Complete regularity is hereditary
From Topospaces
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)
View all topological space metaproperty satisfactions | View all topological space metaproperty dissatisfactions
Get more facts about completely regular space |Get facts that use property satisfaction of completely regular space | Get facts that use property satisfaction of completely regular space|Get more facts about subspace-hereditary property of topological spaces
This article gives the statement, and possibly proof, of a basic fact in topology.
Contents
Statement
Any subset of a completely regular space is completely regular in the subspace topology.
Definitions used
Term | Definition used |
---|---|
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 . |
Related facts
Similar facts
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:
Opposite facts
Facts used
Proof
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 .
Proof (specific):
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
References
Textbook references
- Topology (2nd edition) by James R. Munkres, ^{More info}, Page 211-212, Theorem 33.2, Chapter 4, Section 33