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.