Cauchy sequence where the set of outputs is finite

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!

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 $|a_n-a_m| \ge d>\epsilon.$

• I guess I'm still confused as to how to apply the Cauchy definition to $\epsilon < d$. – lia.riley Sep 16 '18 at 0:26
• @lia.riley I have tried for an explanation in the part called "Added." – coffeemath Sep 16 '18 at 13:42

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.

• Great answer, I was going to suggest the same thing. Two things: the $\frac{1}{2}$ factor in the definition of $\varepsilon$ isn't necessary, and at some point a contradiction will be helpful. – Aweygan Sep 15 '18 at 23:19
• @Aweygan I agree that using contradiction can be useful at the end. I edited my answer. Thanks for commenting. – Ernie060 Sep 15 '18 at 23:24
• If the sequence is constant, then $d=0$. Since we don't want $\epsilon = 0$, we need to be consider this case separately. Thanks to coffeemath for pointing this out in his answer. – Ernie060 Sep 15 '18 at 23:32
• But how does this imply the sequence is constant from some point on? – lia.riley Sep 16 '18 at 23:58
• Let $\epsilon = \frac{d}{2}$. By definition of the Cauchy sequence, there is a $N\in\mathbb{N}$ such that for all $m,n \geq N$ we have that $|a_m - a_n|<\epsilon$. But if $a_m$ and $a_n$ differ less than the minimal distance between different values, then all the $a_n$ have to be constant for $n \geq N$. – Ernie060 Sep 17 '18 at 7:53