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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?

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up vote 5 down vote accepted

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 $x$.

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In the second to last line, also second to last sentence, you claim the existence of $m_1$ but I don't think I understand. Why does it have to be something of the form $i(m_1)$ above which cauchy-ness is in effect? – Jeff Aug 28 '12 at 3:38
@Jeff: Just choose $m_1$ large enough so that $2^{-m_1}<\epsilon/2$; then by construction $d(x_i,x_j)\le 2^{-m_1}<\epsilon/2$ whenever $i,j\ge i(m_1)$. – Brian M. Scott Aug 28 '12 at 3:40
Ah yes, that's the way you chose the sequence. Thanks for your help! – Jeff Aug 28 '12 at 6:13
Why do you need a $2^{-n}$-argument as there appears no sum? Wouldn't an $\frac{\epsilon}{2}$-argument suffice? – Alexander Frei Jun 10 '14 at 4:17

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 $N:=\lceil\frac{\epsilon}{2}\rceil$

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