# Proof that every finite dimensional normed vector space is complete

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.

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Finite dimensional VS over $\mathbb{Q}$ are not complete, when equipped with, e.g. the Euclidean norm. –  user20266 Jul 8 '12 at 15:49
@Thomas Thank you. Is the proof correct now? (see edit) –  Matt N. Jul 8 '12 at 15:53
Looks good! ${}$ –  Cameron Buie Jul 8 '12 at 16:02
@CameronBuie Thank you! –  Matt N. Jul 8 '12 at 16:11
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 8 '12 at 16:31
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