.org/wiki/Stone%E2%80%93%C4%8Cech_compactification