0
$\begingroup$

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?

  • 2
    Your definition of "effectively enumerable" needs the restriction that the function be computable.2012-06-21
  • 0
    Don't you want to require something on the function besides having codomain $S$ (e.g. computable function)2012-06-21
  • 0
    Your argument only shows that $\emptyset$ is a subset of cod(f).2012-06-21

1 Answers 1