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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?

Moishe Kohan
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Casper
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  • It isn’t true without further conditions on $\preceq$: what if $S$ has order type $\omega^*$, i.e., the type of $\left{\frac1n:n\in\Bbb Z^+\right}$? Then the tails are all finite. – Brian M. Scott Apr 11 '20 at 17:08
  • @BrianM.Scott Thanks! What sort of conditions on $\preceq$ would be sufficient? – Casper Apr 11 '20 at 17:36
  • Do you want $d(x,y)<\epsilon$ for all $y\in S$ with $x\preceq y$, or would you settle for $d(x,y)<\epsilon$ for all $y\in S'$ with $x\preceq y$? – Brian M. Scott Apr 11 '20 at 17:52
  • I would settle for the latter. – Casper Apr 11 '20 at 17:53

1 Answers1

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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.

Brian M. Scott
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  • Thanks! Does this argument work if $S$ is uncountably infinite? – Casper Apr 11 '20 at 18:20
  • @Casper: It works for any infinite $S$. Basically it just says that any infinite $\preceq$-increasing sequence has a subsequence that converges in the metric. Note, though, that this does not guarantee that the subsequence converges with respect to the order: in the notation of my answer it’s entirely possible that $x\prec x_n$ for every $n\in N_0$. Thus, it doesn’t really tell you very much and may not really be what you want. – Brian M. Scott Apr 11 '20 at 18:24
  • Also, does this argument work if we require that the subchain $S^\prime$ is of the same cardinality as the original chain $S$? – Casper Apr 11 '20 at 22:06
  • @Casper: That complicates matters. Then you not only need $\omega$ to be embeddable in $\langle S,\preceq\rangle$, but you need it to be embeddable cofinally in a $\preceq$-interval of $S$ whose every tail has cardinality equal to $|S|$. – Brian M. Scott Apr 11 '20 at 22:12
  • Thanks! Do you have any suggestions about how to do this? I was thinking that an infinite chain is a net, and compactness is equivalent to: every net has a convergent subnet. Hence, our chain has a convergent subnet which, by definition of subnet (which includes cofinality), is also a chain with the same cardinality. I am wondering whether this works and whether it could be simplified in metric spaces. – Casper Apr 12 '20 at 14:28
  • @Casper: There are actually three different definitions of subnet; this question, my answer to it, and the links in my answer cover that ground pretty thoroughly. The three are equivalent as far as convergence is concerned, but the definition due to Aarnes and Andenæs makes it easier to see, for instance, that a subnet does not have to have the same cardinality as a net of which it is a subnet. – Brian M. Scott Apr 12 '20 at 15:42