TITLE: Proof that every finite dimensional normed vector space is complete
QUESTION [43 upvotes]: 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.

REPLY [14 votes]: 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.