$AC_\omega(^\omega\omega)$ implies that $\omega_1$ is regular.

Question

So this statement was something I had skipped before, and I tried to prove it, but I am still not sure if I am sneaking in some choice somewhere, so I would be glad if anyone could point an error in my reasoning.

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Claim. $AC_\omega(^\omega\omega)$ implies that $\omega_1$ is regular.

Proof. Since $\operatorname{cf}(\omega_1)$ is a cardinal $\le \omega_1$, and since we know it's not finite, assume towards a contradiction that $\langle \alpha_n: n \lt \omega\rangle$ is a cofinal sequence of countable ordinals below $\omega_1$.

Now since using the Cantor-Bernstein theorem we can show that $|P(\omega\times\omega)| = |{^\omega\omega}|$, we can assume $AC_\omega(P(\omega\times\omega))$.
Now let $T = \{R \in P(\omega\times\omega): R \text{ is a well-ordering}\}$. And set $f:T\rightarrow \omega_1$, $f(R) = \operatorname{otp}(\langle \omega, R \rangle)$. We can easily see that $f$ is surjective.

So let $Z_n = f^{-1}(\{\alpha_n\})$, for each $n$. Now by $AC_\omega(P(\omega\times\omega))$, choose some $R_n \in Z_n$. Now we shall construct a well-ordering on $\omega\times\omega$. Define $\lhd \subseteq {^2(\omega\times\omega)}$, as follows:

$$(n, m) \lhd (p, q) \text{ iff } n \lt p \text{ or } (n = p \text{ and } mR_nq).$$

It's easy to check that $\lhd$ is a well-ordering and it can be seen that $\langle\omega\times\omega, \lhd\rangle \cong \langle\coprod_{n\lt\omega}\alpha_n, \lt^*\rangle$, where $\coprod_{n\lt\omega}\alpha_n$ is the disjoint union of these $\alpha_n$'s and $\lt^*$ is the canonical induced well-ordering. And the reason that the above isomorphism needs no choice, is because that the isomorphisms between the $\langle \omega, R_n \rangle$'s and the $\langle \alpha_n, \in \rangle$'s are unique.

This gives us $|\coprod_{n\lt\omega}\alpha_n| = \omega$. But since $|\bigcup_{n\lt\omega}\alpha_n| \le |\coprod_{n\lt\omega}\alpha_n|$, we have that $|\bigcup_{n\lt\omega}\alpha_n| = \omega$, which is a contradiction.

Is this proof $AC$-free?

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Another small fact that I used here was $|\bigcup_{n\lt\omega}\alpha_n| \le |\coprod_{n\lt\omega}\alpha_n|$.

In general to me it seems if we have a well-ordered index set $I$ and a sequence $\langle A_i: i \in I\rangle$ we can prove $|\bigcup_{i\in I}A_i| \le |\coprod_{i\in I}A_i|$, by just sending each member to the copy in the disjoint union where they appear with least index in $I$.

Do we have $|\bigcup_{i\in I}A_i| \le |\coprod_{i\in I}A_i|$, for general $I$?

Answer 1

Your proof is choice-free to the extent that you had to choose a countable sequence of well-orders, which is doable under your assumptions.

Note that you can also give a direct (and positive) proof of the following claim:

Assume $\sf AC_\omega(\Bbb R)$, if $A\subseteq\omega_1$ is countable, then it is bounded.

This is in fact what you're proving. And the idea is indeed the same. Choose an enumeration of each $\alpha_n\in A$, and create a well-order on $\omega$ which is longer than all of them.

The small fact that you used is true in this case if the union is well-ordered. As it is in this case (as a union of ordinals). But in general there is no reason to think that this is possible, since we might not be able to choose for each $a\in\bigcup A_i$ an index from whence it came.

You can find out in Injection of union into disjoint union why this is in fact equivalent to the Partition Principle. So definitely not choice-free in general.