# Definition of effective enumerability and empty set

Let $S$ be a set. We say that $S$ is effectively enumerable iff (by definition) there exists a function $f \colon N \to N$ which has $S$ as codomain. My question is: is the empty set an effectively enumerable set by this definition? The sentence

for all $x$ in $S$: $x \in \mathrm{cod}(f)$

is equivalent to

there is not $x$ in $S$: $x \notin \mathrm{cod}(f)$,

and this last one is obvious if $S=\emptyset$. So I believe that the statement

There exists a function $f \colon N \to N$ such that there is not $x$ in $S$: $x \notin \mathrm{cod}(f)$

is trivially satisfied for any $f$, so empty set is effectively enumerable. Am I right?

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Your definition of "effectively enumerable" needs the restriction that the function be computable. – Ted Jun 21 '12 at 7:14
Don't you want to require something on the function besides having codomain $S$ (e.g. computable function) – Asaf Karagila Jun 21 '12 at 7:14
Your argument only shows that $\emptyset$ is a subset of cod(f). – Ted Jun 21 '12 at 7:16

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.
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.