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So I am looking through a proof in my lecture notes about the fact that all finite dimensional normed spaces are complete. This result was in the lecture notes seen as corollary to the fact that any two norms on a finite dimensional vector space are equivalent. It was shown that any two norms on a finite dimensional vector spaces are equivalent by considering the two norms $||.||$ and $||.||_{1}$ by showing that $\forall \ x \ \in X \ (fin-dim-vectorspace)$ \begin{align*} ||x|| &\leq C ||x||_{1}\\ K||x||_{1} &\leq ||x|| \iff ||x||_{1} \leq \frac{1}{k} ||x||. \end{align*} So now the fact that a finite dimensional normed space $X$ is complete should be an "easy" corollary. However, there is one step in my lecture notes that I don't see right away. So the proof goes as follows:

Proof:

Take $\{ x_{n} \}$ cauchy in $||.||,$

$\implies ||x_{n} - x_{m}||_{1} \leq \frac{1}{k} ||x_{n} - x_{m}|| \rightarrow 0 \ for \ n,m \rightarrow \infty, $

$\implies \{x_{n}\} \ is \ cauchy \ in \ ||.||_{1}$

SO $x_{n} \rightarrow x \ wrt. ||.||_{1}$ (THIS IS the step I am confused about, why does it follow that if $x_{n}$ is cauchy in the one norm then this sequence $x_{n}$ converges to $x$ in the one norm as well? Is it because all cauchy sequences in the one norm converges?...

The proof then ends with:

$||x_{n} - x ||_{1} \rightarrow 0 \ for \ n \rightarrow \infty$

$\implies ||x_{n} - x|| \leq C||x_{n} - x||_{1} \rightarrow 0 \ for \ n \rightarrow \infty$

So it seems like the proof is just to show that convergence in one norm => convergence in another norm and then from there we know that every finite dimensional vector space is complete by transitivity of norms?.

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1 Answer 1

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Since you are only showing part of the proof, I can only guess. And my guess is that who wrote that had already proved that $\Bbb R^n$ is complete with respect to some norm, possibly the usual norm $\|\cdot\|_2$. So, given a sequence $(x_n)_{n\in\Bbb N}$ of elements of $\Bbb R^n$ which is a Cauchy sequence with respect to some norm $\|\cdot\|$, the idea is:

  • since $(x_n)_{n\in\Bbb N}$ is a Cauchy sequence with respect to $\|\cdot\|$, it is also a Cauchy sequence with respect to $\|\cdot\|_2$;
  • therefore it converges in $\bigl(\Bbb R^n,\|\cdot\|_2\bigr)$;
  • so, it converges in $\bigl(\Bbb R^n,\|\cdot\|\bigr)$.
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  • $\begingroup$ Okay thank you for your clarification. So just to make sure that I understand. A vector space is called finite dimensional if some list of vectors in it spans the space. (Recall by definition every list has a finite length). And then I note that $\mathbb{F}^{n} = \mathbb{R}^{n} \cup \mathbb{C}^{n}$ is a finite dimensional vector space. BUT $\mathbb{R^{n}}$ is complete thus, if $x_{n}$ is cauchy in a finite dimensional vector space then it converges because $\mathbb{R^{n}}$ is complete and $\mathbb{R^{n}}$ is a finite dimensional vector space. $\endgroup$
    – Chengdu
    Oct 14, 2021 at 10:52
  • $\begingroup$ I don't know what you mean when you say that $\Bbb R^n$ is complete. A set neither is nor isn't complete. The term “complete” is about metric spaces not about sets. Given a metric $d$ in $\Bbb R^n$, then the metric space $(\Bbb R^n,d)$ either is or isn't complete. And it turns out that if $d(v,w)=\|v-w\|_2$, then, yes, $(\Bbb R^n,d)$ is complete. $\endgroup$ Oct 14, 2021 at 10:56
  • $\begingroup$ yes you are right, with some metric of course. Thank you! $\endgroup$
    – Chengdu
    Oct 14, 2021 at 10:58

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