Regularity is hereditary

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

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Similar facts

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:

Opposite facts

Facts used

  1. T1 is hereditary


Given: A topological space X, a subset A of X. X is regular.

To prove: A is regular.

Proof: The T1 property for A follows from Fact (1). It thus suffices to show the separation property for A, i.e., that any point and disjoint closed subset in A can be separated by a continuous function. In other words, we want to prove the following.

To prove (specific): For any point x \in A and any closed subset C of A such that x \notin A, there exist disjoint open subsets U_1, U_2 of A such that x \in U_1 and C \subseteq U_2.

Proof (specific):

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 There exists a closed subset D of X such that C= D \cap A. definition of subspace topology C is closed in A.
2 x is not in D x \in A, x \notin C Step (1) [SHOW MORE]
3 There exist disjoint open subsets V_1, V_2 of X such that x \in V_1, D \subseteq V_2. X is regular Steps (1), (2) Step-given combination direct
4 Define U_1 = V_1 \cap A and U_2 = V_2 \cap A.
5 U_1, U_2 are open subsets of A. definition of subspace topology Steps (3), (4) By Step (2), V_1,V_2 are open, so by the definition of subspace topology, U_1, U_2 are open as per their definitions in Step (3).
6 U_1, U_2 are disjoint. Steps (3), (4) follows directly from V_1,V_2 being disjoint
7 x \in U_1, C \subseteq U_2 x \in A, C \subseteq A Steps (3), (4) By Step (4), U_1 = V_1 \cap A. By Step (3), x \in V_1, and we are also given that x\in A, so x_1 \in V_1 \cap A = U_1. Similarly, C \subseteq V_2 and C \subseteq A, so C \subseteq V_2 \cap A = U_2.
8 U_1, U_2 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


Textbook references

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