# Is $\mathsf{ZFC+V=L}$ consistently $\omega$-complete?

This was previously asked and bountied on MSE:

For brevity, let $$T$$ be $$\mathsf{ZFC+V=L}$$.

Say that an extension of $$\mathsf{ZFC}$$ is $$\omega$$-complete iff it has exactly one $$\omega$$-model up to elementary equivalence. While the $$\omega$$-incompleteness of $$T$$ is easily provable in theories only slightly stronger than $$T$$ itself, I don't immediately see how to do it in $$T$$ alone. My question is:

Is the theory $$S:=T+$$ "$$T$$ is $$\omega$$-complete" consistent?

Here are a couple observations:

• If we replace "$$\omega$$-model" by "well-founded model," the answer is obviously yes under standard assumptions. Let $$\alpha$$ be the second-smallest ordinal such that $$L_\alpha\models\mathsf{ZFC}$$. Then $$L_\alpha$$ also satisfies "$$\mathsf{ZFC+V=L}$$ has exactly one well-founded model." Unfortunately, we have no analogous hierarchy of $$\omega$$-models, so this is a non-starter here.

• As to the specific choice of theory in question, the point is that (something like) $$\mathsf{V=L}$$ is needed to block an easy proof of a negative answer via forcing. For example, reasoning in $$\mathsf{ZFC}$$, if $$\mathsf{ZFC}$$ had an $$\omega$$-model $$\mathcal{M}$$ it would have a countable one $$\hat{\mathcal{M}}$$, and we could force over $$\hat{\mathcal{M}}$$ to get a non-elementarily-equivalent $$\omega$$-model $$\hat{\mathcal{N}}$$. (Forcing over ill-founded countable models is no harder really than forcing over well-founded ones.) The key point here is that forcing preserves $$\mathsf{ZFC}$$. This breaks down of course for $$\mathsf{V=L}$$ and so this argument is irrelevant here. Given the paucity of techniques we currently have for building models of $$\mathsf{ZFC}$$ in the first place, this seems to be a real issue.

Ultimately I suspect that the answer is negative, but the above two points between them rule out all the lines of attack I've been able to think of so far.

EDIT: In light of Farmer S's answer below, let me explicitly mention a rule of thumb which I forgot: when thinking about properties which are not too far from first-order definable, always consider the hyperarithmetic hierarchy!

For example, for every $$\mathcal{L}_{\omega_1,\omega}$$-sentence $$\varphi$$, if $$\varphi$$ has a model then it has a model $$M$$ which is countable in $$L$$, and moreover the $$L$$-least (real coding a) model of $$\varphi$$ is hyperarithmetic relative to (any real coding) $$\varphi$$. The property "Is an $$\omega$$-model of $$T$$" is expressible as a computable $$\mathcal{L}_{\omega_1,\omega}$$-sentence, and this drives Farmer S's point that $$T^+\in L_{\omega_1^{CK}}$$.

• Is an $\omega$-model one where the natural numbers are the standard ones? I tried to find a definition, and could only find: math.stackexchange.com/questions/1113639/what-is-an-omega-model Mar 8, 2021 at 20:39
• @PaceNielsen Yup, an $\omega$-model is one where the naturals are standard. In general, in any context where we have a theory $T$ together with a canonical interpretation of some theory of arithmetic into $T$, we say that an $\omega$-model of $T$ is one in which this interpretation yields a structure isomorphic to $\mathbb{N}$ itself. The usual examples are: second-order (or higher-order) arithmetic together with the "first-order part" interpretation, and set theories together with the "$\omega$" interpretation. ("Categoricity" results can justify the privileging of a specific interpretation.) Mar 8, 2021 at 20:39
• So, if I'm understanding correctly, what you'd like is (under the assumption that $T$ has an $\omega$-model) the construction of a structure (in the language of set theory) that satisfies $T$, it has a unique proper substructure satisfying $T$, and $\omega$ is standard in the structure. Is that correct? I ask because I'm trying to figure out what "$T$ is $\omega$-complete" would mean as a statement in some language. I can see how to express it using a language that allows infinite conjunctions, but that goes beyond the language of set theory. Mar 8, 2021 at 20:50
• @PaceNielsen "$T$ is $\omega$-complete" is already a statement in the object language: every model $M$ of $\mathsf{ZFC}$ has a notion of "$\omega$-model of $\mathsf{ZFC}$," namely "model of $\mathsf{ZFC}$ whose $\omega$ is isomorphic to my $\omega$." There's no linguistic or otherwise "meta" issue here. (In particular, note that if $M\models\mathsf{ZFC}$ is not an $\omega$-model itself then the $\omega$-models of $\mathsf{ZFC}$ in the sense of $M$ - assuming $M$ thinks there are any at all - will not be genuine $\omega$-models.) So the $S$ of the OP is a genuine first-order theory. Mar 8, 2021 at 20:55
• @NoahSchweber Using Barwise compactness one can prove an appropriate version of Gödel's second incompleteness for this context to shows that for any r.e. extension $T$ of ZF, if $T$ has an $\omega$-model, then $T$ has an $\omega$-model that satisfies "$T$ has no $\omega$-model", which provides a high level explanation of what's happening in Farmer F.'s recursion-theoretic solution. I will try to the flesh out this as an alternative answer to your question "before long". Mar 10, 2021 at 3:48

Claim: $$T+$$"$$T$$ is $$\omega$$-complete" is inconsistent. For suppose it's consistent and now work in a model $$V$$ of this theory. Let $$T^+$$ be the resulting completion of $$T$$ (i.e. the unique theory of the $$\omega$$-models of $$T$$ in the sense of $$V$$). Then note that $$T^+$$ is a $$\Delta^1_1$$ real, so $$T^+\in L_{\omega_1^{\mathrm{ck}}}$$. But $$L_{\omega_1^{\mathrm{ck}}}\subseteq\mathrm{wfp}(M)$$ whenever $$M\models T$$ is an $$\omega$$-model, and therefore every real $$x\in L_{\omega_1^{\mathrm{ck}}}$$ is such that $$x\leq_{\mathrm{T}} T^+$$ (the sub-$$\mathrm{T}$$ there being "Turing", as opposed to the theory $$T$$). (Given $$x$$, fix a wellorder $$W$$ of $$\omega$$ in ordertype $$\alpha$$ with $$x\in L_\alpha$$. Then (roughly) $$T^+$$ models "$$W$$ is a wellorder", and can recover $$x$$ from $$W$$. (Edited in:) Formally, fix an integer $$e$$ which indexes a recursive wellorder of $$\omega$$ in ordertype $$\alpha$$ with $$x\in L_\alpha$$. Recall $$L_\alpha$$ projects to $$\omega$$, and $$x\leq_{\mathrm{T}} t^{L_\alpha}$$, the first-order theory of $$L_\alpha$$. Fix a Turing reduction $$n$$ of $$x$$ from $$t^{L_\alpha}$$. Then for $$m<\omega$$, we have $$m\in x$$ iff $$T^+$$ contains the statement "Let $$\beta$$ be the ordertype of the wellorder coded by $$e$$, and let $$y\leq_{\mathrm{T}} t^{L_\beta}$$ via the $$n$$th Turing program; then $$m\in y$$".) But with $$T^+\in L_{\omega_1^{\mathrm{ck}}}$$, this gives a contradiction.
• Very nice! However, I think your proof that $x\le_TT^+$ isn't complete. Specifically, we cannot computably build $L_\alpha$ from a copy of $\alpha$, so "[we] can recover $x$ from $W$" isn't really correct on the nose. (In fact, given a well-ordering $A$, the only reals computable from all copies of $A$ are the computable ones.) I think I see how to get around this though. Mar 8, 2021 at 21:12
• Hmm, I guess it also doesn't really make sense to say $T^+$ models "$W$ is a wellorder''; it should be that we let $e$ be some integer which is the index of a wellorder of $\omega$ of the right length in whatever coding we're using, and from that integer, $T^+$ recovers $x$. (That's more formally what I had in mind.) Mar 8, 2021 at 21:15
• I think the following works: grab a whole model instead of just the theory. Specifically, working within $V\models S$, the $L$-least real $r$ coding an $\omega$-model of $\mathsf{ZFC+V=L}$ is hyperarithmetic and hence its jump is an element of the structure it codes, which it consequently computes - an obvious absurdity. Mar 8, 2021 at 21:15
• (I'm not saying to computably build (the theory of) $L_\alpha$ from the wellorder, but that for each $x$, there are integers $e,n$ such that $m\in x$ iff $T^+$ contains the statement "Let $\alpha$ be the ordertype of the wellorder given by $e$, and $y$ be the $n$th real of the theory of $L_\alpha$; then $m\in y$".) Mar 8, 2021 at 21:23