# Does the “special continuum hypothesis” imply the axiom of choice?

In $\mathsf{ZF}$ set theory, does the "special continuum hypothesis" imply the axiom of choice, or is the axiom of choice independent of it?

Here, by the "special continuum hypothesis" we mean the statement that every infinite set of real numbers either has the same cardinality as the set of reals, or the same cardinality as the set of naturals.

There are no implications whatsoever.

As a general rule, if $\varphi$ is a statement which only deals with a single definable set (e.g. the natural numbers, or the real numbers), then it will not imply the axiom of choice.

More correctly, if $\varphi$ is a statement such that for some $\alpha$, $\varphi$ holds if and only if $(V_\alpha,\in)\models\varphi$, then $\varphi$ cannot imply the axiom of choice. Moreover, if $\sf ZFC$ does not prove $\varphi$, then neither will $\sf ZF$ (since we only removed axioms).

The reason is that given a model of $\sf ZFC$, we can always violate choice only above a given rank. So we can violate $\sf AC$ so high, that a statement like "there is an intermediate cardinal between $\aleph_0$ and $2^{\aleph_0}$" (something decided on $V_{\omega+5}$ or so, depending on how you encode functions) cannot possibly imply the axiom of choice, but it will not follow from it either.

More concretely, the continuum hypothesis has many different formulations which are non-equivalent in $\sf ZF$. And stating that there is no intermediate cardinal will not even imply that $2^{\aleph_0}=\aleph_1$.

First of all, note that even the full continuum hypothesis - "Every infinite set which is not countable, admits an injection from $\mathbb{R}$" - does not imply AC. This is because we can have a failure of AC "high up" the set-theoretic hierarchy, so that it just has nothing to do with the reals, at all.
More interestingly perhaps, the "special continuum hypothesis" does not even imply that $\mathbb{R}$ is well-ordered! For example, the Axiom of Determinacy proves that there is no set of reals of intermediate cardinality, but also proves that $\mathbb{R}$ cannot be well-ordered.
• You don't need $\sf AD$ to prove that thing. Solovay model already satisfies that; and in fact you can get it in $\sf ZF$ without large cardinals, as shown by Truss. – Asaf Karagila Aug 12 '15 at 22:04
• That's true - I mentioned AD because it's a neat, philosophically well-motivated axiom, so the fact that AD splits CH and "$\mathbb{R}$ is well-orderable" is especially interesting. But you're quite right that that's overkill. – Noah Schweber Aug 12 '15 at 22:18