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 $T = \left({S, \preceq, \tau}\right)$ 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 $H := \left\{ {\bigcap G: G \subseteq B, G \text{ is finite} }\right\}$

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

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

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