Product of Bernstein sets Remember that a Bernstein set is a set
$B\subseteq \mathbb{R}$ with the property that for any uncountable closed set, $S$, in the real line both
$B\cap S$ and $(\mathbb{R}\setminus B)\cap S$ are non-empty. Some known results are the following: Bernstein sets are Baire spaces, also the Banach-Mazur game played in a Bernstein set is indeterminate. My question is: if $B, D\subseteq\mathbb{R}$ are Bernstein sets then $B\times D$ is a Baire space?
Thanks
 A: There exist some general results which show when a product of Baire spaces is Baire. Specifically, Theorem 2 here states

If $X,Y$ are two Baire spaces and at least one has a locally countable pseudo-base, then $X\times Y$ is a Baire space.

A pseudo-base of a topological space $X$ is a family of nonempty open sets such that every open set in $X$ contains an element of the family. In particular, any base is a pseudo-base. A family is locally countable if any element of the family contains only countably many other ones. In particular, any countable family is locally countable.
Now, any Bernstein set, in fact any subset of a second countable space, is second-countable, which means it has a countable base, which is a locally countable pseudo-base. Therefore, by the theorem, a product of Bernstein sets is Baire.
It is worth noting that not every product of Baire spaces is Baire. Oxtoby (the author of the linked paper) constructs a Baire space whose square is not Baire assuming the continuum hypothesis. Cohen has shown CH is not necessary for that
