1
$\begingroup$

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?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.