Squeeze Theorem for Filter Bases

Theorem
Let $(S,\le,\tau)$ be a Linearly Ordered Space.

Let $F_1$, $F_2$, and $F_3$ be filter bases in $S$.

Suppose that for each $T \in F_1$, $F_2$ has an element $M$ such that all elements of $M$ succeed some element of $T$.

Suppose also that for each $U \in F_3$, $F_2$ has an element $N$ such that all elements of $N$ precede some element of $U$.

Suppose that $F_1$ and $F_3$ each converge to a point $p \in S$.

Then $F_2$ converges to $p$.

Proof
Let $q \in S$, $q < p$.

We will show that $F_2$ has an element which is a subset of ${\uparrow}q$.

Since $F_1$ converges to $p$, it has an element $A \subseteq {\uparrow}q$.

Thus there is an element $k$ in $A$ and an element $M$ in $F_2$ such that all elements of $M$ succeed $k$.

Then by Extended Transitivity, $M \subseteq {\uparrow}q$.

A similar argument using $F_3$ proves the dual statement.

Thus $F_2$ converges to $p$.