Exercise 2.1.8 (Closure of a set)

Proof. We have to demonstrate that

One side is obvious since $\overline{S}$ is itself a closed set containing $S$ and thus the right-hand side is contained in the left-hand side by the properties of intersection.

Now assume for the sake of contradiction that there is a $x$ in the closure $\overline{S}$ of $S$ such that it is not contained in the closed hull of $S$. Since $x$ is an adherent point of $S$, for all open neighbourhoods $U$ of $x$ we have $U\cap S\ne \varnothing$. Since $x$ is not contained in the closed hull, there exists a closed set $V$ containing $S$ such that $x\notin V$. Since $x$ is not an adherent point of $V$, we can find an open neighbourhood $U^{\prime }$ of $x$ such that $U^{\prime }\cap V=\varnothing$. But $U^{\prime }\cap S\subset U^{\prime }\cap V$ - a contradiction to the fact that every open neighbourhood of $x$ has a nonempty intersection with $S$. □