Alexandroff Extension is Topology

Theorem
Let $T = \left({S, \tau}\right)$ be a non-empty topological space.

Let $p$ be a point which is not in $S$.

Let $S^* := S \cup \left\{{p}\right\}$.

Let $T^* = \left({S^*, \tau^*}\right)$ be the Alexandroff extension on $S$.

Then $\tau^*$ is a topology on $S^*$.

Proof
Recall the definition of the Alexandroff extension on $S$:

$U$ is open in $T^*$ :
 * $U$ is an open set of $T$

or
 * $U$ is the complement in $T^*$ of a closed and compact subset of $T$.

Each of the open set axioms is examined in turn:

$O1$: Union of Open Sets
Let $\left \langle{U_i}\right \rangle_{i \mathop \in I}$ be an indexed family of open sets of $T^*$.

Some (perhaps all, perhaps none) of the $U_i$ are open sets of $T$.

The rest of them are each the complement of a closed and compact subset of $T$.

Let $J \subseteq I$ be the subset of $I$ consisting of the indices of the former open sets of $T^*$.

Let $K = I \setminus J$ be the subset of $I$ consisting of the indices of the latter open sets of $T^*$.

Let $\displaystyle \mathcal U_J = \bigcup_{j \mathop \in J} U_j$ be the union of $\left \langle{U_j}\right \rangle_{j \mathop \in J}$.

By definition, each $S \setminus U_j$ is closed in $T$.

From Intersection of Closed Sets is Closed:
 * $\displaystyle \mathcal V_J := \bigcap_{j \mathop \in J} \left({S \setminus U_j}\right)$ is closed in $T$

By De Morgan's Laws:
 * $\displaystyle S \setminus \mathcal U_J = \mathcal V_J = \bigcap_{j \mathop \in J} \left({S \setminus U_j}\right)$

By definition of closed set it follows that $\mathcal U_J$ is open in $T$.

By definition of the Alexandroff extension on $S$, it follows that $\mathcal U_J$ is open in $T^*$.

Let $\displaystyle \mathcal U_K = \bigcup_{k \mathop \in K} U_k$ be the union of $\left \langle{U_k}\right \rangle_{k \mathop \in K}$.

Let $m \in J$ be arbitrary.

Let $\displaystyle \mathcal U_K' = \bigcup_{\substack {k \mathop \in K \\ k \mathop \ne m} } U_k$.

Then by De Morgan's Laws:
 * $\displaystyle S \setminus \mathcal U_K' = \bigcap_{\substack {k \mathop \in K \\ k \mathop \ne m} } \left({S \setminus U_k}\right)$

Let:
 * $\displaystyle \mathcal V_K' := S \setminus \mathcal U_K' = \bigcap_{\substack {k \mathop \in K \\ k \mathop \ne m} } \left({S \setminus U_k}\right)$

Each of $S \setminus U_k$ is closed and compact in $T$.

From Intersection of Closed Sets is Closed:
 * $\mathcal V_K'$ is closed in $T$.

But $S \setminus U_m$ is also closed in $T$.

$S \setminus U_m$ is also compact in $T$.

Let $\mathcal V_K := \mathcal V_K' \cap \left({S \setminus U_m}\right)$.

So from Intersection of Closed Sets is Closed:
 * $\mathcal V_K$ is closed in $T$

and from Intersection of Closed Set with Compact Subspace is Compact:
 * $\mathcal V_K$ is compact in $T$.

But:
 * $\mathcal U_K = S \setminus \mathcal V_K$

and so by definition $\mathcal U_K$ is open in $T^*$.

Finally:

By definition, both $\left({S \setminus \mathcal U_J}\right)$ and $\left({S \setminus \mathcal U_K}\right)$ have been demonstrated to be closed in $T^*$.

So by Intersection of Closed Sets is Closed:
 * $\left({S \setminus \mathcal U_J}\right) \cap \left({S \setminus \mathcal U_K}\right)$ is closed in $T^*$.

Thus $\mathcal U = \mathcal U_J \cup \mathcal U_K$ is an open set of $T^*$.

$O2$: Intersection of Open Sets
Let $U_1$ and $U_2$ be open sets of $T^*$.

$(1): \quad$ Suppose $U_1$ and $U_2$ are both open sets of $T$.

Then as $T$ is a topological space, $U_1 \cap U_2$ is open in $T$.

By definition of the Alexandroff extension on $S$, it follows that $U_1 \cap U_2$ is open in $T^*$.

$(2): \quad$ Suppose that neither $U_1$ and $U_2$ is an open set of $T$.

Then both of their complements $S \setminus U_1$ and $S \setminus U_2$ in $S$ are both closed and compact in $T$.

From Finite Union of Compact Sets is Compact, $\left({S \setminus U_1}\right) \cup \left({S \setminus U_2}\right)$ is compact in $T$.

From Finite Union of Closed Sets is Closed, $\left({S \setminus U_1}\right) \cup \left({S \setminus U_2}\right)$ is closed in $T$.

But by De Morgan's Laws:
 * $S \setminus \left({U_1 \cap U_2}\right) = \left({S \setminus U_1}\right) \cup \left({S \setminus U_2}\right)$

Thus $S \setminus \left({U_1 \cap U_2}\right)$ is both closed and compact in $T$.

Hence by definition $U_1 \cap U_2$ is an open set in $T^*$.

$(3): \quad$ Suppose that either $U_1$ or $U_2$ (but not both) is an open set of $T$.

WLOG suppose $U_1$ is an open set of $T$ and $U_2$ is not.

As $U_1$ is an open set of $T$ it follows that $p \notin U_1$.

Thus it follows that $p \notin U_1 \cap U_2$.

We have that $S^* \setminus U_2$ is not an open set of $T$.

Thus $S^* \setminus U_2$ is both closed and compact in $T$.

From Finite Union of Closed Sets is Closed, $\left({S \setminus U_1}\right) \cup \left({S \setminus U_2}\right)$ is closed in $T$.

But by De Morgan's Laws:
 * $S \setminus \left({U_1 \cap U_2}\right) = \left({S \setminus U_1}\right) \cup \left({S \setminus U_2}\right)$

Thus $S \setminus \left({U_1 \cap U_2}\right)$ is closed in $T$.

But as $p \notin S_1 \cap S^2$ it follows that $U_1 \cap U_2$ is an open set of $T$.

Hence by definition $U_1 \cap U_2$ is an open set of $T^*$.

$O3$: Underlying Set
From Relative Complement with Self is Empty Set:
 * the complement of $S^*$ relative to $S^*$ is $\varnothing$.

From Empty Set is Compact Space‎, $\varnothing$ is a compact subspace of $T^*$.

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

All the open set axioms are fulfilled, and the result follows.