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?

  • $\begingroup$ To show completeness, you need to show that any Cauchy sequence converges, you don't get to construct the sequence. Where did the $\epsilon_i$ come from? $\endgroup$ – copper.hat Mar 24 '15 at 6:24
  • $\begingroup$ @copper.hat yes I understand that we need to show convergence for any Cauchy sequence. The $\epsilon_i$ came from the completeness of $\mathbb{R}$ under the absolute norm $\endgroup$ – MAS Mar 24 '15 at 6:27
  • 1
    $\begingroup$ Well, you have the gist of a correct proof, but given an $\epsilon>0$ you can find an $N_i$ for component $i$, so you then let $N= \max_i N_i$. you don't choose different $\epsilon$s for each component. $\endgroup$ – copper.hat Mar 24 '15 at 6:29

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

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

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


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