January 25, 2012: I have found a mistake in one of the arguments, it is not an important one for the answer (in fact it is flat out irrelevant) but I should rewrite this answer anyway.
The short answer is that $AH(0)$ implies $CH$ under $ZF$, but is unprovable from $CH$ under $ZF$ alone.
$\boxed{ZF\vdash AH(0)\rightarrow CH}$:
Suppose $2^{\aleph_0} = \aleph_1$, suppose $\frak a$ is a set whose cardinality is between $\aleph_0$ and $2^{\aleph_0}=\aleph_1$. $\frak a$ can be well ordered (it can be injected into an ordinal) and therefore has the cardinality of some aleph number. If it is not $\aleph_1$, then it has to be $\aleph_0$.
So in $ZF$ you have that $AH(0)\rightarrow CH$.
From this we have that it is not consistent with $ZF$ that $CH\rightarrow\lnot AH(0)$. We also have that it is consistent relatively to $ZF$ that $AH(0)\land CH$ is true.
$\boxed{ZF\nvdash CH\rightarrow AH(0)}$:
We exhibit a model in which $CH$ is true, but $AH(0)$ is false. Note that this implies that $\aleph_1\nleq2^{\aleph_0}$. Such result can be achieved either when the continuum to be a countable union of countable sets or the presence of an inaccessible cardinal (which is a slight increase in consistency strength).
Surprisingly enough, in the Feferman-Levy model in which the continuum is a countable union of countable sets $CH$ fails. There exists a set of real numbers which is uncountable but there is no injection from $2^\omega$ into the set [1, Remark 3.4].
Consider now the Solovay model of $ZF$, we start with an inaccessible cardinal and end up with a model in which every subset of reals is Lebesgue measurable. It is also a model of the assertion "Every uncountable set of reals has a perfect set", where perfect sets always contain a copy of the Cantor set and therefore have cardinality continuum. In fact, Truss proved in [2] that repeating the Solovay construction from any limit cardinal results in a model where the perfect set property holds, and $AH(0)$ fails, so the inaccessible is redundant for this proof.
Every set of reals, in such model, is either countable or of cardinality continuum. In particular $CH$ holds, but $AH(0)$ not.
Therefore in a model of $ZF$ (without choice) exactly one of the options holds:
- $CH\land AH(0)$,
- $CH\land\lnot AH(0)$ (Solovay's model, and Truss models),
- $\lnot CH\land\lnot AH(0)$ (Feferman-Levy model).
This shows that $CH$ cannot prove or disprove $AH(0)$. Note, by the way, that the first and third options can be found in models of choice, such as Godel's constructible universe and Cohen's construction where the continuum hypothesis fails; the latter can be even shown in models like Cohen's first model where there is a dense Dedekind-finite set of real numbers. However we see more here: the assertion $\aleph_1\leq2^{\aleph_0}$ is unprovable from $ZF$.
Bibliography:
Miller, A. A Dedekind Finite Borel Set. Arch. Math. Logic 50 (2011), no. 1-2, 1--17.
Truss, J. Models of set theory containing many perfect sets. Ann. Math. Logic 7 (1974), 197–219.
You are correct that without the axiom of choice $2^{\aleph_0}\newcommand{\CH}{\mathsf{CH}}$ may not be an $\aleph$. Therefore the continuum hypothesis split into two inequivalent statements:
- $(\CH_1)$ $\aleph_0<\mathfrak p\leq2^{\aleph_0}\rightarrow2^{\aleph_0}=\frak p$.
- $(\CH_2)$ $\aleph_1=2^{\aleph_0}$.
Whereas the second variant implies that the continuum is well-ordered, the first one does not.
You suggested a third variant:
- $(\CH_3)$ $\aleph_0<\mathfrak b\rightarrow 2^{\aleph_0}\leq\mathfrak b$.
Let's see why $\CH_3\implies\CH_2\implies\CH_1$, and that none of the implications are reversible.
Note that if we assume $\CH_3$, then it has to be that $2^{\aleph_0}\leq\aleph_1$ and therefore must be equal to $\aleph_1$. If we assume that $\CH_2$ holds, then every cardinal less or equal to the continuum is finite or an $\aleph$, so $\CH_1$ holds as well.
On the other hand, there are models of $\sf ZF+\lnot AC$, such that $\CH_1$ holds and $\CH_2$ fails. For example, Solovay's model in which all sets are Lebesgue measurable is such model.
But $\CH_2$ does not imply $\CH_3$ either, because it is consistent that $2^{\aleph_0}=\aleph_1$, and there is some infinite Dedekind-finite set $X$, that is to say $\aleph_0\nleq |X|$. Therefore we have that $\aleph_0<|X|+\aleph_0$. Assuming $\CH_3$ would mean that if $X$ is infinite, then either $\aleph_0=|X|$ or $2^{\aleph_0}\leq|X|$. This is certainly false for infinite Dedekind-finite sets (one can make things stronger, and use sets that have no subset of size $\aleph_1$, while being Dedekind-infinite).
One can also think of the continuum hypothesis as a statement saying that the continuum is a certain kind of successor to $\aleph_0$. As luck would have it, there are $3$ types of successorship between cardinals in models of $\sf ZF$, and you can find the definitions in my answer here.
It is easy to see that $\CH_1$ states "$2^{\aleph_0}$ is a $1$-successor or $3$-successor of $\aleph_0$", and $\CH_3$ states that "$2^{\aleph_0}$ is a $2$-successor of $\aleph_0$" -- while not explicitly, it follows from the fact that I used to prove $\CH_3\implies\CH_2$.
So where does $\CH_2$ gets here? It doesn't exactly get here. Where $\CH_1$ and $\CH_3$ are statements about all cardinals, $\CH_2$ is a statement only about the cardinality of the continuum and $\aleph_1$. So in order to subsume it into the $i$-successor classification we need to add an assumption on the cardinals in the universe, for example every cardinal is comparable with $\aleph_1$ (which is really the statement "$\aleph_1$ is a $2$-successor of $\aleph_0$").
All in all, the continuum hypothesis can be phrased and stated in many different ways and not all of them are going to be equivalent in $\sf ZF$, or even in slightly stronger theories (e.g. $\sf ZF+AC_\omega$).
Without the axiom of choice we can have two notions of ordering on the cardinals, $\leq$ which is defined by injections and $\leq^*$ which is defined by surjections, that is to say, $A\leq^* B$ if there is a surjection from $B$ onto $A$, or if $A$ is empty. These notions are clearly the same when assuming the axiom of choice but often become different without it (often because we do not know if the equivalence of the two orders imply the axiom of choice, although evidence suggest it should -- all the models we know violate this).
So we can formulate $\CH$ in a few other ways. An important fact is that $\aleph_1\leq^*2^{\aleph_0}$ in $\sf ZF$, so we may formulate $\CH_4$ as $\aleph_2\not\leq^*2^{\aleph_0}$. This formulation fails in some models while $\CH_1$ holds, e.g. in models of the axiom of determinacy, as mentioned by Andres Caicedo in the comments.
On the other hand, it is quite easy to come up with models where $\CH_4$ holds, but all three formulations above fail. For example the first Cohen model has this property.
All in all, there are many many many ways to formulate $\CH$ in $\sf ZF$, which can end up being inequivalent without some form of the axiom of choice. I believe that the correct way is $\CH_1$, as it captures the essence of Cantor's question.
Interesting links:
- What's the difference between saying that there is no cardinal between $\aleph_0$ and $\aleph_1$ as opposed to saying that...
- Relationship between Continuum Hypothesis and Special Aleph Hypothesis under ZF
Best Answer
The order dimension of the Turing degrees is such a cardinal. See K. Higuchi, S. Lempp, D. Raghavan and F. Stephan, On the order dimension of locally countable partial orderings, Proc. Amer. Math. Soc., Vol. 148 (2020), no. 7, 2823--2833.