Condition for Alexandroff Extension to be T1 Space

Theorem
Let $T = \struct {S, \tau}$ be a non-empty topological space.

Let $p$ be a new element not in $S$.

Let $S^* := S \cup \set p$.

Let $T^* = \struct {S^*, \tau^*}$ be the Alexandroff extension on $S$.

Then $T^*$ is a $T_1$ (Fréchet) space $T$ is a $T_1$ (Fréchet) space.

Necessary Condition
Let $T = \struct {S, \tau}$ be a $T_1$ space.

By definition, $T$ is a $T_1$ space all points of $S$ are closed in $T$.

We have that $S$ is open in $T$ by definition of a topology.

Thus by definition of the Alexandroff extension, $S$ is open in $T^*$.

So as $S = S^* \setminus \set p$ is open in $T^*$, it follows that $\set p$ is closed in $T^*$.

That is, $p$ is a closed point of $T^*$.

Now let $x \in S^*$ such that $x \ne p$.

As $T$ is a $T_1$ space, $\set x$ is closed in $T$.

From Finite Topological Space is Compact, $\set x$ is a compact subset of $T$.

Thus by definition of the Alexandroff extension, $S^* \setminus \set x$ is open in $T^*$.

Thus by definition $\set x$ is closed in $T^*$.

That is, $x$ is a closed point of $T^*$.

Thus it has been shown that all points in $T^*$ are closed points of $T^*$.

Thus by definition $T^*$ is a $T_1$ space.

Sufficient Condition
Let $T^*$ be a $T_1$ space.

By definition of $T$ is a subspace of $T^*$.

By $T_1$ Property is Hereditary, it follows that $T$ is a $T_1$ space.