In the class of ordinals every ordinal $\alpha<\omega^\omega$ is definable over the empty set in terms of $\leq$ alone, because the class of limit ordinals is definable, as is the class of limits of limit ordinals, and limits of limits of limit ordinals, and so on for any finite number of levels. Since the successor function and the 'limit successor' function and the 'limit limit' successor function and so on are all definable you can construct the Cantor normal form of $\alpha$ directly.

Furthermore since ZFC proves $(A,\leq)\equiv(\text{Ord},\leq)$ for any unbounded subclass $A\subseteq\text{Ord}$, we have that for any ordinal $\beta$ that is definable over the empty set, the function $\alpha\mapsto\alpha+\beta$ is definable over the empty set.

The question is: is that essentially it? Defining more complicated ordinal arithmetic seems unclear and anything relying on 'too much ambient set theory' like initial ordinals such as $\omega_1$ must be impossible by forcing absoluteness, right? Relatedly does ZFC prove that $(\omega^\omega,\leq)\equiv(\text{Ord},\leq)$? If not what is the smallest ordinal $\alpha$ such that $(\alpha,\leq)\equiv(\text{Ord},\leq)$

  • $\begingroup$ Have you tried looking into quantifier elimination in the language $\leq, p_n,n \in \mathbb{N}$ where either $\beta$ is "$n$-limit" (limit of limit of ... of limit ordinals) and $p_n(\beta) = \beta$, or it isn't and $p_n(\beta) + \omega^n = \beta$? $\endgroup$
    – nombre
    Jun 18, 2017 at 9:19
  • $\begingroup$ It seems plausible, but I think you need a little more because, for instance, every finite ordinal is definable, but I think you need to use quantifiers to do it in that language. $\endgroup$ Jun 18, 2017 at 22:23

1 Answer 1


Ordinals first order definable in (Ord, <) are precisely those $<ω^ω$. Your argument about definability of ordinals $<ω^ω$ is correct.

The converse follows from the more general statement proved by Itay Neeman in "Monadic Definability of Ordinals": An ordinal is definable in monadic second order logic of (Ord, <) iff it can be obtained using addition, multiplication, and definable regular cardinals (including 0 and 1). Thus, for countable ordinals, only ordinals $<ω^ω$ are definable. For the first order theory, only countable ordinals are definable since the set of definable ordinals is countable and first order properties are preserved under transitive collapse. For the first order theory, there should also be a simpler and more direct proof.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.