Questions about Aleph-Aleph-Null

Question

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.

For the math symbols: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference — Holo, Jan 02 '19 at 15:39
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 $\aleph_\omega$. — pseudocydonia, Jan 02 '19 at 15:40
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! — pseudocydonia, Jan 02 '19 at 15:42

score 21 · Answer 1 · answered Jan 02 '19 at 15:57

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.}

Thanks @Asaf, I never realized Hilbert's hotel only has one bathroom. I think I'll stay at the Holiday Inn instead. — user4894, Jan 02 '19 at 23:31
Well, they have them in their room. But only one at the bar downstairs where they threw the party. — Asaf Karagila, Jan 02 '19 at 23:32
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. — Patrick Malone, Jan 03 '19 at 18:15

score 15 · Answer 2 · answered Jan 02 '19 at 15:45

15

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 $\aleph_\alpha$.

answered Jan 02 '19 at 15:45

David C. Ullrich

92,839

5

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 $\beth_{\alpha}$. – Clive Newstead Jan 02 '19 at 15:50
Thanks Clive, was literally about to comment a question about that. – Patrick Malone Jan 02 '19 at 15:52
@PatrickMalone I was going to reply to the comment you deleted: I sounds like you really don't know much about set theory - giving complete answers to your questions would involve writing a little essay on the topic, which is not really what MSE is supposed to be about. Lucky you, the other answer to the question is an introductory essay. You might get a lot out of https://www.amazon.com/Naive-Theory-Dover-Books-Mathematics/dp/0486814874/ref=sr_1_1_twi_pap_1?ie=UTF8&qid=1546444969&sr=8-1&keywords=halmos+set+theory – David C. Ullrich Jan 02 '19 at 16:07
@CliveNewstead Indeed. But then we have to try to explain why "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?" doesn't show that $\beth_{\omega+1}=\beth_\omega$... – David C. Ullrich Jan 02 '19 at 16:10
The cardinality of the set representation of $\omega$ is $\aleph_0$, but they are not exactly equal. After all, the cardinality of the set representation of $\omega +1$ is also $\aleph_0$. – Acccumulation Jan 02 '19 at 16:39
@Acccumulation: In the usual formalism (which assumes AC) a "cardinal number" means an initial ordinal. By this definition $\aleph_0$ literally equals $\omega$ but only some ordinals (namely, the initial ordinals!) are their own cardinailities. – hmakholm left over Monica Jan 02 '19 at 16:50
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@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 https://en.wikipedia.org/wiki/Cardinal_number – David C. Ullrich Jan 02 '19 at 17:06
Relevant to the discussion with @Accumulation, I think https://math.stackexchange.com/questions/1414845/is-aleph-0-mathbbn – Asaf Karagila Jan 02 '19 at 17:10
@Henning: Even without choice, well-ordered cardinals mean initial ordinals of the right cardinality. So choice is not needed to have $\aleph_0=\omega$, rather it is the fact we use the standard definition for cardinals and not fallback to the equivalence classes definition (modulo Scott's trick if necessary) which is needed. However, the naive approach to cardinals (which was Cantor's original idea) is indeed to use equivalence classes and in that case $\aleph_0$ is not $\omega$. From a type theoretic point of view, since cardinals and ordinals are different, they are also not the same. – Asaf Karagila Jan 02 '19 at 17:11
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@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. – Acccumulation Jan 02 '19 at 17:24
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@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 $1$.) – Misha Lavrov Jan 02 '19 at 18:23
@MishaLavrov I said "equivalence class", not set. – Acccumulation Jan 02 '19 at 18:29
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@Acccumulation: Do you agree that the natural number $2$ is the cardinal of the set ${0,1}$? – Asaf Karagila Jan 02 '19 at 18:42
@Acccumulation Classes are something that can only be spoken of informally in ZFC, so they can't help us come up with an actual definition. – Misha Lavrov Jan 02 '19 at 19:10
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@Acccumulation The first sentence of the section I said you should look at reads "Formally, assuming the axiom of choice, the cardinality of a set X is the least ordinal number α such that there is a bijection between X and α." Hence the cardinality of $\omega$ is $\omega$. Towards the end of the section they mention the notation $\aleph_0$ for the cardinaity of $\Bbb N$. – David C. Ullrich Jan 02 '19 at 21:36
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@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. – David C. Ullrich Jan 02 '19 at 21:49
@DavidC.Ullrich Not sure how I missed that sentence. But if $\aleph_0 = \omega$, and $\omega +\omega \neq \omega$, then is it not true that $\aleph_0+\aleph_0=\aleph_0$? – Acccumulation Jan 03 '19 at 16:20
@Acccumulation As you correctly pointed out, idenitying cardinals with certain ordinals can cause confusion because the operations are different. People use "$+$" for two different things here; which is intended is supposed to be clear from context. When we write $\omega+\omega$ that means $\omega+_o\omega$, while $\aleph_0+\aleph_0$ means $\aleph_0+_c\aleph_0$. Hence $\omega=\aleph_0$ but $\omega+\omega\ne\alpha_0+\aleph_0$. (Cf "operator overloading" in programming...) – David C. Ullrich Jan 03 '19 at 16:30
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@Acccumulation: Wait until you find out that $\omega^\omega$ is countable and $\aleph_0^{\aleph_0}$ is not. Indeed, this is an unfortunate case of operator overloading and the difference between ordinal and cardinal arithmetic. But similarly, if I wrote $f^2(x)$, you might think $f(x)\cdot f(x)$ or $f(f(x))$, and either would be a possible answer (with the former being a bit irresponsible notation, but not unheard of). Context matters when you read mathematics, and notation is hinting for context. – Asaf Karagila Jan 03 '19 at 17:58

score 2 · Answer 3 · answered Jan 02 '19 at 17:38

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 $\aleph_\omega<\aleph_{\omega+1}<\aleph_{\omega+2}<\cdots$.

Questions about Aleph-Aleph-Null

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