Baire property is open subspace-closed

This article gives the statement, and possibly proof, of a topological space property (i.e., Baire space) satisfying a topological space metaproperty (i.e., open subspace-closed property of topological spaces)
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Statement

Verbal statement

Every open subset of a Baire space is itself a Baire space, under the subspace topology.

Definitions used

Baire space

Further information: Baire space

A topological space is termed a Baire space if an intersection of countably many open dense subsets is dense.

Subspace topology

Further information: Subspace topology

Facts used

1. Open subset of open subspace is open

Proof

Given: A Baire space $X$, an open subset $A$. A countable family of open dense subsets, $U_n,n\in \mathbb{N}$ of $A$

To prove: The intersection $T={\bigcap }_{n\in \mathbb{N}}{U}_n$ is dense in $A$

Proof:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	Consider the set $B=X\setminus \overline{A}$, where $\overline{A}$ is the closure of $A$ in $X$. $B$ is an open subset of $X$.				the complement of a closed subset is open by definition.
2	Each $U_n$ is open in $X$.	Fact (1)	$A$ is open in $X$ Each $U_n$ is open in $A$.
3	Each $U_n\cup B$ is open in $X$ with $B$ defined in Step (1).			Steps (1), (2)	Step-combination, and the observation that a union of open subsets is open
4	Each $U_n\cup B$ is dense in $X$. In other words, for any open subset $V$ of $X$, the intersection $(U_n\cup B)\cap V$ is nonempty.				[SHOW MORE] Consider any open subset $V$ of $X$. If the open subset intersects $B$, we are done. Otherwise the open subset is in $\overline{A}$. Hence, by definition, it intersects $A$ in a nonempty open subset, say $W$. Then $W$ is an open subset of $A$. Since $U_n$ is dense in $A$, it intersects $W$ nontrivially, so it intersects $V$ nontrivially, completing the proof of density.
5	Consider the collection of subsets $U_n\cup B$. This is a countable collection of open dense subsets of $X$			Steps (3), (4)	Step-combination direct
6	The intersection ${\bigcap }_{n\in \mathbb{N}}(U_n\cup B)$ is dense in $X$.		$X$ is a Baire space.	Step (5)	Step-given direct
7	${\bigcap }_{n\in \mathbb{N}}(U_n\cup B)=T\cup B$ where $T={\bigcap }_{n\in \mathbb{N}}{U}_n$				pure set theory
8	$T\cup B$ is dense in $X$.			Steps (6), (7)	Step-combination direct
9	$T$ is dense in $A$. In other words, for any nonempty open subset $S$ of $A$, $S\cap T$ is nonempty	Fact (1)	$A$ is open in $X$	Step (8)	[SHOW MORE] Since $S$ is open in $A$, $S$ is also open in $X$ (by Fact (1)). By Step (8), $T\cup B$ is dense in $X$, so $S\cap (T\cup B)=(S\cap T)\cup (S\cap B)$ is nonempty. But since $S\subseteq A$, $S\cap B$ is empty, so $S\cap T$ must be nonempty, completing the proof.

References

Textbook references

• Topology (2nd edition) by James R. Munkres, ^{More info}, Page 297, Lemma 48.4, Chapter 4, Section 48