Let the structure $\mathfrak{A} = (A, R\_1, ..., R\_n)$ be strongly acceptable iff $\mathfrak{A}$ is an acceptable structure (in the sense of Moschovakis' *Elementary Induction on Abstract Structures*), $\mathfrak{A}$ has a countable domain, there is isomorphic copy of the natural numbers $\mathcal{N}^{\mathfrak{A}}$ "living" in $\mathfrak{A}$, and there is an $\mathfrak{A}$-hyperelementary mapping of $A$ into $\mathcal{N}^{\mathfrak{A}}$.


Let $\mathfrak{M}$ be strongly acceptable, and let $\pi$ be a $ℍ\_$-recursive projection of $ℍ\_$ into (we know it exists by the classic results of Barwise in *Admissible Sets and Structures* (II.5.14, V.5.3, and VI.4.11/12)). The domain of , $\_{\pi}$, is $\Sigma$ over $\mathbb{H}YP\_{\mathfrak{M}}$ and cannot be hyperelementary over : if it were, then $D\_π$ would be in $ℍ\_$, and thus, by $\Sigma$-replacement, it would be the case that $ℍ\_$ ∈ $ℍ\_$.


Consider now the structure $(,\_{\pi})$.


1. **Is it the case that $ℍ\_ \in ℍ\_{(,\_π)}$?** My fist intuition is as follows. $\_π$ is in $ℍ\_{(,\_π)}$. But then, since $\forall x \in D\_{\pi}\,.\,\exists y\,.\,x \in \pi(y)$, and $\Sigma$-replacement holds for $ℍ\_{(,\_π)}$, there is a $c = \{y \,:\,\exists x \in D\_{\pi}\,.\,x \in \pi(y)\} \in ℍ\_{(,\_π)}$. But $c = ℍ\_$ itself, and thus $ℍ\_ \in ℍ\_{(,\_π)}$. Additionally, I have had this other thought:


* By theorem 3D.2 of Moschovakis' *Elementary Induction on Abstract Structures*, for any $Q$ that is inductive non-hyperlementary on $\mathfrak{M}$, for any $P \subseteq M^n$, it is the case that $Q$ is hyperelementary on $(\mathfrak{M}, P)$ iff $\kappa^{\mathfrak{m}} < \kappa^{(\mathfrak{M}, P)}$.
Clearly, since $D\_{\pi}$ is $\Sigma$ on $\mathbb{H}YP\_{\mathfrak{M}}$, $D\_{\pi}$ is inductive on $\mathfrak{M}$, and cannot be hyperelementary on $\mathfrak{M}$ (otherwise it would be in $\mathbb{H}YP\_{\mathfrak{M}}$, and we have seen that this is impossible). Thus, we can apply Moschovakis 3D.2 instantiating both $Q$ and $P$ with $D\_{\pi}$. We get that $D\_{\pi}$ is hyperelementary on $(\mathfrak{M}, D\_{\pi})$ iff $\kappa^{\mathfrak{m}} < \kappa^{(\mathfrak{M}, D\_{\pi})}$. Since $D\_{\pi}$ is hyperelementary on $(\mathfrak{M}, D\_{\pi})$, $\kappa^{\mathfrak{m}} < \kappa^{(\mathfrak{M}, D\_{\pi})}$. So, the ordinal of $ℍ\_{(,\_π)}$ must be stricly above the ordinal of $\mathbb{H}YP\_{\mathfrak{M}}$.
* This also implies that the inductive relations on $\mathfrak{M}$ are hyperelementary on $ℍ\_{(,\_π)}$. By result 6D.2 of Moschovakis, we conclude that for any $n \in \omega$, the collection $\mathbb{H}YP^n\_{\mathfrak{M}} = \{X \subseteq M^n\,:\,X \text{ is hyperelementary on }\mathfrak{M}\}$ is a set in $\mathbb{H}YP\_{(\mathfrak{M}, D\_{\pi})}$ since it is inductive on $\mathfrak{M}$. Since $\mathfrak{M}$ is strongly acceptable, then it holds true in $\mathbb{H}YP\_{(\mathfrak{M}, D\_{\pi})}$ that $\forall n \in \mathcal{N}^{\mathfrak{M}}\,.\,\exists a\,.\,a = \mathbb{H}YP^n\_{\mathfrak{M}}$.
Consequently, the collection $b = \{a : \exists n \in \mathcal{N}^{\mathfrak{M}}\,.\,a = \mathbb{H}YP^n\_{\mathfrak{M}}\}$ is a set in $\mathbb{H}YP\_{(\mathfrak{M}, D\_{\pi})}$. The union $\bigcup b$ belongs to $\mathbb{H}YP\_{(\mathfrak{M}, D\_{\pi})}$, too, and $\bigcup b = \mathbb{H}YP\_{\mathfrak{M}}$ according to the definition in 6D of Moschovakis. Hence, when $\mathfrak{M}$ is strongly acceptable, $\mathbb{H}YP\_{\mathfrak{M}} \in \mathbb{H}YP\_{(\mathfrak{M}, D\_{\pi})}$.


2. Let $\mathbb{N}$ be the standard model of arithmetic, and $\pi^{\mathbb{N}}$ be a projection of $\mathbb{H}YP\_{\mathbb{N}}$ into $\omega$, having domain $D\_{\pi}^{\mathbb{N}}$. **What is the ordinal of $ℍ\_{(\mathbb{N},D\_{\pi}^{\mathbb{N}})}$?**. If the reasoning above is correct, it should be strictly above the ordinal of $ℍ\_{\mathbb{N}}$.


Thank you for your help!


(Also asked on Math Stackexchange, link here <https://math.stackexchange.com/q/4742349/1080052>)



Let $\pi$ be a $ℍ_$-recursive projection of $ℍ_$ into . What does $ℍ_{(, Domain(\pi))}$ contain?
The following result will immediately give the answer to your second question:



> 
> **Proposition.** Let $\mathfrak{M}$ be strongly acceptable and $\pi\colon \mathrm{HYP}\_\mathfrak{M}\to\mathfrak{M}$ be a $\mathrm{HYP}\_\mathfrak{M}$-recursive projection of domain $D\_\pi$. Then $\mathrm{HYP}\_{\mathfrak{M},D\_\pi}$ is the next admissible set of $\mathrm{HYP}\_\mathfrak{M}$. That is, $\mathrm{HYP}\_{\mathfrak{M},D\_\pi}=\mathrm{HYP}\_{\mathrm{HYP}\_\mathfrak{M}}$.
> 
> 
> *Proof.* It suffices to claim that if $A$ is an admissible set, then $\mathfrak{M}, D\_\pi\in A$ if and only if $\mathrm{HYP}\_\mathfrak{M}\in A$. If $\mathfrak{M}, D\_\pi\in A$, then by the argument you gave, we can reconstruct $\mathrm{HYP}\_\mathfrak{M}$ inside $A$: More precisely, let $\phi(x,y)$ be a $\Sigma\_1$ formula equivalent to $x\in \pi(y)$ over $\mathrm{HYP}\_\mathfrak{M}$. Since $\mathfrak{M}\in A$, $\mathrm{HYP}\_\mathfrak{M}\subseteq A$. Since
> $$\mathrm{HYP}\_\mathfrak{M}\models \forall x\in D\_\pi \exists y \phi(x,y),$$
> we can see that $\forall x\in D\_\pi \exists y \phi(x,y)$ also holds over $A$.
> Then by $\Sigma$-Collection over $A$ (Theorem I 4.4. of Barwise), we get a set $c\in A$ such that
> $$\forall x\in D\_\pi \exists y\in c\phi(x,y)\land \forall x\in c\exists y\in D\_\pi \phi(x,y).$$
> However, we know that every $x\in D\_\pi$ belongs to $\pi(y)$ for a precisely one $y\in \mathrm{HYP}\_\mathfrak{M}$, in other words, $\forall x\in D\_\pi \exists! y\in\mathrm{HYP}\_\mathfrak{M} \phi(x,y)$ so $c=\mathrm{HYP}\_\mathfrak{M}$.
> 
> 
> The converse is easy. If $\mathrm{HYP}\_\mathfrak{M}\in A$, then clearly $\mathfrak{M}\in A$. Also, since $D\_\pi$ is $\Sigma$ definable over $\mathrm{HYP}\_\mathfrak{M}$, $D\_\pi\in A$ by bounded Separation applied to $A$.
> 
> 
> 


You can see that $\mathrm{HYP}\_\mathbb{N}$ is $L\_{\omega\_1^{CK}}$, where $\omega\_1^{CK}$ is the least admissible ordinal. Thus $\mathrm{HYP}\_{(\mathbb{N},D\_\pi)}$ is $L\_{\omega\_2^{CK}}$, where $\omega\_2^{CK}$ is the second least admissible ordinal.


