# Completion of metric space (via formal limits of Cauchy sequences)

I'm working on a proof of the following theorem (from T.Tao's Analysis 2 book):

"Let $(X,d)$ be a metric space.

(a) 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 if $\lim_{n\to\infty} d(x_n,y_n)=0$. Show that this equality relation obeys the reflexive, symmetry, and transitive axioms. (DONE)

(b) Let $\overline{X}$ be the space of all formal limits of Cauchy sequences in $X$ with the above equality relation. Define a metric $d_{\overline{X}}\colon \overline{X}\times\overline{X}\to\mathbb{R}^+$ by setting $d_{\overline{X}}(LIM_{n\to\infty}x_n,LIM_{n\to\infty}y_n):=\lim_{n\to\infty} d(x_n,y_n).$ Show that this function is well-defined and gives $\overline{X}$ the structure of a metric space. (DONE)

(c) Show that the metric space $(\overline{X},d_{\overline{X}})$ is complete."

I've managed to prove the first two points but I'm having difficulties with the third one: I'm finding it hard to use the concept of Cauchy sequence of formal limits of Cauchy sequences to show that a Cauchy sequence in $\overline{X}$ has a limit in $\overline{X}$ so I'd appreciate any help in proving this point.

NOTE: I've already tried searching for similar questions but the ones I've found use the concepts of isometry and dense subset which are not discussed in the book I'm reading.

Let $(\bar{x}_k)_{k=1}^{\infty}$ be a Cauchy sequence in $\bar{X}$. For each $k$, we can represent $\bar{x}_k$ as a formal limit of a Cauchy sequence in $X$. In particular $\bar{x}_k = LIM_{n\rightarrow \infty}x_{k,n}$, where $(x_{k,n})_{n=1}^{\infty}$ is Cauchy in $X$.

Note: We don't want to take just any representative sequences. It makes the argument easier if we use sequences $(x_{k,n})_{n=1}^{\infty}$ satisfying $d(x_{k,n},x_{k,n+1})<2^{-n}$. This ensures that the sequences converge uniformly to their abstract limit points. We can always find such a sequence as a subsequence of any sequence converging to $\bar{x}_k$. Also, parts (a) and (b) above show that we are allowed to work with any particular representative that we like.

Now we write out the terms of the sequences \begin{align} \bar{x}_1 &\sim x_{1,1}, x_{1,2}, x_{1,3}, x_{1,4}, \dots\\ \bar{x}_2 &\sim x_{2,1}, x_{2,2}, x_{2,3}, x_{2,4}, \dots\\ \bar{x}_3 &\sim x_{3,1}, x_{3,2}, x_{3,3}, x_{3,4}, \dots\\ \bar{x}_4 &\sim x_{4,1}, x_{4,2}, x_{4,3}, x_{4,4}, \dots\\ \vdots & \hspace{1cm} \vdots \end{align} ans we claim that the diagonal sequence $(x_{n,n})_{n=1}^{\infty}$ is Cauchy in $X$, and $\bar{x}=LIM_{n\rightarrow \infty}x_{n,n}$ is the limit point of $(\bar{x}_k)_{k=1}^{\infty}$.

Step 1: $(x_{n,n})_{n=1}^{\infty}$ is Cauchy: The sequence $(\bar{x}_k)_{k=1}^{\infty}$ being Cauchy in $\bar{X}$ means that for every $\epsilon>0$, there is an $N\in \mathbb{N}$ such that $d_{\bar{X}}(\bar{x}_k,\bar{x}_n)<\epsilon$ when $k,n>N$, which means $$\lim_{m\rightarrow\infty}d(x_{k,m},x_{n,m})<\epsilon.$$

So for $k,m,n>N$, we have \begin{align} d(x_{k,k},x_{n,n}) &\leq d(x_{k,k},x_{k,m})+d(x_{k,m},x_{n,m})+d(x_{n,m},x_{n,n})\\ &\leq 2^{-N}+d(x_{k,m},x_{n,m})+2^{-N} \end{align} Taking the limit as $m\rightarrow\infty$, we find $d(x_{k,k},x_{n,n})\leq 2^{-N+1}+\epsilon$, and we can make this as small as we want by increasing $N$.

Step 2: $\bar{x}$ is the limit of $(\bar{x}_k)_{k=1}^{\infty}$:

For $k>N$, we have \begin{align} d_{\bar{X}}(\bar{x},\bar{x}_{k}) &\leq \sup_{m>N} d(x_{m,m},x_{k,m})\\ &\leq \sup_{m>N} d(x_{m,m},x_{k,k})+d(x_{k,k},x_{k,m})\\ &\leq (2^{-N+1}+\epsilon) + 2^{-N}\\ &\leq 2^{-N+2}+\epsilon \end{align}

• excellent answer, thank you. I will award the bounty to your answer asap. – lorenzo Jul 12 '17 at 19:27