In the real line in the usual topology the compact sets are the closed and bounded sets. In the left topology, the topology generated by the left-open intervals $(-,a)$, the compact sets are exactly the closed sets, which are the left-closed intervals $(-,a]$ and also it seems the real line itself - since the only open cover of it is itself.
Now the real line is a totally ordered set - a chain - how far can these results be generalised to this context?
In the ordinal $2\omega=\omega+\omega$, with the interval topology; a straight generalisation of the above would say that $[1,\omega+1]$ is compact because it is closed & bounded. But this can't be right since the interval topology just reduces to the discrete topology here; and thus a compact subset is simply a finite subset. The condition appears to be neccessary but not sufficient.
The left topology appears to generalises - it seems that the compact subsets are again exactly the closed subsets.
Finally, do these results generalise to posets?
In the left topology, the compact sets are not exactly the closed subsets (a counter-example being the disjoint union of a countable number of real lines). They seem to be charactised by having a finite number of maximal elements. Is that right? In the interval topology, a neccessary condition is that they have a finite number of maximal & minimal elements - but given the example of the ordinal above it can't be sufficient.