Complete regularity is hereditary

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)
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This article gives the statement, and possibly proof, of a basic fact in topology.

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 $X$ is completely regular if it is a T1 space and given any point $x\in X$ and closed subset $C\subseteq X$ such that $x\notin C$ , there exists a continuous map $f:X\to [0,1]$ such that $f(x)=0$ and $f(a)=1$ for all $a\in C$ .
subspace topology	For a subset $A$ of the space $X$ , the subspace topology on $A$ is defined as follows: a subset of $A$ is open in $A$ iff it can be expressed as the intersection with $A$ of an open subset of $X$ . Also, a subset of $A$ is closed in $A$ iff it can be expressed as the intersection with $A$ of a closed subset of $X$ .

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

Normality is not hereditary

Facts used

T1 is hereditary

Proof

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

To prove: $A$ is completely 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 exists a continuous function from $A$ to $[0,1]$ such that $f(x)=0$ and $f(a)=1$ for all $a\in C$ .

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 $A$ .		Direct from definition. Note: We can, for concreteness, take $D={\overline {C}}$ , the closure of $C$ in $X$ -- this is the smallest $D$ that works.
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 exists a continuous function $g:X\to [0,1]$ such that $g(x)=0$ and $g(a)=1$ for all $a\in D$ .		$X$ is completely regular	Steps (1), (2)	Step-given combination direct
4	Define $f$ as the restriction of $g$ to $A$ . In other words, if $i:A\to X$ is the inclusion mapping, then $f=g\circ i$ . Then, $f$ is a continuous function from $A$ to $[0,1]$ .	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 $f$ defined in Step (4) has the property that $f(x)=0$ and $f(a)=1$ for all $a\in C$ .			Steps (1), (3), (4)	The condition on $f$ follows directly from the condition on $g$ in Step (3) and the fact that $C=D\cap A$ , so $C\subseteq D$ .