T3 1/2 Property is Hereditary

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space which is a $T_{3 \frac 1 2}$ 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 $T_{3 \frac 1 2}$ space.

Proof
Let $T = \left({S, \tau}\right)$ be a $T_{3 \frac 1 2}$ space.

Then:
 * For any closed set $F \subseteq S$ and any point $y \in S$ such that $y \notin F$, there exists an Urysohn function for $F$ and $\left\{{y}\right\}$.

We have that the set $\tau_H$ is defined as:
 * $\tau_H := \left\{{U \cap H: U \in \tau}\right\}$

Let $F \subseteq H$ such that $F$ is closed in $H$.

Let $y \in H$ such that $y \notin F$.

From Closed Set in Topological Subspace $F$ is also closed in $T$.

Because $T$ is a $T_{3 \frac 1 2}$ space, we have that there exists an Urysohn function for $F$ and $\left\{{y}\right\}$:

That is, there exists a continuous mapping $f: S \to \left[{0 \,.\,.\, 1}\right]$, where $\left[{0 \,.\,.\, 1}\right]$ is the closed unit interval, such that:
 * $f {\restriction_F} = 0, f {\restriction_{\left\{{y}\right\}}} = 1$

where $f {\restriction_F}$ denotes the restriction of $f$ to $F$.

That is:
 * $\forall a \in F: f \left({a}\right) = 0$
 * $\forall b \in \left\{{y}\right\}: f \left({b}\right) = 1$

From Continuity of Composite with Inclusion, as $f$ is continuous on $S$, then $f {\restriction_H}$ is continuous on $H$.

Thus $f {\restriction_H}$ is an Urysohn function for $F$ and $\left\{{y}\right\}$ in $H$.

So the $T_{3 \frac 1 2}$ axiom is satisfied in $H$.