L-stages in ill-founded countable model of ZFC

Question

I apologize that I need a preamble in which I digress into philosophy, but here it goes:

One of the perceived goals of set-theory is to provide a foundation for mathematics. Natural, rational, real numbers and other structures can be defined inside set-theory. However, students reason about these numbers and structures independently of set-theory. Numbers for example are not naturally perceived as sets, they are atomic, and sets are only needed to collect things. Even in the context of model theory, I feel that I am allowed to work with a model of set-theory, put it aside for a moment, and think about non-set-theoretic objects. They all may coexist independently. I may believe that our "real" sets are sufficiently correctly described by ZFC, so that I believe in the existence of ordinals up to and beyond say $\omega_1$, again alongside other non-set-theoretic atomic objects and structures. I may also have a notion of countable and uncountable collections independent of a concrete set-theoretical framework. Countable collections are enumerable, uncountable are not. This is the key idea of Cantors' diagonalization - there is no exhaustive enumeration of the reals.

End of preamble.

Assume a model of ZFC which is presented as a directed graph: The model universe is the collection of vertices, the element relation is the collection of edges. Vertices are atomic, they are not sets, they represent sets. I choose this representation to avoid distracting concepts around real set membership.

Given such presentation of a model, one can identify some exemplary model-sets. The empty set of the model is represented by the vertex with no inbound edges. The set $\{\emptyset\}$ is represented by the vertex with not inbound edges except the one originating from the empty-set-vertex. And so on.

Define an external mapping $\operatorname{Vert}(s)$ which maps a model-set $s$ to the collection of vertices that represent $s$'s model-elements. Whatever $s$ is, $\operatorname{Vert}(s)$ is a countable collection of vertices, if the model is countable.

Assume we have an ill-founded countable model and want to identify various stages of the constructible universe of that model.

Wikipedia defines: $L_\alpha = \bigcup_{\beta < \alpha} \operatorname{Def} (L_\beta)$

It is not clear to me which ordinals I need to use to construct stages. Is it the "real" ones or the ones the model sees? However, I have an issue with both choices.

Choice 1: Use real (i.e. external) ordinals. I assume that the stages $L_\circ$ satisfy $L_\alpha \subsetneq L_\beta$ for $\alpha < \beta$. If this is so, this should translate to $\operatorname{Vert}(L_\alpha) \subsetneq \operatorname{Vert}(L_\beta)$ for $\alpha < \beta$. Then according to [1], with $\omega_1$ begin externally uncountable, $\operatorname{Vert}(L_{\omega_1})$ should be an uncountable collection of vertices, which is impossible.

Choice 2: Use in-model ordinals. According to [2] there is an infinite reverse path of ordinal-vertices: $\alpha_0 \leftarrow \alpha_1 \leftarrow \alpha_2 \cdots$

How is $L_{\alpha_0}$ even well-defined, when there is no base-case for the recursive definition?

[1] limit of uncountable strictly increasing sequence of sets

[2] Infinite decreasing ordinal chain in ill-founded countable model

Answer 1

Every model $M$ of $\mathsf{ZFC}$, ill-founded or not, has its own version $L^M$ of the $L$-hierarchy constructed internally. The levels of $L^M$ are indexed by the $M$-ordinals.

While the $M$-ordinals may be ill-founded externally, they appear well-founded internally. This makes ordinal-indexed constructions work well from $M$'s perspective: $M$ doesn't "see" a descending sequence in its own ordinals, so as far as $M$ is concerned there is (for example) only one possible candidate for $L_{\alpha_0}$. Note that we know this has to be the case ahead of time since $\mathsf{ZFC}$ proves that transfinite recursion works as desired and $M\models\mathsf{ZFC}$.

Note that countability concerns and the cardinality of $M$ don't enter into this. Moreover, nothing special about the $L$-hierarchy is being used here, simply its status as a definable transfinite recursion construction.

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EDIT: Per the comments below, I think it's a good idea for me to give a toy example of an "internal construction."

Let $\Sigma$ be the language consisting of a single constant symbol $0$, a single binary relation symbol $<$, a single unary function symbol $s$, and a single unary relation symbol $U$. Consider the theory $T$ whose axioms say the following:

• $<$ is a linear order with least element $0$, where each element has an immediate successor given by $s$ and each element other than $0$ has an immediate predecessor.

• $U(0)\wedge\neg U(s(0))\wedge \neg U(s(s(0))\wedge\neg U(s(s(s(0))))$, and $\forall x(U(x)\leftrightarrow U(s(s(s(s(x))))))$. (Intuitively, this says that $U$ names the elements divisible by $4$.)

• Finally, the full first-order induction scheme for all $\Sigma$-formulas.

There is an obvious "standard" model of $T$, but - per compactness - $T$ also has lots of "nonstandard" models (= models where $<$ is ill-founded). Let $M$ be a nonstandard model of $T$, and consider the following imagined dialogue between Alice and Bob:

Alice makes a particular claim about the definable ("internal") structure of $M$:

There is exactly one $M$-definable unary relation $E$ such that $E(0^M)$ and for all $x\in M$ we have $E(x)\leftrightarrow \neg E(s^M(x))$.

Bob takes issue with this.

• "Wait a second!" Bob reasonably objects. "Clearly $E$ is meant to denote the set of even numbers. But (in the absence of addition or multiplication) we define evenness recursively, using the ordering alone, and this doesn't work in $M$ since $<^M$ is ill-founded!"

• "Yes," Alice responds, "that's true - the most natural approach to producing such an $E$ doesn't work. However, consider the formula $\eta(x)\equiv U(x)\vee U(s(s(x))).$ Then we have $$M\models \eta(0)\wedge\forall x(\eta(x)\leftrightarrow\neg \eta(s(x)))$$ as desired."

• "Well, that's not really fair," Bob says, "that was sort of a loophole: you've just reconstructed a desired relation from other built-in structure. Isn't that cheating?"

• "It would be cheating if that additional structure weren't part of $M$ itself," Alice answers. "I'm not claiming that we don't need $U$, but it would be silly not to use it if we're given it."

• "... OK, I think I see why I'm dissatisfied," Bob says. "You've used some of the 'extra structure' of $M$ to produce a definable function with the desired properties. But why should it be unique? By making things depend on structure other than $<$ alone, you've raised the possibility of having multiple such definable relations which disagree - which would clearly ruin your day."

• "True," admits Alice. "But I haven't used all the axioms of $T$ yet! Keep in mind I also have the induction scheme (which is first-order, so permits nonstandard models). Suppose $\psi(x)$ were some other formula such that $M\models\psi(0)\wedge\forall x(\psi(x)\leftrightarrow\psi(s(s(x)))$. Consider the new formula $\theta(x)\equiv \eta(x)\leftrightarrow\psi(x)$. We have $\theta(0)$ (trivially) and $\forall x(\theta(x)\rightarrow\theta(s(x)))$ (go by cases). So by our induction scheme we get $\forall x\theta(x)$, or to put it another way $\forall x(\eta(x)\leftrightarrow\psi(x))$. So $\psi$ defines the same relation on $M$ as $\theta$."

The same thing is going on with the $L$-hierarchy internal to a nonstandard model $M$. While ill-foundedness implies that the $M$-ordinals alone are insufficient for such a construction even if we went beyond first-order logic, the $M$-ordinals together with the rest of $M$ give $M$ the power to produce a definable relation with the desired properties, and additional properties of $M$ (e.g. Foundation) ensure that exactly one definable relation with the desired properties exists. Moreover, there is a single formula which defines the desired definable relation within $M$ for every model $M$ (we don't have to use different formulas in different models).