I have two questions, actually. The first is as the title says: how do we know there exists an infinite cardinal such that there exists no other cardinals between it and $ \aleph_0 $? (We would have to assume or derive the existence of such an object before we label it something like $ \aleph_1 $.)

My second question is, can we say for certain if there's any limit to the number of cardinals existing between $ \aleph_0 $ and continuum (i.e. $ 2^{\aleph_0} $)? I mean, how could we know that there's not an infinite number of cardinals between the two - perhaps even more than $ \aleph_0 $?

Asaf Karagila
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    This is not the definition of $\aleph_1$ with which I'm familiar. The definition I've seen is just that it is the cardinality of $\omega_1$ (the first uncountable ordinal), and the proof that $\omega_1$ exists is just what Asaf says below. This doesn't imply, in ZF, that $\aleph_1$ is the smallest possible cardinality of an uncountable set; you need the well-ordering principle to assert this, which is equivalent to AC. Without AC, I think the collection of all possible cardinalities is not even totally ordered, let alone well-ordered. – Qiaochu Yuan Jun 25 '11 at 16:57
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    @Qiaochu: If the cardinalities are totally ordered then they are well-ordered and AC holds. If AC does not hold, well.. all hell breaks loose. – Asaf Karagila Jun 26 '11 at 19:23
  • Linked with this question **Why infinite cardinalities are not “dense”?** http://math.stackexchange.com/questions/374198/why-infinite-cardinalities-are-not-dense – MphLee Apr 30 '13 at 20:45
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    I have to admit that I don't understand why someone would downvote this question. – Asaf Karagila May 17 '13 at 10:58

1 Answers1


To prove the existence of $\aleph_1$ we use the concept of Hartogs number. The question asks, really, why are there uncountable ordinals, since $\aleph_1$ is by definition the least ordinal which is not countable.

Take a set of cardinality $\aleph_0$, say $\omega$. Now consider all the orders on $\omega$ which are well-orders, and consider the order isomorphism as an equivalence relation. The collection of all equivalence classes is a set.

Fact: If $(A,R)$ is a well-ordered set, then there exists a unique ordinal $\alpha$ such that $(A,R)\cong(\alpha,\in)$.

Map every equivalence class to the unique ordinal which is order isomorphic to the members of the class. We now have a set and all its members are ordinals which correspond to possible well-ordering of $\omega$.

Fact: The union of a set of ordinals is an ordinal, it is in fact the supremum of the elements in the union.

Let $\alpha$ be the union of the set defined above. We have that $\alpha$ is an ordinal, and that every ordinal below $\alpha$ is a possible well-ordering of $\omega$ (and therefore countable).

Suppose $\alpha$ was countable too, then $\alpha+1$ was also countable (since $\alpha+1=\alpha\cup\{\alpha\}$), and therefore a possible well ordering of $\omega$. This would contradict the above fact that $\alpha$ is greater or equal to all the ordinals which correspond to well-orderings of $\omega$, since $\alpha<\alpha+1$.

This means that $\alpha$ is uncountable, and that it is the first uncountable ordinal, since if $\beta<\alpha$ then $\beta$ can be injected into $\omega$, and so it is countable. Therefore we have that $\alpha=\omega_1=\aleph_1$.

Note that the above does not require the axiom of choice and holds in $\sf ZF$. The collection of all well-orders is a set by power set and replacement, so is the set of equivalence classes, from this we have that the collection of ordinals defined is also a set (replacement again), and finally $\alpha$ exists by the axiom of union. There was also no use of the axiom of choice because the only choice we had to do was of "a unique ordinal" which is a definable map (we can say when two orders are isomorphic, and when a set is an ordinal - without the axiom of choice).

With the axiom of choice this can be even easier:

From the axiom of choice we know that the continuum is bijectible with some ordinal. Let this order type be $\alpha$, now since the ordinals are well-ordered there exists some $\beta\le\alpha$ which is the least ordinal which cannot be injected into $\omega$ (that is there is no function whose domain is $\beta$, its range is $\omega$ and this function is injective).

From here the same argument as before, since $\gamma<\beta$ implies $\gamma$ is countable, $\beta$ is the first uncountable ordinal, that is $\omega_1$.

As to why there is no cardinals strictly between $\aleph_0$ and $\aleph_1$ (and between any two consecutive $\aleph$-numbers) also stems from this definition.

  1. $\aleph_0 = |\omega|$, the cardinality of the natural numbers,
  2. $\aleph_{\alpha+1} = |\omega_{\alpha+1}|$, the cardinality of the least ordinal number which cannot bijected with $\omega_\alpha$,
  3. $\aleph_{\beta} = \bigcup_{\alpha<\beta}\aleph_\alpha$, at limit points just take the supremum.

This is a function from the ordinals to the cardinals, and this function is strictly increasing and continuous. Its result is well-ordered, i.e. linearly ordered, and every subset has a minimal element.

This implies that $\aleph_1$ is the first $\aleph$ cardinal above $\aleph_0$, i.e. there are no others between them.

Without the axiom of choice, however, there are cardinals which are not $\aleph$-numbers, and it is consistent with $\sf ZF$ that $2^{\aleph_0}$ is not an $\aleph$ number at all, and yet there are not cardinals strictly between $\aleph_0$ and $2^{\aleph_0}$ - that is $\aleph_0$ has two distinct immediate successor cardinals.

For the second question, there is no actual limit. Within the confines of a specific model, the continuum is a constant, however using forcing we can blow up the continuum to be as big as we want.

This is the work of Paul Cohen. He showed that you can add $\omega_2$ many subsets of $\omega$ (that is $\aleph_2\le 2^{\aleph_0}$), and the proof is very simple to generalize to any higher cardinal.

In fact Easton's theorem shows that if $F$ is a function defined on regular cardinals, which has a very limited set of constraints, then there is a forcing extension where $F(\kappa) = 2^\kappa$, so we do not only violate $\sf CH$ but we violate $\sf GCH$ ($2^{\aleph_\alpha}=\aleph_{\alpha+1}$) in a very acute manner.

Asaf Karagila
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    Some minor comments: The set of well-orderings of $\omega$ is not a set of ordinals; I guess you are taking the set of ordinals corresponding to well-orderings of $\omega$. It is a minor point but for a technical issue: This presentation of the argument uses replacement, although one can avoid the use of replacement at the cost of losing $\aleph_1$ *as an ordinal*; one gets instead that there is a well-ordered set of smallest possible order-type that is uncountable. Also, $\aleph_1$ is *a* successor of $\aleph_0$ but, without choice, it is consistent that there are others. – Andrés E. Caicedo Jun 24 '11 at 03:40
  • @Andres: That is a very good point, I will correct the answer. – Asaf Karagila Jun 24 '11 at 05:39
  • @Asaf I don't understand why choosing representatives from equivalence classes is not a choice. Help – Katlus Jul 22 '12 at 11:56
  • @Katlus: I revised that part a bit, I hope it is clearer now. – Asaf Karagila Jul 22 '12 at 13:49
  • But that only proves that among the cardinalities of ordinals, there's a minimal cardinality above $\aleph_0$, right? But then this covers only sets that can be well-ordered, and without AC this doesn't necessary cover all sets, right? – celtschk Jul 22 '12 at 13:56
  • @celtschk: The question was "How can be prove that $\aleph_1$ exists?" which is by definition the existence of an uncountable ordinal (and thus we can prove there is a minimal one with this property). If you wish to show that there is a proper class of cardinals you need to reiterate this process, but this is going to work only up to the least inaccessible (consistency-wise, though, it is enough) and then you are going to have to use the fact that this inaccessible cardinal is itself a set, and "restart". – Asaf Karagila Jul 22 '12 at 14:03
  • Generally speaking in ZF, there are different meanings to "successor cardinal", which in a sense is minimal, we can prove that one of these interpretation always exists, but not the others (without choice that is). Lastly, it is also possible that there are several incomparable "minimal" cardinals for some of the cardinals. – Asaf Karagila Jul 22 '12 at 14:06
  • @AsafKaragila: So you say without choice, cardinalities are not totally ordered, and all non-orderable cardinalities are incomparable with $\aleph_1$, do I understand that right? Of course that naturally leads to the question whether on all "branches" of that partial order there's a minimal uncountable set, or if there can be branches where you always find a cardinal in between $\aleph_0$ and any given cardinal from that branch. But I admit that goes beyond the original question. – celtschk Jul 22 '12 at 15:08
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    @celtschk: Linear order of cardinality implies choice; so without it indeed there are incomparable cardinals. Where they sit depends a lot on the amount of choice holding in the universe. It might be that there are infinite cardinalities incomparable with $\aleph_0$; and sometimes all cardinals are comparable with $\aleph_1$ or higher. There is a complicated structure there, which is something I hope to discuss in my Ph.D. work over the next four-five years. – Asaf Karagila Jul 22 '12 at 15:22
  • @AsafKaragila: So in short, without choice it's complicated :-) Thank you for your explanations. – celtschk Jul 22 '12 at 15:29
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    Guru$^{2}{}{}{}{}$ :) – t.b. Jul 22 '12 at 18:08
  • Isn't it only a theorem of Von Neumann–Bernays–Gödel set theory and not Zermelo-Fraenkel set theory that there is an increasing function from ordinal numbers to cardinal numbers because there is not set of all ordinal numbers? – Timothy Dec 18 '17 at 23:44
  • @Timothy: I have absolutely no idea what you're on about. But in either case, please don't post an answer, and then complain about a similar answer that I posted. – Asaf Karagila Dec 18 '17 at 23:54
  • I can't see why if $\alpha +1$ were countable, then it would necessarily be isomorphic to a well-ordering of $\omega$. Could you please explain this? – CuriousKid7 Jan 20 '18 at 22:45
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    @CuriousKid7: Because being countable means in bijection with $\omega$, so by transport of structure it will be isomorphic to a well-ordering of $\omega$. – Asaf Karagila Jan 21 '18 at 00:22
  • The argument even holds in Z, because you don't need Replacement per se to get the set of well-orderings of a fixed set, just Bounded Separation. And one doesn't really need to work with the von Neumann ordinals, so don't need Replacement there either. – theHigherGeometer Apr 24 '19 at 01:30
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    @David: Indeed. There was a question about uncountable well orders without Replacement somewhere on the site, where I said exactly that. But you do need Replacement if you want $\aleph_1$ to exist as a von Neumann ordinal. – Asaf Karagila Apr 24 '19 at 06:38