The above answers sums it up pretty good, but I wanted to add something from my understanding of this topic.
For the pumping lemma to apply you must have a substring $y$ within your string $s$ that can be pumped, but that requires that $y$ is effectively some string other than the empty string, or else you couldn't be able to pump it in the first place.
Recall that $y$ represents a loop inside the finite deterministic automaton that accepts the language, so if for your initial string $s$ you have $|y| = 0$ then there is no loop at all, which is not the same as avoiding the loop and going straight to the accept state (when $i=0$).
Maybe the order the conditions are given is misleading, because you first gotta check for conditions 2 and 3 in order to pump $y$, i.e. check condition 1.