[proofplan] We prove the inclusion directly from the closure definition and the triangle inequality. If $x$ lies in the closure of the open ball, then every small ball around $x$ meets $B(x_0,r)$, so we can choose points $y$ with $d(y,x_0)<r$ and $d(x,y)$ arbitrarily small. The triangle inequality then gives $d(x,x_0)\le r$. To show the inclusion can be proper, we use the two-point discrete metric, where a point at exact distance $r$ from the center lies in the closed ball but is isolated from the open ball. [/proofplan] [step:Approximate a closure point by points inside the open ball] Let $x \in \overline{B(x_0,r)}$. By the definition of closure in a [metric space](/page/Metric%20Space), for every $\varepsilon>0$ there exists a point $y_\varepsilon \in X$ such that \begin{align*} y_\varepsilon \in B(x_0,r) \cap B(x,\varepsilon). \end{align*} Thus \begin{align*} d(y_\varepsilon,x_0)<r \end{align*} and \begin{align*} d(x,y_\varepsilon)<\varepsilon. \end{align*} [guided] Let $x \in \overline{B(x_0,r)}$. The notation $\overline{B(x_0,r)}$ means the closure of the open ball $B(x_0,r)$ in the metric space $X$; it is not the same notation as the closed ball $\overline{B}(x_0,r)$, which we are trying to reach. By the definition of closure in a metric space, every open ball centered at $x$ must meet the set whose closure contains $x$. Therefore, for each $\varepsilon>0$, the open ball $B(x,\varepsilon)$ intersects $B(x_0,r)$. Hence there exists a point $y_\varepsilon \in X$ such that \begin{align*} y_\varepsilon \in B(x_0,r) \cap B(x,\varepsilon). \end{align*} Unpacking the two ball conditions gives \begin{align*} d(y_\varepsilon,x_0)<r \end{align*} and \begin{align*} d(x,y_\varepsilon)<\varepsilon. \end{align*} This point $y_\varepsilon$ is the approximation to $x$ from inside the open ball. The parameter $\varepsilon$ is arbitrary, so these approximations can be made as close to $x$ as needed. [/guided] [/step] [step:Use the triangle inequality to place the closure point in the closed ball] For every $\varepsilon>0$, the triangle inequality applied to the three points $x,y_\varepsilon,x_0 \in X$ gives \begin{align*} d(x,x_0)\le d(x,y_\varepsilon)+d(y_\varepsilon,x_0)<\varepsilon+r. \end{align*} Since this holds for every $\varepsilon>0$, we must have $d(x,x_0)\le r$. Indeed, if $d(x,x_0)>r$, then choosing $\varepsilon=(d(x,x_0)-r)/2>0$ would contradict $d(x,x_0)<r+\varepsilon$. Therefore $x \in \overline{B}(x_0,r)$. Since $x \in \overline{B(x_0,r)}$ was arbitrary, we conclude \begin{align*} \overline{B(x_0,r)} \subset \overline{B}(x_0,r). \end{align*} [/step] [step:Exhibit a metric space where the inclusion is proper] Let $X=\{a,b\}$, define the metric $d:X\times X\to [0,\infty)$ by setting $d(u,v)=0$ when $u=v$ and $d(u,v)=1$ when $u\ne v$, and take $x_0=a$ and $r=1$. Then \begin{align*} B(a,1)=\{a\}. \end{align*} The point $b$ is not in the closure of $\{a\}$, because \begin{align*} B(b,1/2)=\{b\} \end{align*} and hence $B(b,1/2)\cap \{a\}=\varnothing$. Thus \begin{align*} \overline{B(a,1)}=\{a\}. \end{align*} On the other hand, since $d(a,a)=0\le 1$ and $d(b,a)=1\le 1$, the closed ball is \begin{align*} \overline{B}(a,1)=X=\{a,b\}. \end{align*} Therefore \begin{align*} \overline{B(a,1)}=\{a\}\subsetneq \{a,b\}=\overline{B}(a,1). \end{align*} This proves that the inclusion can be proper. [/step]