The terms of a sequence X_n take on only finitely many values a_1,...,a_k.
That is, for every n, X_n = a_i for some i (the index i depending on n). Prove X_n has a cluster point.
However, $\displaystyle X_n$ will always have a subsequence with a limit point.
Proof: Let $\displaystyle A=\{a_1,...,a_k\}$. Because there are only finitely many points, $\displaystyle A$ is compact, and thus $\displaystyle X_n$ is a sequence in a compact set. So it follows from the Bolzano-Weierstrass Theorem that $\displaystyle X_n$ has a convergent subsequence (and therefore a limit point in that subsequence).
Yes, but that is irrelevant to the question. Any such convergent subsequence is "eventually constant". The definition of "cluster point" is that p is a cluster point if, given any $\displaystyle \epsilon> 0$, there exist an infinite] number of points other than p itself whose distance to p is less than $\displaystyle \epsilon$. If the sequence has only a finite number of points then the set of distances a given point p and points in the sequence, other than p itself, is finite and has a non-zero minimum. Let [itex]\epsilon[/itex] be that minimum distance. Given any p, there exist no other point whose distance is less than $\displaystyle \epsilon$. A finite set does NOT have a cluster point and so no finite valued sequence has a cluster point.
OK. I was treating $\displaystyle x_n$ as a distinct point from $\displaystyle x_m$, even if they both had the same value $\displaystyle a_i$. I was thinking that if $\displaystyle x_n\to x$, then for all $\displaystyle n$ greater than some $\displaystyle N$, $\displaystyle d(x_n,x)=0<\epsilon$ and $\displaystyle \{x_N,x_{N+1},...\}$ was the infinite set (even though all members of that set have the same value). It was a definition mix-up on my part.