Any infinite set of a compact set $K$ has a limit point in $K$?

Solution 1:

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.

Solution 2:

Since for all $q$, $V_q \cap E$ has at most one element, for any finite subset $\{q_1,\ldots,q_n\}$,

$\bigcup_{i=1}^n V_{q_i} \cap E$ has at most $n$ elements, i.e., is finite. But $E$ is assumed to be infinite, so we cannot have $\bigcup_{i=1}^n V_{q_i} \supset E$. Since $K \supset E$, therefore we cannot have $\bigcup_{i=1}^n V_{q_i} \supset K$, contradicting the compactness of $K$.

Solution 3:

Suppose every $ p\in K $ has an open nbhd $ U_p $ such that $ E\cap U_p $ is finite. Then $ C=\{U_p: p\in K\} $ is an open cover of $ K $ but any finite $ D\subset C $ covers only a finite subset of $ E. $ Note that we do not need to assume that $ K $ is a $ T_1 $ space nor even a $ T_0 $ space.

I said "open nbhd" because some people say that a nbhd $ U $ of $ p $ is an open set with $ p\in U, $ while some say that a nbhd $ U $ of $ p $ is any subset of the space, such that $ p\in V\subset U $ for some open $ V.$