While reading a Functional Analysis book, the autor stated, in a somewhat crude manner, that, if $X$ is a Baire space and $H \subset X$ is meagre, then $X \setminus H$ contains a dense $G_\delta$ set. I know that, if $X$ is Baire and $H$ is meagre in $X$, then $X \setminus H$ is dense in $X$, but I was not able to prove the author's claim by using this.
How do we, starting from the fact that $X \setminus H$ is dense in $X$, can conclude that $X \setminus H$ has a dense $G_\delta$ subset?
Any help is appreciated.
Complement of meagre set contains a dense $G_\delta$ set
baire-categoryfunctional-analysisgeneral-topology
Best Answer
Suppose $H$ is meager. Then $H=\bigcup N_i$ is a countable union of nowhere dense sets $N_i$. Let $C_i=\overline{N_i}$ be the closure of $N_i$. Then each $C_i$ is nowhere dense and closed, so $F=\bigcup C_i$ is a meager $F_\sigma$ set and $H\subseteq F$. It follows that $G=X\setminus F$ is a comeager $G_\delta$ set and $G\subseteq X\setminus H$.
Since $X$ has the property of Baire, $G$ is dense. So $X\setminus H$ contains a dense $G_\delta$ set.