[proofplan] We prove the identity $\overline{A} = A \cup A'$ by showing each side is contained in the other. The inclusion $A \cup A' \subset \overline{A}$ follows from the fact that $\overline{A}$ contains $A$ and every limit point of $A$. The reverse inclusion uses the [Neighbourhood Characterisation of Closure](/theorems/1005): if $x \in \overline{A}$ but $x \notin A$, then every open neighbourhood of $x$ meets $A$ in a point other than $x$, placing $x$ in $A'$. [/proofplan] [step:Show $A \cup A' \subset \overline{A}$] The inclusion $A \subset \overline{A}$ holds by extensivity of the closure. For the inclusion $A' \subset \overline{A}$, let $x \in A'$. By definition of the derived set, every open set $U \in \tau$ containing $x$ satisfies $U \cap (A \setminus \{x\}) \neq \varnothing$. In particular, $U \cap A \neq \varnothing$. By the [Neighbourhood Characterisation of Closure](/theorems/1005), $x \in \overline{A}$. [guided] We need two inclusions. First, $A \subset \overline{A}$: this is the extensivity property of closure — the closure of a set always contains the set itself, since $\overline{A}$ is an intersection of closed sets each containing $A$. Second, $A' \subset \overline{A}$: let $x \in A'$. By definition, $x$ is a limit point of $A$, meaning every open neighbourhood $U$ of $x$ contains a point of $A$ different from $x$: \begin{align*} U \cap (A \setminus \{x\}) \neq \varnothing. \end{align*} This implies the weaker condition $U \cap A \neq \varnothing$. By the [Neighbourhood Characterisation of Closure](/theorems/1005), this places $x \in \overline{A}$. [/guided] [/step] [step:Show $\overline{A} \subset A \cup A'$] Let $x \in \overline{A}$ and suppose $x \notin A$. By the [Neighbourhood Characterisation of Closure](/theorems/1005), every open set $U \in \tau$ containing $x$ satisfies $U \cap A \neq \varnothing$. Since $x \notin A$, every point in $U \cap A$ is distinct from $x$, so $U \cap (A \setminus \{x\}) \neq \varnothing$. As $U$ was arbitrary, $x \in A'$. In either case, $x \in A \cup A'$. [guided] Let $x \in \overline{A}$. We consider two cases. **Case 1: $x \in A$.** Then $x \in A \cup A'$ and there is nothing further to show. **Case 2: $x \notin A$.** By the [Neighbourhood Characterisation of Closure](/theorems/1005), every open neighbourhood $U$ of $x$ satisfies $U \cap A \neq \varnothing$. Since $x \notin A$, every point in $U \cap A$ is automatically different from $x$. Hence $U \cap (A \setminus \{x\}) \neq \varnothing$ for every open neighbourhood $U$ of $x$, which means $x$ is a limit point of $A$, i.e., $x \in A'$. In both cases $x \in A \cup A'$, completing the proof. [/guided] [/step]