I'm studying set theory and I'm focusing on von neumann ordinals. I've built an understanding of the reasoning that brings to the set-theoretic construction of the natural numbers whose soundness I'm questioning. I should stress that I find important point 2 to be the starting point of this construction. I'm going to outline it:
- Let's assume that we have already defined ordinals. We know that every successor of an ordinal is still an ordinal.
- Starting from $0 = \emptyset$, which is the smallest ordinal, we can construct some ordinals by finding iteratively (but finitely) the successor ordinal of the previous one. That is $$0 = \emptyset\\ 1 = \{\emptyset\}\\ 2 = \{\emptyset,\{\emptyset\}\}\\ 3 = \{\emptyset,\{\emptyset\},\{\emptyset,\{\emptyset\}\}\}\\\dots\\ n = \{\emptyset,\{\emptyset\},\{\emptyset,\{\emptyset\}\},\dots\}\\\dots$$ we know that these ordinals exist, right?. What we don't know is if there exist a set containing them.
- We define the class of finite ordinals with the following formula: $$FON(x) = ON(x)\wedge\forall y[(y \le x) \wedge (y \neq 0) \Rightarrow \exists z\{y = z \cup \{z\}\}]$$ So finite ordinals are those successor ordinals whose elements are all successor ordinals. We have that each ordinals $n$ we constructed above satisy this formula. Here is the first question: can we say the converse? Can we say that every $x$ such that $FON(x)$ is explicitely definable (with a finite iteration) as for $n$? However our new objective is to identify natural numbers and finite ordinals, and we want to prove that the class finite ordinals is a set $\omega$ (which we will eventually identify with $\mathbb{N}$).
- We introduce the axiom of infinity: $$\exists I(\emptyset \in I \wedge \forall x \in I((x \cup \{x\}) \in I))$$ and we'll call $I$ an inductive set. We want to show that every finite ordinal belongs to every inductive set. If the equivalence, conjectured in the previuos point, between finite ordinals and ordinals iteratively constructed from $\emptyset$ were to be true, then this would follow easly. Now, thanks to the axiom scheme of specification, we can extract from $I$ the subset of all the elements of $I$ that are finite ordinals. So we define $$\omega = \{x \in I : FON(x)\}$$ We then have that, since every finite ordinal belongs to every inductive set, $$FON(x) \Rightarrow x \in I \Rightarrow x \in \omega$$ on the other hand we have $$x \in \omega \Rightarrow FON(x)$$ by the definition of $\omega$. So finally we get: $$\omega = \{x : FON(x)\}$$
Is this line of reasoning solid? Can we fill the holes in it? Thanks