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



Product of Bernstein sets
There exist some general results which show when a product of Baire spaces is Baire. Specifically, Theorem 2 [here](http://matwbn.icm.edu.pl/ksiazki/fm/fm49/fm49113.pdf) states



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> 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.
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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](https://www.jstor.org/stable/2041855?seq=1#page_scan_tab_contents) CH is not necessary for that


