T5 Property is Hereditary

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Theorem

Let $T = \left({S, \tau}\right)$ be a topological space which is a $T_5$ space.

Let $T_H = \left({H, \tau_H}\right)$, where $\varnothing \subset H \subseteq S$, be a subspace of $T$.


Then $T_H$ is a $T_5$ space.


Proof

Let $T = \left({S, \tau}\right)$ be a $T_5$ space.

Then:

$\forall A, B \subseteq S, \operatorname{cl}_S \left({A}\right) \cap B = A \cap \operatorname{cl}_S \left({B}\right) = \varnothing: \exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \varnothing$

where $\operatorname{cl}_S \left({A}\right)$ 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:

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


Let $A, B \subseteq H$ such that $\operatorname{cl}_H \left({A}\right) \cap B = A \cap \operatorname{cl}_H \left({B}\right) = \varnothing$.

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


From Closure in Subspace it follows that $\operatorname{cl}_S \left({A}\right) \cap B = A \cap \operatorname{cl}_S \left({B}\right) = \varnothing$ and so $A$ and $B$ are separated in $S$ as well as in $H$.



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

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

As $F \subseteq H$ and $y \in H, y \notin F$ we have that:

$A \subseteq U \cap H, B \subseteq V \cap H: \left({U \cap H}\right) \cap \left({V \cap H}\right) = \varnothing$

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

$\blacksquare$


Sources