The $\aleph$-numbers are cardinalities which can be well ordered. Since each well-ordered set can be identified with an ordinal, each $\aleph$-number is the cardinality of some ordinal. If we consider the least ordinal of this cardinality we will have an initial ordinal (i.e. an ordinal which cannot be bijected with any ordinal strictly smaller than itself).
The axiom of choice is equivalent to the assertion that every set can be well ordered, therefore every infinite cardinality is an $\aleph$-number.
Suppose that the axiom of choice does not hold, then we have some sets which cannot be well-ordered. Cardinalities which cannot be well-ordered cannot be $\aleph$-numbers.
This gets worse. It is consistent that every set can be linearly ordered, and the axiom of choice still fails. It is even consistent that there is a set that cannot be linearly ordered at all (an amorphous set cannot be linearly ordered, for example). In Thomas Jech's book The Axiom of Choice he has a chapter about cardinal arithmetics without choice. It shows how wild the universe of ZF can be with respect to cardinals, when the axiom of choice is absent.
A few examples:
It is consistent that there is no "canonical" representative to the cardinalities. When the axiom of choice is absent, we are reduced to define cardinalities as equivalent classes in the relation of "there exists a bijection", by using Scott's trick we have that this is a well defined relation. Where as the axiom of choice allows us to pick a representative from each equivalence class (namely, initial ordinals) it is consistent that without the axiom of choice there is no way to choose such representative from all cardinalities at once. (i.e. there is no definable function which returns exactly one element of each cardinality)
We can have that for all infinite sets $|\{0,1\}\times A|=|A|$ but the axiom of choice still does not hold (this in contrast to $|A|\times|A| = |A|$ for every infinite $A$ implies choice)
The continuum hypothesis is independent of choice, that is there is a model in which the axiom of choice does not hold, and there are no sets of cardinality strictly between $\aleph_0$ and $2^{\aleph_0}$.
These are just examples to show how cardinals can behave like a bunch of unsupervised teenagers in a house full of alcohol.
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
This is a consequence of the following theorem:
Since we define $\aleph_0$ to be the cardinality of $\Bbb N$, this means that every infinite subset of a set of size $\aleph_0$ is itself of size $\aleph_0$, and so there cannot be a smaller infinite cardinal.
Note that the above proves that $\aleph_0$ is a minimal element of the infinite cardinals. There is no smaller. To prove that it is in fact the smallest of the infinite cardinals we need to use some other set theoretical assumptions (e.g. every two cardinals are comparable) which are commonly assumed throughout mathematics nowadays.
The proof of the aforementioned theorem is simple, by the way. Suppose that $A$ is infinite, then the map $a\mapsto |\{a'\in A\mid a'<a\}|$ is a bijection between $A$ and $\Bbb N$. The proof of that is by induction.