[proofplan] The proof is an unpacking of the metric topology. In one direction, every open ball $B(x,r)$ is an open neighbourhood of $x$, so the [limit point](/page/Limit%20Point) condition applies directly. In the other direction, an arbitrary open neighbourhood of $x$ contains some metric ball centered at $x$, and the assumed punctured-ball condition supplies a point of $A$ distinct from $x$ inside that neighbourhood. [/proofplan] [step:Apply the limit point definition to each metric ball] Assume that $x$ is a limit point of $A$ with respect to $\tau_d$. Let $r > 0$ be arbitrary. By definition of the topology induced by a metric, the set \begin{align*} B(x,r) := \{y \in X : d(x,y) < r\} \end{align*} is $\tau_d$-open, and $x \in B(x,r)$ because $d(x,x)=0<r$. Since $x$ is a limit point of $A$, every $\tau_d$-open neighbourhood of $x$ intersects $A$ at a point distinct from $x$. Applying this to $B(x,r)$ gives \begin{align*} (B(x,r) \setminus \{x\}) \cap A \ne \varnothing. \end{align*} Since $r > 0$ was arbitrary, the punctured-ball condition holds for every $r > 0$. [/step] [step:Use openness to place a metric ball inside an arbitrary neighbourhood] Assume that for every $r > 0$, \begin{align*} (B(x,r) \setminus \{x\}) \cap A \ne \varnothing. \end{align*} Let $U \subset X$ be a $\tau_d$-[open set](/page/Open%20Set) with $x \in U$. Because $\tau_d$ is the topology induced by $d$, there exists a radius $\rho > 0$ such that \begin{align*} B(x,\rho) \subset U. \end{align*} By the assumed punctured-ball condition applied to this radius $\rho$, there exists a point $a \in X$ such that \begin{align*} a \in (B(x,\rho) \setminus \{x\}) \cap A. \end{align*} Thus $a \in A$, $a \ne x$, and $a \in B(x,\rho) \subset U$. Hence \begin{align*} a \in (U \setminus \{x\}) \cap A, \end{align*} so $(U \setminus \{x\}) \cap A \ne \varnothing$. [guided] We now prove that the punctured-ball condition implies the usual topological definition of a limit point. The topological definition requires checking every open neighbourhood of $x$, not only balls. So let $U \subset X$ be an arbitrary $\tau_d$-open set with $x \in U$. The reason metric balls are enough is that they form neighbourhoods inside every open set in the metric topology. Since $\tau_d$ is induced by $d$ and $U$ is $\tau_d$-open with $x \in U$, there exists a real number $\rho > 0$ such that \begin{align*} B(x,\rho) \subset U. \end{align*} Now we use the hypothesis with this specific radius $\rho$. The assumption says that every punctured ball centered at $x$ meets $A$, so \begin{align*} (B(x,\rho) \setminus \{x\}) \cap A \ne \varnothing. \end{align*} Therefore there exists a point $a \in X$ satisfying \begin{align*} a \in (B(x,\rho) \setminus \{x\}) \cap A. \end{align*} Unpacking this membership gives three facts: \begin{align*} a \in B(x,\rho), \qquad a \ne x, \qquad a \in A. \end{align*} Since $B(x,\rho) \subset U$, the first fact implies $a \in U$. Combining the three facts, we obtain \begin{align*} a \in (U \setminus \{x\}) \cap A. \end{align*} Thus every $\tau_d$-open set $U$ containing $x$ satisfies $(U \setminus \{x\}) \cap A \ne \varnothing$, which is exactly the statement that $x$ is a limit point of $A$ with respect to $\tau_d$. [/guided] [/step] [step:Conclude the equivalence] The first step proves that the limit point condition implies the punctured-ball condition for every $r > 0$. The second step proves that the punctured-ball condition implies the limit point condition. Therefore, for $x \in X$, the point $x$ is a limit point of $A$ with respect to the metric topology $\tau_d$ if and only if \begin{align*} (B(x,r) \setminus \{x\}) \cap A \ne \varnothing \end{align*} for every $r > 0$. [/step]