Regularity is hereditary: Difference between revisions

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{{topospace metaproperty satisfaction}}
{{topospace metaproperty satisfaction|
 
property = regular space|
{{basic fact}}
metaproperty = subspace-hereditary property of topological spaces}}


==Statement==
==Statement==
===Property-theoretic statement===
The [[property of topological spaces]] of being a [[regular space]] is a [[hereditary property of topological spaces]].
===Verbal statement===


Any subset of a [[regular space]] is regular under the [[subspace topology]].
Any subset of a [[regular space]] is regular under the [[subspace topology]].
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==Definitions used==
==Definitions used==


===Regular space===
{{fillin}}
 
{{further|[[Regular space]]}}
 
===Subspace topology===
 
{{further|[[Subspace topology]]}}
 
==Proof==
==Proof==



Revision as of 21:40, 24 January 2012

This article gives the statement, and possibly proof, of a topological space property (i.e., regular space) satisfying a topological space metaproperty (i.e., subspace-hereditary property of topological spaces)
View all topological space metaproperty satisfactions | View all topological space metaproperty dissatisfactions
Get more facts about regular space |Get facts that use property satisfaction of regular space | Get facts that use property satisfaction of regular space|Get more facts about subspace-hereditary property of topological spaces

Statement

Any subset of a regular space is regular under the subspace topology.

Definitions used

Fill this in later

Proof

Proof outline

Any subspace of a T1 space is T1, so we only need to check separation of points and closed sets. We do this as follows:

  • Pick a point, and a closed set not containing it, in the subspace (the set is closed relative to the subspace)
  • By the definition of subspace topology, find a closed set in the whole space, whose intersection with the subspace is the given closed set)
  • Separate the point and this bigger closed set, in the whole space, by disjoint open sets (using regularity of the whole space)
  • Intersect these open sets with the subspace to get a separation by disjoint open sets in the subspace

Note that this proof does not work for normality because we would need to enlarge both closed subsets, and the process of enlarging might lead to intersection.

References

Textbook references

  • Topology (2nd edition) by James R. Munkres, More info, Page 196, Theorem 31.2(b), Chapter 4, Section 31