# Can we make ZF − infinity + “all ordinals are finite” as strong as ZFC?

Consider the theory ZFfin := ZF − axiom of infinity + “all ordinals are finite” (i.e., every ordinal is zero or successor).

Of course, if we add the axiom of choice, this is not very interesting: it's basically first-order arithmetic in disguise. Basically what I'd like to know is whether we can make something combinatorially interesting out of ZFfin (without Choice).

Specifically:

Question: Can we find some axiom (or perhaps axiom scheme) T of a combinatorial flavor which makes ZFfin + T arithmetically at least as strong as ZFC (meaning that all arithmetical statements which are theorems of ZFC should also be theorems of ZFfin+T; of course, we also demand consistency of ZFfin+T)?

I don't have a precise definition of “of combinatorial flavor” but an example of something that is definitely not of that nature is the axiom scheme “all arithmetical theorems of ZFC are true”. I want something which looks like a large cardinal axiom, or a reflection scheme, or something set-theoretic like that, except that you can't do that as such because all ordinals are finite.

(If the question turns out to be too difficult, I might settle for other axioms which make ZFfin much stronger than arithmetic, without going as far as ZFC.)

Motivation: there are various axioms that can be added to the theory NFU (Quine's New Foundations with Urelements) to make it strong in various ways, and can be compared to the large cardinal axioms of ZFC. My question is whether analogous axioms can be found for ZFfin instead of NFU.

• I think ZFfin proves the axiom of choice: ZFfin proves there is no infinite set (i.e., if there is an infinite set, then $\omega$ exists), so every set is finite, and every finite set is well-orderable. – Hanul Jeon Dec 13 '20 at 0:37
• @HanulJeon How do you prove in ZFfin that if there's an infinite set, then $\omega$ exists? – Alex Mennen Dec 13 '20 at 0:46
• @AlexMennen I thought I have proved it, but I may confuse the background theory (especially, the theory that assumes $\in$-induction and $\Delta_0$-Collection.) I will think about it again. – Hanul Jeon Dec 13 '20 at 0:52
• @AlexMennen Take an infinite set $X.$ Then $\mathcal{P}(\mathcal{P}(X))$ has a countably infinite subset, setting $X_n=\{Y \subset X: |Y|=n\}.$ A replacement on that should yield $\omega.$ – Elliot Glazer Dec 13 '20 at 1:12
• Alright. Also, I do think there's an interesting question to be asked here if power set is removed, and we look for strengthenings of ZFfin - Power set which interpret $\text{ZFC}^-.$ – Elliot Glazer Dec 13 '20 at 19:57

## 1 Answer

ZFfin implies every set is finite, and in particular choice holds. Suppose there is an infinite set $$X.$$ By replacing every element of $$\mathcal{P}_{\text{fin}}(X)$$ with its cardinality, we see $$\omega$$ exists, contradiction. So this theory is roughly arithmetic.