Suppose $M$ is a metric space, not necesserily complete.
Given the following definition of Cauchy sequence:
For every $\epsilon > 0$ there exists an open $\epsilon$-ball that contains all but a finite number of elements of the sequence.
consider the following equivalence relation:
def1. Cauchy sequences $\{a_i\}$ and $\{b_i\}$ are equivalent iff $\{a_0, b_0, a_1, b_1, a_2, b_2, \ldots\}$ is a Cauchy sequence.
I came up with a following alternative definition of the equivalence:
def2. $\{a_i\}$ and $\{b_i\}$ are equivalent iff for every $\epsilon > 0$ there exists an open $\epsilon$-ball that contains an infinite number of both $\{a_i\}$ and $\{b_i\}$.
Question: So, are def1 and def2 indeed equivalent?
It seems to me that they are not, but I can neither prove it nor think of a counterexample. Intuitively, def2 fails to address that there's a finite number of elements of $\{a_0, b_0, a_1, b_1, \ldots\}$ outside the ball. This observation helped me to came up with a counterexample to a similar statement:
$\{a_i\}$ and $\{b_i\}$ are equivalent iff (some ball contains infinite number of elements of $\{a_i\}$) always implies (this same ball contains an infinite number of elements of $\{b_i\}$).
The counterexample: $M = \mathbb{R}$, $a_i = (-\frac{1}{2})^i$, $b_i = -\frac{1}{2^i}$, and the ball $B_\epsilon$ could be taken to be $(0, 2\epsilon)$.
An example with multiple limit points could be a problem. However, it is easy to prove that a Cauchy sequence can have no more than 1 limit point, and both def1 and def2 require $\{a_i\}$ and $\{b_i\}$ to be Cauchy sequences.
I think that rather than "some ball contains an infinite number of elements of both ${a_i}$ and ${b_i}$," you want "some ball contains all but a finite number of ${a_i}$ and ${b_i}$. Notice that "containing an infinite number" is a weaker condition than "containing all but a finite number". In the reals, an $\epsilon$-ball around $1$ contains infinitely many elements of the sequence $(-1)^n$, but it certainly does not contain all but finitely many terms.