Cauchy sequence $a_{n}$ that has finitely many different values Prove if Cauchy sequence $a_{n}$ that has finitely different values,
there must exist $n' \in N$ such that whenever $n>n'$, $a_{n}=c$ for some real c.
Approach) I did start by letting $\epsilon=min||a_{n}-a_{m}||$ for $n\neq m$. 
However, this way of defining epsilon does not exclude the case $\epsilon =0$, since there might be some $n \neq m $ such that $a_{n}=a_{m}$.
Can anyone give me a hint ?  
 A: Why not look at the $\max$, and then start trimming the sequence?
Namely, there are finitely many values on the set
$$X_{(a_n)}:=\{|a_i-a_j| \mid i,j \in \mathbb{N}\}.$$
If $\max X_{(a_n)}$ is zero, you are done with $n'=1$. If not, it is greater than zero, and you can pick $\epsilon$ a smaller number than such a maximum, but still greater than $0$. Since the sequence is Cauchy, there is a moment $N$ after which the distances are less than that $\epsilon$, and $X_{(a_n)_{n \geq N}}$ will be your last $X$ with at least one less element. Repeat this until you exhaust all non-zero elements.
If you want to keep with the $\min$ (which is essentially just skipping a lot of the initial iterations of the above argument), then you can consider $\min (X_{(a_n)} \backslash\{0\})$. This is obviously greater than zero. Pick $\epsilon$ between zero and such minimum. Since the sequence is Cauchy, there must be a moment $n'$ such that if $i,j>n'$, then $|a_i-a_j| <\epsilon$. But since $|a_i-a_j|$ belongs to $X_{a_n}$ and by construction of $\epsilon$ we know that $|a_i-a_j|$ is not on $X_{(a_n)}\backslash \{0\}$ (when $i,j>n'$, of course), it must follow that $|a_i-a_j|$ is zero.
A: Let $\{a_n:n\in \Bbb N\}=S$ be a finite set. For $t\in S$ let $V(t)=\{n\in \Bbb N:a_n=t\}.$  Since $S$ is finite  and $\cup_{t\in S}V(t)=\Bbb N,$ at least one $V(t)$ is infinite.
(i).  If $V(t)$ is infinite, and if $V(t')$ is finite for all $t'\in S\setminus \{t\}$ then the set $A=\cup\{V(t'):t\ne t'\in S\}$ is a finite subset of $\Bbb N$ so there exists $M\in \Bbb N$ such that $\forall n>M\;(n\not \in A).$ So $\forall n>M\;(n\in V(t))$ , so $\forall n>M\;(a_n=t).$
(ii). If $V(t')$ and $V(t'')$ are infinite with $t'\ne t''$  let $\epsilon =|t'-t''|/2.$ For any $n\in \Bbb N$ there exists $n'>n$ and $n''>n$ with $n'\in V(t')$ and $n''\in V(t''),$ but also with $|a_{n'}-a_{n''}|>\epsilon,$ so the sequence $(a_n)_{n\in \Bbb N}$ is not Cauchy.
