Finite Ordinals and Natural Numbers

axiomselementary-set-theorylogicnatural numbersordinals

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:

  1. Let's assume that we have already defined ordinals. We know that every successor of an ordinal is still an ordinal.
  2. 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.
  3. 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 $\mathbb{N}$).
  4. 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

Best Answer

Let me try to answer 3. first. It is not even clear that we can state the result you have in mind. Indeed how do you define “finite iteration“ without the finite ordinals under hand ?

And if you already have the finite ordinals, then the question becomes trivially “yes“ because to construct a given finite ordinal as a finite iteration, just use that ordinal as an indexation. However this does not reflect (or so I think) the spirit of your question, which seems to be asking about ”honest” finite iteration, like $1,2, 3$ “and so on“. However the whole problem lies in this “and so on”, which we cannot make precise without appealing to finite ordinals, and then we enter a loop.

Here's one way to answer negatively. Assuming ZF(C) is consistent, it has a model $M$ such that externally, the set of finite ordinals contains a subset on which the membership relation has the order type of $\mathbb Q$. In particular this means that the “finite ordinals“ of this subset cannot be attained by an ”honest finite iteration“. But then, you might say that this model $M$ is artifical, and not the “real“ one : I would agree, but how do you know that from inside of $M$ ?

So the answer to 3. is essentially that you will not be able to prove that in any nontrivial meaningful way.

But all hope is not lost. You can still perform induction on those finite ordinals, and prove that they all belong to $I$, where $I$ is any inductive set (which will allow you to conclude for 4. So your reasoning can be saved !

If you know about transfinite induction, then it should be clear what to do : prove by induction on all ordinals $\alpha$ the formula “$\alpha\in I$ or $\alpha$ is not finite“ which should not be too complicated.

If you do not know about transfinite induction, it's not a problem either, you only need to know that the class of ordinals is itself well-ordered. Once you have that, you can simply ponder : if $I$ is an inductive set, what is the least ordinal that doesn't belong to it ? (If there is any ! - but if there isn't well certainly all finite ordinals belong to it; although if you keep learning about ordinals you'll see that this situation doesn't happen)

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