# Prove the following metric space is complete

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, $\forall i \in \mathbb{N}, \lbrace x_n^i \rbrace _{n=1}^\infty \overset{n \rightarrow \infty}{\longrightarrow} L_i < \infty$.



1. 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.

• Your proof is okay, just remember not to use $i$ as an index right after you have fixed it as in "Let $i\in\mathbb{N}$ be arbitrary...". – SEBASTIAN VARGAS LOAIZA Jul 26 '18 at 22:56

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 $\lim_{n\to\infty}x(n)=(x_k)_{k\in\mathbb N}$.
• The distance $\overline d(a,b)$ is the usual distance if the numbers $a$ and $b$ are close to each other and $1$ otherwise. As far as limits (or being a Cauchy sequence) are concerned, this is the same thing as the usual didtane, but it cann never be greater than $1$. And the distance $D$ between two sequences $(a_n)_{n\in\mathbb N}$ and $(b_n)_{n\in\mathbb N}$ is small if and only if the distance between the first terms of both sequences is small. – José Carlos Santos Jul 24 '18 at 17:19
• @ikoikoia It sure would. For instance, neither $\overline d$ nor $D$ would be metrics. – José Carlos Santos Jul 24 '18 at 17:35