> [!quote] Idea > > Accumulation points of a set are points that are stuck to it. They cannot be separated from the set using neighbourhoods, no matter how small they are. > [!definition] > > Let $(X, \topo)$ be a [[Topological Space|topological space]] and $A \subset X$. A point $x \in X$ is an **accumulation point** of $A$ if $A \cap (U \setminus \bracs{x}) \ne \emptyset$ for every [[Neighbourhood|neighbourhood]] $U$ of $x$. > [!theorem] > > Let $A \subset X$ and $\text{acc}(A)$ be the set of accumulation points of $A$. Then adding them to $A$ results in its [[Topological Closure|closure]], > $ > \ol{A} = A \cup \text{acc}(A) > $ > and $A$ is [[Closed Set|closed]] if and only if it contains all of its accumulation points. > > *Proof*. Let $x \in \ol{A}^c$, then $\ol{A}^c$ is an [[Open Set|open set]] and therefore a neighbourhood of $x$ that does not intersect $A$. So $x \not\in \text{acc}(A)$, $x \not\in A$ and $\ol{A}^c \subset A^c \cap \text{acc}(A)^c$ implies that $\ol{A} \supset A \cup \text{acc}(A)$. > > Now let $x \in A^c \cap \text{acc}(A)^c$. Then there exists an open neighbourhood $V$ of $x$ such that $(V \setminus \bracs{x}) \cap A = \emptyset$. As $x \not\in A$, $V \cap A = \emptyset$ and $V$ is an open neighbourhood of $x$ disjoint from $A$. $V^c$ then is a closed set containing $A$ that does not contain $x$. Therefore $x \not\in \ol{A}$. Since $A^c \cap \text{acc}(A)^c \subset \ol{A}^c$, we have $\ol{A} \subset A \cup \text{acc}(A)$.