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 regular 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:
 * $a \in x$

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

Let
 * $L$ be continuous.

Thus $T$ is locally compact.