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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Statement

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

Definitions used

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Related facts

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

Normality is not hereditary

Facts used

T1 is hereditary

Proof

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] We are given that $x \in A$ , so if $x \in D$ , then $x \in D \cap A$ . By Step (1), $D \cap A = C$ . Thus, if $x \in D$ , then $x \in C$ , which we know is not true. Hence, $x \notin D$ .
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.

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References

Textbook references

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