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?