[proofplan] The proof is a direct comparison of two definitions. Convergence of the sequence $(x_k)_{k\in\mathbb{N}}$ to $x$ means that every positive radius eventually bounds the distances $d(x_k,x)$. Membership in the open ball $B(x,r)$ means exactly the same inequality $d(x_k,x)<r$. Thus each direction follows by replacing the distance inequality with the equivalent open-ball membership statement. [/proofplan]

[step:Rewrite convergence as an eventual distance inequality] By definition, $x_k\to x$ in the [metric space](/page/Metric%20Space) $(X,d)$ means that for every $r>0$ there exists $N\in\mathbb{N}$ such that \begin{align*} d(x_k,x)<r \end{align*} for every $k\in\mathbb{N}$ with $k\ge N$. [/step]

[step:Use the open ball definition to replace the inequality by membership]Fix $r>0$ and $k\in\mathbb{N}$. Since $B(x,r)$ is the open ball \begin{align*} B(x,r)=\{y\in X:d(y,x)<r\}, \end{align*} we have the equivalence \begin{align*} x_k\in B(x,r)\iff d(x_k,x)<r. \end{align*}[/step]

[guided]Fix a radius $r>0$ and an index $k\in\mathbb{N}$. The point $x_k$ belongs to $X$ because $(x_k)_{k\in\mathbb{N}}$ is a sequence in $X$. The open ball centered at $x$ with radius $r$ is defined by \begin{align*} B(x,r)=\{y\in X:d(y,x)<r\}. \end{align*} Therefore, substituting the particular point $y=x_k$ into this defining condition gives \begin{align*} x_k\in B(x,r)\iff d(x_k,x)<r. \end{align*} This is the only conversion needed in the proof: the quantified ball condition is precisely the quantified distance condition in the definition of convergence.[/guided]

[step:Derive the ball condition from convergence] Assume $x_k\to x$ in $(X,d)$. Let $r>0$ be arbitrary. By the convergence definition, there exists $N\in\mathbb{N}$ such that \begin{align*} d(x_k,x)<r \end{align*} for every $k\in\mathbb{N}$ with $k\ge N$. By the open ball equivalence from the previous step, this implies \begin{align*} x_k\in B(x,r) \end{align*} for every $k\in\mathbb{N}$ with $k\ge N$. Since $r>0$ was arbitrary, the stated ball condition holds. [/step]

[step:Derive convergence from the ball condition] Assume that for every $r>0$ there exists $N\in\mathbb{N}$ such that \begin{align*} x_k\in B(x,r) \end{align*} for every $k\in\mathbb{N}$ with $k\ge N$. Let $r>0$ be arbitrary. Choose such an $N$. By the open ball equivalence, for every $k\in\mathbb{N}$ with $k\ge N$ we have \begin{align*} d(x_k,x)<r. \end{align*} This is exactly the definition of $x_k\to x$ in the metric space $(X,d)$. Hence the two conditions are equivalent. [/step]