The natural numbers are well-ordered without the axiom of choice. In fact they still serve as the definition for countability without the axiom of choice assumed. Therefore the basic things we know about the natural numbers hold regardless to the axiom of choice. In particular $\mathbb{N\times N}$ is still countable, and $\{A\subseteq\mathbb N\mid A\text{ is finite}\}$ is also countable.
However when the axiom of choice is negated some other weird things could happen in the power set of the natural numbers, and in other infinite sets:
- There could be a set which is infinite, but has no countably infinite subset.
- It could be that there are no free ultrafilters on the natural numbers.
- It could be that the power set of the natural numbers cannot be well-ordered (it can still be linearly ordered, though).
- Countable unions of general countable sets need not be countable.
- It is possible that there is no linear basis for $\mathbb R$ over $\mathbb Q$.
Let us focus on the first one for a moment, such sets are known as infinite Dedekind-finite sets. Their existence contradicts the axiom of countable choice, so if we assume that we can prove that every infinite set has a countably infinite subset. The fourth one has the same properties, it negates the axiom of countable choice.
Both the second, third and fifth points, however, are compatible with countable choice.
As for why we accept the axiom of choice, historically we did not accept it. People found its consequence strange (regardless to the fact they have used it intuitively). After it was proved that assuming the axiom of choice does not add inconsistencies to ZF, people began using the axiom of choice more and see its wonderful applications. It simply made things easier.
For more reading:
- Advantage of accepting the axiom of choice
- Motivating implications of the axiom of choice?
- Advantage of accepting non-measurable sets
- Why is the axiom of choice separated from the other axioms?
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
For the first one, note that there is a surjection from $X$ onto $X/{\equiv}$. This means that the fibres of this functions define an injection from $X/{\equiv}$ into $2^X$, but you can show that this injective function actually maps $2^{X/{\equiv}}$ into $2^X$ injectively.
For the second one, note that quite literally, $X/{\equiv}$ is a subset of $2^X$, when considering $2^X$ as the power set of $X$. So we trivially get $\leq$. But since there is a surjection from $X$ onto $X/{\equiv}$, this $\leq$ cannot possibly be $=$, since that would contradict Cantor's theorem, which holds without choice (there is no surjection from $X$ onto $2^X$). Therefore $X/{\equiv}<2^X$.