Squeeze Theorem for Filter Bases
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Theorem
Let $\struct {S, \le, \tau}$ 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 $q^\ge$.
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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 $p$.
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$\blacksquare$