Question about a Cauchy sequence. prove that if $a_n$ is a Cauchy sequence, and the set {${ a_n | n\in \mathbb{N} }$} (which means the set of all values the seuqnece $a_n$ can have) is finite, then there is $N_0$ s.t for each $n&gt;N_0$ the sequence $a_n$ is constant. 

can someone give me a hint?
Since the given sequence is a Cauchy sequence, you have that for every positive number $\epsilon$, there is a positive integer $N$ s.t. for all natural numbers $m, \, n &gt; N$:
$$
|a_m - a_n| &lt; \epsilon
$$

Take $\epsilon$ to be the minimum distance between two distinct points, which exists because the set of pairs of distinct points is also finite. Now, take $N$ such that when $m, \, n &gt; N$, it is true that $|a_m - a_n| &lt; \frac{\epsilon}{2}$. But this means that $a_m = a_n$ by the way $\epsilon$ is defined.
Since the given sequence is a Cauchy sequence, you have that for every positive number $\epsilon$, there is a positive integer $N$ s.t. for all natural numbers $m, \, n &gt; N$:
$$
|a_m - a_n| &lt; \epsilon
$$

Take $\epsilon$ to be the minimum distance between two distinct points, which exists because the set of pairs of distinct points is also finite. Now, take $N$ such that when $m, \, n &gt; N$, it is true that $|a_m - a_n| &lt; \frac{\epsilon}{2}$. But this means that $a_m = a_n$ by the way $\epsilon$ is defined.