Are numbers sets?

One of the milestones of contemporary philosophy of mathematics is Benacerraf‘s 1965 article “What Numbers Could Not Be“.1 There he offers a compelling argument according to which numbers cannot be considered as sets — namely, they cannot be metaphysically identified with sets.

The argument is quite simple: if numbers were sets, we should be able to find a unique progression of sets with which numbers can be identified. But this is apparently impossible: there is a lot of ω-series that can serve as well for the aim. For example, we can adopt von Neumann’s series, and say that $0=\emptyset, 1=\{\emptyset\}, 2=\{\emptyset,\{\emptyset\}\}$, and so on, where the successor function $S$ is defined by $S(x)=x \cup \{x\}$. Or we can adopt Zermelo’s series, and say that $0=\emptyset, 1=\{\emptyset\}, 2=\{\{\emptyset\}\}$, and so on, where the successor function $S$ is defined by $S(x)=\{x\}$. Now, the problem is: is $3=\{\{\{\emptyset\}\}\}$ or is $3=\{\emptyset,\{\emptyset\},\{\emptyset,\{\emptyset\}\}\}$? Benacerraf presents then the example of two children, Ernie and John. The first learned that von Neumann’s ordinals are the natural numbers, while the latter that Zermelo’s ordinals are the natural numbers. Now, they will be easily able to learn arithmetic set theoretically via the above constructions, and they will agree on any arithmetical theorem, except that for Ernie it is true that $3 \in 17$, while for John it is false!

It can sound quite odd to ask whether a number belongs or not to another number. It is actually not an arithmetical question. But this is the point: Ernie and John agree on every arithmetical question; they disagree only on non-arithmetical issues, but these issues cannot be considered as essential in order to point out the metaphysical status of numbers. We can do arithmetic either with Zermelo’s ordinals or von Neumann’s ordinals; but if we are going to metaphysically identify numbers with sets, we must choose one of the two. We cannot admit that $\{\emptyset,\{\emptyset\},\{\emptyset,\{\emptyset\}\}\}=\{\{\{\emptyset\}\}\}$. But how can we choose? According to Benacerraf, we cannot choose, since there is no arithmetical reason to prefer one series on the other. Thus, for we can’t have more than one set of natural numbers, we must admit that numbers are not sets.

But then? What are numbers? Benacerraf’s solution is that we have to move from objects to structures. What permits to Ernie and John to agree on arithmetical theorems is not the nature of any single number, but the fact that they are considering two different instantiations of the very same structure — so we can say that numbers are anything that has the right kind of structure.

Benacerraf’s article has been variously discussed over the years but there have been very few attempts to directly challenge his argument. One of the most interesting of these attempts can be found in Eric Steinhart’s 2002 article “Why Numbers are Sets”.2 In this article, Steinhart argues that we actually have reasons to prefer von Neumann’s ordinals over Zermelo’s ordinals (and also over any other alternative solution), and hence we can overcome Benacerraf’s worries and say that yes, numbers are really (metaphysically) sets. Benacerraf poses two conditions on a set-theoretic structure $\alpha= (\omega,f,e,\prec)$ in order to be a good candidate for being the natural numbers:3

• arithmetical condition (AC): $\alpha$ satisfies the AC iff $(\omega,f,e)$ is a model of the Dedekind-Peano Axioms;
• cardinal condition (CC): $\alpha$ satisfies the CC iff it identifies the numerical ‘less than’ relation $<$ with the set-theoretic relation $\prec$ such that the cardinality of a set $S$ is $n$ iff there is a 1-1 correspondence between $S$ and $\{m|n \prec m\}$.

Taken together, AC and CC constitutes the “Natural Number condition” (NNC). According to Benacerraf, NNC suffices to define the natural numbers, so that any other condition we want to eventually add is just over the necessary — that’s what makes the choice between Zermelo’s and von Neumann’s ordinals impossible. Steinhart, on the contrary, claims that actually

there is one set of sets that stands out very clearly for the mathematicians as the natural numbers. The mathematicians standardly identify the natural numbers with the finite von Neumann ordinals. They make the identification because not all apparent reductions satisfy the NN-conditions equally well. (p. 345)

Five reasons, according to Steinhart, justify this preference:

1. the set of von Neumann ordinals is recursively defined;
2. its sets uniquely satisfies certain ordering conditions;
3. it is uniformly extendible to the transfinite ;
4. it is a minimal $\omega$-series;
5. its $n$-th member is the set of all $m$ less than $n$.

1-5 actually specifies five further conditions, beyond NNC, that $\alpha$ has to satisfy in order to be the natural numbers. However, one might still hold that, even if 1-5 justify the mathematicians’ preference for von Neumann ordinals, they still don’t compel us to admit that numbers have to be metaphysically identified with this set of ordinals. Indeed, 1-5 seem to be rather stylistical reasons and seem not to have a metaphysical relevance for our choice. Yet, what is interesting in Steinhart’s article is a mathematical proof he gives to convince us that actually natural numbers are von Neumann ordinals. The proof runs as follows.

CC implies the following axiom (C1) and definition (C2):

• C1: for all $x$, if $x$ is in $\omega$, then there exists a set $x^*=\{z \in \omega | z;
• C2: the cardinality of any set $S$ is $x$ iff there exists some 1-1 correspondence between $x$ and $x^*$.

Now, let’s suppose we choose an $\alpha=(\omega,f,e,\prec)$ to serve as our natural numbers. Because of CC, we must admit that, for each $\alpha$-number $\alpha_n$, the set of $\alpha$-numbers less than $\alpha_n$ is in the NN-universe (i.e., it exists). Thus, if we assume that $\alpha$ is the natural numbers, we must admit that our NN-universe contains the cardinality sets $\{\emptyset\}, \{\alpha_0\}, \{\alpha_0, \alpha_1\},\{\alpha_0, \alpha_1, \alpha_2\}$ and so on. Hence, the NNC entails that $\omega^* = \{x^*|x \in \omega\}$ is in the NN-universe. Similarly, we can define $f^*, e^*$ and $\prec^*$, and show that they are all in our NN-universe. So, we can prove that if $\alpha=(\omega,f,e,\prec)$ satisfies the NNC, then $\alpha^*=(\omega^*,f^*,e^*,\prec^*)$ is in the NN-universe and satisfies the NNC. Therefore, if $\alpha=N$, then $\alpha$ satisfies the NNC; but if $\alpha$ satisfies the NNC, then $\alpha^*$ satisfies the NNC, and hence we can say that $\alpha^*=N$. It follows that $\alpha= \alpha^*$. This means that $\omega=\omega^*$, and therefore that $n=n^*$ for any $n$ in $\omega$. But if $n=n^*$ for any $n$ in $\omega$, then $0=\emptyset$ and $n=\{0,1,2,\ldots,n-1\}=\{n-1 \cup \{n-1\}\}$, and these are precisely the von Neumann ordinals. Then, if $\alpha$ is the natural numbers, then $\alpha$ is the von Neumann finite ordinals.

The argument is very interesting, but I think it is not valid. I am not completely sure about it, so I will present my objection very cautiously. The key point in Steinhart argument consists in the fact that the C1 imposes the existence, for any $x$ in $\omega$, of the set $x^*=\{z \in \omega | z \prec x\}$. According to Steinhart,

If the NN-conditions assert some rule $R$, then the NN-universe contains the domain of $R$, the range of $R$, the extension of $R$, and nothing else. For if we cannot reason to the existence of these objects in the NN-universe, then that rule is meaningless (it plays no role in determining the models of the NN-conditions). (p.353)

This is what warrants Steinhart in saying that $\omega^*$ is in the NN-universe, and that’s precisely what I can’t understand. It seems to me that there is no need to say that the NN-universe has to contain all this stuff. Of course, the domain, the range and the extension of $R$ must be contained in the universe of our set theory (the set theory in which we carry on the reduction), but there’s no need to say that they have to be contained in the NN-universe. Now, if I am right in noticing this, then we are no longer entitled to say that $\omega^*$ is in the NN-universe. It is in the universe of our set theory, and, since it satisfies NNC, this means that we have a different way to identify numbers with sets; but we can no longer conclude that $\alpha=\alpha^*$, and hence the proof is no longer valid. Simply, $\alpha$ implies the existence in our set-theory universe of $\alpha^*$, which is needed in order to say that the cardinality of a set $S$ is $n$ iff $S$ can be put in a 1-1 correspondence with the set $n^*=\{m|m \prec n\}$. It remains a matter of style whether we want adopt this or that progression of sets, but nothing compels us to say that if we can identify numbers with sets, then we cannot identify numbers with anything but von Neumann finite ordinals.

A further consideration can be made, as a conclusion of this post. If I am right in rejecting Steinhart’s proof (and I am not sure I am right in doing this — I restate), then the only way to reject Benacerraf’s conclusion is to hold a strong version of mathematical naturalism: numbers are von Neumann finite ordinals because that’s what working mathematicians do. And, viceversa, if we agree with Benacerraf, we cannot stick ourselves to a too much strong version of mathematical naturalism.

[1] P. Benacerraf (1965), “What Numbers Could Not Be”, The Philosophical Review, vol. 74(1), pp. 47-73. Reprinted in P. Benacerraf and H. Putnam (eds) (1984), Philosophy of Mathematics, Cambridge University Press, New York, pp. 272-295.
[2] E. Steinhart (2002), “Why Numbers Are Sets”, Synthese, vol. 133(3), pp. 343-361.
[3] $\omega$ is a set of sets, $f$ is a one-to-one function from $\omega$ to $\omega$ , $e$ is a particular set belonging to $\omega$, $\prec$ is a set-theoretic relation. The idea is that if we identify $\alpha$ with the natural numbers, we have that $\omega$ is the set of natural numbers, $f$ is the successor function, $e$ is the initial number $0$ and $\prec$ is the ‘less than’ relation.