We know that if $X$ is complete metric space. Then $X$ is of second category in itself by Baire Category theorem. If $X$ is a complete metric space, it can be shown that any open $G \subset X$ is of 2nd category in $X$. My question is:
If $X$ is of 2nd category in itself and $G$ be open in $X$, then is it necessary that $G$ is of 2nd category in $X$?
I think it is not true. But I failed to produce a counterexample. Any help/suggestions?
To elaborate on my comment from earlier, a space $X$ is Baire iff every open $G\subset X$ is of second category in $X$ and Baire spaces are of second category in themselves (both of those assertion pretty much follow from the definitions), so you need a space which is of second category in itself but is not Baire.
Let $X=[0,1]$ (any Baire space will work) and let $Y=X\sqcup\Bbb Q$ be the disjoint union of $X$ and $\Bbb Q$ (both with the standard topology), $Y$ is of second category in itself, it can't be written as a countable union of nowhere dense subsets since $X$ can't, but it's not Baire, $\bigcap\limits_{q\in\Bbb Q} Y\setminus\{q\}$ is a countable intersection of open dense sets which is not dense.
Now is $\Bbb Q$ open in $Y$? Is it of second category in $Y$?