I am working through a Real Analysis book on my own and have come across the proof of the equivalence relation between two Cauchy sequences, and have a question regarding transitivity.
If we let $x_1,x_2,...$,$y_1,y_2,...$ and $z_1,z_2,...$ be Cauchy sequences of rationals such that the $x-y$ and $y-z$ pairs are equivalent, then we can use the triangle inequality as follows:
For some $\epsilon > 0$ we have an $m$ such that for all $k \geq m$ the following holds:
$|x_k - z_k| \leq |x_k - y_k| + |y_k - z_k| < \epsilon + \epsilon = 2\epsilon$
Now, the way I have seen it is to let both $|x_k - y_k|$ and $|y_k - z_k|$ be less than $\frac{\epsilon}{2}$ so that we have:
$|x_k - z_k| \leq |x_k - y_k| + |y_k - z_k| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$
I am confused with how we can freely change between these two methods. Is it because our $\epsilon$ is arbitrary and our $m$ is based off of this said $\epsilon$? Is there a simple algebraic step between the two I am missing, or is it just that our $\epsilon$ can be anything so we have just have to let our $m$ shift accordingly?
Given $\epsilon>0,$ they need $|x_k-z_k|<\epsilon,$ so they made $|x_k-y_k|$ and $|y_k-z_k|$ smaller than $\epsilon/2$ for convenience. This way at the end they get $|x_k-z_k|<\epsilon$ and not $|x_k-z_k|<2\epsilon.$
Here is the justification: For each $\epsilon>0,$ there exists $m$ such that for all $k\geq m$ we have $|x_k-y_k|<\epsilon$ and $|y_k-z_k|<\epsilon.$
Thus, given $\epsilon>0,$ we can let $\epsilon'=\epsilon/2.$ Then $\epsilon'>0$ as well, so there exists $m$ such that for all $k\geq m$ we have $|x_k-y_k|<\epsilon'=\epsilon/2$ and $|y_k-z_k|<\epsilon'=\epsilon/2.$