A symmetric $\beta$-model

TL;DR

This paper constructs a countable $eta$-model where definability between reals coincides with hyperarithmetical reducibility, providing insights into the structure of models in descriptive set theory.

Contribution

It proves the existence of a countable $eta$-model with a unique definability property linking reals and hyperarithmetical reducibility, advancing understanding of $eta$-models.

Findings

Existence of a countable $eta$-model with specific definability properties

Equivalence of definability and hyperarithmetical reducibility within the model

Related results and open questions in the theory of $eta$-models

Abstract

We prove that there exists a countable $\beta$-model in which, for all reals $X$ and $Y$, $X$ is definable from $Y$ if and only $X$ is hyperarithmetical in $Y$. We also obtain some related results and pose some related questions.

Full text

1. 1 The main results
2. 2 Conservation results
3. 3 Recursion-theoretic analogs
4. 4 Some additional results

1 The main results

Our context is the study of $\omega$-models of subsystems of second order arithmetic [7, Chapter VIII]. As in [7, Chapter VII], a $\beta$-model is an $\omega$-model $M$ such that, for all $\Sigma^{1}_{1}$ sentences $\varphi$ with parameters from $M$, $\varphi$ is true if and only if $M\models\varphi$. Theorems 1.1 and 1.3 below are an interesting supplement to the results on $\beta$-models which have been presented in Simpson [7, §§VII.2 and VIII.6].

Let $\mathrm{HYP}$ denote the set of hyperarithmetical reals. It is well known that, for any $\beta$-model $M$, $\mathrm{HYP}$ is properly included in $M$, and each $X\in\mathrm{HYP}$ is definable in $M$.

Theorem 1.1**.**

There exists a countable $\beta$-model satisfying

$\forall X\,($if $X$ is definable, then $X\in\mathrm{HYP})$.

Proof.

Fix a recursive enumeration $S_{e}$, $e\in\omega$, of the $\Sigma^{1}_{1}$ sets of reals. If $p$ is a finite subset of $\omega\times\omega^{<\omega}$, say that $\langle X_{n}\rangle_{n\in\omega}$ meets $p$ if $X_{n_{1}}\oplus\cdots\oplus X_{n_{k}}\in S_{e}$ for all $(e,\langle n_{1},\dots,n_{k}\rangle)\in p$. Let $\mathcal{P}$ be the set of $p$ such that there exists $\langle X_{n}\rangle_{n\in\omega}$ meeting $p$. Put $p\leq q$ if and only if $p\supseteq q$. Say that $\mathcal{D}\subseteq\mathcal{P}$ is dense if for all $p\in\mathcal{P}$ there exists $q\in\mathcal{D}$ such that $q\leq p$. Say that $\mathcal{D}$ is definable if it is definable over the $\omega$-model $\mathrm{HYP}$, i.e., arithmetical in the complete $\Pi^{1}_{1}$ subset of $\omega$. Say that $\langle G_{n}\rangle_{n\in\omega}$ is generic if for every dense definable $\mathcal{D}\subseteq\mathcal{P}$ there exists $p\in\mathcal{D}$ such that $\langle G_{n}\rangle_{n\in\omega}$ meets $p$. We can show that for every $p\in\mathcal{P}$ there exists a generic $\langle G_{n}\rangle_{n\in\omega}$ meeting $p$. (This is a fusion argument, a la Gandy forcing.) Clearly $\{G_{n}:n\in\omega\}$ is a $\beta$-model. We can also show that, if $C$ is countable and $C\cap\mathrm{HYP}=\emptyset$, then there exists a generic $\langle G_{n}\rangle_{n\in\omega}$ such that $C\cap\{G_{n}:n\in\omega\}=\emptyset$.

Let $L_{2}(\langle X_{n}\rangle_{n\in\omega})$ be the language of second order arithmetic with additional set constants $X_{n}$, $n\in\omega$. Let $\varphi$ be a sentence of $L_{2}(\langle X_{n}\rangle_{n\in\omega})$. We say that $p$ forces $\varphi$, written $p\Vdash\varphi$, if for all generic $\langle G_{n}\rangle_{n\in\omega}$ meeting $p$, the $\beta$-model $\{G_{n}:n\in\omega\}$ satisfies $\varphi[\langle X_{n}/G_{n}\rangle_{n\in\omega}]$. It can be shown that, for all generic $\langle G_{n}\rangle_{n\in\omega}$, the $\beta$-model $\{G_{n}:n\in\omega\}$ satisfies $\varphi[\langle X_{n}/G_{n}\rangle_{n\in\omega}]$ if and only if $\langle G_{n}\rangle_{n\in\omega}$ meets some $p$ such that $p\Vdash\varphi$.

If $\pi$ is a permutation of $\omega$, define an action of $\pi$ on $\mathcal{P}$ and $L_{2}(\langle X_{n}\rangle_{n\in\omega})$ by $\pi(p)=\{(e,\langle\pi(n_{1}),\dots,\pi(n_{k})\rangle):(e,\langle n_{1},\dots,n_{k}\rangle)\in p\}$ and $\pi(X_{n})=X_{\pi(n)}$. It is straightforward to show that $p\Vdash\varphi$ if and only if $\pi(p)\Vdash\pi(\varphi)$. The support of $p\in\mathcal{P}$ is defined by $\mathrm{supp}(p)=\bigcup\{\{n_{1},\dots,n_{k}\}:(e,\langle n_{1},\dots,n_{k}\rangle)\in p\}$. Clearly if $p,q\in\mathcal{P}$ and $\mathrm{supp}(p)\cap\mathrm{supp}(q)=\emptyset$, then $p\cup q\in\mathcal{P}$.

We claim that if $\langle G_{n}\rangle_{n\in\omega}$ and $\langle G^{\prime}_{n}\rangle_{n\in\omega}$ are generic, then the $\beta$-models $\{G_{n}:n\in\omega\}$ and $\{G^{\prime}_{n}:n\in\omega\}$ satisfy the same $L_{2}$-sentences. Suppose not. Then for some $p,q\in\mathcal{P}$ we have $p\Vdash\varphi$ and $q\Vdash\lnot\,\varphi$, for some $L_{2}$-sentence $\varphi$. Let $\pi$ be a permutation of $\omega$ such that $\mathrm{supp}(\pi(p))\cap\mathrm{supp}(q)=\emptyset$. Since $\pi(\varphi)=\varphi$, we have $\pi(p)\Vdash\varphi$, hence $\pi(p)\cup q\Vdash\varphi$, a contradiction. This proves our claim.

Finally, let $M=\{G_{n}:n\in\omega\}$ where $\langle G_{n}\rangle_{n\in\omega}$ is generic. Suppose $A\in M$ is definable in $M$. Let $\langle G^{\prime}_{n}\rangle_{n\in\omega}$ be generic such that $M^{\prime}=\{G^{\prime}_{n}:n\in\omega\}$ has $M\cap M^{\prime}=\mathrm{HYP}$. Let $\varphi(X)$ be an $L_{2}$-formula with $X$ as its only free variable, such that $M\models(\exists$ exactly one $X)\,\varphi(X)$, and $M\models\varphi(A)$. Then $M^{\prime}\models(\exists$ exactly one $X)\,\varphi(X)$. Let $A^{\prime}\in M^{\prime}$ be such that $M^{\prime}\models\varphi(A^{\prime})$. Then for each $n\in\omega$, we have that $n\in A$ if and only if $M\models\exists X\,(\varphi(X)$ and $n\in X)$, if and only if $M^{\prime}\models\exists X\,(\varphi(X)$ and $n\in X)$, if and only if $n\in A^{\prime}$. Thus $A=A^{\prime}$. Hence $A\in\mathrm{HYP}$. This completes the proof. ∎

Remark 1.2**.**

Theorem 1.1 has been announced without proof by Friedman [3, Theorem 4.3]. Until now, a proof of Theorem 1.1 has not been available.

We now improve Theorem 1.1 as follows.

Let $\leq_{\mathrm{HYP}}$ denote hyperarithmetical reducibility, i.e., $X\leq_{\mathrm{HYP}}Y$ if and only if $X$ is hyperarithmetical in $Y$.

Theorem 1.3**.**

There exists a countable $\beta$-model satisfying

$(*)\qquad\forall X\,\forall Y\,($if $X$ is definable from $Y$, then $X\leq_{\mathrm{HYP}}Y)$.

The $\beta$-model which we shall use to prove Theorem 1.3 is the same as for Theorem 1.1, namely $M=\{G_{n}:n\in\omega\}$ where $\langle G_{n}\rangle_{n\in\omega}$ is generic. In order to see that $M$ has the desired property, we first relativize the proof of Theorem 1.1, as follows. Given $Y$, let $\mathcal{P}^{Y}$ be the set of $p\in\mathcal{P}$ such that there exists $\langle X_{n}\rangle_{n\in\omega}$ meeting $p$ with $X_{0}=Y$. (Obviously [math] plays no special role here.) Say that $\langle G_{n}\rangle_{n\in\omega}$ is generic over $Y$ if, for every dense $\mathcal{D}^{Y}\subseteq\mathcal{P}^{Y}$ definable from $Y$ over $\mathrm{HYP}(Y)=\{X:X\leq_{\mathrm{HYP}}Y\}$, there exists $p\in\mathcal{D}^{Y}$ such that $\langle G_{n}\rangle_{n\in\omega}$ meets $p$.

Lemma 1.4**.**

If $\langle G_{n}\rangle_{n\in\omega}$ is generic over $Y$, then $G_{0}=Y$, and $\{G_{n}:n\in\omega\}$ is a $\beta$-model satisfying $\forall X\,($if $X$ is definable from $Y$, then $X\leq_{\mathrm{HYP}}Y)$.

Proof.

The proof of this lemma is a straightforward relativization to $Y$ of the proof of Theorem 1.1. ∎

Consequently, in order to prove Theorem 1.3, it suffices to prove the following lemma.

Lemma 1.5**.**

If $\langle G_{n}\rangle_{n\in\omega}$ is generic, then $\langle G_{n}\rangle_{n\in\omega}$ is generic over $G_{0}$.

Proof.

It suffices to show that, for all $p$ forcing $(\mathcal{D}^{X_{0}}$ is dense in $\mathcal{P}^{X_{0}})$, there exists $q\leq p$ forcing $\exists r\,(r\in\mathcal{D}^{X_{0}}$ and $\langle X_{n}\rangle_{n\in\omega}$ meets $r)$.

Assume $p\Vdash(\mathcal{D}^{X_{0}}$ is dense in $\mathcal{P}^{X_{0}})$. Since $p\Vdash p\in\mathcal{P}^{X_{0}}$, it follows that $p\Vdash\exists q\,(q\leq p$ and $q\in\mathcal{D}^{X_{0}})$. Fix $p^{\prime}\leq p$ and $q^{\prime}\leq p$ such that $p^{\prime}\Vdash q^{\prime}\in\mathcal{D}^{X_{0}}$. Put $S^{\prime}=\{X_{0}:\langle X_{n}\rangle_{n\in\omega}$ meets $p^{\prime}\}$. Then $S^{\prime}$ is a $\Sigma^{1}_{1}$ set, so let $e\in\omega$ be such that $S^{\prime}=S_{e}$. Claim 1: $\{(e,\langle 0\rangle)\}\Vdash q^{\prime}\in\mathcal{D}^{X_{0}}$. If not, let $p^{\prime\prime}\leq\{(e,\langle 0\rangle)\}$ be such that $p^{\prime\prime}\Vdash q^{\prime}\notin\mathcal{D}^{X_{0}}$. Let $\pi$ be a permutation such that $\pi(0)=0$ and $\mathrm{supp}(p^{\prime})\cap\mathrm{supp}(\pi(p^{\prime\prime}))=\{0\}$. Then $p^{\prime}\cup\pi(p^{\prime\prime})\in\mathcal{P}$ and $\pi(p^{\prime\prime})\Vdash q^{\prime}\notin\mathcal{D}^{X_{0}}$, a contradiction. This proves Claim 1.

Claim 2: $q^{\prime}\cup\{(e,\langle 0\rangle)\}\in\mathcal{P}$. To see this, let $\langle G^{\prime}_{n}\rangle_{n\in\omega}$ be generic meeting $\{(e,\langle 0\rangle)\}$. By Claim 1 we have $q^{\prime}\in\mathcal{D}^{G^{\prime}_{0}}$. Hence $q^{\prime}\in\mathcal{P}^{G^{\prime}_{0}}$, i.e., there exists $\langle X_{n}\rangle_{n\in\omega}$ meeting $q^{\prime}$ with $X_{0}=G^{\prime}_{0}$. Thus $\langle X_{n}\rangle_{n\in\omega}$ meets $q^{\prime}\cup\{(e,\langle 0\rangle)\}$. This proves Claim 2. Finally, put $q^{\prime\prime}=q^{\prime}\cup\{(e,\langle 0\rangle)\}$. Then $q^{\prime\prime}\leq q^{\prime}\leq p$ and $q^{\prime\prime}\Vdash(q^{\prime}\in\mathcal{D}^{X_{0}}$ and $\langle X_{n}\rangle_{n\in\omega}$ meets $q^{\prime})$. This proves our lemma. ∎

The proof of Theorem 1.3 is immediate from Lemmas 1.4 and 1.5.

2 Conservation results

In this section we generalize the construction of §1 to a wider setting. We then use this idea to obtain some conservation results involving the scheme $(*)$ of Theorem 1.3.

Two important subsystems of second order arithmetic are $\mathsf{ATR}_{0}$ (arithmetical transfinite recursion with restricted induction) and $\Pi^{1}_{\infty}$-$\mathsf{TI}_{0}$ (the transfinite induction scheme). For general background, see Simpson [7]. It is known [7, §VII.2] that $\mathsf{ATR}_{0}\subseteq\Pi^{1}_{\infty}$-$\mathsf{TI}_{0}$, and that every $\beta$-model is a model of $\Pi^{1}_{\infty}$-$\mathsf{TI}_{0}$.

Let $(N,\mathcal{S})$ be a countable model of $\mathsf{ATR}_{0}$, where $\mathcal{S}\subseteq P(|N|)$. Define $\mathcal{P}_{(N,\mathcal{S})}=\{p:(N,\mathcal{S})\models(p$ is a finite subset of $\omega\times\omega^{<\omega}$ and there exists $\langle X_{n}\rangle_{n\in\omega}$ meeting $p)\}$. The notion of $\langle G_{n}\rangle_{n\in|N|}$ being generic over $(N,\mathcal{S})$ is defined in the obvious way. As in §1, the basic forcing lemmas can be proved. Let $\langle G_{n}\rangle_{n\in|N|}$ be generic over $(N,\mathcal{S})$. Put $\mathcal{S}^{\prime}=\{G_{n}:n\in|N|\}$.

Lemma 2.1**.**

$(N,\mathcal{S}^{\prime})$ satisfies the scheme $(*)$ of Theorem 1.3.

Proof.

The proof is a straightforward generalization of the arguments of §1. ∎

Lemma 2.2**.**

Let $\psi$ be a $\Pi^{1}_{2}$ sentence with parameters from $|N|$. If $(N,\mathcal{S})\models\psi$, then $(N,\mathcal{S}^{\prime})\models\psi$.

Proof.

Write $\psi$ as $\forall X\,(X\in S_{e})$ for some fixed $e\in|N|$. If $(N,\mathcal{S})\models\psi$, then for each $n\in|N|$ we have that $\{p\in\mathcal{P}_{(N,\mathcal{S})}:(e,\langle n\rangle)\in p\}$ is dense in $\mathcal{P}_{(N,\mathcal{S})}$, hence $\emptyset\Vdash X_{n}\in S_{e}$. Thus $\emptyset\Vdash\forall X\,(X\in S_{e})$, i.e., $\emptyset\Vdash\psi$, so $(N,\mathcal{S}^{\prime})\models\psi$. ∎

Remark 2.3**.**

Since $(N,\mathcal{S})\models\mathsf{ATR}_{0}$ and $\mathsf{ATR}_{0}$ is axiomatized by a $\Pi^{1}_{2}$ sentence, it follows by Lemma 2.2 that $(N,\mathcal{S}^{\prime})\models\mathsf{ATR}_{0}$. Lemma 2.2 also implies that $(N,\mathcal{S})$ and $(N,\mathcal{S}^{\prime})$ satisfy the same $\Pi^{1}_{1}$ sentences with parameters from $|N|$. From this it follows that the recursive well orderings of $(N,\mathcal{S})$ and $(N,\mathcal{S}^{\prime})$ are the same, and that $\mathrm{HYP}_{(N,\mathcal{S})}=\mathrm{HYP}_{(N,\mathcal{S}^{\prime})}$. It can also be shown that $\mathrm{HYP}_{(N,\mathcal{S})}=\mathcal{S}\cap\mathcal{S}^{\prime}$.

We now have:

Theorem 2.4**.**

$\mathsf{ATR}_{0}+(*)$ is conservative over $\mathsf{ATR}_{0}$ for $\Sigma^{1}_{2}$ sentences.

Proof.

Let $\varphi$ be a $\Sigma^{1}_{2}$ sentence. Suppose $\mathsf{ATR}_{0}\not\vdash\varphi$. Let $(N,\mathcal{S})$ be a countable model of $\mathsf{ATR}_{0}+\lnot\,\varphi$. Let $\mathcal{S}^{\prime}=\{G_{n}:n\in|N|\}$ as above. Since $\lnot\,\varphi$ is a $\Pi^{1}_{2}$ sentence, we have by Lemmas 2.1 and 2.2 that $(N,\mathcal{S}^{\prime})\models\mathsf{ATR}_{0}+(*)+\lnot\,\varphi$. Thus $\mathsf{ATR}_{0}+(*)\not\vdash\varphi$. ∎

In order to obtain a similar result for $\Pi^{1}_{\infty}$-$\mathsf{TI}_{0}$, we first prove:

Lemma 2.5**.**

$(N,\mathcal{S}^{\prime})\models$ “all ordinals are recursive”, i.e., “every well ordering is isomorphic to a recursive well ordering”.

Proof.

Recall from [7, §§VIII.3 and VIII.6] that all of the basic results of hyperarithmetical theory are provable in $\mathsf{ATR}_{0}$. In particular, by [7, Theorem VIII.6.7], the Gandy/Kreisel/Tait Theorem holds in $\mathsf{ATR}_{0}$. Thus for all $p\in\mathcal{P}_{(N,\mathcal{S})}$ we have

$(N,\mathcal{S})\models$ there exists $\langle X_{n}\rangle_{n\in\omega}$ meeting $p$ such that $\forall n\,(\omega_{1}^{X_{n}}=\omega_{1}^{\mathrm{CK}})$.

Now, it is provable in $\mathsf{ATR}_{0}$ that the predicate $\omega_{1}^{X}=\omega_{1}^{\mathrm{CK}}$ is $\Sigma^{1}_{1}$. Thus we have $\emptyset\Vdash\forall X\,(\omega_{1}^{X}=\omega_{1}^{\mathrm{CK}})$, i.e., $\emptyset\Vdash$ “all ordinals are recursive”. This proves our lemma. ∎

Remark 2.6**.**

An alternative proof of Lemma 2.5 is to note that $\mathsf{ATR}_{0}+(*)\vdash$ “$O$ does not exist”, because $O$ would be definable but not hyperarithmetical. And “$O$ does not exist” is equivalent over $\mathsf{ATR}_{0}$ to “all ordinals are recursive”. (Here $O$ denotes the complete $\Pi^{1}_{1}$ set of integers. See [7, §VIII.3].)

Theorem 2.7**.**

$\Pi^{1}_{\infty}$-$\mathsf{TI}_{0}+(*)$ is conservative over $\Pi^{1}_{\infty}$-$\mathsf{TI}_{0}$ for $\Sigma^{1}_{2}$ sentences.

Proof.

By Lemma 2.5 it suffices to prove: if $(N,\mathcal{S})\models$ transfinite induction for recursive well orderings, then $(N,\mathcal{S}^{\prime})\models$ transfinite induction for recursive well orderings. Let $e\in|N|$ be such that $(N,\mathcal{S})\models$ “${<}_{e}$ is a recursive well ordering”. Suppose $p\Vdash\exists n\,(n\in\mathrm{field}({<}_{e})$ and $\varphi(n))$. Put $A=\{n\in\mathrm{field}({<}_{e}):p\not\Vdash\lnot\,\varphi(n)\}$. Clearly $A\neq\emptyset$. By definability of forcing, $A$ is definable over $(N,\mathcal{S})$. Hence there exists $a\in A$ such that $(N,\mathcal{S})\models$ “$a$ is the ${<}_{e}$-least element of $A$”. For each $n<_{e}a$ we have $p\Vdash\lnot\,\varphi(n)$, but $p\not\Vdash\lnot\,\varphi(a)$, so let $q\leq p$ be such that $q\Vdash\varphi(a)$. Then $q\Vdash$ “$a$ is the ${<}_{e}$-least $n$ such that $\varphi(n)$”. Thus $(N,\mathcal{S}^{\prime})\models$ transfinite induction for recursive well orderings. ∎

Remark 2.8**.**

Theorems 2.4 and 2.7 are best possible, in the sense that we cannot replace $\Sigma^{1}_{2}$ by $\Pi^{1}_{2}$. An example is the $\Pi^{1}_{2}$ sentence “all ordinals are recursive”, which is provable in $\mathsf{ATR}_{0}+(*)$ but not in $\Pi^{1}_{\infty}$-$\mathsf{TI}_{0}$.

3 Recursion-theoretic analogs

The results of §§1 and 2 are in the realm of hyperarithmetical theory. We now present the analogous results in the realm of recursion theory, concerning models of $\mathsf{WKL}_{0}$. For background on this topic, see Simpson [7, §§VIII.2 and XI.2].

Let $\mathrm{REC}$ denote the set of recursive reals. It is well known that, for any $\omega$-model $M$ of $\mathsf{WKL}_{0}$, $\mathrm{REC}$ is properly included in $M$, and each $X\in\mathrm{REC}$ is definable in $M$. The recursion-theoretic analog of Theorem 1.1 is:

Theorem 3.1**.**

There exists a countable $\omega$-model of $\mathsf{WKL}_{0}$ satisfying

$\forall X\,($if $X$ is definable, then $X\in\mathrm{REC})$.

Proof.

Use exactly the same construction and argument as for Theorem 1.1, replacing $\Sigma^{1}_{1}$ sets by $\Pi^{0}_{1}$ subsets of $2^{\omega}$. ∎

Remark 3.2**.**

Theorem 3.1 is originally due to Friedman [2, Theorem 1.10, unpublished] (see also [3, Theorem 1.6]). It was later proved again by Simpson [8] (see also Simpson/Tanaka/Yamazaki [9]). All three proofs of Theorem 3.1 are different from one another.

Let $\leq_{T}$ denote Turing reducibility, i.e., $X\leq_{T}Y$ if and only if $X$ is computable using $Y$ as an oracle. The recursion-theoretic analog of Theorem 1.3 is:

Theorem 3.3**.**

There exists a countable $\omega$-model of $\mathsf{WKL}_{0}$ satisfying

$(**)\qquad\forall X\,\forall Y\,($if $X$ is definable from $Y$, then $X\leq_{T}Y)$.

Proof.

Use exactly the same construction and argument as for Theorem 1.3, replacing $\Sigma^{1}_{1}$ sets by $\Pi^{0}_{1}$ subsets of $2^{\omega}$. ∎

The recursion-theoretic analog of Theorems 2.4 and 2.7 is:

Theorem 3.4**.**

$\mathsf{WKL}_{0}+(**)$ is conservative over $\mathsf{WKL}_{0}$ for arithmetical sentences.

Proof.

The proof is analogous to the arguments of §2. ∎

Remark 3.5**.**

Theorems 3.3 and 3.4 are originally due to Simpson [8]. The proofs given here are different from the proofs that were given in [8].

4 Some additional results

In this section we present some additional results and open questions.

Lemma 4.1**.**

Let $\varphi(X)$ be a $\Sigma^{1}_{1}$ formula with no free set variables other than $X$. The following is provable in $\mathsf{ATR}_{0}$. If $\exists X\,(X\notin\mathrm{HYP}$ and $\varphi(X))$, then $\exists P\,(P$ is a perfect tree and $\forall X\,($if $X$ is a path through $P$ then $\varphi(X)))$.

Proof.

This is a well known consequence of formalizing the Perfect Set Theorem within $\mathsf{ATR}_{0}$. See Simpson [7, §§V.4 and VIII.3]. See also Sacks [5, §III.6]. ∎

Lemma 4.2**.**

Let $\varphi(X)$ be a $\Sigma^{1}_{2}$ formula with no free set variables other than $X$. The following is provable in $\mathsf{ATR}_{0}\,+{}$ “all ordinals are recursive”. If $\exists X\,(X\notin\mathrm{HYP}$ and $\varphi(X))$, then $\exists P\,(P$ is a perfect tree and $\forall X\,($if $X$ is a path through $P$ then $\varphi(X)))$.

Proof.

Since $\varphi(X)$ is $\Sigma^{1}_{2}$, we can write $\varphi(X)\equiv\exists Y\,\forall f\,\exists n\,R(X[n],Y[n],f[n])$ where $R$ is a primitive recursive predicate. Let $T_{X,Y}$ be the tree consisting of all $\tau$ such that $\lnot\,(\exists n\leq\mathrm{lh}(\tau))\,R(X[n],Y[n],\tau[n])$. For $\alpha<\omega_{1}^{\mathrm{CK}}$ put

$\varphi_{\alpha}(X)\,\,\equiv\,\,\exists Y\,(T_{X,Y}$ is well founded of height $\leq\alpha)$.

Note that, for each $\alpha<\omega_{1}^{\mathrm{CK}}$, $\varphi_{\alpha}(X)$ is $\Sigma^{1}_{1}$. Reasoning in $\mathsf{ATR}_{0}\,+{}$ “all ordinals are recursive”, we have $\forall X\,(\varphi(X)$ if and only if $(\exists\alpha<\omega_{1}^{\mathrm{CK}})\,\varphi_{\alpha}(X))$. Thus Lemma 4.2 follows easily from Lemma 4.1. ∎

Theorem 4.3**.**

Let $T$ be $\mathsf{ATR}_{0}$ or $\Pi^{1}_{\infty}$-$\mathsf{TI}_{0}$. Let $\varphi(X)$ be a $\Sigma^{1}_{2}$ formula with no free set variables other than $X$. If $T\vdash\exists X\,(X\notin\mathrm{HYP}$ and $\varphi(X))$, then $T\vdash\exists P\,(P$ is a perfect tree and $\forall X\,($if $X$ is a path through $P$ then $\varphi(X)))$.

Proof.

From Friedman [1] or Simpson [7, §VII.2], we have that $T\vdash$ the disjunction (1) all ordinals are recursive, or (2) there exists a countable coded $\beta$-model $M$ satisfying $T\,+{}$ “all ordinals are recursive”. In case (1), the desired conclusion follows from Lemma 4.2. In case (2), we have $M\models\exists X\,(X\notin\mathrm{HYP}$ and $\varphi(X))$, so the proof of Lemma 4.2 within $M$ gives $\alpha<\omega_{1}^{\mathrm{CK}}$ such that $M\models\exists X\,(X\notin\mathrm{HYP}$ and $\varphi_{\alpha}(X))$. It follows that $\exists X\,(X\notin\mathrm{HYP}$ and $\varphi_{\alpha}(X))$. We can then apply Lemma 4.1 to $\varphi_{\alpha}(X)$ to obtain the desired conclusion. ∎

Corollary 4.4**.**

Let $T$ and $\varphi(X)$ be as in Theorem 4.3. If $T\vdash\exists X\,(X\notin\mathrm{HYP}$ and $\varphi(X))$, then $T\vdash\forall Y\,\exists X\,(\varphi(X)$ and $\forall n\,(X\neq(Y)_{n}))$.

Proof.

This follows easily from Theorem 4.3. ∎

Theorem 4.5**.**

Let $T$ and $\varphi(X)$ be as in Theorem 4.3. If $T\vdash(\exists$ exactly one $X)\,\varphi(X)$, then $T\vdash\exists X\,(X\in\mathrm{HYP}$ and $\varphi(X))$.

Proof.

Consider cases (1) and (2) as in the proof of Theorem 4.3. In both cases it suffices to show that, for all $\alpha<\omega_{1}^{\mathrm{CK}}$, if $(\exists$ exactly one $X)\,\varphi_{\alpha}(X)$ then $\exists X\,(X\in\mathrm{HYP}$ and $\varphi_{\alpha}(X))$. This follows from Lemma 4.1 applied to $\varphi_{\alpha}(X)$. ∎

Remark 4.6**.**

Theorems 4.3 and 4.5 appear to be new. Corollary 4.4 has been stated without proof by Friedman [3, Theorems 3.4 and 4.4]. A recursion-theoretic analog of Corollary 4.4 has been stated without proof by Friedman [3, Theorem 1.7], but this statement of Friedman is known to be false, in view of Simpson [8]. A recursion-theoretic analog of Theorem 4.5 has been proved by Simpson/Tanaka/Yamazaki [9].

Question 4.7**.**

Suppose $\mathsf{WKL}_{0}\vdash\exists X\,(X\notin\mathrm{REC}$ and $\varphi(X))$ where $\varphi(X)$ is $\Sigma^{1}_{1}$, or even arithmetical, with no free set variables other than $X$. Does it follow that $\mathsf{WKL}_{0}\vdash\exists X\,\exists Y\,(X\neq Y\land\varphi(X)\land\varphi(Y))$? A similar question has been asked by Friedman [2, unpublished].

Question 4.8**.**

Suppose $\mathsf{WKL}_{0}\vdash(\exists$ exactly one $X)\,\varphi(X)$ where $\varphi(X)$ is $\Sigma^{1}_{1}$ with no free set variables other than $X$. Does it follow that $\mathsf{WKL}_{0}\vdash\exists X\,(X\in\mathrm{REC}$ and $\varphi(X))$? If $\varphi(X)$ is arithmetical or $\Pi^{1}_{1}$ then the answer is yes, by Simpson/Tanaka/Yamazaki [9].