# What are Aleph numbers intuitively?

I cannot get my head around the concept of the `types' of Aleph infinity. What is an easy intuitive way to see when you are given the integer numbers Aleph 0 the Aleph 1 will follow?

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What exactly do you mean? The alephs are not integers. That $\aleph_1$ comes directly after $\aleph_0$ is by definition. For a natural number $n>0$, $\aleph_n$ is the first aleph that comes after $\aleph_{n-1}$. – Michael Greinecker Sep 26 '12 at 9:56
There’s really nothing to see: $\aleph_1$ is by definition the smallest (well-ordered) cardinal greater than $\aleph_0$. – Brian M. Scott Sep 26 '12 at 9:56
@BrianM.Scott : I suggest that the standard definition is that $\aleph_1$ is the cardinality of the set of all countable ordinals. – Michael Hardy Sep 30 '12 at 18:23
@Michael: For me von Neumann cardinals are standard, and $\aleph_\alpha$ and $\omega_\alpha$ are the same thing. – Brian M. Scott Sep 30 '12 at 18:55
So I guess I should learn more about set theory if I want to understand it better... – user29751 Sep 30 '12 at 21:59
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The cardinals are the following ones: $$0,1,2,3,4,5,6,\dots,\aleph_0,\aleph_1,\aleph_2, \aleph_3,\dots,\aleph_\omega,\aleph_{\omega+1},\dots,\aleph_{2\omega},\aleph_{2\omega+1},\dots$$ Where $\aleph_0$ is the first infinity cardinal (the cardinality of each infinite countable set), so $\aleph_0\notin\mathbb N$, is not said to be an integer in the ordinary way. Then $\aleph_1$ is the next cardinal, and so on.. (and this "and so on.." also includes some knowledge about the ordinals).

By Cantor's theorem ($|P(A)| > |A|$ for all sets $A$) we have that for every cardinal there is a bigger cardinal. By the well foundedness and the axiom of choice in ZFC, we have that every cardinal is a cardinal of a well-ordered set (which is in bijection to some ordinal), and it follows that there is always a next cardinal..

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 But is there any simple way to see how you can define the next cardinal? Or should I just view it as an ordered set of some kind? – user29751 Sep 26 '12 at 11:55 Yes, suppose for all ordinals $\gamma <\beta$, the cardinal $\aleph_{\gamma}$ is already defined. Then define $\aleph_{\beta}:=\min\{\kappa\in\mathit{Ord} \mid \forall \gamma<\beta:\ |\kappa| > |\aleph_\gamma| \}$. Since the class of ordinals is well-ordered, it makes a unique sense. – Berci Sep 27 '12 at 9:51 With the usual convention on ordinal arithmetic, $2\omega=\omega$, so you might want to replace it with either $\omega\cdot 2$ or $\omega+\omega$. – tomasz Sep 27 '12 at 23:31

Perhaps you are asking for an example of a set of cardinality $\aleph_1$. Cantor thought the set of all real numbers would be an example. In fact, he thought he had a proof. Then he found a hole in the proof, and restated the question as a hypothesis, which has come down to us as "The Continuum Hypothesis": the cardinality of the reals is $\aleph_1$. No one has been able to prove this, nor to disprove it: In 1940, Gödel published his proof that this hypothesis cannot be disproved on the basis of the axioms of mathematics generally accepted at the time and, in 1963, Cohen published his proof that it cannot be proved either.

The bottom line is, no one has ever been able to present a set provably of cardinality $\aleph_1$ and, on our current understanding of these things, no one ever will.

If that doesn't answer your question, perhaps you'd like to edit your question so someone will be able to get what exactly you're asking.

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I do not understand what you mean, unless you simply mean a set of reals. Because one can exhibit sets of size $\aleph_1$. There are reasonable equivalence relations on the reals that have precisely $\aleph_1$ classes, for example. – Andres Caicedo Sep 27 '12 at 6:59
I think for historical accuracy, one should change the sentence on Cohen. Cohen showed that CH cannot be proven. That CH cannot be disproven has been shown before by Gödel. – Michael Greinecker Sep 27 '12 at 9:42
@Andres, I did have sets of reals in mind, but I confess to being unaware of the kind of example you are talking about. Perhaps you could post an answer giving a set of cardinality $\aleph_1$, as that might be what OP is looking for (and as I might learn something from it). – Gerry Myerson Sep 27 '12 at 10:36
@Michael, I opted for brevity. If it bothers you, feel free to edit. – Gerry Myerson Sep 27 '12 at 10:38
(Gerry, I made the edit.) Anyway, a simple example is the following: We can easily identify a real with a sequence of naturals. For example, an irrational between $0$ and $1$ has an infinite continued fraction. Each natural can be seen as coding two. For example, $2^a(2b+1)$ could code $(a,b)$. This way, each real can be seen as coding a binary relation. Identify two reals iff they do not code well-orderings, or else, they code well-orderings of the same order type. This relation has precisely $\omega_1$ equivalence classes. (On the other hand, finding a "coding free" relation seems harder.) – Andres Caicedo Sep 27 '12 at 22:59

The $\aleph$ numbers are simply ordinals. Informally, ordinals tell you how long is the line to the bathroom is.

In this sense $\aleph_0$ is just the least infinite ordinal, and $\aleph_1$ is the least uncountable ordinal, and so on.

There is a very good analogy between $\aleph_0$ and $\aleph_1$ which might help you understand the difference in size. Note that you cannot reach $\aleph_0$ by a finite process over finite numbers. Namely, every finite sequence of finite numbers is bounded way below $\aleph_0$. Similarly every countable sequence of countable ordinal numbers is bounded way below $\aleph_1$.

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 I'm not sure if this is a very good way of showing it. $\aleph_\omega$ is certainly larger than any $\aleph_j$ for $j\in\omega$, but it is not true that any countable sequence of ordinals smaller than it is bounded way below it. – tomasz Sep 27 '12 at 23:36 @tomasz: But to get to $\aleph_\omega$ you first need to finish processes longer than any $\aleph_n$, for all $n\in\omega$. Even if after you have done all you see that you could have done this in a countable process, you still needed to pass through "checkpoints" which require you for a lot more. – Asaf Karagila Sep 27 '12 at 23:38 Well, I'm not saying that the intuition is completely off, its just that the exact wording does not really generalize as well as you suggest it could -- it looks as if you suggested that all cardinals are regular. – tomasz Sep 27 '12 at 23:42 @tomasz: I don't see how. I never mentioned any cardinal except $\aleph_1$ and $\aleph_0$... – Asaf Karagila Sep 27 '12 at 23:43 You didn't, but this question is about alephs in general, so to someone who doesn't know better, it might appear that that's what you're implying (by transfinite induction on two cases ;) ). – tomasz Sep 27 '12 at 23:48
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The cardinal number $\aleph_1$ is defined as the cardinality of the set of all countable ordinals, i.e. ordinals of cardinality $\le\aleph_0$. If you believe that that set of ordinals exists, then you've got it.

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