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Find all the compact sets of the lower limit topology in $\mathbb{R}$, i.e. the topology given by basis $\{[a,b)|a,b \in \mathbb{R} \}$.

What I have got so far:

necessary conditions: the compact sets of lower limit topology must be countable, also they are closed (in the usual sense) and bounded. Also all of the limit points must be approached above, i.e. for every limit point $L$, there exists $\epsilon$ such that there are no points within $(L-\epsilon,L)$

sufficient condition: countable sets with finitely many limit points, each of the limit points approached above (as defined above) are compact sets in lower limit topology; countable sets with countably many limit points, where the limit points themselves have finitely many limit points, are also compact sets in lower limit topology; as well as countable sets with countably many limit points, the limit points having countably many limit points, the limit points of limit points having finitely many limit points, etc...

As you can see, I am quite close to finding a necessary and sufficient condition, if only I can prove it is impossible to have sets which have countably many limit points, the limit points having countably many limit points,... (descending indefinitely)

Any help is appreciated!

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    $\begingroup$ In this answer Henno Brandsma showed that every compact set is well-ordered with respect to $\ge$; that is, if $C$ is compact, and $S$ is a non-empty subset of $C$, then $S$ has a maximum element. This is your missing ingredient: combine this with the requirement that $C$ be compact in the usual topology, and you have your characterization. $\endgroup$ – Brian M. Scott Jan 21 '16 at 3:46
  • $\begingroup$ I'm confused — $\{0\}\cup\{-1/n:n\in\Bbb N\}$ has at most countably many limit points, but it's not compact in this topology. And yet having countably many limit points is listed as a sufficient condition. $\endgroup$ – Akiva Weinberger Jan 21 '16 at 3:46
  • $\begingroup$ @BrianM.Scott Yeah, that's what I was thinking it would be, but I wasn't sure how to prove it. $\endgroup$ – Akiva Weinberger Jan 21 '16 at 3:47
  • $\begingroup$ The example in your comment violates the requirement, stated earlier, that all of the limit points (in the Euclidean sense) be approached from above. $\endgroup$ – Brian M. Scott Jan 21 '16 at 3:52
  • $\begingroup$ Oh, I thought those were separate conditions, not pieces of a whole big condition. $\endgroup$ – Akiva Weinberger Jan 21 '16 at 3:54
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Thanks @Brian M. Scott for the reference, I have figured out the question, for other's reference, I'll state it here:

S is compact w.r.t. lower limit topology iff S is compact w.r.t. usual topology and S is well ordered with respect to $\geq$ (i.e. every subset of S has a largest element) iff S is compact w.r.t. usual topology, S is countable and every limit point is approached above

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