Is Rice-Shapiro theorem bidirectional?

Question

Rice-Shapiro theorem states that

version A

Let $\Gamma$ be a set of computably enumerable sets, and $I = \{e : W_e \in \Gamma\}$ its index set in some admissible enumeration of c.e sets. If $I$ is c.e, then for all $x$, $x \in I$ if and only if there is $y \in I$ such that $W_y$ is a finite subset of $W_x$.

Or equivalently

$I$ is c.e $\Rightarrow$

$~~~~(1):~ \forall x,y ~ (x \in I \land W_x \subseteq W_y \rightarrow y \in I)$

$\land ~~(2):~ \forall x ~ (x \in I \rightarrow \exists y~(y \in I \land W_y ~\text{is finite} \land W_y \subseteq W_x))$

My question is if it's converse is also true, that is, if (1) and (2) hold, then $I$ is c.e.

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So I found another (weaker) formulation of the theorem in H. Rogers Theory of Recursive Functions and Effective Computability p. 324, which states:

version B

$I$ is c.e $\iff ~(3):~ \forall x ( x \in I \leftrightarrow \exists u (D_{f(u)} \subseteq W_x))$ for some totally computable function $f$

where $D$ is the canonical enumeration of the finite sets (see definition 2.4).

(3) implies that $I$ is c.e, for it is just a $\Sigma_1$ representation of $I$. On the other hand, if $I$ is c.e, by s-m-n theorem take the totally computable function $g$ to be defined such that $W_{g(x)} = D_x$. Now the set $g^{-1}(I) = \{ x : \exists y (g(x) = y \land y \in I)\}$ is also c.e, thus there must be a totally computable function $f$ such that $Range(f) = g^{-1}(I)$. So by version A of the theorem, we also have (1) and (2), which imply that this $f$ is the function that satisfies (3).

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I also found these two answers which state the theorem in bidirectional form, but with an additional condition:

version C

$I$ is c.e $\iff (1) \land (2)$

$~~~~~~~~~~~~~~~~\land (4):~ \{ e : D_e \in \Gamma \}$ is c.e

Now this set in (4) is just the $g^{-1}(I)$ defined above, where we just saw that it is indeed c.e. by the sole fact that $I$ is c.e. So my primitive guess is that (4) is the ingredient for proving the other direction.

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Considering all the observations above, my question is whether the right to left direction holds generally, i.e. does (1) and (2) imply $I$ is c.e? And is there a counterexample if that fails? Assuming (4), can we deduce that $I$ is c.e? And does that mean the wanted counterexample should refute (4)?

Answer 1

No, the converse direction does not hold:

Pick some non-c.e. set $B \subseteq \mathbb{N}$. Now consider $\Gamma = \{A \ c.e. \mid A \cap B \neq \emptyset\}$. By construction, $\Gamma$ satisfies the conclusion of Rice-Shapiro (it suffices to consider singleton sets rather than all finite sets). However, for the resulting index set $I$ we find that $B \leq_m I$, so $I$ cannot be c.e..

Essentially, Rice-Shapiro says that the Markov-effectively open subsets of the subspace of ce subsets of $P(\omega)$ (with the Scott topology) are open, but for already for cardinality reasons alone, not all open subsets can be Markov-effectively open.