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Does there exist a first-order set-theoretic formula $\phi(x)$ using only bounded quantifiers such that, in ZF without assuming the Axiom of Foundation/Regularity, $\phi(x)$ is true if & only if $x$ is (the von Neumann construction of) a natural number? The only predicates for defining the naturals that I can find are variations on $\phi(x) := \forall I(\emptyset\in I \land (\forall y\in I)(y\cup\{y\}\in I) \to x\in I)$ (which isn't bounded) and $\phi(x) := x=\emptyset \lor ((\exists k\in x)(k\cup\{k\}=x) \land (\forall y\in x)(y=\emptyset \lor (\exists k\in x)(k\cup\{k\}=y)))$ (which needs not be satisfied by only the naturals in non-well-founded theories; e.g., the Quine atom satisfies this predicate).

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up vote 3 down vote accepted

I don't think this is possible, for the following reason:

The characteristic property of a formula with bounded quantification $\phi(x)$ is that the truth of $\phi(x)$ for some $x$ in a model $M$ depends only on how the membership relation behaves on the transitive closure of $x$, not on what is in the rest of the model.

Therefore, if we can find two models $M$ and $N$ of $\text{ZF}-\text{Reg}$ with elements $x\in M$ and $y\in N$ such that $x$ is a natural in $M$ but $y$ is not one in $N$, yet the transitive closures of $x$ in $M$ and $y$ in $N$ are isomorphic, then there cannot exist a bounded-quantifier formula that defines the naturals.

Here's a slightly convoluted way to do that (assuming, of course, that ZF is consistent):

Let $N$ be a model of $\text{ZF}-\text{Reg}$ with Aczel's anti-foundation axiom.

Inside $N$, carry out the usual compactness argument that there is a model $M$ of ZF which contains a non-standard integer $x$.

Still working inside $N$, look at the membership structure of the transitive closure of $x$. Because of the anti-foundation axiom, $N$ contains some $y$ whose transitive closure is exactly isomorphic to $x$. But $y$ cannot be a natural in $N$, because $(M,x)$ was explicitly constructed such that $x$ would be a non-standard natural when seen from $N$.

Now, even though $(M,x)$ was constructed inside $N$, we can easily lift it out of $N$ to become a model that lives in our universe side-by-side with $N$. It is still a model of $\text{ZF}-\text{Reg}$ (because everything is first-order, so the conditions for being a model inside $N$ guarantee that the lifted model is a model at our metalevel too), and similarly the isomorphism between $\bar x$ and $\bar y$ that exists in $N$ induces an isomorphism at the metalevel. This completes the proof.

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The method below fails because we need at least one unbounded quantifier in order to obtain the power set of $x$, however I am keeping this answer here so that others attempting to solve this problem will have a starting point.

To define $\omega$ we either need to find a way to express its minimality as an inductive set, or its minimality as an infinite ordinal; or we need to find a way to define a finite ordinal and write $\omega$ as the set of finite ordinals.

  • $x$ is transitive, that is every element of an element of $x$ is also an element of $x$.
  • $x$ is linearly ordered by $\in$, that is given two distinct elements of $x$, one is the element of the other.
  • $x$ is inductive, that is the set with no elements is in $x$, and for every $k$, element of $x$, there is another element of $x$ which is equal to $k\cup\{k\}$.
  • $x$ is well-founded, that is for every element of $\mathcal P(x)$ which is non-empty there is a $\in$-minimal member.
  • $x$ contains only finite sets, that is every $k$, element of $x$, has the property that every collection of subsets of $k$ is either empty or contains a $\subseteq$-maximal element.

Note that I use the power set twice, once for well-foundedness and one for finiteness. I also use Tarski's definition of finite (Dedekind's definition would also be applicable and would require quantification over elements of $\mathcal P(x\times x)$, so we use power sets again as well as products).

It should be remarked that finite-ness and well-foundedness are both second-order properties, i.e. there are models of ZF in which there are uncountably many [internally] finite sets. Spooky but true. It seems a bit unlikely to find a way to define them without appealing to unbounded quantifiers, or at least bounding the quantifiers at the power set or so.

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Wouldn't "$x$ is transitive and well-ordered by $\in$" and "$x$ is Dedekind-finite" be enough? – Hagen von Eitzen Dec 20 '12 at 10:10
@Hagen: What about $\omega+1$? :-) – Asaf Karagila Dec 20 '12 at 10:11
I meant to formalize "$x\in\omega$ directly, not $x=\omega$, therefore I used "$x$ Dedekind-finite". – Hagen von Eitzen Dec 20 '12 at 10:17
@Hagen: Oops! I didn't see... I just woke up and my BAC is still not low enough. :-) Either way, that would define a finite ordinal, but not the collection of all finite ordinals. – Asaf Karagila Dec 20 '12 at 10:20
@Asaf Karagila. How do you write "$x$ is well-ordered by $\in$" as a bounded formula over ZF-Foundation? – Ali Kare Dec 20 '12 at 11:21

This is too long to write in comments so let me note here that if we can express "$x$ is an ordinal" as a bounded formula $\mbox{ord}(x)$ over ZF-Foundation then we can also express "$x$ is a natural number". First define (as in Jech, 3rd ed, p. 164) "$x$ is a limit ordinal" as

$$ \lim(x) \mbox{ iff } \mbox{ord}(x) \wedge (\forall u\in x) (\exists v\in x) u\in v.$$


$$x \mbox{ is a natural number iff } \mbox{ord}(x) \wedge (\neg \lim(x) \vee x=0) \wedge (\forall u\in x) (u=0 \vee \neg\lim(u) ).$$

So if it is the case that "$x$ is a natural number" is not equivalent to a bounded formula over ZF-Foundation then neither is "$x$ is an ordinal". Since the usual definition of "$x$ is an ordinal" employs "$x$ is well-founded", it also follows that "$x$ is well-founded" is not equivalent to a bounded formula over ZF-Foundation.

$$ \mbox{ }$$

Note: If the axiom of foundation holds, then it is easy to define "$x$ is an ordinal" as a bounded formula over ZF. There are different ways to do it: $$\mbox{ord}(x) \:\: \mbox{ iff } \:\: x \mbox{ is a transitive set } \wedge x \mbox{ is totally ordered by } \in.$$ Or $$\mbox{ord}(x) \:\: \mbox{ iff } \:\: x \mbox{ is a transitive set } \wedge \: \in \! \mbox{ satisfies trichotomy on }x.$$ Or $$\mbox{ord}(x) \:\: \mbox{ iff } \:\: x \mbox{ is a transitive set } \wedge (\forall u \in x)(u \mbox{ is a transitive set}).$$

The above equivalences are in Kunen 2011, p.34, p.38, p.39.

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