I am studying axiomatic (ZF) set theory and in particular the construction of the natural numbers from axiomatic set theory. Let me start by giving some introduction to my question.
The axiom of infinity says that there exists at least one inductive set. An inductive set is defined to be a set $I$ for which $\emptyset\in I$ and $(a\in I \Rightarrow s(a)\in I)$, where $s(a) = a\cup\{a\}$. One can then prove that there also exists a minimal inductive set $S$ such that $S\subseteq I$ for all inductive sets $I$ and that $S = \cap\{A\subseteq I : \text{A is a inductive set}\}$. This definition does not depend on the choice of $I$. The set $S$ then is claimed to satisfy the Paeno axioms, by which it can act as a model for the natural numbers. I'd like to prove this, and my question is about one of the first steps of this proof, I think.
What I am trying to prove is that $S = E$, where $E = \{\emptyset, s(\emptyset), s(s(\emptyset)),s(s(s(\emptyset))),…\}$. I think it's clear that, if $E$ exists, $E$ is the minimal inductive set. But I am not sure how to prove that $E$ exists. I do have an idea, but I am wondering if it is valid. It goes simply like this.
We can describe $E$ as $E=\{x\in S : x\in \{\emptyset, s(\emptyset), s(s(\emptyset)),s(s(s(\emptyset))),…\}$. Then, by the separation axiom, E exists.
My main question is basically: does this work? Is this (part of the) proof valid? The reason that I doubt this is because the condition $x\in \{\emptyset, s(\emptyset), s(s(\emptyset)),s(s(s(\emptyset))),…\}$ might not be a valid condition that can be used together with the separation axiom. A difficulty is for example that it is not always clear if a given set $B$ is an element of $E$, as we need to check if $B$ is any of the elements in $E$, of which there are infinite.
Another question, closely related to the former, is whether or not the separation axiom implies that all 'subcollections' of a given set are in fact themselves sets. If this is true, then indeed this guarantees that $E$ exists.
EDIT:
I think I found an anwser to my first question. We can prove that the minimal inductive set is $E = \{\emptyset, s(\emptyset), s(s(\emptyset)),s(s(s(\emptyset))),…\}$ by using the separation axiom. We just need to describe $E$ in terms of first order logical propositions. We can do this as follows.
$E = \{n \in S : (\neg \exists m(m\in S \wedge(s(m)=n)))\rightarrow n=\emptyset \}$. Basically, it says that each element $n$ of $E$ either has a 'predecessor' $m$ (such that $s(m)=n$) or is the empty set. This makes sure the empty set is the only 'starting point' and thus all other elements are related to the empty set by repeated succession. This is correct ZF language, I believe, so this should do the trick. By the separation axiom $E$ exists. And since $C$ is itself an inductive set, and is contained in the minimal inductive set, it should itself be the minimal inductive set.
If anyone agrees or disagrees, I would love to hear it!
Best Answer
To have any hope of proving, in axiomatic set theory, that $S=\{\emptyset,s(\emptyset),s(s(\emptyset)),\dots\}$, you'd first have to express $\{\emptyset,s(\emptyset),s(s(\emptyset)),\dots\}$ in the language of axiomatic set theory. The difficulty here is the ellipsis "$\dots$". I don't see any way to define that without a prior notion of "natural number" or "finite" or some such concept. I think you'll be much better off of you try to prove directly from the definition of $S$ that it satisfies the Peano axioms.
You're right that the separation axioms won't give you $E$, at least not until you handle the issue of expressing $E$ in the language of axiomatic set theory. The separation axioms apply only to formulas in that language. In particular, they do not say that all "subcollections" of a given set are again sets.