'Completeness' of ordered topological space

Question

This is a follow-up question to an answer given by Henno Brandsma in this thread How to prove ordered square is compact.

In the answer it is shown that:

A non-empty LOTS (linearly ordered topological space) $(X,<)$ is compact iff every subset $A\subseteq X$ has a supremum $sup(A)$.

Since $\mathbb{R}$ is a locally compact LOTS , with the completeness property, i.e., that every bounded subset $A\subseteq \mathbb{R}$ has a supremum $\sup(A)$, can a similar argument be made for general LOTS?

More precisely, can we something along the lines of:

A LOTS $(X,<)$ is locally compact if and only if every bounded subset $A\subseteq X$ has a supremum $\sup(A)$.

Or perhaps some other condition can characterize such a 'completeness' property for ordered spaces.

Answer 1

If a LOTS $X$ obeys

(order completeness:) For every non-empty subset $A$ that has an upperbound, $\sup(A)$ exists.

then $X$ is locally compact: If $x \in X$ does not equal the min or max of $X$ (if it exists), pick any $a,b$ with $a < x, x < b$ and note that $Y=[a,b]$ is closed (true in any LOTS) and even compact using the characterisation I showed (for any order convex subset of a LOTS the subspace topology and induced-order topology are the same, so we can see $Y$ as a LOTS too): let $A \subseteq Y$ be any subset: $A=\emptyset$ is OK: $\sup(A) = \min(Y)=a$ in that case. If $A \neq \emptyset$, $b$ is always an upperbound and the assumption tells us $\sup(A)$ exists in $X$, but as $\sup(A)\in \overline{A}$ (true in any LOTS), and $Y$ is closed, $\sup(A) \in Y$ and we're done showing $Y$ compact. A similar argument can be held for $\min(X), \max(X)$ if they exist, and we see all points of $X$ have a compact neighbourhood (and as any LOTS is Hausdorff, $X$ obeys all equivalent definitions of local compactness).

The lexicographically ordered plane (which is locally compact) has sets like $A=\{0\} \times \Bbb R$ that have upper (and lower) bounds but not a sup. So the completeness condition is quite a bit stronger than local compactness. It's customary (in "LOTS theory") to define the equivalence relation $a \sim b$ iff $[a,b]$ is compact, for locally compact LOTS, and study the equivalence classes, see this paper for an example. I'm currently not aware of a natural completeness condition on $(X,<)$ characterising that the corresponding LOTS is locally compact.

So the order-completeness of $\Bbb R$ has indeed its local compactness (and connectedness) as a consequence. But the reverse is not so clear.

Final remark: the compactness characterisation for LOTS also implies that a connected LOTS with min and max is compact. It's (I think) fun to see that where in general compactness and connectedness can be quite independent properties, and in LOTS they are quite closely linked. It's an interesting class.