A net $(v_\alpha)_{\alpha\in I}$ in a topological vector space (=TVR) $(V, \mathcal{T})$ is called Cauchy if
$$\forall U \in \mathcal{V}_V(0): \exists \alpha_0 \in I : \forall \alpha, \beta \geq \alpha_0: v_\alpha-v_\beta \in U$$
A TVR is called complete if every Cauchynet converges.
If $(V,\Vert \cdot \Vert)$ is a Banach space, then is it true that $V$ is a complete topological vector space? Clearly every Cauchysequence in $V$ converges, but also every Cauchynet?
Let $(V,\|\cdot\|)$ be a Banach space. Every Cauchy sequence converges.
Let $(v_\alpha)_{\alpha \in I}$ be a Cauchy net. To show: $(v_\alpha)$ converges.
We recursively construct a sequence $\alpha_n \in I$, $n=1,2,3,\dots$. We use the fact that $(v_\alpha)$ is a Cauchy net. There is $\alpha_n \in I$ so that,
(i) for all $\alpha,\beta \in I$ with $\alpha,\beta \ge \alpha_n$ we have $\|v_\alpha-v_\beta\| < \frac1n$, and
(ii) $\alpha_n \ge \alpha_j$ for all $j < n$.
We claim the sequence $(v_{\alpha_n})_{n \in \mathbb N}$ is a Cauchy sequence. Indeed, let $\epsilon > 0$ be given. There is $n \in \mathbb N$ so that $\frac{1}{n} < \epsilon$. Then for all $j,k \in \mathbb N$ with $j,k \ge n$, we have by (ii) that $\alpha_j \ge \alpha_n$ and $\alpha_k \ge \alpha_n$, so by (i) $\|v_{\alpha_j}-v_{\alpha_k}\| < \frac1n < \epsilon$. Thus $(v_{\alpha_n})$ is a Cauchy sequence.
Therefore, the sequence $(v_{\alpha_n})$ converges. Say $\lim_{n\to\infty} v_{\alpha_n} =u$. We claim the net $(v_\alpha)$ also converges to $u$. Let $\epsilon > 0$. There is $n_1 \in \mathbb N$ so that $\frac{1}{n_1} < \frac{\epsilon}{2}$. There is $n_2 \in \mathbb N$ so that for all $n \ge n_2$ we have $\|v_{\alpha_n} - u\| < \frac{\epsilon}{2}$. Let $n = \max\{n_1,n_2\}$. Then for all $\alpha \ge \alpha_n$ we have:
$n \ge n_1$, so $\alpha \ge \alpha_{n_1}$ and $\alpha_n \ge \alpha_{n_1}$ so that $\|v_\alpha-v_{\alpha_n}\| < \frac{1}{n_1} \le \frac{\epsilon}{2}$, and
$n \ge n_2$ so $\|\alpha_n - u\| \le \frac{\epsilon}{2}$
From the triangle inequality, we get $\|v_\alpha - u\| < \epsilon$. Thus, $\lim_{\alpha\in I} v_\alpha = u$ .