TITLE: What's the difference between saying that there is no cardinal between $\aleph_0$ and $\aleph_1$ as opposed to saying that...
QUESTION [17 upvotes]: ... $\aleph_1$ is the immediate successor of $\aleph_0$?
I was reading the wiki article on $\aleph_1$ where a distinction is made between the two. If there's isn't a cardinal between $\aleph_1$ and $\aleph_0$, the implication is that they must follow one another so why must one need AC to assert that $\aleph_1$ is the smallest after $\aleph_0$?

REPLY [3 votes]: Generally $\aleph_1$ is the cardinal notation of the smallest infinite ordinal not in one-to-one correspondence with $\omega = \mathbb{N}$.  As such, there is nothing between $\aleph_0$ and $\aleph_1$:
Suppose that $X$ were a set and $\aleph_0 \leq | X | < \aleph_1$.  Then there is a one-to-one function $f : X \to \omega_1$, and we may use such a function to define a well-ordering of $A$:  $x \leq y$ iff $f(x) \leq f(y)$.  Then the order-type of this well-ordering must be strictly less than $\omega_1$, and so by definition of $\omega_1$ there must be a bijection from $X$ onto $\omega$, so actually $\aleph_0 = |X|$.  (Note that the above argument did not use Choice anywhere.)  Therefore $\aleph_1$ is an immediate successor of $\aleph_0$.
However, if you do not assume the Axiom of Choice, I do not think you can  turn the an into the: i.e., it might be possible that $\aleph_0$ has an immediate successor different from $\aleph_1$.  (It is certainly possible that there is a cardinal $\mathfrak{p} > \aleph_0$ such that neither $\mathfrak{p} \leq \aleph_1$ nor $\aleph_1 \leq \mathfrak{p}$ hold, but I am turning myself into knots trying to get an immediate successor of $\aleph_0$.)  This is because if you do not assume the Axiom of Choice, then not all sets can be well-ordered, and the cardinality of a non-well-orderable set need not be comparable with a specific aleph, i.e., the cardinality of an ordinal.

A common example (though not exactly the focus of this question) is an infinite Dedekind-finite set: a set $D$ which is not equinumerous to any finite set, but yet there is no injection $\omega \to D$.  It follows that the cardinals $\aleph_0$ and $| D |$ are not comparable, since the required witnessing functions do not exist.  (That $\aleph_0 \not\leq |D|$ is by definition of Dedekind-finite; that $|D| \not\leq \aleph_0$ follows because if there was an injection $f : D \to \omega$, the range of $f$ must be infinite, and we could use that to construct an injection $\omega \to D$.)
Another common example is that $\mathcal{P} ( \mathbb{N} )$ need not be well-orderable.  By Cantor's Theorem we know that $\aleph_0 = | \mathbb{N} | < | \mathcal{P} ( \mathbb{N} ) |$, but in this situation the cardinals $\aleph_1$ and $| \mathcal{P} ( \mathbb{N} ) |$ need not be comparable.  Thus there must be at least two incomparable immediate successors of $\aleph_0$.