Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics II / 4 / a / ii / Solution

Suppose $E$ is a closed set and $p\notin E$. Then $p$ is not a limit point of $E$, so some punctured neighbourhood of $p$ misses $E$. As $p\notin E$, the entire neighbourhood misses $E$. Thus $\mathbb{R}\setminus E$ is an open set.

Conversely, suppose every point $p\notin E$ has a neighbourhood disjoint from $E$. No such point can be a limit point of $E$, so every limit point lies in $E$. This proves the equivalence between containing all limit points and having an complement that is an open set.