Baire property is open subspace-closed: Difference between revisions

From Topospaces
No edit summary
No edit summary
Line 1: Line 1:
{{topospace metaproperty satisfaction}}
{{topospace metaproperty satisfaction|
property = Baire space|
metaproperty = open subspace-closed property of topological spaces}}


==Statement==
==Statement==

Revision as of 00:06, 25 January 2012

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)
View all topological space metaproperty satisfactions | View all topological space metaproperty dissatisfactions
Get more facts about Baire space |Get facts that use property satisfaction of Baire space | Get facts that use property satisfaction of Baire space|Get more facts about open subspace-closed property of topological spaces

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, Un,n∈N of A is dense in A

Proof: First, consider the set B=X∖A¯, where A¯ is the closure of A in X. Clearly B is open in X.

Since A is open in X, and each Un is open in A, each Un is open in X. Hence, each Un∪B is open in X.

Next, we want to argue that each Un∪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 A¯. Hence, by definition, it intersects A in a nonempty open subset, say W. Then W is an open subset of A. Since Un is dense in A, it intersects W nontrivially, completing the proof of density.

Thus, consider the collection of subsets Un∪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∪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∩T is nonempty. Since S is open in A, S is also open in X, so S∩(T∪B) is nonempty, because T∪B is dense in X. But since S⊂A, S∩B is empty, so S∩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