[proofplan] We prove boundedness by replacing the finitely many individual bounds with one global bound. Empty sets cause no difficulty because the ambient [metric space](/page/Metric%20Space) is nonempty, so an arbitrary center is available in the vacuous case. We then restrict attention to the nonempty members of the family. For each nonempty $A_i$, boundedness gives a center $c_i$ and a finite radius $R_i$ controlling all points of $A_i$. We then choose one common center from the union and use the triangle inequality, followed by the existence of a maximum over finitely many [real numbers](/page/Real%20Numbers), to obtain a single bound for the whole union. [/proofplan] [step:Discard the vacuous case where every set is empty] Let \begin{align*} U:=\bigcup_{i=1}^m A_i \end{align*} denote the finite union. If $U=\varnothing$, choose a point $c_\varnothing\in X$, which exists because $X$ is nonempty, and set $M_\varnothing:=1$. Then the inequality \begin{align*} d(x,c_\varnothing)\le M_\varnothing \end{align*} holds for every $x\in U$ because there is no point $x\in U$ at which it can fail. Hence $U$ is bounded, and the desired conclusion holds in this case. Assume from now on that $U\neq\varnothing$. [/step] [step:Choose individual bounds and one common center] Define the nonempty index set \begin{align*} I:=\{i\in\{1,\dots,m\}:A_i\neq\varnothing\}. \end{align*} Since $U\neq\varnothing$, the set $I$ is nonempty. Choose an index $i_0\in I$ and choose a point $c\in A_{i_0}$. For each $i\in I$, since $A_i$ is bounded, there exist a point $c_i\in X$ and a real number $R_i>0$ such that \begin{align*} d(x,c_i)\le R_i \end{align*} for every $x\in A_i$. [/step] [step:Use the triangle inequality to compare every set to the common center] For each $i\in I$, define the real number \begin{align*} M_i:=R_i+d(c_i,c). \end{align*} If $x\in A_i$, then the triangle inequality in the metric space $(X,d)$ gives \begin{align*} d(x,c)\le d(x,c_i)+d(c_i,c). \end{align*} Using the bound $d(x,c_i)\le R_i$, we obtain \begin{align*} d(x,c)\le R_i+d(c_i,c)=M_i. \end{align*} [guided] The reason for introducing the common point $c$ is that boundedness of the union requires one bound measured from one fixed reference point, rather than a different center for each $A_i$. For a fixed nonempty set $A_i$, boundedness gives a center $c_i\in X$ and a radius $R_i>0$ such that \begin{align*} d(x,c_i)\le R_i \end{align*} for every $x\in A_i$. Now take any point $x\in A_i$. To estimate the distance from $x$ to the common center $c$, insert the intermediate point $c_i$. The triangle inequality for the metric $d:X\times X\to\mathbb{R}$ states that \begin{align*} d(x,c)\le d(x,c_i)+d(c_i,c). \end{align*} The first term is controlled by the boundedness of $A_i$, so \begin{align*} d(x,c)\le R_i+d(c_i,c). \end{align*} Thus every point of $A_i$ lies within the finite distance \begin{align*} M_i:=R_i+d(c_i,c) \end{align*} of the single common center $c$. [/guided] [/step] [step:Take the maximum of the finitely many bounds] The set $\{M_i:i\in I\}$ is a nonempty finite subset of $\mathbb{R}$, so it has a maximum. Define \begin{align*} M:=\max_{i\in I} M_i. \end{align*} Then $M\ge M_i$ for every $i\in I$. Let $x\in U$. By definition of $U$, there exists $i\in\{1,\dots,m\}$ such that $x\in A_i$. Since $x\in A_i$, the set $A_i$ is nonempty, so $i\in I$. From the previous step, \begin{align*} d(x,c)\le M_i\le M. \end{align*} Therefore every $x\in U$ satisfies $d(x,c)\le M$. Hence $U$ is bounded. Since \begin{align*} U=\bigcup_{i=1}^m A_i, \end{align*} the finite union $\bigcup_{i=1}^m A_i$ is bounded. [/step]