0
$\begingroup$

Let $V$ be a finite dimensional vector space with a norm $| \ |,$ and let $d(x,y)=|x-y|$ be the corresponding metric. Show that the metric space $(V,d)$ is complete.

Do I start off the proof by saying let $( x_{n_1}, x_{n_2}, \ldots, x_{n_m})$ be a Cauchy sequence then show $\forall \epsilon >0: \exists N: \forall m: m, n \ge N: d \left({x_n, x_m}\right) < \epsilon$?

$\endgroup$
1
  • 1
    $\begingroup$ The steps are: (1) $\mathbb R$ is complete (with absolute value as distance). (2) Guess the limit of your Cauchy sequence. (3) Prove that it is the limit (and that it exists). $\endgroup$
    – user42761
    Nov 10, 2013 at 0:02

2 Answers 2

4
$\begingroup$

Close. The definition of a complete metric space is one where all Cauchy sequences converge. So, take an arbitrary Cauchy sequence $\{x_n\}$ and show that for all $\epsilon>0$ $\exists N\in\mathbb{N}$ such that $\forall n,m > N$, $|x_n-x_m|<\epsilon$ implies that $\{x_n\}$ converges. What you have written there is the definition of a Cauchy sequence. You want to use this definition to show that there is an $x \in V$ such that $|x_n-x|<\epsilon$. Hint: every Cauchy sequence is bounded.

$\endgroup$
0
2
$\begingroup$

I think you need that in $V$ (finite dimension), all norms are equivalents, then you can reduce the problem to : $\mathbb R^m$ is complete.

$\endgroup$
1
  • $\begingroup$ But I did say finite dimension. $\endgroup$
    – user104235
    Nov 10, 2013 at 4:09

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .