[proofplan] We prove the result directly from the metric definition of convergence. Membership of each $x_k$ in the closed ball gives the uniform bound $d(x_k,x_0) \leq r$. Convergence lets us choose terms $x_k$ arbitrarily close to $x$, and the triangle inequality transfers the radius bound from $x_k$ to $x$ up to an arbitrary error. Letting that error vanish gives $d(x,x_0) \leq r$, which is exactly $x \in \overline{B}(x_0,r)$. [/proofplan] [step:Convert closed ball membership into a distance bound] For every $k \in \mathbb{N}$, the hypothesis $x_k \in \overline{B}(x_0,r)$ and the definition of $\overline{B}(x_0,r)$ give \begin{align*} d(x_k,x_0) \leq r. \end{align*} Thus the whole sequence has distance at most $r$ from the center $x_0$. [/step] [step:Use convergence and the triangle inequality to bound $d(x,x_0)$ up to an arbitrary error] Let $\varepsilon > 0$. Since $x_k \to x$ in $(X,d)$, there exists $N_\varepsilon \in \mathbb{N}$ such that for every $k \geq N_\varepsilon$, \begin{align*} d(x_k,x) < \varepsilon. \end{align*} Choose $k_\varepsilon := N_\varepsilon$. By symmetry of the metric and the triangle inequality applied to the triple $x,x_{k_\varepsilon},x_0 \in X$, \begin{align*} d(x,x_0) \leq d(x,x_{k_\varepsilon}) + d(x_{k_\varepsilon},x_0). \end{align*} Using $d(x,x_{k_\varepsilon}) = d(x_{k_\varepsilon},x) < \varepsilon$ and $d(x_{k_\varepsilon},x_0) \leq r$, we obtain \begin{align*} d(x,x_0) < r + \varepsilon. \end{align*} [guided] Fix an arbitrary number $\varepsilon > 0$. The purpose of introducing $\varepsilon$ is to measure how close a sequence term must be to the [limit point](/page/Limit%20Point) $x$. By the definition of convergence in the [metric space](/page/Metric%20Space) $(X,d)$, there exists an index $N_\varepsilon \in \mathbb{N}$ such that for every $k \geq N_\varepsilon$, \begin{align*} d(x_k,x) < \varepsilon. \end{align*} We now select one such index, namely $k_\varepsilon := N_\varepsilon$. This gives a concrete sequence term $x_{k_\varepsilon}$ close to $x$. Since the metric is symmetric, \begin{align*} d(x,x_{k_\varepsilon}) = d(x_{k_\varepsilon},x) < \varepsilon. \end{align*} The triangle inequality for the metric $d: X \times X \to [0,\infty)$, applied to the points $x,x_{k_\varepsilon},x_0 \in X$, gives \begin{align*} d(x,x_0) \leq d(x,x_{k_\varepsilon}) + d(x_{k_\varepsilon},x_0). \end{align*} The first term on the right is less than $\varepsilon$ by convergence. The second term is at most $r$ because $x_{k_\varepsilon} \in \overline{B}(x_0,r)$. Therefore \begin{align*} d(x,x_0) < \varepsilon + r. \end{align*} Equivalently, \begin{align*} d(x,x_0) < r + \varepsilon. \end{align*} This is the key estimate: the limit point $x$ is no farther than $r$ from $x_0$, except for an error $\varepsilon$ that we may choose arbitrarily small. [/guided] [/step] [step:Remove the arbitrary error and conclude membership in the closed ball] We have shown that for every $\varepsilon > 0$, \begin{align*} d(x,x_0) < r + \varepsilon. \end{align*} Set $a := d(x,x_0) \in [0,\infty)$. We claim that $a \leq r$. If instead $a > r$, define $\varepsilon_0 := (a-r)/2 > 0$. Then the preceding estimate with $\varepsilon = \varepsilon_0$ gives \begin{align*} a < r + \varepsilon_0 = r + \frac{a-r}{2} = \frac{a+r}{2}. \end{align*} This implies $2a < a+r$, hence $a < r$, contradicting $a > r$. Therefore $d(x,x_0) = a \leq r$. By the definition of the closed ball, \begin{align*} x \in \overline{B}(x_0,r). \end{align*} This proves the theorem. [/step]