Recall the definition of the Alexandroff extension on $S$: $U$ is open in $T^*$ {{iff}}: :$U$ 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 in Topological Space: :$\displaystyle \VV_J := \bigcap_{j \mathop \in J} \paren {S \setminus U_j}$ 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 in Topological Space: :$\VV_K'$ 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 in Topological Space: :$\VV_K$ 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: {{begin-eqn}} {{eqn | l = \displaystyle \UU = \bigcup_{i \mathop \in I} U_i       | r = \UU_J \cup \UU_K       | c = Definition of $\UU_J$ and $\UU_K$ }} {{eqn | ll= \leadsto       | l = S \setminus \UU       | r = S \setminus \paren {\UU_J \cup \UU_K}       | c =  }} {{eqn | r = \paren {S \setminus \UU_J} \cap \paren {S \setminus \UU_K}       | c =  }} {{end-eqn}} 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 in Topological Space: :$\paren {S \setminus \UU_J} \cap \paren {S \setminus \UU_K}$ is closed in $T^*$. Thus $\UU = \UU_J \cup \UU_K$ is an open set of $T^*$. {{qed|lemma}} === $\text O 2$: 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 in Topological Space, $\paren {S \setminus U_1} \cup \paren {S \setminus U_2}$ 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 $T$. {{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 in Topological Space, $\paren {S \setminus U_1} \cup \paren {S \setminus U_2}$ 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 $T^*$. {{qed|lemma}} === $\text O 3$: 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 $T^*$. {{qed|lemma}} All the open set axioms are fulfilled, and the result follows. {{qed}}