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
The cardinality of the continuum is $2^{\aleph_0}$, the argument is roughly as you describe, although there are some subtleties to be dealt with (e.g. some numbers would have two distinct binary expansions, but we can correct for that).
The problem is that assuming only the axioms of $\sf ZFC$, we are unable to determine whether or not $2^{\aleph_0}=\aleph_1$ or $2^{\aleph_0}=\aleph_2$, or maybe it is actually much larger.
The only thing we can prove that that the cofinality of $2^{\aleph_0}$ is uncountable. This is a technical concept, but in this case we can state it simply: if $F\colon\Bbb{R\to N}$ is any function, then there is some $n\in\Bbb N$ such that $\{r\in R\mid F(r)=n\}$ has the same cardinality as the continuum. This mean that $2^{\aleph_0}\neq\aleph_\omega$, where $\aleph_\omega$ is the first infinite cardinal which has infinitely many infinite cardinals below it.
But we know that this is the only limitation. We can have the continuum as large or as small as we want as long as it satisfies the above requirement.
As a side note, nobody proved recently that $2^{\aleph_0}\geq\aleph_2$. The work of Asperó and Schindler show that assuming a technical axiom called $\sf MM^{++}$, which implies $2^{\aleph_0}=\aleph_2$, we can prove an even more technical axiom known as $(*)$.
But $\sf MM^{++}$ itself is not provable. You can assume it, or you can assume its negation. And whether or not there is an absolute mathematical truth, so far as a mathematical society we seem more keen to assume that it is something like $\sf ZFC$, and so we do not have enough information to determine whether or not $\sf MM^{++}$, $(*)$ or $\sf CH$ are true.
Best Answer
Your logic is wrong.
You are conflating truth and provability. While working in a given model of $\sf ZFC$ either $\sf CH$ is true or it's false, but one of them must be the case. Even if you don't know which one.
There are sets of reals that we can perfectly define, and their cardinality is either $\aleph_0$ or $\aleph_1$, regardless to the cardinality of the continuum, and so they may serve as a counterexample to the Continuum Hypothesis in some models and not in others.
In the case of $\sf AC$ your solution is entirely wrong. It is consistent without $\sf AC$ that every set which cannot be well-ordered is the union of two sets of smaller cardinality, and that in addition $\Bbb R$ cannot be well-ordered (so its power set cannot either).
Nevertheless, the two propositions that you're stating are true.
In the case of $(0,1)$ we can define a bijection from $\Bbb R$ into $(0,1)$ (for example $x\mapsto 1/e^x$). Or we can define an injection from $\mathcal P(\Bbb N)$ into $(0,1)$.
In the case of discontinuous functions, we can either define an injection from $\mathcal P(\Bbb R)$ into the set. Or we can appeal to an abstract theorem: If $|A|=|A|+|A|$, and $B\subseteq\mathcal P(A)$ such that $|A|=|B|$, then $|\mathcal P(A)|=|\mathcal P(A)\setminus B|$. The proof is not trivial, and you can find it elsewhere on this very site.