Regularity is hereditary
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)
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|Property||Proof that is is subspace-hereditary|
|Hausdorff space||Hausdorffness is hereditary|
|completely regular space||complete regularity is hereditary|
|Urysohn space||Urysohn is hereditary|
|T1 space||T1 is hereditary|
Other similar facts:
Given: A topological space , a subset of . is regular.
To prove: is 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 exist disjoint open subsets of such that and .
|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 .|
|2||is not in||Step (1)||[SHOW MORE]|
|3||There exist disjoint open subsets of such that .||is regular||Steps (1), (2)||Step-given combination direct|
|4||Define and .|
|5||are open subsets of .||definition of subspace topology||Steps (3), (4)||By Step (2), are open, so by the definition of subspace topology, are open as per their definitions in Step (3).|
|6||are disjoint.||Steps (3), (4)||follows directly from being disjoint|
|7||Steps (3), (4)||By Step (4), . By Step (3), , and we are also given that , so . Similarly, and , so .|
|8||are the desired open subsets.||Steps (5)-(7)||Step-combination, this is exactly what we want to prove.|
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 196, Theorem 31.2(b), Chapter 4, Section 31