# Please help me prove $K \cap A' \neq \varnothing$

If $A$ is an infinite subset of a compact set $K$, then $K \cap A' \neq \varnothing$.

Definition: Point $x\in X$ is a accumulation point to the metric space $A \subseteq (X,d)$ if: $$T(x,r) \cap (A \setminus \{x\}) \neq \varnothing, \quad \forall r>0.$$

Set of accumulation points to the set of $A$ called derivatives set and denoted by $A'$.

Previously, thank you very much.

If $K\cap A'=\emptyset$ then, by definition, for each point $p\in K$ there would exist an open ball $B_p$ such that $B_p$ contains at most one point of $A$ (this case would occur if $p$ also belongs to $A$). Take the open cover $\mathcal{O}=\{B_p:p\in K\}$. Note that we can not get a finite subcover since all the points of $A$ are in different balls and $A$ is infinite. This is a contradiction to the compactness of $K$.