Claim: Let $X$ be a complete metric space or a locally compact Hausdorff space. Let $A\subseteq X$ be a $G_{\delta}$ set. Prove that $A$ is a Baire space.
I tried to prove the claim, and got stuck. That's what I got so far:
$A$ is $G_{\delta}$ therefore there exist $A_{1},A_{2},…$ open sets in $X$ such that $A=\bigcap _{n=1} ^{\infty} A_{n}$.
We will show that $A$ is a Baire space by showing that every countable intersection of open dense sets in $A$, is dense. So, let $G_{1},G_{2},…$ be open dense sets in $A$, and define $G=\bigcap _{n=1} ^{\infty} G_{n}$.
Assume (by contradiction) that exists some open set $U$ in $A$ such that $U\cap G=\emptyset$.
$U$ is open in $A$ therefore there exists some $U'$ open set in $X$ such that $U=U'\cap A$, and also there exists some $G_{n}'$ open in $X$ such that $G_{n}=A\cap G_{n}'$ for every $n$.
Then: $U\cap G = (U'\cap A)\cap (\bigcap_{n=1} ^{\infty} A\cap G_{n}')=A\cap U' \cap \bigcap _{n=1} ^ {\infty} G_{n}'$ ….
I cannot see how to proceed from here.
Best Answer
We can suppose $A$ is dense in $X$. Indeed, the closure $\overline A$ is complete metric or locally compact Hausdorff as $X$ is, and we replace $X$ by that closure. Thus, from now on, $\overline A=X$, which is a Baire space. Since $A$ is a $G_\delta$ set, we have $A=\bigcap G_n$, for some sets $G_n$ open in $A$, and dense in $X$ as $A$ is. Now let $\{V_m\}$ be open dense subsets of $A$ and let us see that $\bigcap V_m$ is dense in $A$. For each $m$ pick an open set $W_m$ of $X$ with $V_m=A\cap W_m$; since $V_m$ is dense in $A$, which is dense in $X$, the set $W_m$ is dense in $X$. Since $X$ is Baire, the intersection $\bigcap_{n,m}G_n\cap W_m$ is dense in $X$. But that intersection is $$ \bigcap_{n,m}G_n\cap W_m=\bigcap_m\big(\bigcap_n G_n)\cap W_m=\bigcap_m A\cap W_m=\bigcap_mV_m, $$ and we are done.
This is somehow an adaptation of more general arguments. I suggest to look at Engelking's General Topology, this book is a jewel.