I'm reading principles of mathematical analysis and have a question about a theorem 2.37.
Theorem 2.37
If $E$ is an infinite subset of a compact set $K$, then $E$ has a limit point in $K$.
The proof is
If no point of $K$ were a limit point of $E$, each $q \in K$ would have a neighborhood $V_q$ which contains at most one point of $E$.
It is clear that no finite subcollection of $\{V_q\}$ can cover $E$.
The same is true of $K$, since $E \subset K$.
This contradicts the compactness of $K$.
I understand the first part that states that no finite subcollection of $\{V_q\}$ can cover $E$, in other words, $E$ is not compact. But I don't understand why it means that no finite subcollection can cover $K$. Is the author saying that if a subset of a set K is not compact, then $K$ is not compact? If that's the case, a compact set may have open subsets which are not compact, so I'm confused.
Thanks in advance.
Best Answer
A set with no limit points is necessarily a closed set. Being a closed subset of a compact space, it is compact. On the other hand, you're looking at a proof that $E$ is not compact, so you've got a contradiction.
Alternatively, look at this set of neighborhoods that cover $E$ and add one more open set to this collection: the complement of $E$. That set is open, since as noted above, $E$ is closed. Now you've got an open cover of $K$. It must therefore have a finite subcover. But every finite subset of this cover fails to cover all of $E$, so again you have a contradiction.