# Which sets are “clompact”?

This is an exercise taken verbatim from Abbott's Understanding Analysis:

Let’s call a set clompact if it has the property that every closed cover (i.e., a cover consisting of closed sets) admits a finite subcover. Describe all of the clompact subsets of $$\mathbf R$$.

I am unable to fully resolve the problem. So far, I have been able to see that singleton sets are always clompact, because the single element must be in at least one set belonging to the closed cover, and that one set is a sufficient finite subcover. The null set is also clompact for obvious reasons. I know that every non-singleton interval (regardless of if they are open, closed, or half-open) is not clompact. As an example, the closed cover $$\{0\}\cup\bigcup_1^\infty\left[\frac1{n+1},\frac1n\right]$$ for $$[0,1]$$ does not have a finite subcover. Similar constructions of closed covers show that $$[a,b]$$, $$(a,b]$$, $$[b,a)$$ and $$(a,b)$$ are not clompact as well. In addition, $$\mathbf R$$ itself and any unbounded interval is also not clompact.

Is anyone able to help in solving this problem? Any assistance is appreciated.

Hint: Singletons are closed, so if $$K\subset\mathbb R$$, then $$\{\{x\}:x\in K\}$$ is a closed cover of $$K$$.