Problem:
Prove that if $X$ is a Baire space and $U\subset X$ is an open set, then the subspace $U$ is a Baire space.
My proof:
Let $ \{A_n \} $ be any countable collection of closed sets of $ U $, each of which has empty interior. Since $ U $ is open, all the sets $ A_n $ is closed in $ X $ as well. Since $ X $ is a Baire space, $ \bigcup A_n $ is has empty interior in $ X $. Since $ A_n\subset U $ for each $ n $, $ \bigcup A_n\subset U $ as well, and hence $ A_n $ has empty interior in $ U $. It follows that $ U $ is a Baire space.
Question: Is my proof correct? I am claiming that a set closed in an open set is closed in $X$ as well. I that true?
Best Answer
A closed (in the subspace topology) subset $A$ of $U$ is generally not closed in $X$. Consider the subset $A = [0,1)$ of $U = (-1,1) \subset \mathbb{R} = X$.
But only a small modification is needed. Since the boundary of an open set has empty interior, it follows that for subsets of $U$, being nowhere dense in $U$ implies being nowhere dense in $X$. For, if $A \subset U$ is not nowhere dense in $X$, i.e. $V := \operatorname{int}_X(\operatorname{cl}_X(A)) \neq \varnothing$, then $W = V\cap U$ is a nonempty open set contained in $\operatorname{cl}_U(A) = U \cap \operatorname{cl}_X(A)$.
Thus, if $(A_n)$ is a sequence of relatively closed subsets of $U$ with empty interior, then $(\overline{A}_n)$ is a sequence of closed (in $X$) sets with empty interior, and $A_n = U \cap \overline{A}_n$ for all $n$.