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I know that one formulation of Baire's theorem is:

If $(M,d)$ is a complete metric space and $\{A_n\}_{n\in\mathbb{N}}$ is a sequence of closed subsets such that $M=\bigcup_{n\in\mathbb{N}}A_n$, then there exists $n_*\in\mathbb{N}$ such that $\operatorname{int}(A_{n_*})\neq\emptyset$.

This is in fact equivalent to the stronger-seeming statement:

If $(M,d)$ is a complete metric space and $\{A_n\}_{n\in\mathbb{N}}$ is a sequence of closed subsets such that $\operatorname{int}(A_{n})=\emptyset$ for all $n\in\mathbb{N}$, then $\operatorname{int}\left(\bigcup_{n\in\mathbb{N}}A_n\right)=\emptyset$.

This inspires the following definition:

A topological space $(X,\mathcal{T})$ is a Baire space if and only if for every sequence $\{A_n\}_{n\in\mathbb{N}}$ of closed subsets such that $\operatorname{int}(A_{n})=\emptyset$ for all $n\in\mathbb{N}$, then $\operatorname{int}\left(\bigcup_{n\in\mathbb{N}}A_n\right)=\emptyset$.

Is it true that if for a topological space $(X,\mathcal{T})$ we have that for every sequence $\{A_n\}_{n\in\mathbb{N}}$ of closed subsets with $X=\bigcup_{n\in\mathbb{N}}A_n$ there exists $n_*\in\mathbb{N}$ such that $\operatorname{int}(A_{n_*})\neq\emptyset$, then $X$ is a Baire-space?

Redundant Aunt
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  • http://math.stackexchange.com/questions/105834/definitions-of-baire-first-and-second-category-sets plus answers might shed some light. – Henno Brandsma Jan 22 '17 at 20:24
  • The first is equivalent to : every countable intersection of open dense sets is dense, the second to: every first category subset of $X$ has empty interior ,or no open subset of $X$ is first category. – Henno Brandsma Jan 23 '17 at 05:25
  • Also see http://math.stackexchange.com/questions/714504/equivalence-of-baire-space-definitions?rq=1 – Henno Brandsma Jan 23 '17 at 05:27

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