Why is the infinite set from the axiom of infinity the natural numbers?

Question

Why is the infinite set from the axiom of infinity the natural numbers?

Is there any reason such set was chosen? Couldn't the axiom yield a set that looks like $\Bbb R$ for example?

Answer 1

Simplicity.

The language of set theory contains exactly one symbol, which is a binary relation symbol. That symbol is $\in$ and it is used to represent the membership relation.

In fact in sufficiently old treatments, you will find that even $=$ is dropped out of the language and is added via the conservative addition as it is definable from $\in$ when assuming the axiom of extensionality.

Now you also need to observe that the infinite set guaranteed by the axiom of infinity is not $\Bbb N$.

1. It is an inductive set, which may or may not be $\omega$ (the least infinite ordinal).
2. $\omega$ can be used to model the natural numbers internally to set theory. But we can also choose different interpretations. The important thing is that we can prove they are all "essentially the same" (read: isomorphic).

So what is $\Bbb R$? You could argue that $\Bbb R$ is the unique linear order which is both Archimedean and complete. But how would you express this in the language of set theory? You will have to express what is a linear order, what does it mean to be Archimedean (which will include invariably talking about $\Bbb N$, or $\omega$, thus postulating its existence). And what does it mean that it is unique?

And you will have to do all this work in your axioms. Axioms are supposed to be simple as possible. They should be almost definitions (and some argue that they are definitions). From the axioms of $\sf ZFC$ we can prove that we can internalize first-order logic, define semantics in a sound way, and that there exists a unique-up-to-isomorphism such and such structure or another.

Now compare this to the axiom of infinity which simply postulates the existence of an inductive set. Not even "there is a least inductive set". Just the existence of one inductive set.

This is simplicity. This is beauty.

Andres remarked that we can postulate other type of "infinite sets" existing. We can postulate that there exists a set which is not Dedekind-finite, or that there is a set which is not finite using other formulations that do not refer to the natural numbers. But usually these formulations require something like "injection" or "power set" or "maximal element". All those things require us to first interpret ordered pairs, and what does it mean for something to have certain properties. But ordered pairs can be interpreted in several different ways, not just the Kuratowski interpretation. Note that the Replacement axioms, which ultimately talk about functions, don't refer to functions as objects. They talk about formulas which define a "functional relation", without talking about the set or class of ordered pairs they can be used to define.

So again you end up with something which is more contrived, or more complex to state, and requires an extra mile of work. And again, compare this to the simplicity of the following axiom: $$\exists A\bigg(\exists z\big(z\in A\land\forall y(y\notin z)\big)\land\forall x\big(x\in A\rightarrow(\exists z(z\in A\land\forall y(y\in z\leftrightarrow(y\in x\lor y=x))))\big)\bigg)$$ Or, if you prefer the simpler formulation after adding $\varnothing$ to the language, and using $\{\}$ notation, $$\exists A(\varnothing\in A\land\forall x(x\in A\rightarrow x\cup\{x\}\in A)).$$

Answer 2

No reason other than canonicity. It is not hard to see in the presence of the other axioms of set theory that if there is an infinite set, then there is a countably infinite one. (And you do not need the axiom of choice for this to be the case, see here for a short sketch.)

Answer 3

EDIT. I missed the paragraph where Asaf comments briefly on inductive sets not necessarily being $\omega$, but I'd like to keep the observation that that (infinite) cardinals are inductive.

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Andrés' and Asaf's respective answers are nice, but I would like to add a minor, though not trivial remark.

The set $A$ (in the notation of Asaf's answer) that the Axiom of Infinity gives you is not necessarily the set of the natural numbers, or more correctly, the first infinite ordinal $\omega$. You actually need to separate the natural numbers out of this set by using comprehension (or if you like, taking the intersection of the class of all inductive sets, i.e., sets satisfying the property ascribed to $A$ by the axiom).

Actually, $2^{\aleph_0}$ (the cardinal of $\mathbb{R}$, under AC) is inductive: it is an example of $A$ satisfying the Axiom of Infinity.

Answer 4

To get a set that "looks like" $\mathbb R$ (in the sense of having that cardinality), you can take $2^A = \text{the set of all subsets of }A$, where $A$ is countably infinite. To show that that exists, you first need to know that a countably infinite set exists.