Is it possible to prove the axiom of infinity from the real number axioms?

Question

What I mean by the title is whether, given that there is a class $\Bbb R$ and operations $+,\cdot,<$ that satisfy the ordered field axioms and the least upper bound axiom, you can prove the existence of an infinite set in $\sf ZF-Inf$. My intuition says this is possible, because those axioms are sufficient to prove that $|\Bbb R|=|2^\Bbb N|$ in $\sf ZF$, and most (all?) models that satisfy the negation of infinity are countable, which means that classes like this $\Bbb R$ cannot possibly fit in the model. (Note that I am not assuming that $\Bbb R\in V$, which would make the question trivial.)

Answer 1

The question is a bit tricky to formalize. If we work in $\mathsf{ZF}-\text{Infinity} + \neg \text{Infinity}$ then any ordered field is necessarily a proper class, so the least upper bound property is really an axiom schema and not a single axiom. However, I think that in any reasonable formulation the answer to the question is "yes".

Work in $\mathsf{ZF}-\text{Infinity}$. If $\omega$ is a set then we're done, so assume that $\omega$ is a proper class (in which case it is the class of all ordinals.) Then there is a definable surjection $\omega \to V$.

Assume that $\mathcal{R} = (R,+^\mathcal{R},\times^\mathcal{R},<^\mathcal{R})$ is an ordered field. We can either formalize what we are proving as a theorem schema with one theorem for every possible quadruple of formulas defining an ordered field, or we can formalize it as a theorem in the language of set theory expanded by predicate symbols $R$, $+$, $\times$, and $<$. In any case, we will show that $\mathcal{R}$ is not complete.

Recall that $V$ is countable, so in particular $R$ is countable. Let $f : \omega \to R$ be a surjection. We may then use Cantor's proof of the uncountability of the reals to define a $<^\mathcal{R}$-bounded subclass of $R$ with no least upper bound (if $\mathcal{R}$ were complete, we would get an element of $R$ not in the range of $f$, but every element of $R$ is in the range of $f$, so instead the argument tells us that $\mathcal{R}$ is not complete.)

More details: we recursively define a strictly $<^\mathcal{R}$-increasing sequence $(a_n : n<\omega)$ and a strictly $<^\mathcal{R}$-decreasing sequence $(b_n : n<\omega)$ such that for every $n < \omega$ we have $a_n <^\mathcal{R} b_n$ and either $f(n) <^\mathcal{R} a_n$ or $b_n <^\mathcal{R} f(n)$. (We don't need the Axiom of Choice to choose $a_n$ and $b_n$ because the universe is definably well-ordered in order type $\omega$ via $f$.) Then the set $\{x \in R : \exists n \in \omega\,(x <^\mathcal{R} a_n)\}$ is a bounded subset of $\mathcal{R}$ with no least upper bound.

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EDIT: In summary, your intuition is essentially correct. However, we need more than the fact that "most" models (e.g. the intended model $V_\omega$) of $\mathsf{ZF} - \text{Infinity} + \neg\text{Infinity}$ are seen to be countable from the outside. We need to use the fact that the theory $\mathsf{ZF} - \text{Infinity} + \neg\text{Infinity}$ proves "$V$ is countable," which allows us, working in this theory, to use Cantor's argument to rule out the existence of a complete ordered field of the same size as $V$.