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Given a Gödel numbering $\phi_n$ of the partial recursive unary functions, the Rice-Shapiro theorem states

Let $\mathcal A$ be a family of partial-recursive unary functions such that its index set $A = \{ x \in \mathbb N\ |\ \phi_x \in \mathcal A \}$ is r. e. Then for any partial recursive unary $f,$ $$f \in \mathcal A \iff \exists \theta \in \mathcal A \text{ finite function with } \theta \subseteq f.$$

Let $\mathcal B = \{U_\theta\ |\ \theta \text{ finite function}\},$ for $$ U_\theta = \{f \text{ partial recursive } |\ \theta \subseteq f \}. $$ Then $\mathcal B$ is the basis of a topology in the set of all partial recursive functions (in fact, this extends to the set of all partial unary functions). Then the content of the Rice-Shapiro theorem as stated is that any such $\mathcal A$ is open with respect to this topology. The corresponding Wikipedia page makes a stronger claim: in this notation,

$\mathcal A$ has r. e. index set iff it is a recursively enumerable union of $U_\theta$.

I'm not $100\%$ sure of what a r. e. union means in this context. I first assumed that for a family $\{ U_\theta\ |\ \theta \in \Theta \}$ it means something like:

There is a (total) recursive function $k(x)$ such that {$\phi_{k(x)}\ |\ x \in \mathbb N\} = \Theta.$

However that does not seem enough, because there must also be an effective way to know what the domain of $\phi_{k(x)}$ is.

So I guess that an appropriate bijective enumeration $\theta_n$ of the set of all finite functions must be used, such that the predicate “$\phi_x \text{ extends } \theta_y$” is partially decidable. Then r. e. subsets of finite functions correspond to r. e. subsets of $\mathbb N.$

If I use that definition I think I can finish the proof for both implications. So my question is, is this the correct definition? If so, I can intuitively see that such enumeration exists, but is there a reference where it's constructed explicitely? For example in Hartley-Roger's book there is a version of the theorem for sets, but not for finite functions.

Keplerto
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Let $\mathscr F$ be the set of all finite functions from $\mathbb N$ to $\mathbb N,$ and think of any $\theta\in\mathscr F$ as a code for the set $U_\theta = \{\varphi\mid\varphi\text{ is partial recursive and }\theta\subseteq\varphi\}.$ The Wikipedia article you linked to calls these sets frusta. $U_\theta$ is the frustum coded by $\theta.$

Having a recursively enumerable union of frusta really means: (1) having a recursively enumerable set of codes for frusta, and (2) taking the union of those particular frusta. (This is all using the specific coding $\theta\mapsto U_\theta$ above.)

In other words, it means that we have an recursively enumerable $W\subseteq\mathscr F$ and we're looking at $\bigcup_{\theta\in W} U_\theta.$

You can look at this in terms of recursive functions (as you were suggesting), but it's just a bit more complicated: You'd start with a recursive function $k:\mathbb N\to\mathscr F,$ and you'd then look at $\bigcup_{n\in\mathbb N} U_{k(n)}.$ However, you'd need to treat the empty set as a special case because the empty set, although r.e., isn't the range of a recursive function.

Mitchell Spector
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  • I see, but you still have to define what a "recursively enumerable" subset of $\mathscr F$ is. Perhaps this is standard, but I don't know. I guess you do it via a coding? Like a function $\theta$ defined on $x_1 < \dots < x_n$ with respective values $y_1, \dots, y_n$ might be coded by $2^{\pi(x_1, y_1)}+ 2^{\pi(x_1, y_1) + \pi(x_2-x_1, y_2)} + \dots + 2^{\pi(x_1, y_1) + \pi(x_2-x_1, y_2) + \dots \pi(x_n - x_{n-1}, y_n)}$ for some coding $\pi$ of $\mathbb N \times \mathbb N$. Or is it another type of definition? – Keplerto Jun 29 '23 at 06:52
  • Yes, you can do this with a coding of finite functions as natural numbers, as you suggested. (I didn't check out the details of your coding, but that sort of thing would work.) Alternatively, since the finite functions here are finite objects, you can code them straightforwardly as strings, and then define partial recursive functions via Turing machines on the alphabet you're using. – Mitchell Spector Jun 29 '23 at 07:00
  • Ok, thanks then! Now I think I can complete the proof for the harder implication to the right. First show that $\theta \subseteq \phi_x$ is partially decidable in the variables $\theta, x$, so $\exists x\ (\theta \subseteq \phi_x \land \phi_x \in \mathcal A)$ is also partially decidable. Therefore, by Rice-Shapiro, the set of all finite functions in $\mathcal A$ (which is the set of all $\theta$ satisfying that property) is r. e., and $\mathcal A$ is the r. e. union of their frustra. – Keplerto Jun 29 '23 at 07:36
  • But I've realised there can be functions with that property not in A. – Keplerto Jun 29 '23 at 08:05
  • I think one still needs to find some recursive function $f(\theta)$ with $\phi_{f(\theta)} = \theta$ and then $\phi_{f(\theta)} \in \mathcal A$ is the needed partially decidable predicate. – Keplerto Jun 29 '23 at 08:14