Originally Posted by
redsoxfan325 d.) Assuming this sequence is in the metric space $\displaystyle (M,d)$, if $\displaystyle \{b_n\}$ converges, then it is bounded. ($\displaystyle \exists x_0\in M, R\in\mathbb{R}$ such that $\displaystyle d(x_0,b_n)<R~\forall n$)
So $\displaystyle d(a_n,x_0)-R<d(a_n,b_n)~\forall n$. (Triangle Inequality)
But since $\displaystyle \{a_n\}$ is unbounded, so is $\displaystyle d(a_n,x_0)$, because $\displaystyle x_0$ is a fixed point. Therefore $\displaystyle d(a_n,b_n)$ is unbounded as well.
@ Drexel: $\displaystyle x_n$ needs to converge.