Alexandroff Extension is Topology

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 $\tau^*$ 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:

$\text O 1$: Union of Open Sets
Let $\family {U_i}_{i \mathop \in I}$ 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 $\displaystyle \UU_J = \bigcup_{j \mathop \in J} U_j$ be the union of $\family {U_j}_{j \mathop \in J}$.

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

From Intersection of Closed Sets is Closed:
 * $\displaystyle \VV_J := \bigcap_{j \mathop \in J} \paren {S \setminus U_j}$ is closed in $T$

By De Morgan's Laws:
 * $\displaystyle S \setminus \UU_J = \VV_J = \bigcap_{j \mathop \in J} \paren {S \setminus U_j}$

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

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

Let $\displaystyle \UU_K = \bigcup_{k \mathop \in K} U_k$ be the union of $\family {U_k}_{k \mathop \in K}$.

Let $m \in J$ be arbitrary.

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

Then by De Morgan's Laws:
 * $\displaystyle S \setminus \UU_K' = \bigcap_{\substack {k \mathop \in K \\ k \mathop \ne m} } \paren {S \setminus U_k}$

Let:
 * $\displaystyle \VV_K' := S \setminus \UU_K' = \bigcap_{\substack {k \mathop \in K \\ k \mathop \ne m} } \paren {S \setminus U_k}$

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

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

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

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

Let $\VV_K := \VV_K' \cap \paren {S \setminus U_m}$.

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

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

But:
 * $\UU_K = S \setminus \VV_K$

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

Finally:

By definition, both $\paren {S \setminus \UU_J}$ and $\paren {S \setminus \UU_K}$ have been demonstrated to be closed in $T^*$.

So by Intersection of Closed Sets is Closed:
 * $\paren {S \setminus \UU_J} \cap \paren {S \setminus \UU_K}$ is closed in $T^*$.

Thus $\UU = \UU_J \cup \UU_K$ is an open set of $T^*$.

$\text O 2$: 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, $\paren {S \setminus U_1} \cup \paren {S \setminus U_2}$ is compact in $T$.

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

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

Thus $S \setminus \paren {U_1 \cap U_2}$ 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 $T$.

, 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, $\paren {S \setminus U_1} \cup \paren {S \setminus U_2}$ is closed in $T$.

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

Thus $S \setminus \paren {U_1 \cap U_2}$ 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 $T^*$.

$\text O 3$: Underlying Set
From Relative Complement with Self is Empty Set:
 * the complement of $S^*$ relative to $S^*$ is $\O$.

From Empty Set is Compact Space‎, $\O$ 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.