Since this question was tagged as computability, I am assuming that in the definition of effectively enumerable, the function $f : \omega \rightarrow \omega$ is a total computable function.
In computability theory, the usual definition of a computably enumerable (c.e.) set is the domain of a partial computable function.
It can be shown by so called "Listing Theorem" that a set $A$ is c.e. if and only if $A = \emptyset$ or $A$ is the range of a total computable function.
So the $\emptyset$ is a c.e. set. However, it is not the range of any total computable function $f: \omega \rightarrow \omega$, since the range of such function can not be empty. Therefore, the "Listing Theorem" characterization of c.e. sets separately includes the $\emptyset$ as a c.e. set. (Of course, $\emptyset$ is c.e. if you use the original definition of being the domain of a partial computable function.)
Intuitively, computable set are those that whose membership can be computed, i.e. there is an algorithm that halt on all inputs and tells you correctly the membership. Computably enumerable sets are those that possess that an algorithm that will eventually halt if $n \in A$ and not halt if $n \notin A$. So in some sense, the definition using domain of partial computable functions adhere more closely to this intuition than "listable" since $\emptyset$ does not quite work. Of course, no one would deny that $\emptyset$ is computable, and every computable set should be c.e.