T5 Property is Hereditary

Theorem

Let $T = \struct {S, \tau}$ be a topological space which is a $T_5$ space.

Let $T_H = \struct {H, \tau_H}$, where $\O \subset H \subseteq S$, be a subspace of $T$.

Then $T_H$ is a $T_5$ space.

Proof

Let $T = \struct {S, \tau}$ be a $T_5$ space.

Then:

$\forall A, B \subseteq S, \map {\cl_S} A \cap B = A \cap \map {\cl_S} B = \O: \exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \O$

where $\map {\cl_S} A$ denotes the closure of $A$ in $S$.

That is:

For any two separated sets $A, B \subseteq S$ there exist disjoint open sets $U, V \in \tau$ containing $A$ and $B$ respectively.

We have that the set $\tau_H$ is defined as:

$\tau_H := \set {U \cap H: U \in \tau}$

Let $A, B \subseteq H$ such that $\map {\cl_H} A \cap B = A \cap \map {\cl_H} B = \O$.

That is, $A$ and $B$ are separated in $H$.

Then:

\(\ds \map {\cl_S} A \cap B\)

\(=\)

\(\ds \map {\cl_S} A \cap \paren {H \cap B}\)

as $B \subseteq H$

\(\ds \)

\(=\)

\(\ds \paren {\map {\cl_S} A \cap H} \cap B\)

Intersection is Associative

\(\ds \)

\(=\)

\(\ds \map {\cl_H} A \cap B\)

Closure of Subset in Subspace

\(\ds \)

\(=\)

\(\ds \O\)

Assumption

Similarly:

$A \cap \map {\cl_S} B = \O$

So $A$ and $B$ are separated in $S$.

Because $T$ is a $T_5$ space, we have that:

$\exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \O$

It follows that:

$\exists U \cap H, V \cap H \in \tau_H : A \subseteq U \cap H, B \subseteq V \cap H, \paren {U \cap H} \cap \paren {V \cap H} = \O$

and so the $T_5$ axiom is satisfied in $H$.

$\blacksquare$

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Functions, Products, and Subspaces