[proofplan] We prove both implications directly from the metric definition of a [limit point](/page/Limit%20Point). If $x$ is a limit point of $A$, then every punctured ball around $x$ meets $A$, so choosing one point in each ball of radius $1/n$ produces a sequence in $A \setminus \{x\}$ converging to $x$. Conversely, if such a sequence converges to $x$, then every ball around $x$ contains some tail element different from $x$, so every punctured ball meets $A$. [/proofplan] [step:Choose points in shrinking punctured balls around the limit point] Assume first that $x$ is a limit point of $A$. By definition, for every $r > 0$, \begin{align*} B(x,r) \cap (A \setminus \{x\}) \neq \varnothing, \end{align*} where $B(x,r) := \{y \in X : d(y,x) < r\}$. For each $n \in \mathbb{N}$, apply this definition with $r = 1/n$. Choose \begin{align*} a_n \in B(x,1/n) \cap (A \setminus \{x\}). \end{align*} Then $(a_n)_{n=1}^{\infty}$ is a sequence in $A \setminus \{x\}$, and for every $n \in \mathbb{N}$, \begin{align*} 0 \le d(a_n,x) < \frac{1}{n}. \end{align*} Since $1/n \to 0$ in $\mathbb{R}$, the squeeze property for real sequences gives \begin{align*} \lim_{n \to \infty} d(a_n,x) = 0. \end{align*} Thus $a_n \to x$ in the [metric space](/page/Metric%20Space) $(X,d)$. [guided] Assume that $x$ is a limit point of $A$. The definition means that no matter how small a radius $r > 0$ we choose, the open ball around $x$ of radius $r$ contains a point of $A$ different from $x$. In symbols, for every $r > 0$, \begin{align*} B(x,r) \cap (A \setminus \{x\}) \neq \varnothing, \end{align*} where $B(x,r) := \{y \in X : d(y,x) < r\}$. We want to build a sequence converging to $x$, so we force the chosen points to lie in balls whose radii go to $0$. For each $n \in \mathbb{N}$, the number $1/n$ is positive, so the limit point property applied with $r = 1/n$ gives a point \begin{align*} a_n \in B(x,1/n) \cap (A \setminus \{x\}). \end{align*} This defines a sequence $(a_n)_{n=1}^{\infty}$, and every term lies in $A \setminus \{x\}$ by construction. Because $a_n \in B(x,1/n)$, the distance from $a_n$ to $x$ satisfies \begin{align*} 0 \le d(a_n,x) < \frac{1}{n} \end{align*} for every $n \in \mathbb{N}$. The real sequence $(1/n)$ converges to $0$, and distances are always non-negative. Therefore the squeeze property for real sequences yields \begin{align*} \lim_{n \to \infty} d(a_n,x) = 0. \end{align*} By the definition of convergence in a metric space, this is exactly the statement that $a_n \to x$. [/guided] [/step] [step:Use a convergent sequence to meet every punctured neighbourhood] Conversely, suppose there exists a sequence $(a_n)_{n=1}^{\infty}$ such that $a_n \in A \setminus \{x\}$ for every $n \in \mathbb{N}$ and \begin{align*} \lim_{n \to \infty} d(a_n,x) = 0. \end{align*} Let $r > 0$ be arbitrary. By convergence to $0$ of the real sequence $(d(a_n,x))$, there exists $N \in \mathbb{N}$ such that \begin{align*} d(a_N,x) < r. \end{align*} Since $a_N \in A \setminus \{x\}$, and since $d(a_N,x) < r$ means $a_N \in B(x,r)$, we have \begin{align*} a_N \in B(x,r) \cap (A \setminus \{x\}). \end{align*} Thus \begin{align*} B(x,r) \cap (A \setminus \{x\}) \neq \varnothing. \end{align*} Because $r > 0$ was arbitrary, every punctured ball around $x$ meets $A$. Hence $x$ is a limit point of $A$. [/step]