Exercise 3.1 in Baby Rudin states:
Prove that convergence $\{s_n\}$ implies convergence of $\{|s_n|\}$. Is the converse true?
My question is not on how to do this, but on interpreting what Rudin was asking.
First, the chapter deals with sequences of complex numbers, so it seems safe to safe that $s_n \in \mathbb{C}$. Of course, the sequence of modulu will lie in $\mathbb{R} \subset \mathbb{C}$. For complex numbers, we do have an earlier fact from chapter $1$ that states that $$ ||x|-|y|| \leq |x-y|. $$ However, the more general definition of sequence converge doesn't use absolute value to denote distance, but the more general $d(x,q)$. That is, it states that for any $\epsilon > 0$, there exists $N$ so that $n \geq N$ implies $d(s_n, s) < \epsilon$.
The proofs of these fact seem to rely on using this absolute-value/Euclidean metric.
Am I missing something? Is this the only available metric on $\mathbb{C}$? Or is there a more general proof that does not rely on the fact I mentioned?