Completion of metric space (Terence Tao Analysis II Exercise 1.4.8 part (c)) $\def\LIM{\operatorname{LIM}}$
Let $(X,d)$ be a metric space and given any cauchy sequence $(x_n)_{n=1}^{\infty}$ in $X$ we introduce the formal limit $\LIM_{n\to \infty}x_n$. We say that two formal limits $\LIM_{n\to \infty}x_n$ and $\LIM_{n\to \infty}y_n$ are equal iff $\lim_{n \to \infty}d(x_n,y_n)=0$. We then define $\bar{X}$ to be set of all the formal limits of Cauchy sequences in $X$. We define the metric $d_{\bar{X}}$ as follows: $$d_{\bar{X}}(\LIM_{n\to \infty}x_n,\LIM_{n\to \infty}y_n)= \lim_{n \to \infty} d(x_n,y_n)$$
I have proved that $(\bar{X},d_{\bar{X}})$ is indeed a metric space that that the definition of metric is well defined. But I am stuck to prove that $(\bar{X},d_{\bar{X}})$ is a complete metric space. This problem could be resolved without taking into account topological spaces as that concept in later in the book. Any suggestion on how to go about this problem without using machinery of topology would be highly invaluable. Thanks in advance.
 A: Generally, if you have a sequence of sequences, one thing to consider is the diagonal sequence:
\begin{array}{ccccc}
\color{purple}{x_{1,1}} & x_{1,2} & x_{1,3} & x_{1,4} & \dots \\
x_{2,1} & \color{purple}{x_{2,2}} & x_{2,3} & x_{2,4} & \dots \\
x_{3,1} & x_{3,2} & \color{purple}{x_{3,3}} & x_{3,4} & \dots \\
x_{4,1} & x_{4,2} & x_{4,3} & \color{purple}{x_{4,4}} & \dots \\
\vdots & \vdots & \vdots & \vdots & \ddots
\end{array}
You should be able to show that if $(\operatorname{LIM}_n x_{i,n})_{i = 1}^\infty$ is Cauchy then the limit is $\operatorname{LIM}_n x_{n,n}$.
A: Let $(LIM_{n\to\infty} x_{in})_{i=1}^\infty$ be a Cauchy sequence in $(\overline{X},d_{\overline{X}})$, where $(x_{in})_{n=1}^\infty$ is a Cauchy sequence in (X, d) for each i. Define the sequence $n(i)_{i=1}^\infty$ recursively as follows:$ n(1) := min\{N \geq 1: d(x_{1i},x_{1j}) \leq 1, \forall i,j \geq N\}. $ $ \forall k > 1, n(k) := min\{N > n(k-1): d(x_{ki}, x_{kj}) \leq 1/k, $ $\forall i, j \geq N\}.$ (That is, we take the smallest term in the first sequence after which the  distance of any pair is less than 1, then we take the smallest term from that term on in the second sequence after which the distance of any pair is less than 1/2 ,..., etc.) $n(i)_{i=1}^\infty$ is an increasing sequence. Let $(x_{k})_{k=1}^\infty := (x_{kn(k)})_{k=1}^\infty$.
Step 1: $(x_{k})_{k=1}^\infty$ is Cauchy: Let $N \geq 1$. By construction, $\forall i , j \geq N$, $d(x_{in(i)}, x_{jn(j)}) \leq d(x_{in(i)}, x_{in(j)}) $ $+$ $d(x_{in(j)}, x_{jn(j)}) \leq 1/N$ $+$ $o$(N), which can be bounded as small as desired by taking N sufficiently large.
Step 2: $LIM_{k\to\infty} (x_{k})_{k=1}^\infty$ is the limit: For any i, j, note that: $d_(x_{ij}, x_{jn(j)}) \leq d(x_{ij}, x_{in(j)})$ $+$ $d(x_{in(j)}, x_{in(i)})$ $+$ $d(x_{in(i)}, x_{jn(i)})$ $+$ $d(x_{jn(i)}, x_{jn(j)})$. Each of the four terms on the RHS can be bounded arbitrarily small by taking taking i, j sufficiently large. Hence we have: $lim_{i\to\infty} d_{\overline{X}}(LIM_{j\to\infty} x_{ij}, LIM_{j\to\infty} x_{j}) = lim_{i\to\infty} lim_{j\to\infty} d(x_{ij}, x_{jn(j)}) = 0$, as desired.
