[proofplan] We argue by contradiction. If some neighbourhood of $x$ contained only finitely many points of $A$, then after passing to an open neighbourhood inside it, the finitely many points of $A$ different from $x$ could be removed one at a time. The $T_1$ axiom makes each singleton closed, so the complement of each such point remains open and still contains $x$. The resulting neighbourhood of $x$ avoids $A \setminus \{x\}$, contradicting that $x$ is a [limit point](/page/Limit%20Point) of $A$. [/proofplan] [step:Assume one neighbourhood contains only finitely many points of $A$] Suppose, for contradiction, that there exists a neighbourhood $N \subset X$ of $x$ such that $A \cap N$ is finite. Since $N$ is a neighbourhood of $x$, there exists an [open set](/page/Open%20Set) $U \in \tau$ such that $x \in U \subset N$. Define the finite set \begin{align*} F := (A \cap U) \setminus \{x\}. \end{align*} Since $A \cap U \subset A \cap N$ and $A \cap N$ is finite, the set $F$ is finite. [/step] [step:Remove the finitely many points of $A$ different from $x$] For each $y \in F$, we have $y \neq x$. Since $(X,\tau)$ is $T_1$, the singleton $\{y\}$ is closed, so $X \setminus \{y\}$ is open and contains $x$. Define \begin{align*} V := U \cap \bigcap_{y \in F} (X \setminus \{y\}). \end{align*} Because $F$ is finite, $V$ is a finite intersection of open sets, hence $V \in \tau$. Also $x \in V$, since $x \in U$ and $x \neq y$ for every $y \in F$. By construction, no point of $F$ lies in $V$. Therefore \begin{align*} A \cap V \subset \{x\}. \end{align*} Equivalently, \begin{align*} V \cap (A \setminus \{x\}) = \varnothing. \end{align*} [guided] The goal is to turn the assumed finite set of nearby points into an actual contradiction with the definition of limit point. The finite set we must eliminate is \begin{align*} F := (A \cap U) \setminus \{x\}. \end{align*} These are exactly the points of $A$ lying in the open neighbourhood $U$ other than possibly $x$ itself. For each point $y \in F$, we know $y \neq x$. The $T_1$ axiom says that singletons are closed, so $\{y\}$ is closed. Hence its complement $X \setminus \{y\}$ is open. Since $x \neq y$, this open set contains $x$. Thus removing the point $y$ does not destroy the fact that we still have an open neighbourhood of $x$. Because $F$ is finite, we may remove all points of $F$ at once by taking the finite intersection \begin{align*} V := U \cap \bigcap_{y \in F} (X \setminus \{y\}). \end{align*} This set is open because it is a finite intersection of open sets. It contains $x$ because $x \in U$ and $x \notin \{y\}$ for every $y \in F$. Now let $a \in A \cap V$. Since $V \subset U$, we have $a \in A \cap U$. If $a \neq x$, then $a \in (A \cap U) \setminus \{x\} = F$. But $V$ was constructed to avoid every point of $F$, a contradiction. Hence every point of $A \cap V$ must be $x$, and therefore \begin{align*} A \cap V \subset \{x\}. \end{align*} This is the same as saying \begin{align*} V \cap (A \setminus \{x\}) = \varnothing. \end{align*} [/guided] [/step] [step:Contradict the limit point property] The set $V$ is an open neighbourhood of $x$. Since $x$ is a limit point of $A$, every neighbourhood of $x$ intersects $A \setminus \{x\}$. Thus \begin{align*} V \cap (A \setminus \{x\}) \neq \varnothing. \end{align*} This contradicts the conclusion from the previous step that \begin{align*} V \cap (A \setminus \{x\}) = \varnothing. \end{align*} Therefore no neighbourhood $N$ of $x$ can have $A \cap N$ finite. Hence every neighbourhood of $x$ contains infinitely many points of $A$. [/step]