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 ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() |
subspace topology | For a subset ![]() ![]() ![]() ![]() ![]() ![]() ![]() Also, a 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 ![]() ![]() ![]() |
definition of subspace topology | ![]() ![]() |
Direct from definition. Note: We can, for concreteness, take ![]() ![]() ![]() ![]() | |
2 | ![]() ![]() |
![]() |
Step (1) | [SHOW MORE] | |
3 | There exists a continuous function ![]() ![]() ![]() ![]() |
![]() |
Steps (1), (2) | Step-given combination direct | |
4 | Define ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() |
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 ![]() ![]() ![]() ![]() |
Steps (1), (3), (4) | The condition on ![]() ![]() ![]() ![]() |
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