=== Condition $(1)$ === Let $T$ be locally compact. By Topology forms Complete Lattice: :$L$ is complete lattice. Thus by Way Below Closure is Directed in Bounded Below Join Semilattice: :$\forall x \in \tau: x^\ll$ is directed Thus by definition: :$L$ is up-complete. Let $x \in \tau$. We will prove that :$x \subseteq \sup \left({x^\ll}\right)$ Let $a \in x$. By definition: :$x$ is open set in $T$ By definition of locally compact: :there exists local basis $\mathcal B$ of $a$ such that all elements are compact. By definition of local basis: :$\exists y \in \mathcal B: y \subseteq x$ Then :$y$ is compact. By Way Below if Between is Compact Set in Ordered Set of Topology: :$y \ll x$ By definition of way below closure: :$y \in x^\ll$ By Set is Subset of Union/Set of Sets: :$\displaystyle y \subseteq \bigcup \left({x^\ll}\right)$ By definition of subset: :$\displaystyle a \in \bigcup \left({x^\ll}\right)$ Thus by proof of Topology forms Complete Lattice: :$a \in \sup \left({x^\ll}\right)$ {{qed|lemma}} According to definition of set equality it remains to prove that: :$\sup \left({x^\ll}\right) \subseteq x$ Let $a \in \sup \left({x^\ll}\right)$. By proof of Topology forms Complete Lattice: :$\displaystyle a \in \bigcup \left({x^\ll}\right)$ By definition of union: :$\exists y \in x^\ll: a \in y$ By definition of way below closure: :$y \ll x$ By Way Below implies Preceding: :$y \preceq x$ Then :$y \subseteq x$ Thus by definition of subset: :$a \in x$ By definition: :$L$ satisfies the axiom of approximation. Thus by definition :$L$ is continuous. {{qed|lemma}} === Condition $(2)$ === Let $T$ be a $T_3$ space. Let $L$ be continuous. Let $x \in S, A \in \tau$ such that: :$x \in A$ By definition of continuous: :$L$ satisfies the axiom of approximation. {{begin-eqn}} {{eqn | l = A       | r = \sup \left({A^\ll}\right)       | c = {{Defof|Axiom of Approximation}} }} {{eqn | r = \bigcup \left({A^\ll}\right)       | c = Topology forms Complete Lattice }} {{end-eqn}} By definition of union: :$\exists y \in A^\ll: x \in y$ By definition of way below closure: :$y \ll A$ By definition of Definition:T3 Space/Definition 2: :$\exists N_x: \complement_S \left({N_x}\right) \in \tau: \exists V \in \tau: x \in V \subseteq N_x \subseteq y$ By definition of interior: :$x \in N_x^\circ$ where $N_x^\circ$ denotes the interior of $N_x$. We will prove that $N_x$ is topologically compact. Let $\mathcal F$ be a set of open subsets of $S$ such that: :$\mathcal F$ is a cover of $N_x$ Define a set of open subsets of $S$: :$\mathcal F' := \mathcal F \cup \left\{ {\complement_S\left({N_x}\right)}\right\}$ {{begin-eqn}} {{eqn | l = \bigcup \mathcal F'       | r = \left({\bigcup \mathcal F}\right) \cup \bigcup \left\{ {\complement_S\left({N_x}\right)}\right\}       | c = Union Distributes over Union/Sets of Sets }} {{eqn | r = \left({\bigcup \mathcal F}\right) \cup \complement_S\left({N_x}\right)       | c = Union of Singleton }} {{eqn | o = \supseteq       | r = N_x \cup \complement_S\left({N_x}\right)       | c = {{{Defof|Cover of Set}} }} {{eqn | r = S       | c = {{Defof|Relative Complement}} }} {{eqn | o = \supseteq       | r = A }} {{end-eqn}} Then :$A \subseteq \bigcup \mathcal F'$ By Way Below in Ordered Set of Topology: :there exists a finite subset $\mathcal G'$ of $\mathcal F'$: $y \subseteq \bigcup \mathcal G'$ Define a finite subset of $\mathcal F$: :$\mathcal G := \mathcal G' \setminus \left\{ {\complement_S\left({N_x}\right)}\right\}$ By definition of union: :$N_x \subseteq \bigcup \mathcal G$ Thus by definition: :$N_x$ has finite subcover of $\mathcal F$ Thus by definition: :$N_x$ is compact. {{qed|lemma}} Thus by Locally Compact iff Open Neighborhood contains Compact Set; :$T$ is locally compact. {{qed}}