1
$\begingroup$

Theorem: If $E$ is an infinite subset of a compact set $K$, then $E$ has a limit point in $K$.

If no point of $K$ were a limit point of $E$, then each $q ∈ K$ would have a neighborhood $V_q$ which contains at most one point of $E$ (namely, $q$, if $q∈E$). It is clear that no finite subcollection of ${Vq}$ can cover $E$; and the same is true of $K$, since $E⊂K$. This contradicts the compactness of $K$.

I understand the theorem's proof. But I need to understand if $E$ has a limit point in $K$, isn't it still we can't find a finite subcollection of ${Vq}$ that can cover $E$, which contradicts the compactness of $K$?

$\endgroup$
  • 1
    $\begingroup$ Try to write down a cover that has no finite sub-cover. $\endgroup$ – uniquesolution Jan 11 '18 at 18:16
  • 1
    $\begingroup$ $\{V_q\}_q$ is a cover, and it suppose to have a finite sub-cover for $K$ but it does not. $\endgroup$ – Yanko Jan 11 '18 at 18:17
2
$\begingroup$

"if $E$ has a limit point in $K,$ isn't it still we can't cover $K$ by a finite subcover"

No, it's not.

Consider the set $S = \left\{\, \dfrac 1 n : n\in\{\,1,2,3,\ldots\,\} \,\right\}.$

We can put a neighborhood around each member of this set $S$ that excludes the other members and also excludes the limit point, which is $0,$ and that is an open cover of $S$ that has no finite subcover. Thus $S$ is not compact. However, $0$ is a limit point that is (as you assumed) in $K.$ Thus an open cover of $K$ contains an open set that covers $0.$ Every open set that covers $0$ also covers all except finitely many members of $S,$ so it only takes finitely many to cover $0$ and all members of $S.$

$\endgroup$

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.