Every compact chain has a convergent subchain Suppose that (1) $S$ is an infinite chain (i.e. an infinite set with a relation $\preceq$ which is antisymmetric, transitive, and complete), (2) $S$ is endowed with some metric, and (3) $S$ is compact in the corresponding metric topology.
Does it then follow that $S$ has a 'convergent' 'subchain'?
I am putting these in scare quotes since I am not sure how to define them myself. But I was thinking something like: subchain = a cofinal subset of a chain, a subchain $S^\prime$ of $S$ is convergent = there is some $z$ in $S$ and for all $\epsilon>0$ there is an $x$ in the subchain such that, for all $y$ with $x\preceq y$, $d(z,y)<\epsilon$.
Does this make sense? Is it true? If so, how to prove it?
If it doesn't make sense, can you direct me to mathematically legit ideas in the same ballpark which do make sense?
 A: In effect you have a compact metric space $S$ endowed with a linear order $\preceq$. $S$ is sequentially compact, so every infinite sequence in $S$ has a convergent subsequence. Let $\langle x_n:n\in\omega\rangle$ be a sequence in $S$, and suppose that there is an infinite $N\subseteq\omega$ such that the subsequence $\langle x_n:n\in N\rangle$ is $\preceq$-increasing. Then there is an infinite $N_0\subseteq N$ such that $\langle x_n:n\in N_0\rangle$ converges to some $x\in S$. Let $A=\{x_n:n\in N_0\}$, and note that if $x_n,x_m\in A$, then $x_n\preceq x_m$ iff $n\le m$. For each $\epsilon>0$ there is an $m_\epsilon\in N_0$ such that $d(x,x_n)<\epsilon$ for all $n\in N_0$ such that $n\ge m_\epsilon$, but this just says that $d(x,y)<\epsilon$ for all $y\in A$ such that $x_m\preceq y$.
Thus, you want to rule out all order types in which $\omega$ cannot be embedded, i.e., all order types in which every strictly increasing sequence is finite. In all other order types there is a subsequence of the desired kind.
