Why if the set of values of the sequence terms is finite then the sequence is eventually a constant?

Question

I am reading a proof of theorem that states that the sequence is Cauchy only if it's convergent. At the beginning of the proof it is stated that if the set of the values of sequence is finite then the sequence is ultimately a constant so from there it is convergent as well. For example a sequence ${(-1)^n}$ forms a finite set of values, but for a sequence to be Cauchy, shouldn't inequality $|a_n-a_m|<\epsilon$ be true for whatever epsilon? I don't see how this sequence is ultimately constant it if every second term is opposite of the first one.

Answer 1

There are sequences with only finitely many values that aren't eventually constant, as $(-1)^n$ shows.

But Cauchy sequences with only finitely many values are eventually constant. Indeed let $(u_n)$ be such a Cauchy sequence and let $x_1,...,x_k$ be the values it takes, we take the $x_i$ to be distinct. Let $e= \min_{i\neq j} |x_i-x_j|>0$ and $\epsilon = e/2$.

Then by Cauchy-ness, there is $N$ such that for $n,m\geq N$, $|u_n-u_m|<\epsilon$.

Let $i$ be such that $u_N = x_i$. Then for $n\geq N$, $|u_n - u_N|< \epsilon < e$. Let $k$ be such that $u_n = x_k$. Then if $k\neq i$, $|u_n-u_N| = |x_k-x_i| \geq e > |u_n-u_N|$, a contradiction. Therefore $x_k = x_i$. Therefore for all $n\geq N, u_n = x_i$ : $(u_n)$ is eventually constant.