What exactly is $\lim_{n\to\infty}a_n$? Is $\lim_{n\to\infty}a_n$ a term of the cauchy sequence $\{a_i\}$, or the limit? I suppose it can't be both. I am leaning towards limit because if we select any $a_k\in\{a_i\}$, we have $a_j\in\{a_i\},j>k$. And any member of the sequence can be selected thus.
Motivation: 
In the metric space of equivalence classes of cauchy sequences, [$\{a_i\}\sim\{b_i\}$ iff $\forall\epsilon\in\Bbb{R}, \exists N\in\Bbb{N}$ such that $d(a_p,b_q)<\epsilon\forall p,q>N$] the metric $d^*$ is defined thus: $d^*(\{a_i\},\{b_i\})=\lim_{n\to\infty}d(a_n,b_n)$. If $\lim_{n\to\infty}a_n$ and $\lim_{n\to\infty}b_n$ are indeed limits and not points of the sequence, how can we even be sure that such limits exist in the metric space?
Note that it has nowhere been mentioned that the metric space is complete. 
Thanks in advance!
 A: In this definition there is nowhere used a limit of $a_n$ nor $b_n$ which indeed might not exist in given space. Instead it uses $\lim_{n\to\infty}d(a_n,b_n)$, and $d(a_n,b_n)$ isn't an element of given metric space, but element of $\mathbb R$ (as metric is defined as a function from given space to $\mathbb R$). As $a_n$ and $b_n$ are Cauchy sequences, this limit is well defined.
A: Note that $\{d(a_n,b_n)\}$ is a sequence in $\Bbb R$, which is a complete metric space, so to see that $\lim_{n\to\infty}d(a_n,b_n)$ exists it suffices to prove that the sequence is Cauchy.
Choose $N$ such that for $n, m \gt N$, $$d(a_n, a_m) \lt {\varepsilon \over 2}, d(b_n, b_m) \lt {\varepsilon \over 2}.$$ Then by the triangle inequality we get: $$d(a_n, b_n) \le d(a_n, a_m) + d(b_n, b_m) + d(a_m, b_m).$$
Swapping $n$ and $m$ will give us $$|d(a_n, b_n) - d(a_m, b_m)| \le d(a_n, a_m) + d(b_n, b_m) \lt \varepsilon,$$ hence $\{d(a_n,b_n)\}$ is Cauchy in $\Bbb R$ and $\lim_{n\to\infty}d(a_n,b_n)$ exists.
Now, if you can show (and I will leave this up to you) that $\{c_i\} \sim \{a_i\}, \{d_i\} \sim \{b_i\}$ implies $d^*(\{a_i\}, \{ b_i\}) = d^*(\{c_i\}, \{ d_i\})$, then $d^*$ is well-defined.
