Let $\left( a_{n}\right) _{n}$ be a Cauchy sequence such that $\left\{ a_{n}:n\in \mathbb{N} \right\}$ is finite.

Question

Let $\left( a_{n}\right) _{n}$ be a Cauchy sequence such that $\left\{ a_{n}:n\in \mathbb{N} \right\}$ is finite. Show that $\left( a_{n}\right) _{n}$ is constant.

Answer 1

Let us denote $A=\{|a_n-a_k|:n,k\in \mathbb{N}\}\setminus\{0\}$. Since $\{a_n:n\in \mathbb{N}\} $ is finite $A$ is finite as well which follows there is a minimum in $A$. Denote $m=min\{a\in A\}$. Note that $m>0$ because $0\not \in A$. $\{a_n\}$ is cauchy which implies that there exists $n_0\in \mathbb{N}$ s.t. $\forall n>n_0$ $0\leq|a_{n+1}-a_n|<m$. If $0<|a_{n+1}-a_n|<m$ it's a contradiction to the definition of $m$. So we have $\forall n>n_0$ $a_n=a_{n_0}$.

Answer 2

We assume the space is complete e.g. $a_n \in \mathbb{R}$

if $\left\{ a_{n}:n\in \mathbb{N} \right\}^2$ is finite then so is $\left\{ |a_{n} - a_{m}|:n,m\in \mathbb{N} \right\}$

hence $\left\{ a_{n}:n\in \mathbb{N} \right\}$ has a minimum separation $\epsilon$

$\left\{ a_{n}:n\in \mathbb{N} \right\}$ is Cauchy and so eventually a tail will all be within $\epsilon/2$ of each other, so this tail is constant $\square$