Complete regularity is hereditary
This article gives the statement, and possibly proof, of a topological space property (i.e., completely regular space) satisfying a topological space metaproperty (i.e., subspace-hereditary property of topological spaces)
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This article gives the statement, and possibly proof, of a basic fact in topology.
|completely regular space||A space is completely regular if it is a T1 space and given any point and closed subset such that , there exists a continuous map such that and for all .|
|subspace topology|| For a subset of the space , the subspace topology on is defined as follows: a subset of is open in iff it can be expressed as the intersection with of an open subset of .|
Also, a subset of is closed in iff it can be expressed as the intersection with of a closed subset of .
|Property||Proof that is is subspace-hereditary|
|Hausdorff space||Hausdorffness is hereditary|
|regular space||regularity is hereditary|
|Urysohn space||Urysohn is hereditary|
|T1 space||T1 is hereditary|
Other similar facts:
Given: A topological space , a subset of . is completely regular.
To prove: is completely regular.
Proof: The T1 property for follows from Fact (1). It thus suffices to show the separation property for , i.e., that any point and disjoint closed subset in can be separated by a continuous function. In other words, we want to prove the following.
To prove (specific): For any point and any closed subset of such that , there exists a continuous function from to such that and for all .
|Step no.||Assertion/construction||Facts used||Given data used||Previous steps used||Explanation|
|1||There exists a closed subset of such that||definition of subspace topology||is closed in .||Direct from definition. Note: We can, for concreteness, take , the closure of in -- this is the smallest that works.|
|2||is not in||Step (1)||[SHOW MORE]|
|3||There exists a continuous function such that and for all .||is completely regular||Steps (1), (2)||Step-given combination direct|
|4||Define as the restriction of to . In other words, if is the inclusion mapping, then . Then, is a continuous function from to .||definition of subspace topology (the inclusion mapping from a subset with the subspace topology is continuous), a composite of continuous maps is continuous||Step (3)||Step-given combination direct|
|5||The function defined in Step (4) has the property that and for all .||Steps (1), (3), (4)||The condition on follows directly from the condition on in Step (3) and the fact that , so .|
This proof uses a tabular format for presentation. Learn more about tabular proof formats|View all pages on facts with proofs in tabular format
- Topology (2nd edition) by James R. Munkres, More info, Page 211-212, Theorem 33.2, Chapter 4, Section 33