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I'm interested in whether ordinal numbers can be described by a first-order theory without presupposing ZFC or any particular set theory. Such a theory might look like Peano arithmetic, but generalized to include all the infinite ordinals that appear in standard treatments. A similar question is asked here, but the top answer there assumes a "background" set theory, and it even explicitly claims that "unlike the Peano axioms, these axioms require a larger theory of sets."

I understand that the standard treatment of ordinals lives within set theory, where the set/class distinction is important. We might like to say something like, "for any set—but not class—of ordinals, there exists a supremum," but logical predicates extend to classes, not sets, so there are clearly some difficult issues that need to be handled carefully. But I would still be surprised if it were impossible to capture the structure of the class of all ordinals and their arithmetic as a standalone first-order theory.

Initial stab:

It seems natural to begin with the axioms of a strictly well-ordered class. We have a built-in binary relation $<$, and we can define $\leqslant,\nless,>,$ etc. in the usual ways.

  1. $\forall x:x\nless x$
  2. $\forall x,y,z:(x<y)\wedge(y<z)\to(x<z)$
  3. $\forall x,y:(x\leqslant y)\vee(y\leqslant x)$
  4. (Schema) $\exists x\varphi(x)\to\exists m\Big(\varphi(m)\wedge\big(\forall y\varphi(y)\to m\leqslant y\big)\Big)$

We should also have a constant $0$ and unary function $s$ obeying the same axioms as in Peano arithmetic, with the added axiom that $\forall x:x<s(x)$.

But this is not enough the guarantee the existence of $\omega$, and we will probably need a special axiom for this, as ZFC also needs an axiom for $\mathbb N$. We could include a unary relation in our language called $f$ (for "finite") and add the axioms $f(0)$ and $\forall x:f(x)\to f(s(x))$. Then we could add an "axiom of infinity": $\exists y\forall x:f(x)\to x<y,$ and from this we can prove the existence of a unique smallest element that is greater than all the finite elements.

But this is where I get stuck/confused. I think we need some analog of replacement to get to $\omega\cdot2$ and higher limit ordinals. But I can't even imagine what we need in order to get to $\omega_1$, since this will require the language having a way of comparing cardinalities.

WillG
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    What does "capture the structure of the class of all ordinals and their arithmetic as a standalone first-order theory" mean precisely? – Noah Schweber Mar 18 '22 at 19:32
  • Couldn't we just have $\omega$ as a constant? "$\omega$ is the least limit" can be written down in your language: $$ 0 < \omega \land \forall x (s(x) \ne \omega) \land \forall y (0 < y \land (\forall x(s(x) \ne y)) \to \omega \le y) $$ – martini Mar 18 '22 at 19:40
  • @martini Good point, I think that works just fine. – WillG Mar 18 '22 at 21:25
  • @NoahSchweber I suppose for one thing, it means you can prove the existence of things like $\omega\cdot2$, $\omega^2$, $\omega^\omega$, $\epsilon_0$, etc, and prove that the arithmetic relations between them (+, •, ^) agree with those of ZFC. But then ideally we could also have a way to define the cardinals and show that something analogous to the aleph numbers must exist. – WillG Mar 18 '22 at 21:30
  • $@NoahSchweber but unfortunately I’m not familiar enough to know what other key properties of ordinals and cardinals are worth calling out and trying to preserve in the logical formalism. – WillG Mar 18 '22 at 21:34
  • The answer is yes to a relevant question https://math.stackexchange.com/q/1702207/1030967. – C7X Mar 18 '22 at 22:52
  • In one of my Coq developments, I produced a definition of ordinals that essentially says: an ordinal is either the successor of another ordinal, or the supremum of some set of ordinals (encoded as a function from some index type $I$ to the ordinals), and is constructed as the smallest type with these constructors. I then defined an equivalence relation to determine which ordinals should be considered equal, as well as a closely related order relation, and proved it's well-ordered. – Daniel Schepler Mar 19 '22 at 00:47
  • (There are some subtleties you need to beware of in order to avoid running into the Burali-Forti paradox: mostly restrictions on the index types $I$ allowed in such a way that $I$ cannot be an index of the type of all ordinals. As it happens, Coq's type hierarchy automatically took care of that in the way I formulated the inductive type.) – Daniel Schepler Mar 19 '22 at 00:49
  • Have you already seen this question https://mathoverflow.net/questions/230504/direct-axiomatization-of-ordinal-and-cardinal-numbers – Cristian Gratie Mar 19 '22 at 20:36
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    Maybe useful: Gaisi Takeuti, Construction of the set theory from the theory of ordinal numbers (1954) – Mauro ALLEGRANZA Mar 22 '22 at 12:49

1 Answers1

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I'm not an expert but I'm very interested in both ordinals and in doing math without ZFC so I'll have a go at it. I'll start by rewriting your axioms to use the same notation throughout and to only use $<$ (personal preference to not write the axioms with symbols that are not part of the signature).

  1. $\forall x: \lnot (x < x)$
  2. $\forall x: \forall y: \forall z: x < y \land y < z \to x < z$
  3. $\forall x: \forall y: x < y \lor y < x \lor x = y$
  4. $(\exists x: \varphi(x)) \to \exists m: \phi(m) \land \forall y : \varphi(y) \to m < y \lor m = y$

Axioms 1 through 4 ensure that $<$ is a well ordering, but put no other constraints on the universe, which can even be empty. Rather than adding $0$ $s$ and the limit ordinal as constants to the signature of the language, I want to try forcing their existence via axioms.

Axiom 4 is a good candidate for identifying unique elements of our language because, once we can prove that $\varphi$ is satisfied by at least one element, the $m$ from the axiom can be proved to be unique. Now, we can add an axiom to make sure our universe is not empty:

  1. $\exists x: x = x$

And then define $0$ from axiom 4 applied to $\varphi(x) := x = x$. For a given $n$ we can also define $s(n)$ from axiom 4 with $\varphi(x) := n < x$ provided that we add the following axiom as well:

  1. $\forall x: \exists y : x < y$

Now, with the above it should be possible to construct $\mathbb N$, but we need more for $\omega$. One obvious candidate is an axiom that postulates the existence of limit ordinals as ordinals without a predecessor.

  1. $\forall x: \exists y: x < y \land \forall z: z < y \to \exists t: z < t \land t < y$

This is for example presented as the infinity axiom for a theory of ordinals and sets of ordinals in [1]. I have changed it to postulate the existence of a limit ordinal larger than any other ordinal, in the hopes that it will help construct the other ordinals, but I'm not sure if it's safe from the Burali-Forti paradox.

In any case, with this axiom it should be possible to get $\omega$ using $$ \varphi(y) : = (\exists x: x < y) \land \forall z: z < y \to \exists t: z < t \land t < y $$ It seems to me also that we could get $\omega \cdot n$ too, but not $\omega^2$.

An alternative for axiom 7 could be an axiom scheme for the existence of limit ordinals, something along the lines of

7'. $ (\exists x: \varphi(x) \land (\forall y: \varphi(y) \to x < y) \land \forall z: \varphi(z) \to z = x \lor \exists t: \varphi(t) \land t < z \land \lnot \exists u: \varphi(u) \land t < u \land u < z) \\ \to \exists v: \forall w: \varphi(w) \to w < v $

which says that an ordinal larger than all elements $x$ satisfying $\varphi$ exists if the set of these elements is such that it has a first element and all other elements have a predecessor in $\varphi$. The problem with this is that it's not obvious how to write a formula $\varphi$ for $\mathbb N$ in order to get $\omega$ from it via axioms 7 and 4.

At least one more axiom would be needed here for transfinite induction, but its exact formulation depends on how limit ordinals are added.

What I would like about an axiom system like this (assuming it actually works, or course) is that each axiom serves a clear purpose related to the ordinals.

I hope this makes sense and contributes ideas for your question.

[1] https://arxiv.org/pdf/math/0502265.pdf

Cristian Gratie
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  • Thanks, this is a good start. I think depending on taste, your axiom 5 can be dropped, since typical formulations of first-order logic allow proving $\exists x:x=x$ from scratch, i.e., they implicitly assume a nonempty domain of discourse. – WillG Mar 21 '22 at 15:09