Models of second-order arithmetic closed under relative constructibility

Question

I know little to nothing about second-order arithmetic and its subsystems. However, I would like to understand when a model of (a subsystem of) second-order arithmetic ($\mathsf{Z}_2$) is downward closed with respect to the relative constructibility preorder. More precisely, consider this statement:

Suppose that $M$ is a ($\beta$-?) model of $\mathsf{T}\subseteq \mathsf{Z}_2$, let $x,y \in \mathbb{R}$ such that $x \in M$ and $y \le_c x$ (i.e. $y \in L[x]$). Then $y \in M$.

The questions are

• What can $\mathsf{T}$ be to make the statement true? Would $T = \varPi_1^1$-$\mathsf{CA}_0$ suffice?
• Does $M$ need to be a $\beta$-model, for the statement to be true?

References would be more than appreciated.
Thanks

Answer 1

First, let me remark that every $\beta$-model can be identified with a transitive model of $\mathsf{ATR_0^{set}}$ (the detail is available in Simpson's book on the second-order arithmetic.) Thus let me work with transitive models of a set theory instead.

The following argument shows your first statement negatively:

Claim. Suppose that $M$ is a transitive model of $\mathsf{ATR_0^{set}}$ and closed under the relative constructibility. If $a\in M$ is a real, then $\omega_1^{L[a]}\subseteq M$.

Proof. Suppose that $\xi<\omega_1^{L[a]}$, then we can find a well-order $X$ in $L[a]$ whose field is $\omega$ and isomorphic with $\xi$. Clearly $X\le_c a$, so the assumption implies $X\in M$. By the axiom Beta over $M$, $\xi\in M$.

On the other hand, the least $\beta$-model of $\Pi^1_1\text{-}\mathsf{CA}_0$ has the height $\omega_\omega^\mathsf{CK}$ (the limit of the first $n$ admissible ordinals), which is strictly below than $\omega_1^L$. Thus not every model of $\Pi^1_1\text{-}\mathsf{CA}_0$ is closed under relative constructibility. The answer is negative even in the case $T=\mathsf{Z}_2$ since there is a transitive model of $\mathsf{Z}_2$ of height less than $\omega_1^L$.

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Also, your second question has a partial positive answer by the following argument:

Claim. Every $\omega$-model closed under the relative constructibility is a $\beta$-model.

Proof. Let $M$ be an $\omega$-model of $\mathsf{ATR_0^{set}}$ (that is, $\omega^M=\omega$.) We claim that $M$ reflects every true $\Sigma^1_1$-sentence.

Let $a\in M$ be a real, and $\exists x \phi(x,a)$ be a true $\Sigma^1_1$ sentence with parameter $a$ and an arithmetical formula $\phi$. By Shoenfield absoluteness, $L[a]$ also thinks $\exists x \phi(x,a)$ holds, so we can find $b\in L[a]$ such that $\phi(b,a)$. From $b\le_c a$ we get $b\in M$ and so $M\vDash \phi(b,a)$.

Answer 2

Elaborating on Hanul's answer, the issue is that relative constructibility as such is given for free really big ordinals. Letting $M^\mathsf{set}$ be the $\{\in\}$-analogue of a $\beta$-model $M$, the issue is that if $M^{\mathsf{set}}$'s height is less than $\omega_1^L$ then no matter how "correct" $M$ is we will never have $M^\mathsf{set}$ closed under $\le_c$. To be precise, by condensation every (constructible) first-order $\{\in\}$-theory $T$ which has a well-founded model of height $\ge\omega_1^L$ will also have a well-founded model of height $<\omega_1^L$.

We can fix this by replacing $\le_c$ with a height-bounded version: given a "reasonably closed" (e.g. primitive recursively closed) ordinal $\alpha$, let $x\le_{c,\alpha}y$ iff $x\in L_\alpha[y]$. This then suggests the following version of your question:

If $M$ is a $\beta$-model and $\alpha=ht(M^\mathsf{set})$, must $M$ be closed under $\le_{c,\alpha}$?

The answer to this question is yes: every $\beta$-model satisfies $\mathsf{ATR}_0$, and that's enough to "internally" develop the $L$-hierarchy in the appropriate way.