# $G_{\delta}$ sets in locally compact Hausdorff or complete metric space are Baire spaces

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.

• A subspace $Y$, of a complete metric space $X$, is a completely metrizable space iff $Y$ is $G_{\delta}$ in $X.$ – DanielWainfleet Dec 29 '16 at 18:41

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.
• Thanks! But, why is $\overline {A}$ is a baire space? I mean, as far as I know, hausdorff and locally compact is baire space only when it's also regular, isn't it? – Astro Nauft Feb 4 '15 at 14:20
• I don't understand how $G_n \cap W_m$ is dense in $X$ in the third sentence from the very last line. – izimath Apr 11 '19 at 3:59
• I got it. It holds since they are both dense, open and $X$ is Baire. – izimath Apr 11 '19 at 4:03