# The purpose of the $\sf ZFC$ Axiom of Infinity

The purpose of the $\sf ZFC$ Axiom of Infinity seems to me to have no other purpose than to provide a set from which the natural numbers can be extracted. Is this correct? If so, do other $\sf ZFC$ axioms, e.g. Pairing, likewise have no other purpose than to enable this extraction process?

• The purpose of the majority of the axioms of ZFC is to allow us to construct set-theoretic versions of the standard types of objects that we use in mathematics. "Unordered pair" is not directly useful, but in combination with other axioms it lets us construct a set-theoretic version of ordered pair, which is of great importance. – André Nicolas Aug 25 '13 at 20:49
• ??? The purpose of the axioms taken as a whole is to describe the universe of sets as (we think) we understand it. – Brian M. Scott Aug 25 '13 at 20:49
• @DanChristensen: We want a framework within which all standard mathematical constructions can be done. Definitely the ZFC construction of ordered pair is artificial, that's not what ordered pairs really are. And again, if we are introducing individual natural numbers, the convoluted construction starting from the empty set and using sickeningly many braces is not a nice one. But it works. More elaborately, the set of reals can be constructed within ZFC, by say formalizing the Dedekind cut approach. When we do analysis, we can largely forget about the set-theoretic background. (Cont.) – André Nicolas Aug 26 '13 at 2:32
• As asked, this is very naive. Set theory is a theory of infinite sets. What is remarkable of the axiom of infinite is not that it provides us with a formal surrogate for the natural numbers, but rather that this suffices, when combined with the other axioms, to give us the rich landscape that follows. – Andrés E. Caicedo Aug 26 '13 at 3:53
• Still makes no sense, but anyway, under what ought to be the natural interpretation of your question, you literally do not get anything new. These "axioms" are provable in $\mathsf{ZF}$ without the axiom of infinity. In this theory, you cannot prove that $\omega$ is a set, but you can prove that as a (perhaps proper) class, it satisfies both first and second order $\mathsf{PA}$. Yes, you want the redundancy here, because in this theory you cannot prove the existence of infinite sets, so the second order version may be vacuously true, while the first order version has content regardless. – Andrés E. Caicedo Aug 28 '13 at 5:17

Set theory is a theory of infinite sets, one could say that this is the point (that it serves us as a foundation for mathematics is extra, the cherry on top). What is remarkable about the axiom of infinity is not that it provides us with a formal surrogate for the natural numbers (I mean, we better do have something in our axioms that allows us to find such a surrogate, else, this would be a terrible theory of infinity, and an even worse foundation), but rather that this suffices, when combined with the other axioms, to give us the rich landscape that follows.

That said, the axiom of infinity is definitely used to prove many results beyond the construction of the naturals. "There are dense linear orders without end points" is an example. "There is an $\omega_1$-Aronszajn tree" is another. "Every Goodstein sequence terminates", etc. (Note that the last is an example of a statement about the natural numbers.)

Assuming the other axioms, the axiom of infinity is trivially equivalent to the statement "$\omega$ exists". In this sense, any use of the axiom of infinity is just using that there is a "set of natural numbers". But, as the examples above indicate, these uses can lead in many different directions, well beyond anything resembling the natural numbers (and, remarkably, also give us theorems about the natural numbers, as an additional bonus). One can wonder whether using the axiom is truly necessary for some of these results. The answer is yes and, not only that, but we in fact have a very good understanding of what results precisely need the use of the axiom of infinity. Namely, as Peter Smith's answer indicates, the theory resulting from replacing the axiom of infinity with its negation is just first order Peano arithmetic. Any theorem that goes beyond this framework needs the axiom of infinity. (This is not to say that any result which is not explicitly about natural numbers requires the axiom of infinity. We can code and discuss certain infinite objects in this setting, but not everything we would like. A precise formalization of this is carried out in the context of subsystems of second order arithmetic. In particular, the theory known as $\mathsf{ACA}_0$ allows us to prove some explicit results about some infinite sets, without requiring any commitments beyond the resources of first order Peano Arithmetic. See here for a brief introduction, and this book for details.)

In particular, the axiom of infinity goes well beyond the Peano axioms (and not simply in terms of consistency strength or expressive power). The Peano axioms are provable in $\mathsf{ZF}$ without the axiom of infinity. In this theory, you cannot prove that $\omega$ is a set, but you can prove that as a (perhaps proper) class, it satisfies both first and second order $\mathsf{PA}$. Typically, the second order formulation of the axioms subsumes the first order formulation, but not here, since in this theory one cannot prove the existence of infinite sets, so the second order version may be vacuously true, while the first order version still has content.

• Thanks Andres. I have accepted your answer, but I wonder what you think of an alternative to the AOI that I proposed a few days ago at math.stackexchange.com/questions/472045/… Do agree with posters there that it is equivalent to AOI? – Dan Christensen Aug 28 '13 at 19:26
• I'll take a look later today, thanks for the link. – Andrés E. Caicedo Aug 28 '13 at 19:28

Here's a well-known bit of folklore. The theory "ZFC - the axiom of infinity + the negation of the axiom of infinity" is equivalent to Peano Arithmetic in the sense that each theory is interpretable in the other and the interpretations are inverse to each other. (Roughly speaking, anyway. The details involved in spelling this out accurately as a tight result are a bit tricky. See the paper by Kaye and Wong.)

Hence there's one good sense in which ZFC with the negation of the axiom of infinity gets you arithmetic -- so, in exactly what sense is the axiom of infinity "needed to get the natural numbers"?

• Well, you could do arithmetic on natural numbers without the axiom. But you wouldn't have the SET of all natural numbers, which is needed to construct the other number systems such as the real numbers. Please correct me if I'm wrong. – Brusko651 Aug 25 '13 at 21:05
• Sure, there won't be an infinite set! But as I recall the OP was originally asking about whether the axiom was "needed to get the natural numbers", though the question has since been edited (I think!). – Peter Smith Aug 25 '13 at 21:08
• @PeterSmith I'm not saying the axiom of infinity is necessary for constructing the natural numbers, but that it seems to be used for this and only this purpose. But I stand to be corrected. Are you saying that the natural numbers can be constructed in ZFC without Infinity? That would be interesting. – Dan Christensen Aug 26 '13 at 2:29
• @Dan: In a model of $\sf ZFC$ with the negation of the axiom of infinity instead, the natural numbers are just the ordinals of the universe. – Asaf Karagila Aug 26 '13 at 3:26
• See also here. – Andrés E. Caicedo Aug 26 '13 at 3:31