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$ZFC+V=L$ implies that $P(\mathbb{N})$ is a subset of $L_{\omega_1}$. But I’m wondering what layer of the constructible Universe contains a smaller set.

My question is, what is the smallest ordinal $\alpha$ such that for all formulas $\phi(n)$ in the language of second-order arithmetic, the set $\{n\in\mathbb{N}:\phi(n\}\in L_\alpha$? Does this depend on whether we assume $V=L$?

I’m guessing $\alpha>\omega_1^{CK}$, and that $\alpha$ is greater than the ordinal $\beta_0$ discussed in my question here. But can we say anything more about it?

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It is consistent that there is no such $\alpha$.

More precisely, it is consistent with ZFC that there is a formula $\varphi$ in the language of second-order arithmetic such that $\{x:\varphi(x)\}$ is not constructible. For example, $0^\sharp$ if it exists has this property (it's $\Delta^1_3$-definable if it exists).


EDIT: Of course, if V = L then such an $\alpha$ trivially exists. Throughout the rest of this answer we assume V=L.

The key point is that there is a "definable translation" between first-order formulas over $L_{\omega_1}$ and second-order formulas of arithmetic:

  • One direction is immediate: any second-order arithmetic formula can be rephrased in $L_{\omega_1}$ since sets of naturals are already elements of $L_{\omega_1}$.

  • The other direction is the interesting one. Given a well-founded tree $T\subset\omega^{<\omega}$ (note that we can definably conflate subsets of $\omega$ and subsets of $\omega^{<\omega}$, and that the set of well-founded trees is second-order definable), we recursively define a map $Set_T$ from nodes of $T$ to sets, by setting $$Set_T(\sigma)=\{Set_T(\sigma^\smallfrown \langle k\rangle): k\in\omega, \sigma^\smallfrown\langle k\rangle\in T\};$$ for example, if $\sigma$ is a leaf of $T$ then $Set_T(\sigma)=\emptyset$. We then let $Set(T)=Set_T(\langle\rangle)$ be the set assigned to the empty string (= the root of $T$). It's easy to check that the relations "$Set(T_0)=Set(T_1)$" and "$Set(T_0)\in Set(T_1)$" are definable in second-order arithmetic, and this gives us an interpretation of $L_{\omega_1}$ into $\mathcal{P}(\omega)$.

The projectively-definable reals are precisely the parameter-freely definable elements of the first-order structure $(\omega,\mathcal{P}(\omega); +,\times,\in)$, and the translation above identifies these with the set $M$ of parameter-freely definable elements of the first-order structure $(L_{\omega_1}; \in)$ (which I'll conflate with $L_{\omega_1}$).

The final point is that since $L$ has definable Skolem functions, $M$ is in fact an elementary submodel of $L_{\omega_1}$ and hence$^1$ $M=L_\eta$ for some $\eta$. This $\eta$ is exactly our $\alpha$. That is:

Assuming V=L, $\alpha$ is the height of the smallest elementary submodel of $L_{\omega_1}$.

In particular, this is massively bigger than $\beta_0$, since $\beta_0$ is parameter-freely definable in $L_{\omega_1}$.


$^1$This is a cute fact. The Condensation Lemma alone doesn't kill this off: in order to apply Condensation we need to know that $M$ is transitive. But a priori, it's not clear that it needs to be - for example, a countable elementary submodel of $L_{\omega_2}$ obviously can't be transitive, since it must contain $\omega_1$ as an element.

So what's special about $\omega_1$ here? The trick here is the following:

Suppose $A$ is a "sufficiently closed" transitive set (= contains $\omega$ and such that eveyr countable element of $A$ is countable within $A$) - for example, $A=L_{\omega_1}$ - and $B$ is an elementary substructure of $A$ (conflating a transitive set with the corresponding $\{\in\}$-structure as usual). Then the set of countable ordinals in $A$ is closed downwards.

Rough proof: Suppose $\theta$ is a (WLOG infinite) countable ordinal in $A$ and $\gamma<\theta$. Since $A$ computes countability correctly we have in $A$ an $f: \omega\cong\theta$. By elementarity "going down," $B$ contains some $g$ which $B$ thinks is a bijection from $\omega$ to $\theta$; by elementarity "going up," $A$ also thinks $g$ is. So (working in $A$) there is some $n\in\omega$ such that $g(n)=\gamma$; but since $n\in\omega$ we have $n\in B$ (we can't "lose" natural numbers!) and so $g(n)=\gamma\in B$ as well. $\Box$

We can generalize the above observation using further closedness assumptions: e.g. if $B$ is an elementary submodel of a sufficiently closed transitive set $A$ with $\omega_1\subseteq B$ then $B\cap\omega_2$ is closed downwards (running the above argument, we only need that $dom(g)\subset B$).

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  • $\begingroup$ What If we assume $V=L$? Then what is $\alpha$? $\endgroup$ – Keshav Srinivasan May 14 at 3:30
  • $\begingroup$ Does $V=L$ imply that $\alpha=\beta_0$, or $\alpha>\beta_0$, or what? $\endgroup$ – Keshav Srinivasan May 14 at 13:47
  • $\begingroup$ Does the fact that $L_{\beta_0}\cap P(\mathbb{N}$ is a $\beta$-model for second-order arithmetic shed any light on whether $V=L$ implies $\alpha=\beta_0$? $\endgroup$ – Keshav Srinivasan May 16 at 7:06
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    $\begingroup$ @KeshavSrinivasan Addressed. $\endgroup$ – Noah Schweber May 19 at 6:28
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What you want is the least $\delta \in \mathsf{On}$ such that $L_\delta \vDash "\mathsf{ZFC}^{-}+V=HC"$ (where $V=HC$ is the assertion that every set is hereditarily countable.) Since the theory "$\mathsf{ZFC}^{-} + V=HC$" is bi-interpretable with second-order arithmetic, you'll get a relativized version of second-order comprehension.

If the relativzed version of $\mathsf{SOA}$ is acceptable to you, then it's possible for $\delta$ to be countable; that said, if you want full comprehension, then as @Noah pointed out, it's consistent that no such $\delta$ exists (countable or otherwise.)

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  • $\begingroup$ What if we assume $V=L$? What does that imply about the smallest $\alpha$ such you get full comprehension in $L_\alpha$? $\endgroup$ – Keshav Srinivasan May 14 at 22:02
  • $\begingroup$ I don't think this is right - I think they want the relative-to-$L$ version. I don't see that $L_\delta$ needs to compute projective truth correctly (in fact, it won't, since there is a projectively definable real coding it). $\endgroup$ – Noah Schweber May 14 at 22:06
  • $\begingroup$ @NoahSchweber I don't understand. Are you saying that $(\omega, L_\delta \cap \mathcal{P}(\omega), \ldots)$ won't provide a Henkin model for the many-sorted analogy of $\mathsf{SOA}$? (This is what I intended to convey with the emphasis on it being relativized.) $\endgroup$ – Not Mike May 15 at 0:02
  • $\begingroup$ @NoahSchweber Oh! Wow I'm dense. Yeah, you're right. $\endgroup$ – Not Mike May 15 at 0:46
  • $\begingroup$ @NotMike So what’s the conclusion? What does $V=L$ imply about all this? $\endgroup$ – Keshav Srinivasan May 15 at 22:17

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