show that the normed linear vector space $\mathbb{R}^n$ with norm $||x|| = \max{(|x_1|, |x_2|, ... , |x_n|})$ is complete.

Question

I am trying to show that the normed linear vector space $\mathbb{R}^n$ with norm $||x|| = \max{(|x_1|, |x_2|, ... , |x_n|})$ is complete.

My approach was as follows: First, construct a Cauchy sequence ${x_n}$ such that $x_n\in\mathbb{R}^n$ and $\forall \epsilon>0 \exists N\text{ such that }||x_m-x_n||\leq\epsilon \forall m,n\geq N$

We also know that $\mathbb{R}$ is complete, hence, the every Cauchy sequence in $\mathbb{R}$ converges to a scalar in $\mathbb{R}$

Hence, the individual components of my nth vector converges with respect to the absolute value norm. Therefore the $\lim_{n\to\infty} x_n = x$ i.e.

$$\forall \epsilon \exists N\text{ such that }||x_n-x||=(|x_{n1}-x|, |x_{n2}-x|, ... , |x_{nk}-x|)\leq \epsilon \forall n\geq N$$ (1)

Then I claim that that my Cauchy sequence {x_n} will converge under the max-norm to the same vector x, hence I need to show that $\exists N>0 s.t. ||x_n-x||\leq \epsilon$

$$||x_n-x||=\max{(|x_{n1}-x|, |x_{n2}-x|, ... , |x_{nk}-x|})$$ by (1) this is $$||x_n-x||=\max{(\epsilon_1, \epsilon_2, ... , \epsilon_k})=\epsilon$$

is my solution correct?

Answer 1

Let $x_m = (x_{1m},x_{2m},\dots, x_{nm} )\in \mathbb{R^n}$ be a Cauchy sequence and assume it converges to a point $x=(x_1,x_2,\dots,x_n)$. Now see this

$$ || x_m - x ||= \big|\big| (x_{1m}-x_1,x_{2m}-x_2,\dots, x_{n m}-x_n ) \big|\big|$$

$$ = \max \left\{ | x_{1 m}-x_1|,| x_{2 m}-x_2|, \dots, | x_{n m}-x_n| \right\} \leq||x_m-x||_2 <\epsilon. $$