Past exam of the options II course of the University of Oxford / 2012 / E / E3 / a

The closure of $A$ is the intersection of every closed subset containing it:The collection is nonempty because it contains $X$. Arbitrary intersections of closed sets are closed, so this is the smallest closed set containing $A$.

Every subset is contained in its closure, giving $\overline{A}\subseteq \overline{\overline{A}}$. On the other hand, $\overline{A}$ is already closed by its definition as an intersection of closed sets. It is therefore one of the closed supersets used when taking its own closure, so $\overline{\overline{A}}\subseteq \overline{A}$. HenceThis is the idempotence of the closure operation.

If $x\notin \overline{A}$, some closed superset $F$ of $A$ misses $x$. Its complement is an open set containing $x$ and disjoint from $A$. Conversely, if an open set $U$ contains $x$ and misses $A$, the closed set $X\setminus U$ contains $A$ and misses $x$, so $x\notin \overline{A}$.

Negating this equivalence gives the neighbourhood characterization

An accumulation point of $A$ is a point $x\in X$ such that every open neighbourhood of $x$ contains a point of $A$ different from $x$:The point itself need not belong to $A$. Requiring a different point distinguishes accumulation from mere membership in the closure.