Squeeze Theorem/Sequences/Linearly Ordered Space

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

Let $\left\langle{x_n}\right\rangle$, $\left\langle{y_n}\right\rangle$, and $\left\langle{z_n}\right\rangle$ be sequences in $S$.

Let $p \in S$.

Let $\left\langle{x_n}\right\rangle$ and $\left\langle{z_n}\right\rangle$ both converge to $p$

For each $n$, let $x_n \le y_n \le z_n$.

Then $\left\langle{y_n}\right\rangle$ converges to $p$.


Let $m \in S$ and $m < p$.

Then $\left\langle{x_n}\right\rangle$ eventually succeeds $m$.

Thus by Extended Transitivity, $\left\langle{y_n}\right\rangle$ eventually succeeds $m$.

A similar argument using $\left\langle{z_n}\right\rangle$ proves the dual statement.

Thus $\left\langle{y_n}\right\rangle$ is eventually in each ray containing $p$, so it converges to $p$.