# 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 $\aleph_0$?

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Without the axiom of choice, it is conceivable that there is a cardinal $\kappa$ such that $\aleph_0 < \kappa$ but neither $\aleph_1 \le \kappa$ nor $\kappa \le \aleph_1$. – Zhen Lin Dec 5 '12 at 11:50

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.

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 $\aleph_0$.

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 You wrote "it could also be the case that there are smaller cardinalities but none which are strictly between $\aleph_0$ and the countable of such set." It's not clear to me what those cardinalities are smaller compared to. And also, is the last line a typo and you were going for $\aleph_1$ in place of 'the countable of such set'? – Mark Dec 9 '12 at 15:31 @Mark: It is possible to have a cardinal $A$ which is a $1$-successor of $\aleph_0$, but there is some $B show 2 more comments Generally$\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$. - In your last paragraph it's not clear to me that "$\mathcal{P}(\mathbb{N})$is incomparable with$\aleph_1$" implies that that$\aleph_0$has an immediate successor besides$\aleph_1$". (Although it is consistent that$\mathcal{P}(\mathbb{N})$is such a successor.) – Trevor Wilson Dec 5 '12 at 16:38 @TrevorWilson: My intuitions here a quite a bit weaker than Asaf's, and his input would likely (certainly?) be better; as it stands, I am half ready to delete. My original thought would be to take a infinite Dedekind-finite set$D \subseteq \mathbb{R}$. Of course,$\aleph_0 \not\leq | D |$, but we would have$\aleph_0 < |D| + \aleph_0$. By hope would be that some such$D$could be found to make this sum an immediate successor which would have to be different than$\aleph_1$. But I must admit I am having trouble squaring this circle. – Arthur Fischer Dec 6 '12 at 10:41 @Arthur Fischer: In Set Theory by Thomas Jech, the author writes that "one cannot prove without the Axiom of Choice that$\omega_1$is not a countable union of countable sets." That is to say, without AC, the uncountability of$\omega_1$is unprovable. It negates what you wrote in the second paragraph that nowhere was AC used, doesn't it? – Mark Dec 6 '12 at 10:44 @Matt: NO! Some amount of Choice (Countable Choice) is required to show that countable unions of countable sets are countable. There are models of ZF (e.g.,the Feferman-Levy model) in which$\mathbb{R}$is a countable union of countable sets. But the uncountability of$\mathbb{R}$is a ZF theorem. The uncountability of$\omega_1$is a definition. – Arthur Fischer Dec 6 '12 at 10:48 Arthur, you should make it more explicit. If$\mathbb R\$ cannot be well-ordered it does not mean that CH holds. You need to require more (e.g. Solovay's model or Truss' variants which does not require inaccessible). – Asaf Karagila Dec 6 '12 at 11:14