I am wondering if in a totally ordered topological space (order topology) that is connected (hence linear continuum) and metrizable, does every sequence have a monotone subsequence?
Can we generalize this theorem for every totally ordered topological space?
This has nothing to do with topology, the question is simply whether every sequence $\Bbb N\to P$, where $P$ is a totally ordered set, has a monotone subsequence, and the usual proof done for sequences $\Bbb N\to\Bbb R$ works without changes.
Namely let $(p_i)_{i\in\Bbb N}$ be a sequence in $P$. We say that $p_k$ is a peak of the sequence if $l>k$ implies $p_k\geq p_l$. We now distinguish different cases:
If there are infinitely many peaks then the subsequence of peaks is an infinite nonincreasing sequence.
If there are finitely many peaks let $i_1$ be such that $p_{i_1}$ is the successor of the last peak. Since $p_{i_1}$ is not a peak we can find $i_2>i_1$ such that $p_{i_2}\geq p_{i_1}$. Since $p_{i_2}$ is not a peak we can find $i_3>i_2$ such that $p_{i_1}\leq p_{i_2}\leq p_{i_3}$. Continue this construction inductively to get a nondecreasing subsequence.