Note: I apologize in advance for not using proper notation on some of these values, but this is literally my first post on this site and I do not know how to display these values correctly.

I recently was looking up facts about different cardinalities of infinity for a book idea, when I found a post made several years ago about $ℵ_{ℵ_0}$

If the infinite cardinals aleph-null, aleph-two, etc. continue indefinitely, is there any meaning in the idea of aleph-aleph-null?

In this post people are talk about the difference between cardinal numbers and how $ℵ_{ℵ_0}$ should instead be $ℵ_ω$. The responses to the post then go on to talk about $ℵ_{ω+1}$, $ℵ_{ω+2}$, and so on.

Anyways, my understanding of the different values of ℵ was that they corresponded to the cardinalities of infinite sets, with $ℵ_0$ being the cardinality of the set of all natural numbers, and that if set X has cardinality of $ℵ_a$, then the cardinality of the powerset of X would be $ℵ_{a+1}$.

With this in mind, I always imagined that if a set Y had cardinality $ℵ_0$, and you found its powerset, and then you found the powerset of that set, and then you found the powerset of THAT set, and repeated the process infinitely you would get a set with cardinality $ℵ_{ℵ_0}$.

So, I guess my question is, in the discussion linked above, when people are talking about $ℵ_{ω+1}$, how is that possible? Because if you take a powerset an infinite number of times, taking one more powerset is still just an infinite number of times, isn't it?

I hope I worded this question in a way that people will understand, and thanks in advance into any insight you can give me about all this.

  • $\begingroup$ For the math symbols: math.meta.stackexchange.com/questions/5020/… $\endgroup$
    – ℋolo
    Jan 2, 2019 at 15:39
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    $\begingroup$ Here is maybe a better way to think of it: $\aleph_\omega$ is the smallest cardinal which is strictly greater than $\aleph_n$ for all $n$. Then, it's perfectly reasonable to say that $2^{\aleph_\omega}$ is bigger than $\aleph_\omega$. $\endgroup$ Jan 2, 2019 at 15:40
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    $\begingroup$ However, in general you can't assume that the power set of $\aleph_a$ is equal to $\aleph_{a+1}$ - that amounts to asserting the full version of the Generalized Continuum Hypothesis! $\endgroup$ Jan 2, 2019 at 15:42

3 Answers 3


Ordinals are not cardinals.

Recall Hilbert's hotel. Where you have infinitely many rooms, one for each natural number, and they are all full. And there's a party, with all the guests invited. At some point, after so many drinks, people need to use the restroom.

So someone goes in, and immediately after another person comes and stands in line. They only have to wait for the person inside to come out, so they have $0$ people in front of them, and then another person comes and they only have to wait for $1$ person in front of them, and then another and another and so on. That's fine. But the person in the bathroom had passed out, unfortunately, and everyone is so polite, so they just wait quietly. And the queue gets longer.

Let's for concreteness sake, point out that only people who stay in rooms with an even room number go to the toilet. The others are just fine holding it in. Now for every given $n$, there is someone in the queue which needs to wait for at least $n$ people. The queue is infinite. But it's fine, since each person has only to wait a finite amount of time for their turn.

But what's this now? The person in room $3$ has to use the toilet as well. But they cannot cut in the line, that would be impolite. So they stand at the back. Well. There were $\aleph_0$ people in the queue, that's how many, and we added just one more, so there are still $\aleph_0$ people waiting in line. But now we have one person who has to wait for infinitely many people to go before them. So the queue is ordered in a brand new way. If they were lucky and someone decided to let them cut in line, then the queue would have looked the same, just from some point on people would have to wait just one more person to go first.

This is not what happened, though. So the queue looks different. Well, now we continue, all the people in room numbers which are powers of $3$ start to follow. And at some point we get to a queue which looks like two copies of the natural numbers stitched up. And then the bloke from room $5$ joins the line, and he has to two for two infinite queues before their turn. And so on and so forth.

Okay, what's the point of all that?

The point is that for finite queues the question of "how many people" and "how is the queue ordered" are the same question. So adding one person does not matter where this person was added to the queue. But when the queue was infinite, adding one person at the end or adding it to the middle would very much change the queue's order. So "how many" is no longer the same as "how long is the queue".

When iterating an operation transfinitely many times, e.g. by taking power sets or cardinal successors, we work successively. This creates a queue-like structure of cardinals. The first, the second, etc., which are ordinal numbers, they talk about order.

So once you go through the finite ones, you have to move to infinite ordinals, not to infinite cardinals. As such $\omega$ is the appropriate notation, since it denotes an ordinal, rather than $\aleph_0$ which denotes a cardinal.

Between $\aleph_\omega$ and $\aleph_{\omega+1}$ there are similarities: both have infinitely many [infinite] cardinals smaller than themselves. But it is not the same, exactly because we are dealing with the question "how are these ordered" rather than "how many are there".

Note that $\aleph$ numbers are not defined by power sets, these are $\beth$ numbers (Beth is the second letter of the Hebrew alphabet, whereas Aleph is the first one). But this is irrelevant to your actual question.

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    $\begingroup$ Thanks @Asaf, I never realized Hilbert's hotel only has one bathroom. I think I'll stay at the Holiday Inn instead. $\endgroup$
    – user4894
    Jan 2, 2019 at 23:31
  • $\begingroup$ Well, they have them in their room. But only one at the bar downstairs where they threw the party. $\endgroup$
    – Asaf Karagila
    Jan 2, 2019 at 23:32
  • $\begingroup$ I remember something about this when I was studying infinite sets in high school for a project, there was something about omega+1, omega+2, and so on, and it was described in a way similar to what you are saying. I just also remembered something about power sets and thought that was the only way to make infinity bigger. $\endgroup$ Jan 3, 2019 at 18:15
  • $\begingroup$ Cute.${{{{}}}}$ $\endgroup$ Apr 21, 2020 at 22:49

The reason it should really be $\aleph_\omega$ instead of $\aleph_{\aleph_0}$ is that we think of $\aleph_\alpha$ for ordinals $\alpha$. Yes, $\aleph_0=\omega$, but we write $\omega$ when we think of it as an ordinal instead of a cardinal.

The rest of your question is based on a fundamental misunderstanding: $\aleph_{\alpha+1}$ is not the cardinality of the power set of $\aleph_\alpha$; what it is is the smallest cardinal larger than $\aleph_\alpha$.

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    $\begingroup$ It might also be worth noting that the cardinalities obtained from $\mathbb{N}$ by iterating power sets, rather than successor cardinals, are the beth numbers $\beth_{\alpha}$. $\endgroup$ Jan 2, 2019 at 15:50
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    $\begingroup$ @Acccumulation Really? Can you tell me exactly what $\aleph_0$ is? Hint: "the cardinality of the natural numbers" doesn't answer the question - what is a cardinal? Hint: $\aleph_0=\omega$. See "Formal Definition" at en.wikipedia.org/wiki/Cardinal_number $\endgroup$ Jan 2, 2019 at 17:06
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    $\begingroup$ @DavidC.Ullrich One definition is the equivalence class of all sets that have a bijection with the natural numbers. I don't see where your link says that $\aleph_0$ is equal to $\omega$. Cardinal numbers are a separate number system with different operations. Saying that $\aleph_0$ = $\omega$ is a bit like saying that the identity matrix is equal to one. Yes, you can represent the real numbers as a subgroup of GL$_n$(R), but there are contexts where the distinction is important. $\endgroup$ Jan 2, 2019 at 17:24
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    $\begingroup$ @Acccumulation One problem with your definition is that "the set of all sets that have a bijection with $\mathbb N$" does not exist. Instead, we can try to construct what is informally a representative of that equivalence class. We can always make that representative an ordinal (assuming choice) and then the natural thing to do is to choose the least ordinal that works. Thus, $\aleph_0$ is the least ordinal which has a bijection with $\mathbb N$, and that is $\omega$. (I don't disagree with your second point, though here the link is a bit closer than between the identity matrix and $1$.) $\endgroup$ Jan 2, 2019 at 18:23
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    $\begingroup$ @Acccumulation Regardless of whether you believe or approve, in fact the standard definition of $\aleph_0$ is $\aleph_0=\omega$. You can disapprove if you want (although asking why the definition is a good idea would probably make a better impression than explaining why it's a bad idea), but insisting that a standard definition is wrong is just silly. Definitions are what they are. $\endgroup$ Jan 2, 2019 at 21:49

Further to existing answers, let's clarify how transfinite recursion defines $\aleph_\alpha$ for arbitrary ordinals $\alpha$. Fr simplicity I'll identify each aleph with the least ordinal of the intended cardinality. We have $\aleph_0:=\omega$, and for any limit ordinal $\gamma\ne 0$ we have $\aleph_\gamma:=\bigcup_{\beta<\gamma}\aleph_\gamma$. Finally, $\aleph_{\alpha+1}$ is the Hartogs number of $\aleph_\alpha$, i.e. the least ordinal that cannot be injected into $\aleph_\alpha$. I recommend convincing yourself $\aleph_\alpha$ can be injected into $\aleph_\beta$ only if $\alpha=\beta\lor\alpha\in\beta$ (for ordinals we usually write this as $\alpha\le\beta$), and in particular understanding the proof that a Hartogs number exists. Hence $\aleph_\omega<\aleph_{\omega+1}<\aleph_{\omega+2}<\cdots$.


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