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