Way Below Compact is Topological Compact

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

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

Let $x \in \tau$.

Then
 * $x$ is compact in $L$ (it means: $x \ll x$)


 * $T_x$ is compact (topologically)

where $T_x = \left({x, \tau_x}\right)$ denotes the topological subspace of $x$.

Sufficient Condition
Let
 * $x \ll x$

Let $F \subseteq \tau_x$ be a open cover of $x$.

By definition of cover:
 * $x \subseteq \bigcup F$

By definition of topological space:
 * $\forall y \in \tau: x \cap y \in \tau$

By definition of subset:
 * $\tau_x \subseteq \tau$

By Subset Relation is Transitive:
 * $F \subseteq \tau$

By Way Below in Ordered Set of Topology:
 * there exists a finite subset $G$ of $F$: $x \subseteq \bigcup G$

By definition:
 * $G$ is cover of $x$

Thus by definition:
 * $x$ has finite subcover.

Thus by definition:
 * $T_x$ is compact.