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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!

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By its very definition, it is a limit. – MathApprentice Aug 14 '13 at 9:06
So the metric function defined is incorrect? – Ayush Khaitan Aug 14 '13 at 9:08
Consider your base space $M$. Then your equivalence classes of Cauchy sequences will be the points of the completion $M^*$ of $M$, and $d^*$ will be the inherited metric. This is, for instance, one of the two most common ways to construct the real numbers from the rationals. – Arthur Aug 14 '13 at 9:19
The metric function $d^\ast$ (it does not mention $\lim_{n\to\infty}a_n$ or $\lim_{n\to\infty}b_n$) is correct. – GEdgar Aug 14 '13 at 13:22
up vote 5 down vote accepted

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.

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oh ok. So basically $\lim_{n\to\infty}d(a_n,b_n)\neq d(\lim_{n\to\infty}a_n,\lim_{n\to\infty}b_n)$ – Ayush Khaitan Aug 14 '13 at 9:31
It is, but only if the RHS exists. – walcher Aug 14 '13 at 9:32

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.

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