A set is closed if and only if it contains all its limit points.
Proof in book: Suppose $S$ is "not closed". We must show that $S$ does not contain all its limit points. Since
$S$ is "not closed", $S^c$ is "not open". Therefore there is at least one element $x$ of $S^c$ such that every ball $B(x,\epsilon)$ contains at least one element of $S$ …
Why is there at least one element $x\in S^c$ such that every open ball contains at least one element of the open set $S$?
Best Answer
After the edit, everything is fine. Since $$A \text{ open} \Leftrightarrow A^C \text{ closed}$$ $S^C$ is not open, but an open set is characterised by $$\forall\ x \in A\ \exists\ \epsilon > 0 \ : \ B_\epsilon (x) \subset A$$ The contraposition is $$\exists\ x \in A\ : \forall\ \epsilon > 0 : B_\epsilon(x) \cap A^C \neq \emptyset$$ Which is what is stated in the book.