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Can you read my proof and tell me if it's correct? Thanks.

Let $V$ be a vector space over a complete topological field say $\mathbb R$ (or $\mathbb C$) with $\dim(V) = n$, base $e_i$ and norm $\|\cdot\|$. Let $v_k$ be a Cauchy sequence w.r.t. $\|\cdot\|$. Since any two norms on a finite dimensional space are equivalent, $\|\cdot\|$ is equivalent to the $l^1$-norm $\|\cdot\|_1$ which means that for some constant $C$, $\varepsilon > 0$, $k,j$ large enough, $$ \varepsilon > \|v_j - v_k\| \geq C \|v_j - v_k\|_1 e_i= C \sum_{i=1}^n |v_{ji} - v_{ki}| \geq |v_{ji} - v_{ki}|$$ for each $1 \leq i \leq n$. Hence $v_{ki}$ is a Cauchy sequence in $\mathbb R$ (or $\mathbb C$) for each $i$. $\mathbb R$ (or $\mathbb C$) is complete hence $v_i = \lim_{k \to \infty} v_{ki} $ is in $\mathbb R$ (or $\Bbb C$) for each $i$. Let $v = (v_1, \dots , v_n) = \sum_i v_i e_i$. Then $v$ is in $V$ and $\|v_k - v\| \to 0$:

Let $\varepsilon > 0$. Then $$ \|v_k - v\| \leq C \|v_k - v\|_1 = C \sum_{i=1}^n |v_{ki} - v_i| \leq C^{'}n \varepsilon$$

for $k$ large enough.

0x90
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Rudy the Reindeer
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    Finite dimensional VS over $\mathbb{Q}$ are not complete, when equipped with, e.g. the Euclidean norm. –  Jul 08 '12 at 15:49
  • @Thomas Thank you. Is the proof correct now? (see edit) – Rudy the Reindeer Jul 08 '12 at 15:53
  • Looks good! ${}$ – Cameron Buie Jul 08 '12 at 16:02
  • @CameronBuie Thank you! – Rudy the Reindeer Jul 08 '12 at 16:11
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    The proof is correct, there's a litle detail that is incorrect, the first equality in the final equation should be an inequality and possibly a different constant $C'$ instead of $C$ depending on how you define these constants (you can get away with one constant.) – Olivier Bégassat Jul 08 '12 at 16:31
  • @OlivierBégassat Yes you are right of course. I was aware of that. – Rudy the Reindeer Jul 08 '12 at 17:39
  • I kind of think that "Any two norms on a finite dimensional vector space are equivalent" is the key to this proof though. Otherwise you're just proving that $\mathbb{R}^n$ with the usual norm is complete. (Or $\mathbb{C}^n$) – Alex Zorn Aug 03 '15 at 15:05
  • This theorem is just valid for normed spaces what about metric spaces Let consider $(R,d)$ where $d(x,y)=|arctan(x)-arctan(y)|$ it is obvious that $R$ is finite dimensional but this space is not completed . Where is the wrong ? –  Dec 13 '17 at 02:45
  • @user344474: I think this is topological completeness: there is an equivalent metric d'(x,y) ( giving rise to the same topology) under which it is complete. – MSIS Jan 02 '23 at 03:02

2 Answers2

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Yes, your proof is correct. Here, I will just reword it to slightly improve clarity and precision .

Let $V$ be a vector space over $\mathbb R$ (or $\mathbb C$) with $\dim(V) = n$ and norm $\|\cdot\|$. Let $\{e_i\}_{i=1,\cdots , n}$ be a base of $V$. Suppose $v_k$ be a Cauchy sequence w.r.t. $\|\cdot\|$.

Since any two norms on a finite dimensional space are equivalent, $\|\cdot\|$ is equivalent to the $l^1$-norm $\|\cdot\|_1$. So, there are $C,D>0$ such that, for all $w\in V$, $C \|w\|_1 \leq \|w\| \leq D \|w\|_1$.

So, we have, for all $\varepsilon > 0$, there is $N$ such that, if $k,j>N$, $$ \varepsilon > \|v_j - v_k\| \geq C \|v_j - v_k\|_1 = C \sum_{i=1}^n |v_{ji} - v_{ki}| \geq C |v_{ji} - v_{ki}|$$ for each $1 \leq i \leq n$. Hence $v_{ki}$ is a Cauchy sequence in $\mathbb R$ (or $\mathbb C$) for each $i$. Since $\mathbb R$ (or $\mathbb C$) is complete, there is $u_i$ in $\mathbb R$ (or $\mathbb C$) such that $u_i = \lim_{k \to \infty} v_{ki} $, for each $i$. Let $u = (u_1, \dots , u_n) = \sum_i u_i e_i$. Then, it is clear that, $u$ is in $V$.

Let us prove $\lim_{k \to +\infty} \|v_k - u\| = 0$:

$$ \lim_{k \to +\infty} \|v_k - u\| \leq D \lim_{k \to +\infty} \|v_k - u\|_1 = D \lim_{k \to +\infty} \sum_{i=1}^n |v_{ki} - u_i| = D \sum_{i=1}^n \lim_{k \to +\infty} |v_{ki} - u_i|=0$$

Ramiro
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    This is nice. It would be worth relabeling the elements of the limit vector $v$ because currently the vectors $v_j,v_k$ could be confused with the scalar $i^\text{th}$ element $v_i$ of the limit vector $v$. – JS1204 Feb 17 '16 at 16:53
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I think there is a simple way to demonstrate this.

Let $ \left(E, \| \| \right)$ be a $\mathbb{K}$ finite dimensionial vector space. Consider the following application:

\begin{aligned} \mathbb { K }^ { n } & \rightarrow E \\ \left( \lambda _ { 1 } , \ldots , \lambda _ { n } \right) & \mapsto \sum _ { i = 1 } ^ { n } \lambda _ { i } e _ { i } \end{aligned} which is a linear isometric bijection (thus, a homeomorphism) between $ \left(\mathbb{ K }^ { n }, \| \| _\infty\right)$ and $ \left(E, \| \|_{\infty} \right)$.

The pre-image of a complete space by a uniformly continuous and bijective function being complete and $ \left(\mathbb{ K }^ { n }, \| \| _\infty\right)$ being a complete space, then $ \left( E, \| \| _\infty\right)$ is complete. Since all norms on a finite dimensional space are equivalent, it folows that $ \left(E, \| \| \right)$ is complete.

Amir Dib
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