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?


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 $x$.

  • $\begingroup$ 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? $\endgroup$ – Jeff Aug 28 '12 at 3:38
  • $\begingroup$ @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 $i,j\ge i(m_1)$. $\endgroup$ – Brian M. Scott Aug 28 '12 at 3:40
  • $\begingroup$ Ah yes, that's the way you chose the sequence. Thanks for your help! $\endgroup$ – Jeff Aug 28 '12 at 6:13
  • $\begingroup$ Why do you need a $2^{-n}$-argument as there appears no sum? Wouldn't an $\frac{\epsilon}{2}$-argument suffice? $\endgroup$ – C-Star-W-Star 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 $N:=\lceil\frac{\epsilon}{2}\rceil$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.