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