[proofplan] We compare Euclidean boundedness with coordinatewise boundedness using the elementary coordinate inequalities in $\mathbb{R}^n$. If $A$ is contained in a Euclidean ball, then every coordinate is bounded by the same radius because $|x_i|\le |x|$. Conversely, if each coordinate is bounded, then the Euclidean norm is bounded by the square root of the sum of the squared coordinate bounds. The empty-set case is included by the standard convention that the empty subset of a [metric space](/page/Metric%20Space) is bounded. [/proofplan]

[step:Extract coordinate bounds from a Euclidean bound] Assume first that $A$ is bounded in $\mathbb{R}^n$. By the definition of boundedness in the [Euclidean metric](/page/Euclidean%20Metric), there exists $R>0$ such that \begin{align*} |x|\le R \qquad \text{for every } x\in A. \end{align*} Fix $i\in\{1,\dots,n\}$. Let $t\in C_i$. By the definition of $C_i$, there exists $x=(x_1,\dots,x_n)\in A$ such that $t=x_i$. Since the Euclidean norm satisfies $|x_i|\le |x|$, we obtain \begin{align*} |t|=|x_i|\le |x|\le R. \end{align*} Thus every element of $C_i$ has absolute value at most $R$, so $C_i$ is bounded in $\mathbb{R}$. Since $i$ was arbitrary, each coordinate image $C_i$ is bounded. [/step]

[step:Combine coordinate bounds into one Euclidean bound]Assume conversely that $C_i$ is bounded in $\mathbb{R}$ for every $i\in\{1,\dots,n\}$. For each $i\in\{1,\dots,n\}$, choose a number $M_i>0$ such that \begin{align*} |t|\le M_i \qquad \text{for every } t\in C_i. \end{align*} Define \begin{align*} M:=\left(\sum_{i=1}^n M_i^2\right)^{1/2}. \end{align*} Let $x=(x_1,\dots,x_n)\in A$. For each $i\in\{1,\dots,n\}$, the coordinate $x_i$ belongs to $C_i$, hence $|x_i|\le M_i$. Therefore \begin{align*} |x|^2=\sum_{i=1}^n |x_i|^2\le \sum_{i=1}^n M_i^2=M^2. \end{align*} Since both $|x|$ and $M$ are nonnegative, it follows that $|x|\le M$. Thus \begin{align*} |x|\le M \qquad \text{for every } x\in A. \end{align*} By the definition of boundedness in the Euclidean metric, $A$ is bounded.[/step]

[guided]The goal is to turn separate one-dimensional bounds into a single bound in $\mathbb{R}^n$. For each coordinate image $C_i$, boundedness in $\mathbb{R}$ means that there is a number $M_i>0$ controlling the absolute value of every element of $C_i$: \begin{align*} |t|\le M_i \qquad \text{for every } t\in C_i. \end{align*} These constants may depend on the coordinate index $i$, so we need one number that controls all coordinates at once. Define \begin{align*} M:=\left(\sum_{i=1}^n M_i^2\right)^{1/2}. \end{align*} Now take an arbitrary point $x=(x_1,\dots,x_n)\in A$. Because $x\in A$, each coordinate $x_i$ belongs to the corresponding coordinate image $C_i$. Therefore the chosen coordinate bound gives \begin{align*} |x_i|\le M_i \qquad \text{for every } i\in\{1,\dots,n\}. \end{align*} The Euclidean norm is defined by summing the squares of the coordinates, so substituting these coordinate bounds gives \begin{align*} |x|^2=\sum_{i=1}^n |x_i|^2\le \sum_{i=1}^n M_i^2=M^2. \end{align*} Since $|x|\ge 0$ and $M\ge 0$, the inequality $|x|^2\le M^2$ implies $|x|\le M$. The point $x\in A$ was arbitrary, so \begin{align*} |x|\le M \qquad \text{for every } x\in A. \end{align*} This is exactly the definition that $A$ is bounded in $\mathbb{R}^n$ with respect to the Euclidean norm.[/guided]

[step:Account for the empty set and finish the equivalence] If $A=\varnothing$, then every coordinate image $C_i$ is also empty, and the empty subset of $\mathbb{R}$ and the empty subset of $\mathbb{R}^n$ are bounded by the standard convention for bounded subsets of metric spaces. If $A\ne\varnothing$, the two previous steps prove both implications. Hence $A$ is bounded in $\mathbb{R}^n$ if and only if each coordinate image $C_i$ is bounded in $\mathbb{R}$. [/step]