- check $A$ is semantically closed
- ($\Rightarrow$) pick $f \in \mathcal{F}$ s.t. $\nexists g \subseteq f . g \in \mathcal{F} \wedge \mathsf{dom}(g) \textit{ is finite}$
- ($\Leftarrow$) pick $f \notin \mathcal{F}$ s.t. $\exists g \subseteq f . g \in \mathcal{F} \wedge \mathsf{dom}(g) \textit{ is finite}$
Some examples.
- $A = \{ n | \mathsf{dom} (\varphi_n) \backslash \mathsf{ran} (\varphi_n) \neq \varnothing \}$
- $B = \{ n | \mathsf{dom} (\varphi_n) \text{ is finite} \}$
- $C = \{ n | \mathsf{dom} (\varphi_n) \text{ is infinite} \}$
$A \notin \mathcal{RE}$. By R-S $\Leftarrow$, we provide $g(0)=1 \in \mathcal{F}$, also we scheme $f(0)=1,f(1)=0$, s.t. $g \subseteq f \wedge f \notin \mathcal{F}$, both $f$ and $g$ are finite (let the otherwise case undefined).
$B$ and $C$ are both not $\mathcal{RE}$ (directly from R-S theorem). This says, the set of indices of all finite domain functions are not recursively enumerable. So is the set of indices of all infinite domain ones.
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