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

Question

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?

Answer 1

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 $n$.

Answer 2

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.

Answer 3

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 $U$.