Squeeze Theorem for Filter Bases

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Theorem

Let $\left({S, \le, \tau}\right)$ be a linearly ordered space.

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

Let:

$\forall T \in F_1: \exists M \in F_2: \forall x \in M: \exists y \in T: y \le x$

That is:

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

Similarly, let:

$\forall U \in F_3: \exists N \in F_2: \forall x \in N: \exists y \in U: x \le y$

That is:

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

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


Then $F_2$ converges to $p$.


Proof

Let $q \in S$ such that $q < p$.

We will show that $F_2$ has an element which is a subset of $q^\ge$.



Since $F_1$ converges to $p$, it has an element:

$A \subseteq q^\ge$.

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:

$M \subseteq q^\ge$

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

Thus $F_2$ converges to $p$.



$\blacksquare$