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 $K$?

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    $\begingroup$ Try to write down a cover that has no finite sub-cover. $\endgroup$ – uniquesolution Jan 11 '18 at 18:16
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    $\begingroup$ $\{V_q\}_q$ 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

"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 $S.$


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