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:

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 $A$. 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]
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 $C \subseteq D$.
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