Difficulty showing a dense $G_{\delta}$ subset of a Baire space is Baire

Question

I have been working on showing that the irrationals is a Baire space. So far I have shown that the irrationals can be expressed as a $G_{\delta}$ set and I know that if this set was to be Baire then it cannot have any open subset of the first category. I am having some trouble showing this. My approach so far is to assume it does have an open subset of the first category but I cannot seem to reach a contradiction.

Answer 1

It is known that any open subset of a complete metric space is of second category. Here is a sketch: If $U$ is open in a complete metric space $(X,d)$ then $D(x,y)=d(x,y)+|\frac 1 {d(x, U^{c})}-\frac 1 {d(y, U^{c})}|$ defines an equivalent metric which makes $U$ complete, so BCT can be applied.

Now let $U$ be open in the space $S$ of all irrationals. Then $U=V\cap S$ for some open set $V$ in $\mathbb R$. If $U$ is a first catoegory so is $V$ because $V= (V \cap S) \cup (V\cap \mathbb Q)$ and $V\cap \mathbb Q$ is a countable union of singletons. This contradicts the fact that $V%$ is of second category.