Consider $(X,d)$ a pseudometric space and $T$ as the set of all Cauchy's sequences in $X$. We can define $\hat{d}:=\lim_{n\rightarrow\infty}d(x_n,y_n)$ for $x,y\in T$, whichs makes to $T$ be a pseudometric space. I need to prove that $(T,\hat{d})$ is complete. So, consider $\{x^n\}_{n\in\omega}$ be a Cauchy's sequence in $T$ (that is, $x^n=(x^n_1,x^n_2,\dots,x^n_n,\dots)$ is a Cauchy's sequence for each $n\in \omega$). I think that the sequence must converger to a subsequence of the sequence $\{x_n^n\}_{n\in \omega}$ but I can't obtein any satisfactory way to show it.
First, am I correct trying to show that the sequence converge to a subsequence of the diagonal? If the answer is positive, can someone give a hint for to show it?
If you know the construction of the completion of a metric space, using equivalence classes of Cauchy sequences, then you can just mimic that proof.
And you'll see that although it is a variant of a diagonal argument, it is not strictly speaking the diagonal sequence $(x^n_n)$ that one uses: you have to pick the right entry from $x^1$; then the right entry from $x^2$; and so on.
In the link provided they do use $(x^n_n)$, but only after passing to a subsequence of the sequence $(x^n)$.