Baire property is open subspace-closed

This article gives the statement, and possibly proof, of a topological space property satisfying a topological space metaproperty
View all topological space metaproperty satisfactions | View all topological space metaproperty dissatisfactions
|

Property "Page" (as page type) with input value "{{{property}}}" contains invalid characters or is incomplete and therefore can cause unexpected results during a query or annotation process.Property "Page" (as page type) with input value "{{{metaproperty}}}" contains invalid characters or is incomplete and therefore can cause unexpected results during a query or annotation process.

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

Proof

Given: A Baire space $X$, an open subset $A$

To prove: An intersection of a countable family of open dense subsets, $U_n,n\in \mathbb{N}$ of $A$ is dense in $A$

Proof: First, consider the set $B=X\setminus \overline{A}$, where $\overline{A}$ is the closure of $A$ in $X$. Clearly $B$ is open in $X$.

Since $A$ is open in $X$, and each $U_n$ is open in $A$, each $U_n$ is open in $X$. Hence, each $U_n\cup B$ is open in $X$.

Next, we want to argue that each $U_n\cup B$ is dense in $X$. For this, 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, completing the proof of density.

Thus, consider the collection of subsets $U_n\cup B$. This is a countable collection of open dense subsets of $X$, so their intersection is dense in $X$. This intersection is of the form $B\cup T$, where $T$ is an open subset of $A$. We now need to argue that $T$ is a dense subset of $A$.

Let $S$ be a nonempty open subset of $A$; we need to show that $S\cap T$ is nonempty. Since $S$ is open in $A$, $S$ is also open in $X$, so $S\cap (T\cup B)$ is nonempty, because $T\cup B$ is dense in $X$. But since $S\subset A$, $S\cap B$ 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