5
$\begingroup$

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.

$\endgroup$
0
11
$\begingroup$

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

$\endgroup$
3
$\begingroup$

Result: A set is ''clompact'' if and only if it is finite.

Suppose A is infinite then consider the closed covers {{a}: a $\in$ A} given by singleton set of A. Then it has no finite subcover. Hence, if A has finite subcover then it must be finite.

Conversly, if A={$a_1,a_2,\cdots, a_n$} is finite and {$O_\lambda:\lambda \in \Lambda $} be it's closed cover. By definition of cover, there exist $\lambda_i \in \Lambda $ such that $a_i \in O_{\lambda_i}.$ Then take covers {$O_{\lambda_i},i=1,2,...,n$}. this covers A. hence A must be clompact.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.