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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

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    $\begingroup$ Every Bernstein set is second-countable (as is any subset of $\mathbb R$), so has a locally countable pseudo-basis. By Theorem 2 here, this implies $B\times D$ is a Baire space. $\endgroup$ – Wojowu Jun 10 at 19:32
  • $\begingroup$ Thanks @Wojowu, I did not know that great result. $\endgroup$ – Gabriel Medina Jun 10 at 19:43
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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

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    $\begingroup$ In fact, later using Cohen's idea (using stationary sets in $\omega_1$), were constructed two Baire metric spaces whose product is not Baire in ZFC. (Applications of stationary sets in topology - William G.Fleissner in Surveys in General Topology 1980, Pages 163-193) $\endgroup$ – Gabriel Medina Jun 10 at 22:40
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    $\begingroup$ "Weak basis" is also used (at least by Kechris) for pseudo-base. $\endgroup$ – Pedro Sánchez Terraf Jun 11 at 15:23
  • $\begingroup$ @PedroSánchezTerraf Thanks for this remark. It seems that neither term is particularly standard though. $\endgroup$ – Wojowu Jun 11 at 15:38
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    $\begingroup$ @PedroSánchezTerraf, I would say the most standard name is $\pi$-base. $\endgroup$ – Ramiro de la Vega Jun 11 at 23:26
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    $\begingroup$ In modern terminology a pseudo-base is a collection of open sets such that each singleton is the intersection of all elements of the family that contain it. $\endgroup$ – Ramiro de la Vega Jun 11 at 23:35

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