Double limit in metric space 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.
 A: 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 .
