Let us work in $\mathsf{ZF-Inf}$, i.e. Zermelo-Frankel without the Axiom of Choice and without the axiom of infinity. For each set $A$, say that $F$ is an inductive family of subsets of $A$ iff $\varnothing \in F$ and, if $x \in F$ and $y \in A$, then $x \cup \{y\} \in F$. Define a set to be RW-finite iff it is contained in every inductive family of its subsets. It is RW-infinite iff it is not RW-finite.
Question: Suppose there is an RW-infinite set. Is it the case that $\mathsf{ZF-Inf}$ can prove, given this hypothesis, that there is a set $W$ such that $\varnothing \in W$ and, if $x \in W$, then $x \cup \{x\} \in W$? In other words, does the existence of an RW-infinite set imply the axiom of infinity in its usual formulation?
I can prove that the axiom of infinity implies the existence of an RW-infinite set, but I'm not having much luck proving this converse… any help would be greatly appreciated.
Best Answer
Yes, this is provable.
Specifically, suppose $A$ is RW-infinite. Let $f:\mathcal{P}(A)\rightarrow Ord$ send $X$ to the unique ordinal in bijection with $X$ if such exists, and to $0$ otherwise. By replacement, $ran(f)$ is a set; by RW-infiniteness of $A$, it contains every finite ordinal. From this we can extract $\omega$.
Note that this crucially used replacement. I believe this is necessary, but I don't immediately see how to build a witnessing model (that is, a model of $\mathsf{Z-Inf+\neg Inf+RWInf}$).