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Let be

  • $(M,d)$ a metric space,
  • $(x_{k\ell})_{\ell\in\mathbb{N}}$ Cauchy sequences in $M$ (one for each $k\in\mathbb{N}$),
  • for any $\varepsilon>0$ there is $N$ such that $k,\ell\ge N\to d(x_{k\ell},x_{\ell\ell})<\varepsilon$,
  • NEW $(x_{\ell\ell})_{\ell\in\mathbb{N}}$ is Cauchy (the diagonal sequence)

How do I deduce formally the following double limit? $$\tag{*} \lim_{k\to\infty}\lim_{\ell\to\infty}d(x_{k\ell},x_{\ell\ell})=0. $$ I could just state the limit in (*), but I'm not totally convinced.

Mohamed Ali
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  • did you try, for fixed $k$, to study the convergence of the sequence $\big(d(x_{kl},x_{ll})\big)_{l \in \mathbb{N}}$ ? – Olivier Roche Sep 27 '19 at 08:54
  • @OlivierRoche Do you mean that I can show that it's a Cauchy sequence in $\mathbb{R}$ and thus converge to lets say $(\alpha_k)$? – Mohamed Ali Sep 27 '19 at 09:28
  • Yes, that's what I was hoping for. I didn't check, though. – Olivier Roche Sep 27 '19 at 09:31
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    What’s the definition of a double limit? Is it $\lim_{k \to \infty} \left( \lim_{l \to \infty} d(x_{kl}, x_{ll}) \right) = 0$ if and only if $ \forall \epsilon>0, \exists N \in \mathbb{N}, \forall k>N, \lim_{l \to \infty} d(x_{kl}, x_{ll}) < \epsilon$ – Joe Sep 27 '19 at 10:23
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    @Joe I think so, but this doesn't make a sense unless the sequences $\big(d(x_{kl},x_{ll})\big)_{l \in \mathbb{N}}$, $k \in \mathbb{N}$ converge. – Olivier Roche Sep 27 '19 at 11:32
  • @Olivier Roche, oh, I see, the issue is that the distance are between elements of different Cauchy sequences, right? – Joe Sep 27 '19 at 13:29
  • @Joe Yes, that's the problem. – Olivier Roche Sep 27 '19 at 13:34
  • This smells to me like it requires some sort of a triangle inequality, where for any $\epsilon>0$ you pick N based on statement 3 so that the distance is less than $\epsilon/2^i$, then for each $k>N$ say the $x_{kl}$ converges to something since it is Cauchy, and show that the limit of $x_{kl}$ can be made arbitrarily close to the limit of $x_{ll}$ for large enough k. Unfortunately, I don’t have time to try to see if that idea is fruitful. – Joe Sep 27 '19 at 13:50
  • I added a condition and started an answer (still incomplete) as wiki. The afternoon in the mountains brought some fresh air. For today that's it. – Mohamed Ali Sep 27 '19 at 18:10

1 Answers1

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I forgot to mention that $(x_{\ell\ell})_{\ell\in\mathbb{N}}$ is Cauchy.

Then we have $$ d(x_{k\ell},x_{\ell\ell})\le d(x_{k\ell},x_{k\ell'})+d(x_{k\ell'},x_{\ell'\ell'})+d(x_{\ell'\ell'},x_{\ell\ell}),\\ \to\qquad d(x_{k\ell},x_{\ell\ell})-d(x_{k\ell'},x_{\ell'\ell'})\le d(x_{k\ell},x_{k\ell'})+d(x_{\ell'\ell'},x_{\ell\ell}) $$ Similarly $$ d(x_{k\ell'},x_{\ell'\ell'})-d(x_{k\ell},x_{\ell\ell})\le d(x_{k\ell'},x_{k\ell})+d(x_{\ell\ell},x_{\ell'\ell'}) $$ thus $$ \mid d(x_{k\ell'},x_{\ell'\ell'})-d(x_{k\ell},x_{\ell\ell})\mid \le d(x_{k\ell'},x_{k\ell})+d(x_{\ell\ell},x_{\ell'\ell'}) < \tfrac{\varepsilon}{2}+\tfrac{\varepsilon}{2}=\varepsilon, $$ for $\ell,\ell'>N$ since both are Cauchy.

We have proved that $(d(x_{k\ell},x_{\ell\ell}))_{\ell\in\mathbb{N}}$ is Cauchy in $\mathbb{R}$, thus converges to some $\alpha_k$.

Now we have to prove that $$ \lim_{k\to\infty}\alpha_k=0, $$ where $$ \alpha_k=\lim_{\ell\to\infty}d(x_{k\ell},x_{\ell\ell}). $$

As per comment of @Joe: Given the third condition, for any $\varepsilon>0$ there is $N$ such that $k,\ell\ge N\to d(x_{k\ell},x_{\ell\ell})<\varepsilon$; therefore we have for $k\ge N$ $$ \forall \ell\ge N,\ d(x_{k\ell},x_{\ell\ell})<\varepsilon \quad\to\quad \lim_{\ell\to\infty} d(x_{k\ell},x_{\ell\ell})<\varepsilon \quad\to\quad \alpha_k<\varepsilon, $$ and the claim, $\lim_{k\to\infty}\alpha_k=0$, follows.

Note 1. The forth condition follows from the first 3, which are thus sufficient.

Note 2. The fourth consition is easier to see if we go back to the theorem of which this lemma is part of, the completion of a metric space theorem; I proven that $(x_{\ell\ell})$ is Cauchy in a previous step because I needed to show that $[(x_{\ell\ell})]$ is an element of the completion $(M',d')$ of $(M,d)$.

Note 3. The inequality used above should be the quadrilateral inequality, see Completing metric space .

Mohamed Ali
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  • Condition 3 finishes your proof, right? $\forall \epsilon > 0, \exists N \in \mathbb{N}, \forall k,\ell>N, d(x_{k\ell}, x_{\ell\ell}) < \epsilon$, which implies $\alpha_k < \epsilon$. – Joe Sep 28 '19 at 05:30
  • @Joe Thank you, I added your comment to the answer – Mohamed Ali Sep 28 '19 at 06:38
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    nice proof! I was going to try something different, but I couldn’t do it without the metric space being complete. I thought about trying to form the completion to use in the proof, but your proof is much better. – Joe Sep 28 '19 at 06:47