# Is the set of aleph numbers countable?

If I write the set of aleph numbers in this way $\{\aleph_0, \aleph_1, \aleph_2, \aleph_3, \dots\}$ it seems obvious to me that this set is countable, because aleph numbers have integer coefficients. However, maybe we are using the wrong notation for aleph numbers: how do we know that aleph numbers are really countable?

I think my question can be rephrased like this: what is the cardinality of the set $\{\mathbb{N}, \mathscr{P}(\mathbb{N}), \mathscr{P}(\mathscr{P}(\mathbb{N})), \mathscr{P}(\mathscr{P}(\mathscr{P}(\mathbb{N}))), \dots\}$, where $\mathscr{P}(\mathbb{N})$ represents the power set of $\mathbb{N}$? How do I prove it?

• The aleph numbers are indexed by ordinals, and therefore are not even a set : they are all the infinite cardinals. – Captain Lama Jul 22 '16 at 11:31
• This is a good question, which has been asked several times before. – Asaf Karagila Jul 22 '16 at 11:38
• I am certain that I missed a handful of other questions, e.g. questions talking about power sets and not directly about $\aleph$ numbers. But they are out there. – Asaf Karagila Jul 22 '16 at 11:49
• Wikipedia: Aleph number – Martin Sleziak Jul 22 '16 at 17:18

If by $\{ \aleph_0, \aleph_1, \aleph_2, \dots \}$ you mean $\{ \aleph_n : n \in \mathbb{N} \}$, then this set is countable because it's indexed by the natural numbers. However, this set doesn't contain $\aleph_{\omega}$ or any larger alephs.

If by $\{ \aleph_0, \aleph_1, \aleph_2, \dots \}$ you mean 'the set[sic] of all (well-orderable) cardinals', then it's certainly not countable; in fact, it's too big to be a set! (Hence the 'sic'.) There is an aleph for each ordinal $\alpha$, and there are class-many ordinals.

It should also be noted that, unless you accept the generalised continuum hypothesis, it is not necessarily the case that $\aleph_{\alpha} = |\mathscr{P}^{\alpha}(\mathbb{N})|$ for all $\alpha$, as you suggest in your question.

As Clive already pointed out the "set" of $\{\aleph_0,\aleph_1,\aleph_2,\cdots\}$ need not have almost anything to do with the "set" $\{\omega,\mathscr{P}(\omega),\mathscr{P}(\mathscr{P}(\omega)),\cdots\}$. Though the second expression seems more likely to be interpreted as an actual set i.e. $\{\mathscr{P}^n(\omega);\;n\in\omega\}$ due to the fact that you have to introduce a new type of operation to interpret it as anything else. You can't iterate the power set operation more then an arbitrary finite number of times. At some point you have to take a union.
This is a standard convention, where it makes sense, to define $\mathscr{P}^\alpha$ for $\alpha\geq\omega$ by the standard recursion approach if $\alpha$ is a limit then $\mathscr{P}^\alpha(\omega)=\bigcup_{\beta<\alpha}\mathscr{P}^\beta(\omega)$ and when $\alpha=\beta+1$ you just have $\mathscr{P}^\alpha(\omega)=\mathscr{P}(\mathscr{P}^\beta(\omega))$.
Post script: Becoming quite curious as to how much distance you can put between the two sets you mention I asked a question of my own to which Asaf Karagila kindly provided an answer. To summarize assuming ZFC the second class is a subset of the first class but can easily be quite "thin" in it (for example you can make $\mathscr{P}(\omega)> \aleph_\kappa$ where $\kappa$ is the first ordinal such that $\aleph_\kappa=\kappa$).
If you drop choice, then it is consistent that the only thing the two classes have in common is $\omega=\aleph_0$.