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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?

user93089
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Novo
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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 09:56
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    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 09:56
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    @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
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    @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
  • von Neumann ordinals are merely a method of encoding something; they shouldn't be confused with the thing they encode. – Michael Hardy Sep 30 '12 at 19:00
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    So I guess I should learn more about set theory if I want to understand it better... – Novo Sep 30 '12 at 21:59

3 Answers3

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The cardinals are the following ones: $$0,1,2,3,4,5,6,\dots,\aleph_0,\aleph_1,\aleph_2, \aleph_3,\dots,\aleph_\omega,\aleph_{\omega+1},\dots,\aleph_{\omega2},\aleph_{\omega2+1},\dots $$ Where $\aleph_0$ is the first infinite 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..

Asaf Karagila
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Berci
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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? – Novo 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 09:51
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    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
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    so is there a $\aleph_{\aleph_0}$? – tox123 Jul 24 '15 at 22:41
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    @tox123 I think the subscripts have to be ordinals. There is an $\aleph_\omega$. There's also an $\aleph_{\omega+1}$, which is bigger. – Akiva Weinberger Apr 28 '16 at 07:23
  • @tox123 Yes, there is an $\aleph_\alpha$ for each ordinal $\alpha$, in particular for the ordinal $\aleph_0=\omega$. It is customary, though, to write subindices in ordinal notation, though it is not (as the previous comment suggests) that "they have to be" so. The (somewhat artificial) distinction stops making much sense anyway when you notice that there are fixed points: Cardinals $\kappa$ such that $\kappa=\aleph_\kappa$. On the other hand, for small concrete ordinals (such as $\omega$) it is a common convention. – Andrés E. Caicedo Jan 08 '18 at 17:15
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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 countable ordinals exists, then you've got it.

Anders Kaseorg
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Michael Hardy
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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 [EDIT: but see clarification in the comments] 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.

Gerry Myerson
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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. – Andrés E. Caicedo Sep 27 '12 at 06:59
  • 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 09:42
  • @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
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    (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.) – Andrés E. Caicedo Sep 27 '12 at 22:59
  • @Andres, thanks. At the risk of further exposing my ignorance, let me ask: suppose you now form a set $S$ of reals with one element from each of those $\omega_1$ equivalence classes. What can you say about the cardinality of $S$? – Gerry Myerson Sep 27 '12 at 23:24
  • 1940? I thought that Godel proved that in 1938, but then again I might be remembering wrong... – Asaf Karagila Sep 27 '12 at 23:35
  • @AsafKaragila Thanks. Meant to say "published". – Andrés E. Caicedo Sep 27 '12 at 23:52
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    Gerry: The set $S$ would be in bijection with $\omega_1$, so it would have size $\omega_1$. The point is that while the set of equivalence classes is a fairly explicit object, picking a representative from each is a "non-constructive" process. So, of course $S$ has size $\omega_1$, but it is not clear that we have "exhibited" it (in some sense). Also, another example is to take the set of all countable ordinals, but this requires the possibility of identifying order types with ordinals, which needs the axiom of replacement, in general considered strong. I also avoid this in the example above. – Andrés E. Caicedo Sep 27 '12 at 23:58
  • @Andres, thanks again. So now that we have a set of reals of size $\omega_1$, I guess we know that set can't be put in one-one correspondence with the natural numbers, but we are unable to tell (in ZFC) whether it can be put in one-one correspondence with the reals. – Gerry Myerson Sep 28 '12 at 00:21
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    Hi Gerry. Exactly. – Andrés E. Caicedo Sep 28 '12 at 00:22
  • You say no one has presented a set provably of cardinality $\aleph_1,$ so what about the set of all countable ordinals? Do you mean that's not "provably" a set, or that it has not been "presented" (whatever that would mean)? Cantor defined $\aleph_1$ as the cardinality of the set of all countable ordinals, and as far as I know, that's been the standard definition every since. $\qquad$ – Michael Hardy Jan 07 '18 at 17:30
  • @MichaelH, are you talking to me? I explained earlier in the comment stream that what I had in mind when I wrote "set" was "subset of the reals", and then I had a fruitful discussion with Andres, so I think all is well. – Gerry Myerson Jan 07 '18 at 19:03
  • @GerryMyerson : Your first sentence just says "Perhaps you are asking for an example of a set of cardinality $\aleph_1.$" I think few people would guess from that that "set" was intended to mean "subset of the reals". – Michael Hardy Jan 07 '18 at 20:46
  • @MichaelH, few people who read the first comment and my reply to it would have to guess. In any event, take it in the context of an answer to someone with no demonstrated knowledge of set theory. It seemed reasonable to me to assume that OP was thinking in terms of subsets of the reals, and not sets of countable ordinals. (But why are you bringing this up now, five years after everyone else has forgotten it?) – Gerry Myerson Jan 07 '18 at 23:15
  • @GerryMyerson : Most people reading this answer would not have seen those comments. (This came up in my notifications, so I noticed it now.) – Michael Hardy Jan 08 '18 at 16:50
  • @MichaelH, made an edit. Better? – Gerry Myerson Jan 08 '18 at 17:08
  • @GerryMyerson : Better, but better still would be making it self-contained, e.g. "The bottom line is, no one has ever been able to present a set of real numbers provably of cardinality $\aleph_1$ and, on our current understanding of these things, no one ever will." – Michael Hardy Jan 08 '18 at 19:03