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Maybe there are some textbooks which might treat cauchyness by taking double limits...

My question:
Is it sufficient and necessary to consider the double limit: $$x_n\quad \text{cauchy}\quad \Leftrightarrow \quad \lim_{(m,n)}d(x_m,x_n)=0$$ Moreover, if so, does it suffice as well to consider the limits successively and interchanged: $$\lim_m\lim_n d(x_m,x_n)=0 \quad\text{and}\quad \lim_n\lim_m d(x_m,x_n)=0$$ Thanks in advance. Cheers, Alex.

My Answer:
Ok, so far, interchanging limits give the same since: $$\lim_{m}\lim_{n}d(x_m,x_n) \stackrel{d(x,y)=d(y,x)}{=} \lim_{m}\lim_{n}d(x_n,x_m) \stackrel{m\leftrightarrow n}{=} \lim_{n}\lim_{m}d(x_m,x_n)$$ So there are the following two hypothesis still open: $$x_n\quad \text{cauchy}\quad \Leftrightarrow \quad \lim_{(m,n)}d(x_m,x_n)=0$$ $$\lim_{(m,n)}d(x_m,x_n)=0 \quad \Leftrightarrow \quad \lim_{m}\lim_{n}d(x_m,x_n)$$

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  • $\begingroup$ instead of saying "many" you should say "exist" $\endgroup$
    – janmarqz
    Commented Jan 9, 2014 at 17:51
  • $\begingroup$ Hmm, as far as I remember I saw it some times in some proofs, but I will correct it just be sure ;-) $\endgroup$ Commented Jan 9, 2014 at 18:04
  • $\begingroup$ ok.. for your problem you also should use the $\varepsilon-N$-definition of "Cauchy-ness" $\endgroup$
    – janmarqz
    Commented Jan 9, 2014 at 18:12
  • $\begingroup$ well im considering uniform space and nets ...but in principle the definition is basically the same (as far as I know) as for metric spaces and sequences together with the $\epsilon-N$-definition $\endgroup$ Commented Jan 10, 2014 at 9:31
  • $\begingroup$ Oh I just realized I was treating a metric space in my question, so lets for simplicity consider metric space - uniform space comes next ;-) $\endgroup$ Commented Jan 10, 2014 at 9:39

1 Answer 1

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If $(x_\alpha)_{\alpha\in A}$ is a Cauchy net, for every $\varepsilon > 0$, there is an $\alpha(\varepsilon)$ such that $\beta,\gamma \geqslant \alpha(\varepsilon) \Rightarrow d(x_\beta, x_\gamma) < \varepsilon$. Then for each $\beta$, the net $\gamma \mapsto d(x_\beta,x_\gamma)$ is a Cauchy net in $\mathbb{R}$, hence its limit exists. Also the net $\beta \mapsto \lim\limits_{\gamma} d(x_\beta,x_\gamma)$ is a Cauchy net in $\mathbb{R}$ hence its limit exists as well: For $\beta \geqslant \alpha(\varepsilon)$, we have $0 \leqslant \lim\limits_\gamma d(x_\beta,x_\gamma) \leqslant \varepsilon$, hence

$$\lim_\beta\lim_\gamma d(x_\beta,x_\gamma) = 0.$$

Conversely, if $(x_\alpha)_{\alpha\in A}$ is a net such that for every (large enough) $\beta\in A$ the limit $\lim\limits_\gamma d(x_\beta,x_\gamma)$ exists, and $\lim\limits_\beta \lim\limits_\gamma d(x_\beta, x_\gamma) = 0$, then given $\varepsilon > 0$, there is a $\beta(\varepsilon)$ such for all $\beta \geqslant \beta(\varepsilon)$ we have $\lim\limits_\gamma d(x_\beta,x_\gamma) < \varepsilon/2$. In particular, $\lim\limits_\gamma d(x_{\beta(\varepsilon)},x_\gamma) < \varepsilon/2$, so there is a $\gamma(\varepsilon,\beta(\varepsilon))$ such that $d(x_{\beta(\varepsilon)},x_\gamma) < \varepsilon/2$ for all $\gamma \geqslant \gamma(\varepsilon,\beta(\varepsilon))$. Choose $\alpha(\varepsilon) = \gamma(\varepsilon,\beta(\varepsilon))$. Then, for $\beta,\gamma \geqslant \alpha(\varepsilon)$ we have

$$d(x_\beta,x_\gamma) \leqslant d(x_{\beta(\varepsilon)},x_\beta) + d(x_{\beta(\varepsilon)},x_\gamma) < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon,$$

so $(x_\alpha)_{\alpha\in A}$ is a Cauchy net.

To complete everything, we need to see that $\lim\limits_{(\beta,\gamma)} d(x_\beta,x_\gamma) = 0$ is equivalent to $(x_\alpha)$ being a Cauchy net.

If $(x_\alpha)_{\alpha\in A}$ is a Cauchy net, then for $(\beta,\gamma) \geqslant (\alpha(\varepsilon),\alpha(\varepsilon))$ we have $d(x_\beta,x_\gamma) < \varepsilon$, therefore $\lim\limits_{(\beta,\gamma)} d(x_\beta,x_\gamma) = 0$. Conversely, if $\lim\limits_{(\beta,\gamma)} d(x_\beta,x_\gamma) = 0$, then for every $\varepsilon > 0$ there is a pair $(\beta(\varepsilon),\gamma(\varepsilon))$ such that $(\beta,\gamma) \geqslant (\beta(\varepsilon),\gamma(\varepsilon)) \Rightarrow d(x_\beta,x_\gamma) < \varepsilon$. Then we can choose (by directedness) an $\alpha(\varepsilon)$ with $\alpha(\varepsilon) \geqslant \beta(\varepsilon)$ and $\alpha(\varepsilon) \geqslant \gamma(\varepsilon)$, and for $\beta,\gamma \geqslant \alpha(\varepsilon)$ it follows that $d(x_\beta,x_\gamma) < \varepsilon$, hence $(x_\alpha)$ is a Cauchy net.

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  • $\begingroup$ Very nice -especially the sscond assertion very elegant solved - Thanks alot!!! =) $\endgroup$ Commented Jan 10, 2014 at 11:14
  • $\begingroup$ Hope you don't mind if edit your answer. $\endgroup$ Commented Jan 10, 2014 at 12:06
  • $\begingroup$ I don't mind that, but I disagree with your last one. Unless I'm missing something, we also need $\alpha(\varepsilon) \geqslant \beta(\varepsilon)$ there, and we need the directedness for the existence of an $\alpha$ that is $\geqslant$ both, $\beta$ and $\gamma$. $\endgroup$ Commented Jan 10, 2014 at 12:11
  • $\begingroup$ Just read the line before the definition of $\alpha$. The definition of $\beta(\varepsilon)$ is only depending on $\varepsilon)$ $\endgroup$ Commented Jan 10, 2014 at 12:16
  • $\begingroup$ Ah, right, we can choose $\alpha(\varepsilon)$ as $\gamma$ indeed. Good spot. $\endgroup$ Commented Jan 10, 2014 at 12:22

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