Why proof of axiom of choice on $\omega+1$ fails?

Question

I came up with a proof of Michael Greinecker about the truth of "axiom" of choice in finite case (the 3rd answer). Do We Need the Axiom of Choice for Finite Sets?

I understand that he was doing induction on $\omega$ (set of all natural numbers).

As far as I know, in general, we can do induction on a well-founded set, therefore we can do induction on the set $\omega+1$ (sucessor set of $\omega$). I see that if I apply the same logic as the proof of Michael to the set $\omega+1$, I can prove that the axiom of choice is true for all elements of $\omega+1$ which contains $\omega$. It means that there is a proof of axiom of choice for the infinite case (i.e. when there are $\omega$ sets). However, clearly it shouldn't be true. But I fail to understand why the proof of Michael can't be applied to the set $\omega+1$ whereas it was successfully applied to the set $\omega$ given that the 2 sets are well-ordered and they are very alike (i.e. $\omega+1$ has 1 more element than $\omega$ and the other elements are the same).

Could you please help me to figure it out ? Many thanks for your support!

Answer 1

The fact that "we can do induction" doesn't mean that every induction argument holds.

Say that a set is Kuratowski finite if it lies in the smallest class of sets satisfying: $\varnothing$ is Kuratowski finite; if $x$ is Kuratowski finite, then for every set $y$, $x\cup\{y\}$ is Kuratowski finite.

By induction on $\omega$ we can prove that every natural number is Kuratowski finite. And we can also prove that $\omega$ itself isn't. So the induction fails for $\omega+1$.

Even more than that, the statement "Every set is empty" holds, by induction for $1$, but it does not hold for $2$, let alone $\omega$ or $\omega+1$.

There reason is that a proof by induction still requires a proof. Namely, we need to prove the induction step (and base case, if you're doing it with a base case and a step). It might be that we simply cannot prove the step in general for a given situation (e.g., the "every set is empty" example), or what holds in the case of induction is that limit cases require a slightly different formulation which we might not be able to prove (e.g. the case of "Every set is finite" failing for $\omega+1$ while holding for $\omega$).

The latter is the situation with the Axiom of Choice. We can prove that if $A$ is a family of sets which admits a choice function, then $A\cup\{a\}$ will admit a choice function for any non-empty set $a$. This, you might notice, is very close to the concept of Kuratowski finiteness, and those two are indeed related. The consequence is that we can certainly get the Axiom of Choice for families of finite sets. But if a set is infinite, removing a single point does not land us in a place where we already have some induction hypothesis to work with. In other words, if $\{A_n\mid n<\omega\}$ is a family of non-empty sets, and every finite family admits a choice function, the only way to get a finite family is to remove an infinite amount of members. Not just one.

Answer 2

Writing $\mathrm{AC}(X)$ for the axiom of choice for a family indexed by $X$, the limit case for transfinite induction on $\omega + 1$ requires you to prove that $\mathrm{AC}(\omega)$ follows from $\forall n. \mathrm{AC}(n)$, which isn't true.