# 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.

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.

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 $$\operatorname{LIM}_n x_{n,n}$$.

• I have considered the exact thing but can't get to anywhere. Feb 13, 2020 at 18:39
• @Anonymous What would it mean to converge to this diagonal sequence? What does it mean that the sequence of sequences is Cauchy? Feb 13, 2020 at 19:37
• Oh and don't forget to show that $(x_{n,n})$ is in $\overline{X}$. The details of this have been worked out elsewhere on this site if you're really stuck but you should be perfectly capable of working out the details. Just start with the definitions and work slowly. @Anonymous Feb 13, 2020 at 19:48