Topology is Locally Compact iff Ordered Set of Topology is Continuous

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

Let $L = \left({\tau, \preceq}\right)$ be an ordered set where $\preceq \mathop = \subseteq\restriction_{\tau \times \tau}$

Then
 * $(1): \quad T$ is locally compact implies $L$ is continuous
 * $(2): \quad T$ is $T_3$ space and $L$ is continuous implies $T$ is locally compact

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:
 * $y \subseteq \bigcup \left({x^\ll}\right)$

By definition of subset:
 * $a \in \bigcup \left({x^\ll}\right)$

Thus by proof of Topology forms Complete Lattice:
 * $a \in \sup \left({x^\ll}\right)$

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:
 * $a \in \bigcup \left({x^\ll}\right)$

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 axiom of approximation.

Thus by definition
 * $L$ is continuous.

Condition $(2)$
Let
 * $T$ be $T_3$ space.

Let
 * $L$ be continuous.

Let $x \in S, A \in \tau$ such that
 * $x \in A$

By definition of continuous:
 * $L$ satisfies axiom of approximation.

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 $S$
 * $\mathcal F' := \mathcal F \cup \left\{ {\complement_S\left({N_x}\right)}\right\}$

Then
 * $A \subseteq \bigcup \mathcal F'$

By Way Below in Ordered Set of Topology:
 * there exists 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.

Thus by Locally Compact iff Open Neighborhood contains Compact Set;
 * $T$ is locally compact.