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,nN of A is dense in A

Proof: First, consider the set B=XA¯, 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 UnB is open in X.

Next, we want to argue that each UnB 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 UnB. This is a countable collection of open dense subsets of X, so their intersection is dense in X. This intersection is of the form BT, 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 ST is nonempty. Since S is open in A, S is also open in X, so S(TB) is nonempty, because TB is dense in X. But since SA, SB is empty, so ST 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