Why is the Axiom of Infinity necessary?

I am having trouble seeing why the Axiom of Infinity is necessary to construct an infinite set. According to a professor of who's mine teaching a class on "infinity," the Peano axioms are only adequate to establish the existence of all of the natural numbers, but not also that there is an infinite set consisting of them. To do so, we must stipulate not only the Axiom of Induction, but that there also exists an inductive set (via the Axiom of Infinity).

So, why does the existence of an infinite set of the natural numbers not just follow from the existence of all of the natural numbers?

Solution 1:

BrianO's answer is spot-on, but it seems to me you may not be too familiar with models and consistency proofs, so I'll try to provide a more complete explanation. If anything it may better steer you towards what you need to study, as admittedly I'm about to gloss over a lot of material.

Why do we need the axiom of infinity? Because we know (and can prove) that the other axioms of ZFC cannot prove that any infinite set exists. The way this is done is roughly by the following steps:

• Remember a set of axioms $\Sigma$ is inconsistent if for any sentence $A$ the axioms lead to a proof of $A \land \neg A$. This can be written as $\Sigma \vdash A \land \neg A \to \neg Con(\Sigma)$
• If $Inf$ is the statement "an infinite set exists", then $\neg Inf$ is the statement "no infinite sets exist".
• The axiom of infinity is essentially the assumption that $Inf$ is true and hence $\neg Inf$ is false.
• If we don't need the axiom of infinity, then with the other axioms $ZFC^* = ZFC - Inf$, we should be able to prove $Inf$ as a theorem, in other words we'll posit that $ZFC^* \vdash Inf$
• We assume that $ZFC$, and hence the subset $ZFC^*$, are consistent.
• We then add $\neg Inf$ as an axiom to $ZFC^*$, which we'll call $ZFC^+$
• By showing that $(ZFC - Inf) + \neg Inf$ has a model (a set in which all the axioms are true when quantifiers range only over the elements of the set), we can prove the relative consistency $Con(ZFC) \to Con(ZFC^+)$. In other words we're basically just proving $ZFC^+$ is consistent, but we need to be explicit that this proof assumes $ZFC$ is consistent.
• The model we want is $HF$, the set of all hereditarily finite sets. I'll leave it you to verify all the axioms of $ZFC^+$ hold in this set. But the important point is $HF \models ZFC^+$, and our relative consistency is proven. (This follows from Godel's completeness theorem)
• We are assuming that $ZFC^* \vdash Inf$, but because $ZFC^+$ is an extension of $ZFC^*$ it must also be the case that $ZFC^+ \vdash Inf$. But then we have $ZFC^+ \vdash Inf \land \neg Inf$ and is thus inconsistent, a contradiction.

Thus we must conclude that our hypothesis $ZFC^* \vdash Inf$ is false and there is no proof of $Inf$ from the other axioms of ZFC. $Inf$ must be taken as an axiom to be able to prove that any infinite set exists.

Solution 2:

The existence of each natural number follows from the other axioms of set theory, but if you drop the Axiom of Infinity (AxInfinity), the resulting theory ZFC-AxInfinity has a (transitive) model consisting of the hereditarily finite sets, which contains no infinite sets. The axioms of ZFC-AxInfinity provide no way to gather all the natural numbers into a single set.