Prove the following metric space is complete

Question

I've been struggling with this question for many, hours. Even reading the professor's proof outline, and understanding parts of it, I'm struggling with completely reproducing the proof on my own (pun unintended). His outline is in a different language, so I am translating and filling in all the technical details. My goal in this exchange is proof verification.

Set $X= \lbrace (a_n)_{n=1}^{\infty} \mid \forall n \in \mathbb{N} :a_n \in \mathbb{R} \rbrace $, the set of all infinite real sequences.

Define a metric $D$ on $X$:

$D((a_n)_{n=1}^{\infty},(b_n)_{n=1}^{\infty}) := \underset{n}{\sup} \lbrace\bar d(a_n,b_n)/n\rbrace$

Where $\bar d(a,b):=\min \lbrace |a-b|,1\rbrace $, for any $a,b \in\mathbb{R}$.

Prove that $(X,D)$ is a complete metric space.

I know that I need to show that every Cauchy sequence in $(X,D)$ converges in $(X,D)$.

So, given a Cauchy sequence, $\bar x_n = \lbrace x_n^i \rbrace _{i=1} ^\infty$, I imagine setting up an infinite square matrix, where the $n^{th}$ column is the $n^{th}$ sequence in $\bar x_n$ and the $i^{th}$ row is a sequence of all the $i^{th}$ elements in each sequence.

1. First, I want to show that every row sequence $\lbrace x_n^i \rbrace _{n=1}^\infty $, is Cauchy. I'm not sure if I did this correctly.

Let $i \in \mathbb{N}$ be arbitrary. Since $\bar x_n$ is Cauchy, for every $\epsilon > 0$, there exists $N\in \mathbb{N}$ such that for any $m>n>N$:

$D(\bar x_n,\bar x_m)= \underset{i}{\sup} \lbrace\bar d(x_n^i , x_m^i)/i\rbrace = \underset{i}{\sup} \lbrace \min(|x_n^i-x_n^i|,1)/i\rbrace<\epsilon /i$.
It follows that $\forall m>n>N: |x^i_n - x^i_m|/i<\epsilon /i$.
$\implies |x^i_n - x^i_m|<\epsilon$, proving that $\lbrace x_n^i \rbrace _{n=1}^\infty $, is Cauchy in $\mathbb{R}$.
Since $(\mathbb{R},|\cdot|)$ is complete, $\forall i \in \mathbb{N}, \lbrace x_n^i \rbrace _{n=1}^\infty \overset{n \rightarrow \infty}{\longrightarrow} L_i < \infty$.

$ $

2. Next, define $L= \lbrace L_i \rbrace _{i=1}^\infty$. Showing that $ \underset{n \rightarrow \infty}{\lim {\bar x_n}=L}$ completes the proof.

Let $\epsilon >0$ be arbitrary. Set $K > \epsilon^{-1}.$ Since $\forall i \in \mathbb{N} : \underset{n \rightarrow \infty}{\lim} x_n^i=L_i $, there exists $N$ such that $\forall n>N$:

$\forall 1\leq i \leq K : |x_n^i-L_i| < i \epsilon \implies |x_n^i-L_i|/ i < \epsilon \implies \underset{ 1\leq i \leq K}{\sup} \lbrace \bar d(x_n^i , L_i)/i\rbrace < \epsilon$
$\forall i > K : i > \epsilon^{-1} \implies 1/i < \epsilon \implies \underset{i>K}{\sup} \lbrace \bar d(x_n^i , L_i)/i\rbrace < \epsilon$

Finally implying $D(\lbrace x_n^i \rbrace _{i=1}^\infty, \lbrace L_i \rbrace _{i=1}^\infty) < \epsilon \implies D(\bar x_n, L) < \epsilon$

Q.E.D
Wow, this took a very long time to write. ANY feedback would be appreciated.

Answer 1

Here's an outline:

1. Let $\bigl(x(n)\bigr)_{n\in\mathbb N}$ be a Cauchy sequence of elements of your space. In particular, for each $n\in\mathbb N$, $x(n)$ is a sequence $\bigl(x(n)_k\bigr)_{k\in\mathbb N}$ of real numbers.
2. For each $k\in\mathbb N$, prove that the sequence $\bigl(x(n)_k\bigr)_{n\in\mathbb N}$ is a Cauchy sequence of real numbers. Let $x_k$ be its limit.
3. Prove that $\lim_{n\to\infty}x(n)=(x_k)_{k\in\mathbb N}$.