# Baire property is open subspace-closed

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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]
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]

## References

### Textbook references

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