Equivalent forms of the Baire property

Given: The objects, hypotheses, and choice principles stated above.

Let $X$ be a topological space and let $A\subseteq X$. The set $A$ is nowhere dense when $\mathrm{int}(\bar{A})=\mathrm{\varnothing }$ (def-interior-closure-boundary-top). It is meagre when there is a sequence $(N_n{)}_{n\in \mathbb{N}}$ of nowhere dense subsets of $X$ with $A\subseteq {\bigcup }_n{N}_n$. It is residual, or comeagre, when $X\setminus A$ is meagre. The empty union shows that $\mathrm{\varnothing }$ is meagre, including when $X=\mathrm{\varnothing }$. (Nowhere dense, meagre, residual, and comeagre subsets of a topological space).

A topological space $(X,\mathcal{T})$ (def-topological-space) is a Baire space when for every sequence $(U_n{)}_{n\in \mathbb{N}}$ of subsets of $X$ that are open and dense in $X$ (def-dense-top, def-sequence-convergence-top, def-natural-numbers), the intersection ${\bigcap }_{n\in \mathbb{N}}{U}_n$ is dense in $X$. (Baire space: a topological space in which every countable intersection of dense open subsets is dense).

For all sets $X$, $a$ and $b$, $X\setminus (a\cup b)=(X\setminus a)\cap (X\setminus b),X\setminus (a\cap b)=(X\setminus a)\cup (X\setminus b)\mathrm{.}$ Let $F$ be a set with $F\mathrm{\ne }\mathrm{\varnothing }$. Then $\{X\setminus a:a\in F\}$ is a nonempty set and $X\setminus \bigcup F=\bigcap \{X\setminus a:a\in F\},X\setminus \bigcap F=\bigcup \{X\setminus a:a\in F\}\mathrm{.}$ ($X\setminus (a\cup b)=(X\setminus a)\cap (X\setminus b)$ and $X\setminus (a\cap b)=(X\setminus a)\cup (X\setminus b)$; and for a nonempty set $F$, $X\setminus \bigcup F=\bigcap \{X\setminus a:a\in F\}$ and $X\setminus \bigcap F=\bigcup \{X\setminus a:a\in F\}$).