[proofplan] We prove both implications directly from the definitions. If $A$ is closed, then $X \setminus A$ is open, so a point outside $A$ has an open neighbourhood disjoint from $A$ and therefore cannot be a [limit point](/page/Limit%20Point) of $A$. Conversely, if every limit point of $A$ belongs to $A$, then each point outside $A$ is not a limit point, which gives an open neighbourhood meeting $A$ only possibly at that point; since the point is outside $A$, this neighbourhood is disjoint from $A$. Thus $X \setminus A$ is open, so $A$ is closed. [/proofplan] [step:Show that closedness forces every limit point to lie in $A$] Assume that $A$ is closed in $X$. By definition of closedness, $X \setminus A \in \tau$. Let $x \in A'$. Suppose, for contradiction, that $x \notin A$. Then $x \in X \setminus A$, and since $X \setminus A$ is open, the set \begin{align*} U := X \setminus A \end{align*} is an open neighbourhood of $x$. But \begin{align*} U \cap (A \setminus \{x\}) = (X \setminus A) \cap (A \setminus \{x\}) = \varnothing, \end{align*} contradicting the defining property of $x \in A'$. Hence $x \in A$. Since $x \in A'$ was arbitrary, $A' \subset A$. [guided] Assume that $A$ is closed in $X$. The definition of closedness in a [topological space](/page/Topological%20Space) says exactly that the complement $X \setminus A$ is open, so \begin{align*} X \setminus A \in \tau. \end{align*} We want to prove $A' \subset A$, so let $x \in A'$ be arbitrary. The goal is to show $x \in A$. Suppose instead that $x \notin A$. Then $x$ lies in the complement $X \setminus A$. Since this complement is open, it is an open neighbourhood of $x$. Now compare this neighbourhood with the definition of a limit point. Because $x \in A'$, every open neighbourhood of $x$ must meet $A \setminus \{x\}$. Taking the particular open neighbourhood \begin{align*} U := X \setminus A, \end{align*} we compute \begin{align*} U \cap (A \setminus \{x\}) = (X \setminus A) \cap (A \setminus \{x\}) = \varnothing. \end{align*} This contradicts the condition that $x \in A'$. Therefore the assumption $x \notin A$ is impossible, and hence $x \in A$. Since the argument applies to every $x \in A'$, we conclude that \begin{align*} A' \subset A. \end{align*} [/guided] [/step] [step:Show that containing all limit points forces closedness] Assume that $A' \subset A$. We prove that $X \setminus A$ is open. Let $x \in X \setminus A$. Since $x \notin A$ and $A' \subset A$, we have $x \notin A'$. By the definition of $A'$, there exists an [open set](/page/Open%20Set) $U_x \in \tau$ such that $x \in U_x$ and \begin{align*} U_x \cap (A \setminus \{x\}) = \varnothing. \end{align*} Because $x \notin A$, we have $A \setminus \{x\} = A$, so \begin{align*} U_x \cap A = \varnothing. \end{align*} Thus $U_x \subset X \setminus A$. We have shown that for every $x \in X \setminus A$, there exists $U_x \in \tau$ with \begin{align*} x \in U_x \subset X \setminus A. \end{align*} Therefore $X \setminus A$ is open. Hence $A$ is closed in $X$. [/step] [step:Combine the two implications] The first step proves that if $A$ is closed in $X$, then $A' \subset A$. The second step proves that if $A' \subset A$, then $A$ is closed in $X$. Therefore $A$ is closed in $X$ if and only if \begin{align*} A' \subset A. \end{align*} This proves the theorem. [/step]