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


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)

  • $\begingroup$ Perfect. So, just to be clear, the problem with ordinals defined by "finite iteration" is that they cannot be formalized (enclosed in a formula) without assuming already the existance of finite ordinals, right? By the way, what is the model you are referring to? $\endgroup$ – Lorenzo Jun 12 '19 at 6:23
  • $\begingroup$ Yes, without already having a notion of finite. As for the model it's not one specific model, you can prove its existence using the compatcness theorem for instance (assuming ZFC is consistent) $\endgroup$ – Maxime Ramzi Jun 12 '19 at 8:34

We have the following

Proposition: An ordinal $\alpha$ is a natural number if, and only if $\alpha=0$ or is a successor ordinal and for each ordinal $\beta<\alpha$, $\beta=0$ or $\beta$ is a successor ordinal.


$\Longrightarrow$) By induction over the natural numbers. Let

$$A=\{n\in\omega\;|\;(n=0\text{ or }(n\text{ is a successor ordinal}\text{ and }\text{for each }\beta<n,\;\beta=0\text{ or }\beta\text{ is successor})\}$$

  • It is evident that $0$ belongs to $A$
  • Let $n$ be a natural number that belongs to $A$. Then $S(n)$ is a successor ordinal. If $\beta<S(n)$, then $\beta\in n$ or $\beta=n$. In any of those cases $\beta=0$ or $\beta$ is a successor ordinal.

$\Longleftarrow$) Let $\alpha$ be an ordinal which satisfies the property of the statement. By the trichotomy property, exactly one of the following possibilities is given: $\alpha<\omega,\;\alpha=\omega$ or $\omega<\alpha$. If $\alpha=\omega$, then $\alpha$ is not equal to $0$ nor a successor ordinal. If $\omega<\alpha$, then $\omega$ is an ordinal less than $\alpha$ that is not $0$ nor a successor ordinal. The only remaining possibility is for $\alpha$ to be less than $\omega$, that is, $\alpha\in\omega$ so $\alpha$ is a natural number.

So now to you questions. For $3$ I must agree with Max's answer, nothing else should be added.

If you are curious about the actual construction of the set of the natural numbers, please check my answer to this question. After constructing the natural numbers, proving the principle of induction over $\omega$ and the principle of trichotomy of the ordinal numbers, you can immediately give an answer to $4$ via the proposition I just proved before.

Note that I'm also supposing that $\omega$ is not a successor ordinal. Indeed, $\omega$ is different from $0$, and if it were a successor ordinal, then there would exist a natural number $n$ such that $\omega=S(n)$, which cannot be possible, for various reasons: one of them, is that $S(n)$ would be a natural number that is an element of itself. Therefore, $\omega$ is a limit ordinal. But also it is the first limit ordinal, since we have just proved that every element in $\omega$ is $0$ or a successor ordinal.

Hope my answer, although late, is still helpful.


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