Baire property is open subspace-closed: Difference between revisions
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 , an open subset
To prove: An intersection of a countable family of open dense subsets, of is dense in
Proof: First, consider the set , where is the closure of in . Clearly is open in .
Since is open in , and each is open in , each is open in . Hence, each is open in .
Next, we want to argue that each is dense in . For this, consider any open subset of . If the open subset intersects , we are done. Otherwise the open subset is in . Hence, by definition, it intersects in a nonempty open subset, say . Then is an open subset of . Since is dense in , it intersects nontrivially, completing the proof of density.
Thus, consider the collection of subsets . This is a countable collection of open dense subsets of , so their intersection is dense in . This intersection is of the form , where is an open subset of . We now need to argue that is a dense subset of .
Let be a nonempty open subset of ; we need to show that is nonempty. Since is open in , is also open in , so is nonempty, because is dense in . But since , is empty, so 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