[proofplan] We prove that no point of $X$ is a [limit point](/page/Limit%20Point) of an arbitrary subset $A \subset X$. In the discrete topology, each singleton $\{x\}$ is open, so it is an open neighbourhood of $x$. This neighbourhood contains no point of $A$ distinct from $x$, which violates the defining condition for being a limit point. [/proofplan] [step:Use the singleton neighbourhood to exclude each point from the derived set] Let $A \subset X$ be arbitrary, and let $x \in X$. Since $\tau = \mathcal{P}(X)$, the singleton set $\{x\}$ belongs to $\tau$, so $\{x\}$ is an open neighbourhood of $x$. By definition, $x \in A'$ means that every open neighbourhood $U \in \tau$ of $x$ satisfies \begin{align*} U \cap (A \setminus \{x\}) \neq \varnothing. \end{align*} Taking the open neighbourhood $U := \{x\}$, we compute \begin{align*} \{x\} \cap (A \setminus \{x\}) = \varnothing. \end{align*} Therefore $x \notin A'$. [guided] Fix an arbitrary subset $A \subset X$ and an arbitrary point $x \in X$. To prove that $A' = \varnothing$, it is enough to show that this arbitrary point $x$ is not a limit point of $A$. The key feature of the discrete topology is that every subset of $X$ is open. Since $\{x\} \subset X$, we have \begin{align*} \{x\} \in \tau. \end{align*} Thus $\{x\}$ is an open neighbourhood of $x$. Now recall the definition of the derived set: a point $x$ lies in $A'$ exactly when every open neighbourhood $U \in \tau$ of $x$ contains a point of $A$ different from $x$. Equivalently, \begin{align*} x \in A' \end{align*} requires \begin{align*} U \cap (A \setminus \{x\}) \neq \varnothing \end{align*} for every open neighbourhood $U$ of $x$. We test this condition using the particular open neighbourhood $U := \{x\}$. This neighbourhood contains only the point $x$, while $A \setminus \{x\}$ contains only points of $A$ that are not equal to $x$. Hence the two sets are disjoint: \begin{align*} \{x\} \cap (A \setminus \{x\}) = \varnothing. \end{align*} So there exists an open neighbourhood of $x$ that contains no point of $A$ distinct from $x$. This is exactly the negation of the condition $x \in A'$, and therefore $x \notin A'$. [/guided] [/step] [step:Conclude that the derived set is empty] Since $x \in X$ was arbitrary, no point of $X$ belongs to $A'$. Hence \begin{align*} A' = \varnothing. \end{align*} Because $A \subset X$ was arbitrary, the conclusion holds for every subset of $X$. [/step]