# Prove the intersection of a compact set and a set with no accumulation points is finite

Let $S\subset\mathbb{C}$. We say that $z_0$ is an accumulation point of $S$ if for every $r>0$, the intersection $D(z_0,r)\cap S$ is an infinite set. Let $U\subset\mathbb{C}$ be an open set such that $S\subset U$. Suppose that $S$ does not have any accumulation point contained in $U$. Prove that for any compact set $K\subset U$, the intersection $S\cap K$ is finite.

First, if $S$ does not have an accumulation point contained in $U$, that means for every $z \in U$, there exists some $r>0$ such that $D(z,r)\cap S$ is finite.

Now consider a compact set $K \subseteq U$. Since both $S$ and $K$ are subsets of $U$, we can consider two possible cases: either these subsets of $U$ are disjoint or they have a nonempty intersection. If $S$ and $K$ are disjoint subsets of $U$, then $S \cap K = \emptyset$ which is certainly finite. Now consider the case that $S \cap K \ne \emptyset$. We will show $S \cap K$ is finite.

And this is where I'm stuck. I can think of possible directions to go:

One way I could go is to use the fact that $S$ has no accumulation points in $U$, so for every $z \in S \cap K$ there exists $r_z$ such that $D(z,r_z) \cap S$ is finite. However, this is not the same as the number of elements in $S \cap K$ being finite.

Another direction I could go is to use the fact that since $K$ is compact, then any covering (which would also cover $S \cap K$) would be finite. But this does not guarantee the number of elements in $S \cap K$ is finite, only that the cover is finite.

• Sequential compactness could be of help - C is a metric space, so it's no cheating – nelv Jan 28 '15 at 9:21

Your approach works if you combine your two ideas: Form an open set around each point in $K$, such that it contains only finitely many points of $S$. These open sets cover $K$, and since $K$ is compact, extract a finite subcover, each element of which only contains finitely many points of $S$.

Let $X$ be the compact set and $S$ be a set with no accumulation points.

Let $T = X \cap S$.

If $T$ is infinite:

Let $x = ( x_k : k \in \mathbb{N} )$ be an infinite sequence of distinct points in $T$.

Then $x$ has a subsequence $( y_k : k \in \mathbb{N} )$ that converges in $X$.

Can you see that we get a contradiction?

An infinite subset of a compact set has a limit point (Rudin).