What's the difference between saying that there is no cardinal between $\aleph_0$ and $\aleph_1$ as opposed to saying that...

... $\aleph_1$ 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 $\aleph_0$?

Solution 1:

It is consistent without the axiom of choice that there are uncountable sets which cannot be well-ordered, and every infinite set whose cardinality is smaller is necessarily countable. It could also be the case that there are smaller cardinalities, but none which are strictly between $\aleph_0$ and the uncountable set.

Indeed there are three different definitions of successor cardinals when the axiom of choice fails (although the existence of one [for every set] implies choice; there is another which is provable without choice; and a third which is independent).

It is consistent, if so, that there are several successors to $\aleph_0$. It is always true that $\aleph_1$ is the successor of $\aleph_0$, and that it is the minimal aleph above it.

In particular it is consistent that the real numbers form such set.

See, for example, Relationship between Continuum Hypothesis and Special Aleph Hypothesis under ZF

Edit: I have some free time so here are different variants of "successor". This is taken from Jech The Axiom of Choice (p. 163), the original definition is due to Tarski.

Let $\frak p$ and $\frak q$ be cardinals such that $\frak p<q$.

1. $\frak q$ is the $1$-successor of $\frak p$ if whenever $\frak m$ is such that $\frak p\leq m\leq q$ then $\frak p=m$ or $\frak q=m$.
2. $\frak q$ is the $2$-successor of $\frak p$ if whenever $\frak m$ is such that $\frak p < m$ then $\frak q\leq m$.
3. $\frak q$ is the $3$-successor of $\frak p$ if whenever $\frak m$ is such that $\frak m < q$ then $\frak m\leq p$.

Now we see that it is always true that $\aleph_1$ is a $1$- and $3$-successor of $\aleph_0$. However it is consistent that so are the real numbers themselves.

The assertion that $\aleph_1$ is a $2$-successor of $\aleph_0$ is equivalent to every uncountable set $X$ has an injection from $\omega_1$ into $X$. In fact just requiring that $\aleph_0$ has a $2$-successor is enough.

It is consistent that there are several $1$-successors to $\aleph_0$, and that there is a $1$-successor which is not $3$-successor.

For the real numbers it holds that if their cardinal is a $1$-successor of $\aleph_0$ then it is a $3$-successor of $\aleph_0$.

Interestingly, it is consistent that there is a proper class of $1$-successors to $\aleph_0$.

Solution 2:

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 $\aleph_0$.