[proofplan] We prove both implications directly from the Euclidean norm. If $A$ is bounded, then all points of $A$ lie within a fixed Euclidean distance of some center $c\in\mathbb{R}^n$; the triangle inequality converts this into a bound from the origin, and each coordinate is bounded by the Euclidean norm. Conversely, if every coordinate of every point of $A$ lies between $-M$ and $M$, then the Euclidean norm is bounded above by $\sqrt{n}M$, so $A$ lies in a Euclidean ball. [/proofplan]

[step:Convert metric boundedness into coordinate bounds]For $y=(y_1,\dots,y_n)\in\mathbb{R}^n$, write $|y|:=\left(\sum_{j=1}^{n}y_j^2\right)^{1/2}$ for the Euclidean norm, so the [Euclidean metric](/page/Euclidean%20Metric) is $d(u,v)=|u-v|$ for $u,v\in\mathbb{R}^n$. Assume that $A$ is bounded in the Euclidean metric. By the definition of boundedness, there exist a point $c\in\mathbb{R}^n$ and a number $R>0$ such that $|x-c|\le R$ for every $x\in A$. Define $M:=|c|+R+1$. Then $M>0$. Let $x=(x_1,\dots,x_n)\in A$ and let $i\in\{1,\dots,n\}$. Since the absolute value of a coordinate is bounded above by the Euclidean norm, $|x_i|\le |x|$. By the triangle inequality in $\mathbb{R}^n$, we have \begin{align*} |x| \le |x-c|+|c| \end{align*} and therefore \begin{align*} |x| < M. \end{align*} Thus $|x_i|\le M$ for every $i\in\{1,\dots,n\}$. Therefore $x\in[-M,M]^n$. Since $x\in A$ was arbitrary, $A\subset[-M,M]^n$.[/step]

[guided]For $y=(y_1,\dots,y_n)\in\mathbb{R}^n$, write $|y|:=\left(\sum_{j=1}^{n}y_j^2\right)^{1/2}$ for the Euclidean norm, and define the Euclidean metric by $d(u,v)=|u-v|$ for $u,v\in\mathbb{R}^n$. Assume that $A$ is bounded in the Euclidean metric. This means that $A$ is contained in a ball of finite radius, but the center of that ball need not be the origin. More precisely, there exist a point $c\in\mathbb{R}^n$ and a number $R>0$ such that $|x-c|\le R$ for every $x\in A$. The box $[-M,M]^n$ is centered at the origin, so we first turn the bound around $c$ into a bound around $0$. Define $M:=|c|+R+1$. The added $1$ is not mathematically essential, but it guarantees at once that $M>0$ and gives a strict margin. Let $x=(x_1,\dots,x_n)\in A$ and fix an index $i\in\{1,\dots,n\}$. The Euclidean norm controls each coordinate because $|x|^2=\sum_{j=1}^{n}x_j^2$. Since the summand $x_i^2$ is one of the nonnegative terms in the sum, we have $x_i^2\le |x|^2$, hence $|x_i|\le |x|$. Now apply the triangle inequality to the decomposition $x=(x-c)+c$: $|x|\le |x-c|+|c|$. Because $x\in A$, the boundedness hypothesis gives $|x-c|\le R$. Therefore $|x_i|\le |x|\le |x-c|+|c|\le R+|c|<M$. Thus every coordinate of $x$ satisfies $|x_i|\le M$, so $x\in[-M,M]^n$. Since $x\in A$ was arbitrary, we conclude that $A\subset[-M,M]^n$.[/guided]

[step:Bound the Euclidean norm from coordinate bounds] Conversely, assume that there exists $M>0$ such that $A\subset[-M,M]^n$. Let $x=(x_1,\dots,x_n)\in A$. Then $|x_i|\le M$ for every $i\in\{1,\dots,n\}$. Hence \begin{align*} |x|^2 = \sum_{i=1}^{n}x_i^2 \le \sum_{i=1}^{n}M^2 = nM^2. \end{align*} Taking square roots gives $|x|\le \sqrt{n}M$. Hence $|x|<\sqrt{n}M+1$, and every point of $A$ lies in the Euclidean ball centered at $0$ with positive radius $\sqrt{n}M+1$. Therefore $A$ is bounded. This proves the converse implication and completes the proof. [/step]