# Prove that an open subset of a Baire space is a Baire space

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?

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 $n$.

After help from @DanielFischer, I think this proof is correct.

Let $\{A_n \}$ be any countable collection of closed sets of $U$, each of which has empty interior. Since each $A_n$ is closed in $U$, they are nowhere dence in $U$. Since the boundary of an open set has empty interior, and $U$ is open, it follows that each $A_n$ is nowhere dence in $X$ as well. For if $B\subset$ is not nowhere dense set in $X$, $V:=(\overline{B})^\circ\neq \emptyset$, both closure and interior taken in $X$, and $V\cap U$ is a non-empty open set containd in the closure of $B$, so $B$ is not nowhere dense in $U$. Since $A_n$ is nowhere dense in $X$, they have empty interior in $X$, and 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.

Here is a possibly easier proof in terms of the alternative definition for Baire spaces: a space in which the intersection of any countable family of open dense sets is dense.

Suppose we are given $$O_n$$ open in $$U$$ and dense in $$U$$. The set $$G_n=O_n\cup(X\setminus\overline{U})$$ is open in $$X$$ and dense in $$X$$, since its closure contains $$X\setminus\overline U$$ and contains $$\overline{O_n}\supseteq\overline{U}$$. Then $$\bigcap_n G_n=\bigcap_n O_n\cup(X\setminus\overline U)$$ is dense in $$X$$ by the Baire property of $$X$$. So any nonempty open subset of $$U$$ must meet $$\bigcap_n G_n$$, and since it does not meet $$X\setminus\overline U$$, it must meet $$\bigcap_n O_n$$. This means that $$\bigcap_n O_n$$ is dense in $$U$$.