[proofplan] We prove that the complement of the closed ball is open. Given a point outside the closed ball, its distance from the centre exceeds $r$, leaving a positive margin. Choosing a ball around this exterior point with radius half that margin, the triangle inequality for the [metric space](/page/Metric%20Space) shows that every point in this smaller ball still has distance greater than $r$ from $x_0$. Hence every exterior point is interior to the complement, so the complement is [open](/page/Open%20Set) and the closed ball is [closed](/page/Closed%20Set). [/proofplan]

[step:Show that every exterior point has a neighbourhood outside the closed ball]Let \begin{align*} C := \overline{B}(x_0,r) = \{z \in X : d(z,x_0) \le r\}. \end{align*} We show that $X \setminus C$ is [open](/page/Open%20Set). Let $x \in X \setminus C$. By the definition of $C$, this means \begin{align*} d(x,x_0) > r. \end{align*} Define \begin{align*} \varepsilon := \frac{d(x,x_0)-r}{2}. \end{align*} Then $\varepsilon > 0$. Let $y \in B(x,\varepsilon)$, where \begin{align*} B(x,\varepsilon) := \{z \in X : d(z,x) < \varepsilon\}. \end{align*} Using the triangle inequality from the definition of a [metric space](/page/Metric%20Space), in the form \begin{align*} d(x,x_0) \le d(x,y) + d(y,x_0), \end{align*} we obtain \begin{align*} d(y,x_0) \ge d(x,x_0) - d(x,y). \end{align*} Since $d(x,y) < \varepsilon$, it follows that \begin{align*} d(y,x_0) > d(x,x_0) - \varepsilon. \end{align*} By the definition of $\varepsilon$, \begin{align*} d(x,x_0) - \varepsilon = \frac{d(x,x_0)+r}{2}. \end{align*} Because $d(x,x_0) > r$, we have \begin{align*} \frac{d(x,x_0)+r}{2} > r. \end{align*} Therefore $d(y,x_0) > r$, so $y \notin C$. Hence $B(x,\varepsilon) \subset X \setminus C$.[/step]

[guided]Let \begin{align*} C := \overline{B}(x_0,r) = \{z \in X : d(z,x_0) \le r\}. \end{align*} To prove that $C$ is [closed](/page/Closed%20Set), it is enough to prove that its complement $X \setminus C$ is [open](/page/Open%20Set). This means that for every point $x \in X \setminus C$, we must find a radius $\varepsilon > 0$ such that the open ball \begin{align*} B(x,\varepsilon) := \{z \in X : d(z,x) < \varepsilon\} \end{align*} is contained in $X \setminus C$. Fix $x \in X \setminus C$. Since $x$ is not in the closed ball centred at $x_0$ with radius $r$, the defining inequality for membership in $C$ fails, so \begin{align*} d(x,x_0) > r. \end{align*} The positive number $d(x,x_0)-r$ is the margin by which $x$ lies outside the closed ball. We choose half of this margin: \begin{align*} \varepsilon := \frac{d(x,x_0)-r}{2}. \end{align*} Then $\varepsilon > 0$ because $d(x,x_0) > r$. Now take any $y \in B(x,\varepsilon)$. By definition of $B(x,\varepsilon)$, this gives $d(x,y) < \varepsilon$. The triangle inequality from the definition of a [metric space](/page/Metric%20Space), applied to the three points $x$, $y$, and $x_0$, gives \begin{align*} d(x,x_0) \le d(x,y) + d(y,x_0). \end{align*} Rearranging this inequality gives \begin{align*} d(y,x_0) \ge d(x,x_0) - d(x,y). \end{align*} Since $d(x,y) < \varepsilon$, we get \begin{align*} d(y,x_0) > d(x,x_0) - \varepsilon. \end{align*} Substituting the definition of $\varepsilon$ yields \begin{align*} d(x,x_0) - \varepsilon = d(x,x_0) - \frac{d(x,x_0)-r}{2}. \end{align*} Thus \begin{align*} d(x,x_0) - \varepsilon = \frac{d(x,x_0)+r}{2}. \end{align*} Because $d(x,x_0) > r$, the average of $d(x,x_0)$ and $r$ is still strictly larger than $r$: \begin{align*} \frac{d(x,x_0)+r}{2} > r. \end{align*} Therefore $d(y,x_0) > r$, so $y$ is not in $C$. Since the point $y \in B(x,\varepsilon)$ was arbitrary, we have proved \begin{align*} B(x,\varepsilon) \subset X \setminus C. \end{align*}[/guided]

[step:Conclude that the complement is open and the closed ball is closed] For every $x \in X \setminus C$, we have found $\varepsilon > 0$ such that $B(x,\varepsilon) \subset X \setminus C$. Hence $X \setminus C$ is [open](/page/Open%20Set) in the metric space $X$. Therefore $C = \overline{B}(x_0,r)$ is [closed](/page/Closed%20Set) in $X$. [/step]