Prove that if $a_{n}$ and $b_{n}$ are equivalent Cauchy sequences as well as $b_{n}$ and $c_{n}$, then $a_{n}$ and $c_{n}$ are equivalent Cauchy sequences.
MY ATTEMPT
Since $a_{n}$ and $b_{n}$ are equivalent, for every positve rational $\epsilon/2$, there is a natural number $N_{1}\geq 0$ such that $|a_{n} - b_{n}| \leq \epsilon/2$ for $n\geq N_{1}$. Similarly, there is a natural number $N_{2}\geq0$ such that $|b_{n} - c_{n}|\leq \epsilon/2$ for every $n\geq N_{2}$. Consequently, for every positive rational $\epsilon$, there in an $N = \max\{N_{1},N_{2}\}$ such that \begin{align*} |a_{n} - c_{n}| = |(a_{n} - b_{n}) - (b_{n} - c_{n})| \leq |a_{n} - b_{n}| + |b_{n} - c_{n}| \leq \epsilon/2 + \epsilon/2 = \epsilon \end{align*} for every $n\geq N$, and we are done.
Any comments or contributions to my attempt?
That is fine.
My only comment is that you do what I call "epsilon fiddling". In other words, to make your final bound exactly $\epsilon$, you use some multiple of $\epsilon$ in your proof so the final total is exactly $\epsilon$.
I prefer to just use $\epsilon$ at each stage and them combine them. In this case, if the bound in each case is $\epsilon$, the resulting bound is $2\epsilon$.
As far as I am concerned, any reasonable function of $\epsilon$ is good enough.