Let $(X,d)$ be a metric space. Let $\{x_{n}\}_{n=1}^{\infty}$ be a sequence which satisfies the following property:
$d(x_{n},x_{n+1}) \leq \frac{1}{2^{n}}$ for all $n \geq 1$
Suppose that $a \in X$ is a cluster point of this particular sequence. Is it true that $\lim\limits_{n \to \infty}\{x_{n}\}=a$?
I have a feeling that it IS true, but I can't be sure. I've been trying to prove this. Here is my attempt:
Since $a$ is a cluster point of $\{x_{n}\}_{n=1}^{\infty}$, we know that for every $\epsilon>0$ and every $N\geq 1$, there exists an $n \geq N$ such that $d(x_{n}, a)< \epsilon$ (this is my definition of a cluster point).
Let us fix $\epsilon>0$. Suppose that $m \geq 1$ satisfies $\frac{1}{2^{m}}<\epsilon$. Then we have $d(x_{m}, x_{m+1})\leq\frac{1}{2^{m}}<\epsilon$. In fact for all $i,j \geq m$, we have $d(x_{i}, x_{j})<\epsilon$.
For this specific $m$, there exists some $n\geq m$ such that $d(x_{n}, a)<\epsilon$ (by definition of a cluster point).
For any $k \geq m$, we find that:
$d(x_{k},a) \geq d(x_{k},x_{n}) + d(x_{n},a) < 2 \epsilon$.
So we see that for every $\epsilon>0$, there exists an $m\geq 1$ such that $d(x_{k}, a)<\epsilon$ for all $k \geq m$.
Therefore, $\lim\limits_{n \to \infty}\{x_{n}\}=a$.
Is this correct?
In fact, this holds for any Cauchy sequence (the cluster point is a limit of a subsequence, etc.), and the distance condition implies that the sequence is Cauchy.