I am having trouble understanding aleph numbers. I understand $\aleph_0$ is a countable infinity, but after that, I'm lost. What are $\aleph_1,\aleph_2,\aleph_3$, etc. to $\aleph_n$? Is there an infinite amount of $\aleph$ numbers or do they stop at some finite number?

  • $\begingroup$ In a couple of hours I'll be done with this assignment and I'll have a go at explaining them. $\endgroup$ – goblin GONE Oct 24 '14 at 3:37

To understand the $\aleph$ numbers properly you need to understand the ordinal numbers first, at least a little bit.

The idea of an ordinal is to model the notion of a length of a queue to the bathroom in a party. There is an empty queue, then there is one person waiting, then another, and so on. But here we can talk about infinite queues as well. At some point we have an infinite queue, but everyone on that queue is just a finite number of people away from the bathroom. This is the same as the natural numbers, there are infinitely many of them, but each is preceded by just a finite number.

Then comes the poorest schmoe you had the luck to meet. And he has to wait for all those infinitely many people to use the bathroom before he can. Poor guy. But it could be worse, someone comes and takes place after this poor schmoe. And then another and another and soon enough we have two copies of the natural numbers stacked one on top of another.

And we can continue in this fashion. At some point it becomes impossible to properly visualize. It's just a very very long long to the bathroom, and you really take pity on whoever is coming to stand in this queue next.

But after you've done this for every countable queue possible, you have an uncountable queue, and each person in that queue is just a countable number of people away. But here comes a new person, and she has to wait uncountably many people before she can see the inside of that bathroom. That girl is standing on the first uncountable point of the queue. And sure enough, we can continue again, and extend and enlarge and so on and so forth.

Okay, so how do we get from this to the $\aleph$ numbers? The $\aleph$ numbers tell us how many people are in the queue, not how long it is. For finite queues the two notions coincide, but for infinite queues they do not. It is clear that the queue described by $\Bbb N$, and the queue described by adding one more after that are both countable. If that new person had cut in line, he would piss a lot of folks, but the length of the queue remains the same.

So $\aleph$ numbers come to tell us how many people are standing in that line. But in order to make this a well-defined notion, we use the shortest queue that can hold that many people. So $\aleph_0$ is the size of the shortest queue that is infinite, namely the queue where each person has only finitely many predecessors; $\aleph_1$ is the size of the queue where there are uncountably many people, but each has at most $\aleph_0$ predecessors; $\aleph_2$ is the size of the shortest queue where each person has at most $\aleph_1$ predecessors.

This goes on, and we can easily see how a queue holding more than $\aleph_1$ and more than $\aleph_2$ and more than $\aleph_n$, for every finite $n$, is formed. But we can continue, and the next cardinal is exactly that of a line where each person has at most $\aleph_n$, for some $n$, predecessors. This is a limit cardinal and it is denoted by $\aleph_\omega$, since $\omega$ is the first infinite ordinal (it denotes the natural numbers).

And so we can continue, and show that if $\alpha$ is an ordinal, then there is a cardinal with exactly $\alpha$ cardinals preceding it, and we call this cardinal $\aleph_\alpha$.

So how many are there? Well, this mapping is a bijection between the ordinals and the $\aleph$ numbers. Since the collection of all ordinals is too large to even be considered a set, the collection of all cardinals is too large to be a set as well. And we call these collections "proper classes" to say that they are not sets, but collections which while large enough, are still collections we can talk about.

  • $\begingroup$ I think I understand. Using your example, aleph 0 is an infinite amount of people, but made up using finite persons. The aleph after that would be having to wait for a countably infinite amount of people before you. The aleph after that would be waiting for an infinite amount of aleph ones. But, that seems hard to imagine. Couldn't I just say aleph 2, for example, is an extension of aleph 0? If you're just adding more people, it doesn't. Seem like you're changing from one aleph to other. $\endgroup$ – Julian Jefko Oct 24 '14 at 3:55
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    $\begingroup$ Well, you just add one after another, but you do it uncannily many times. $\aleph_1$ is an extension of $\aleph_0$. The change from one cardinal to another happens after "a long time have passed". If you add one person each second, you reach $\aleph_1$ only after $\aleph_1$ seconds. But yes, other than that, you just go from one to the other. $\endgroup$ – Asaf Karagila Oct 24 '14 at 3:59
  • $\begingroup$ Is it somewhat subjective, then, or is just a limit of the example? Also, on the last few paragraphs, since The collection of all alephs can be mapped to the natural numbers, is the collection of all types of infinity the smallest type of infinity possible? $\endgroup$ – Julian Jefko Oct 24 '14 at 4:05
  • $\begingroup$ @JulianJefko: No, $\aleph_1$ has to wait for $\aleph_1$ people, each of whom have to wait for countably many people. $\endgroup$ – Robert Israel Oct 24 '14 at 4:05
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    $\begingroup$ @Joe: I'm not big on believing in things. $\endgroup$ – Asaf Karagila Oct 24 '14 at 4:11

This is an issue about which mathematicians who haven't studied set theory beyond what they actually use are often confused.

$\aleph_1$ is the cardinality of the set of all countable ordinal numbers.

$\aleph_2$ is the cardinality of the set of all ordinal numbers of cardinality $\le\aleph_1$.

$\aleph_3$ is the cardinality of the set of all ordinal numbers of cardinality $\le\aleph_2$.


$\aleph_\omega$ is the cardinality of the set of all ordinal numbers of cardinality no bigger than some $\aleph_n$ for finite $n$.

$\aleph_{\omega+1}$ is the cardinality of the set of all ordinal numbers of cardinality $\aleph_\omega$.


There exists an $\aleph_\alpha$ for every ordinal number $\alpha$.

If every set can be well-ordered --- a proposition equivalent to the axiom of choice --- then every infinite cardinality is one of these.

$2^{\aleph_0}$ is the cardinality of the set of all subsets of a set of cardinality $\aleph_0$. If the continuum hypothesis is true, then there is no cardinality between $\aleph_0$ and $2^{\aleph_0}$. If every cardinality is one of the alephs, then the continuum hypothesis is equivalent to $2^{\aleph_0}=\aleph_1$. It is consistent with the Zermelo--Fraenkel axioms plus the axiom of choice that $2^{\aleph_0} = \aleph_n$ for some finite $n>1$. It is not consistent that $2^{\aleph_0} = \aleph_\omega$ since it can be shown that $2^{\aleph_0}$ is not the supremum of any countable set of cardinalities. But it is consistent that $2^{\aleph_0} >\aleph_\omega$.

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    $\begingroup$ This isn't very helpful if someone doesn't know much about ordinals. $\endgroup$ – Asaf Karagila Oct 24 '14 at 3:36
  • $\begingroup$ True. That is a prerequisite to the answer to the posted question. $\endgroup$ – Michael Hardy Oct 24 '14 at 3:39

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