Cauchy nets in a metric space Say that a net $a_i$ in a metric space is cauchy if for every $\epsilon > 0$ there exists $I$ such that for all $i, j \geq I$ one has $d(a_i,a_j) \leq \epsilon$.  If the metric space is complete, does it hold (and in either case why) that every cauchy net converges?  
 A: Consider a Cauchy net:
$$\forall \lambda,\lambda'\geq\lambda_n:\quad d(x_\lambda,x_\lambda')<\frac{1}{n}$$
Extract a Cauchy sequence:
$$x_n:=x_{\lambda(n)}\quad\lambda(n):=\lambda_1\wedge\ldots\wedge\lambda_n$$
Apply completeness:
$$d(x_\lambda,x)\leq d(x_\lambda,x_{n_0})+d(x_{n_0},x)<\frac{N}{2}+\frac{N}{2}\leq\epsilon$$
where to choose the meet $n_0:=N\wedge n(N)$ with $N:=\lceil\frac{\epsilon}{2}\rceil$
A: Yes, it’s true.
Suppose that the metric space $\langle X,d\rangle$ is complete, and let $\langle x_i:i\in I\rangle$ be a Cauchy net in $X$. Pick $i(0)\in I$ such that $d(x_i,x_j)\le 1$ whenever $i,j\ge i(0)$. Given $i(n)\in I$ such that $d(x_i,x_j)\le 2^{-n}$ whenever $i,j\ge i(n)$, choose $i(n+1)\in I$ such that $i(n+1)\ge i(n)$ and $d(x_i,x_j)\le 2^{-(n+1)}$ whenever $i,j\ge i(n+1)$. Then the sequence $\langle x_{i(k)}:k\in\Bbb N\rangle$ is $d$-Cauchy and therefore converges to some $x\in X$. Fix $\epsilon>0$; there is an $m_0\in\Bbb N$ such that $d(x_{i(n)},x)<\epsilon/2$ whenever $n\ge m$, and there is an $m_1\in\Bbb N$ such that $d(x_i,x_j)<\epsilon/2$ whenever $i,j\ge i(m_1)$. Let $m=\max\{m_0,m_1\}$; then $d(x_i,x)<\epsilon$ whenever $i\ge i(m)$, so $\langle x_i:i\in I\rangle$ converges to $x$.
