Proving $\mathbb{R}^n$ bounded implies $-M\leqslant x_i\leqslant M$

Question

Exercise(Genia 7.2.8): Let $n$ be a positive integer, $d$ the Euclidian metric on $\mathbb{R}^n$, and $X$ a subset of $\mathbb{R}^n$. Prove that $X$ is a bounded in $(\mathbb{R},d)$ if and only if exists a positive number $M$ such that for all $(x_1,x_2,...,x_n)\in X$, $-M\leqslant x_i\leqslant M$, $i=1,2,...n$.

$\Leftarrow$ For $x,y$ such that $x\neq y$ $d(x,y)=\sqrt{\sum_\limits{i=1}^{n}|x_i-y_i|^2}\leqslant \sqrt{\sum_\limits{i=1}^{n}M^2}=\sqrt{nM^2}=\sqrt{n}M$.

So $(X,d)$ is limited.

$\Rightarrow$

If X is bounded it exists a ball that contains $X$ with a radius of suppose M.

Questions:

Is my proof insofar right?

How should I prove the reverse inequality?

Thanks in advance!

Answer 1

If $X$ is bounded, there exists $M\ge0$ such that $\left|\left|x\right|\right|\le M$ for all $x\in X$. If this is the case, then for all $i\in\{1,\dots,n\}$ $$\left|x_i\right|^2\le \sum_{j=1}^n \left|x_j\right|^2 \le M$$ This comes from the fact that $\max_{1\le i\le n} \left|x_i\right| \le \sqrt{\sum_{j=1}^n \left|x_j\right|^2}$ (inequalities between norms, but I don't know how to typeset norms without my personal macros ;-).