In Koepke's paper, "Turing Computations On Ordinals", one has the following (well-known) result:

>A set $x$ is ordinal computable from a finite set of ordinal parameters if an only if it is an element of the constructible universe.


In Sacks' survey article, "E-Recursive Intuitions", one finds these (possibly related) results:

>Proposition 3.1.  $V=L$ iff $L$ is E-recursively enumerable...
>Proposition 2.3.  There exists a $\Delta_1$ definable class that is not E-recursively enumerable.


Now, since the above theorem of Koepke's characterizes the constructible sets as those ordinal computable from a finite set of ordinal parameters (note that the _pure sets_ are sets of ordinals), one should be able to characterize the axiom $V=L$ as

'Every set of ordinals is ordinal computable from a finite set of ordinal parameters.'

Substituting this characterization for $V=L$ in Sacks' theorem, one has 

'Every set of ordinals is ordinal computable from a finite set of ordinal parameters iff $L$ is E-recursively enumerable.'


It is interesting to note that Sacks' proof of Theorem 3.1 from his paper gives an indication of what must happen if $V\neq L$ (most set theorists do not believe that $V$=$L$ is an 'acceptable' axiom):

>Proof of 3.1.  Suppose $\forall x(x\in L\leftrightarrow\{e\}(x)\downarrow)$ and $V\neq L$.  Then for some $b\notin L$, $\{e\}(b)\uparrow$.  By Proposition 2.2 ["If $A$ is E-recursively enumerable, then $A$ is $\Delta_1$ definable."--my comment], divergence [$\uparrow$--my comment] is $\Sigma_1$ definable.  Then by Levy-Shoenfield absoluteness, $\{e\}(x)\uparrow$ for some $x\in L$.

Hence the question asked in the title.


Three other questions:

1.  Can one 'force' $L$ to be not E-recursively enumerable?

2.  If $V\neq L$, are there constructible, non-E-recursively enumerable sets?

3. Since $L$ is usually considered a 'class', is E-recursion defined for classes (since it is based on Normann's "Set recursion")? (Hope this is not too silly a question....)