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My definition of Baire Space is the modern one

A topological space $X$ is a Baire Space iff the intersection of a countable family of open dense everywhere subsets of $X$ is dense everywhere.

There is an old definition that states

A topological space $X$ is a Baire Space iff every non empty open subset of $X$ is of second category.

I'm trying to prove that the old definition implies the modern one

My attempt: Suppose that $X$ is not a Baire Space in the modern sense, then there is a family $\{A_n\}_{n \in \mathbb{N}}$ of open everywhere dense subsets of $X$ and an open set $U \subseteq X$ such that $$U \cap \big( \bigcap_{n \in \mathbb{N}}A_n \big) = \emptyset,$$

so $$U = U - \big( \bigcap_{n \in \mathbb{N}}A_n \big) = \bigcup_{n \in \mathbb{N}} \big( U - A_n \big) $$

I'm tempted to think that for every $n \in \mathbb{N}$, $U - A_n $ is nowhere dense, which would give the result, but I'm not sure.

How do I see that $\mathrm{int}_X ( \mathrm{cl}_X (U-A_n)) = \emptyset???$

Thanks, any help would be appreciated.

Martin Sleziak
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Luis Vera
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1 Answers1

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Let $A$ be a dense open set in $X$, let $U$ be open in $X$, and let $S=U\setminus A$; you want to show that $S$ is nowhere dense in $X$. Clearly $S\subseteq X\setminus A$, and $X\setminus A$ is closed, so $\operatorname{cl}_XS\subseteq X\setminus A$. $A$ is dense, so $X\setminus A$ does not contain any non-empty open set, and in particular $\operatorname{int}_X\operatorname{cl}_XS=\varnothing$.

Brian M. Scott
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