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:

• Normality is weakly hereditary
• Compactness is weakly hereditary

Opposite facts

• Normality is not hereditary

Facts used

1. 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 $A$.
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 $C\subseteq D$.

References

Textbook references

• Topology (2nd edition) by James R. Munkres, ^{More info}, Page 211-212, Theorem 33.2, Chapter 4, Section 33