If Infimum of Filtered Subset belongs to Element of Sub-Basis then Subset and Element Intersect implies Infimum of Subset belongs to Closure of Subset

Theorem
Let $T = \struct {S, \preceq, \tau}$ be a complete topological lattice with lower topology.

Let $B$ be an analytic sub-basis of $T$.

Let $F$ be a filtered subset of $S$ such that
 * $\forall A \in B: \inf F \in A \implies F \cap A \ne \varnothing$

Then $\inf F \in F^-$

where $F^-$ denotes the topological closure of $F$.

Proof
We will prove that
 * $\forall A \in B, x \in F \cap A, y \in F: y \preceq x \implies y \in A$

Let $A \in B$, $x \in F \cap A$, $y \in F$.

By definition of sub-basis:
 * $A$ is open.

By Open Subset is Lower in Lower Topology:
 * $A$ is lower.

By definition of intersection:
 * $x \in A$.

Thus by definition of lower set:
 * $y \preceq x \implies y \in A$.

Define $\ds H := \set {\bigcap G: G \subseteq B, G \text { is finite} }$

By definitions of sub-basis and basis:
 * $H$ is basis of $T$.

We will prove that
 * $\forall A \in H: \inf F \in A \implies F \cap A \ne \O$

Let $A \in H$ such that
 * $\inf F \in A$

By definition of $H$:
 * $\exists G \subseteq B: A = \bigcap G \land G$ is finite.

By definition of intersection:
 * $\forall C \in G: \inf F \in C \in B$

By assumption:
 * $\forall C \in G: F \cap C \ne \O$

By definition of non-empty set:
 * $\forall C \in G: \exists x: x \in F \cap C$

By Axiom of Choice:
 * $\exists f: G \to F: \forall C \in G: \map f C \in F \cap C$

By Image of Mapping from Finite Set is Finite:
 * $f \sqbrk G$ is finite.

By Filtered iff Finite Subsets have Lower Bounds:
 * $\exists h \in F: \forall x \in f \sqbrk G: h \preceq x$

Then
 * $\forall C \in G: h \in C$

By definition of intersection:
 * $h \in \bigcap G$

Thus by definitions of intersection and non-empty set:
 * $F \cap A \ne \O$

Thus by Characterization of Closure by Basis:
 * $\inf F \in F^-$