# Bounded in $\Bbb R^n$

Let $n$ be a positive integer, $d$ the Euclidean metric on $\Bbb R^n$, and $X$ any subset of $\Bbb R^n$. Prove that $X$ is bounded in $(\Bbb R^n, d)$ iff there exists an $M > 0$ such that for all $y =<x_1,x_2,...,x_n>$ $\in X$, $-M \le x_i \le M, i=1, 2, ..., n$.

I start $(\Rightarrow)$ by assuming $(\exists r >0)(d(y_1,y_2) \le r)(\forall y_1,y_2 \in X)$. My idea is to construct an open ball $B_{r+1}(y_i)$ so that $X \subseteq B_{r+1}(y_i)$ then choosing a point outside of $B_{r+1}(y_i)$ that I can call $M$ and compare each individual point to. But I'm not too sure how to do this.

You have to fix one point, say $x_0$, then consider the distance between it and any other point $x\in X$. So by your assumption, for all $x\in X$, $d(x_0,x)\le r$. Then applying triangle inequality yields that for all $x\in X$, $$d(0,x)\le r+d(x_0, 0)\le N<\infty.$$ where $N$ can be just chosen to be $r+d(x_0,0)$. Now can you find an $M$ so large that the closed box defined by $-M\le x_i\le M$ contains the closed ball $\{x\mid d(0,x)\le N\}$ altogether?