Exercise 2.1.5 (Limit points and isolated points)

Proof. The trick is to consider the closure $\overline{\mathit{\lambda S}}$ of the limit points of $S$ and to notice that by definition of an adherent point, any point $x\in \overline{\mathit{\lambda S}}$ must have a limit point $x^{\prime }\in \mathit{\lambda S}$ in every one of its neighbourhoods $U\ni x$ and thus meet $S$ as well (e.g. intersect neighbourhoods of both $x$ and $x^{\prime }$). Since $U\setminus \{x\}$ is open (as a set difference of an open and a closed set), we have proven that $x$ is a limit point of $S$, i.e., $\overline{\mathit{\lambda S}}=\mathit{\lambda S}$. By Theorem 1.3, $\mathit{\lambda S}$ is closed. □

Proof. Let $x$ be from the right-hand side. Then $x$ is either an isolated point or a limit point (clearly not both, hence the disjointness). If $x$ is an isolated point, then $x\in S\subseteq \overline{S}$ by definition; if $x$ is a limit point, then $x$ is obviously adherent und thus $x\in \overline{S}$.
Conversely, suppose that $x\in \overline{S}$. We have two cases: either $x\in S$ or $x\in \overline{S}\setminus S$. Notice how the definitions of limit and isolated points are complementary in $S$; a point $x\in S$ would either be a limit point or an isolated point. In case that $x\in \overline{S}\setminus S$, $x$ must be on the outside of $S$ and cannot be an isolated point of $S$. It is, however, adherent to $S$ and must thus be a limit point of $S$. □