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

A point is in the closure $\overline{E}$ exactly when every neighbourhood meets $E$. This is equivalent to belonging to $E$ or being a limit point of $E$.

If $p\notin \overline{E}$, choose $\epsilon >0$ with $(p-\epsilon ,p+\epsilon )\cap E=\varnothing$. Every $q$ with $|q-p|<\epsilon /2$ has a smaller neighbourhood contained in this interval and disjoint from $E$. Thus $(p-\epsilon /2,p+\epsilon /2)$ misses $\overline{E}$, and the complement of $\overline{E}$ is open. Hence $\overline{E}$ is closed.

Every closed set $F$ containing $E$ contains all limit points of $E$: a limit point outside $F$ would have a neighbourhood missing $F$, and hence missing $E$. Therefore $\overline{E}\subseteq F$. Since $\overline{E}$ is itself one of these closed sets,