Cauchy sequence where the set of outputs is finite

Question

If $\{a_n\}$ is a Cauchy sequence, and $S = \{a_n |n\in\mathbb{N}\}$ is finite, then $\{a_n\}$ is constant from some point on.

The statement makes sense, but I'm not quite sure how to start. I feel like maybe contradiction, saying if it wasn't constant from some point on, $S$ couldn't be finite.

What I have so far:

Let $\{a_n\}$ be a Cauchy sequence, so for each $\epsilon > 0$, there exists an $n^{*}\in\mathbb{N}$ such that $| a_m -a_n| < \epsilon$ for all $m \geq n^*$ and all $n \geq n^*$.

Any help would be appreciated!

Answer 1

Consider $d:=\min\{|a_n - a_m|\mid \forall n,m\in \mathbb{N}\}$. Since $S$ is finite, the sequence $\{a_n\}$ only takes on finitely many different values, and so this minimum exists. Then let $\epsilon= \frac{d}{2}$ and use the definition of a Cauchy sequence together with the fact that $S$ is finite.

Edit: I should have added that the argument above holds if the sequence is not constant. If the sequence is constant, then there is nothing to prove.

Answer 2

If S has only one member the sequence is already constant. Otherwise:

Let $d$ be the minimum of $|a_n-a_m|$ over all pairs $n,m$ with $a_n \neq a_m$ [This exists and is positive since $S$ is finite.] Now apply Cauchy to some $\epsilon < d.$

Added: About applying Cauchy: Fix $0< \epsilon <d.$ Then by Cauchy there is $n_0$ such that for $n,m \ge n_0$ we have $|a_n-a_m|<\epsilon <d.$ By the way $d$ was defined, we cannot have[with $n,m \ge n_0$] $a_n \neq a_m,$ else $|a_n-a_m| \ge d>\epsilon.$