I am wondering if there are ways of defining "structure" on infinite sets that generate sequences of cardinals that cannot be proved to have the same cardinality as either the $\aleph$ or $\beth$ numbers?

Let me explain:

Start by defining $\beth_0=|\mathbb N|$, which then is also the cardinality of any other countably infinite set. Next use the power set axiom to define a sequence of sets: P($\mathbb N$), P(P($\mathbb N$)), P(P(P($\mathbb N$))), ...

So the first kind of "structure" we introduce is that of subsets.

By Cantor's Theorem, $|A|<|$P($A$)$|$. So define the $\beth$ numbers as the cardinality of these successive power sets: $\beth_1=|$P($\mathbb N$)$|$, $\beth_2=|$P(P($\mathbb N$))$|$, ...

Other kinds of structure on infinite sets would generate the same sequence of $\beth$s. For example, if we define F($A$) to be the set of all functions $A\rightarrow$ {0,1}. Then we will have $|$F($\mathbb N$)$|=|$P($\mathbb N$)$|=\beth_1$ and further $|$F(F($\mathbb N$))$|=\beth_2$, ...

A second kind of structure uses ideas of well ordered sets to build up ordinals. Define $\aleph_0=|\omega_0|$ and $\aleph_{\gamma^+}=$ {$\alpha:\alpha$ is an ordinal and $|\alpha|\le\aleph_{\gamma}$}. This will get us the sequence of $\aleph$s.

Again other kinds of structure (e.g. building ordinals using sets with a maximum element instead of using wellorderd sets) would generate the same sequence of $\aleph$s.

The independence of the (generalized) continuum hypothesis means that after $\aleph_0=\beth_0$ there is not much (other then some weak restrictions) we can prove about the relative size of $\aleph$s and $\beth$s in ZFC. And to me it seems like this is because once we get to uncountable sets there is just not enough "structure" avalaible to construct one-to-one functions and nail down their relative sizes.

So my question is are there other ways of defining "structure" on infinite sets to generate another sequence of cardinals (say $\gimel_0,\gimel_1, \gimel_3$ ...), which, at least within ZFC, then can't be shown to have the same cardinality as either some $\aleph$s or some $\beth$s?

Or alternatively, is there a proof that within ZFC that defining cardinals using either the method of $\aleph$s or $\beth$s somehow exhausts possible ways of cardinal definition?

Clarification: I want be clear that I understood comments that in ZFC all cardinals are $\aleph$s, since any set can be well-ordered. But it is undecidable which $\aleph$s the $\beth$s correspond to, right? Even though we know there exists a well-ordered set with cardinality $\beth_1$, we cannot say which $\aleph$ that is without deciding the continuum hypothesis. So the question is are there other ways of defining "structure" to generate cardinals that are similarly undecidable in ZFC? Or are the $\aleph$s or $\beth$s all there is?

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    $\begingroup$ All cardinals are $\aleph$s in ZFC, so maybe we should consider just ZF instead. I believe we can have incomparable cardinals without choice but I don't know of any alternative method for generating them like you are looking for. $\endgroup$ Jan 7, 2018 at 22:35
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    $\begingroup$ @DanBrumleve Small correction. All infinite cardinals are $\aleph$'s in ZFC. $\endgroup$ Jan 7, 2018 at 23:29
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    $\begingroup$ Jon, there are many other such cardinals. Google "cardinal invariants of the continuum" for some such cardinals that live between $\aleph_0$ and $\mathfrak{c}$. $\endgroup$ Jan 8, 2018 at 2:20
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    $\begingroup$ @Jon I think (for example) setting $\gimel_\alpha=\aleph_{\alpha^2}\cdot\beth_\alpha$ will give you an example of what you're looking for: we can't compute $\gimel_\alpha$ in ZFC alone, either in terms of $\aleph$s or in terms of $\beth$s. $\endgroup$ Jan 8, 2018 at 10:56
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    $\begingroup$ In this case, I'm not sure there is any question here. After all, any definition of the "structure" giving a notion we are denoting $\gimel_\alpha$, there is a definable function $F$ on the ordinals such that $\gimel_\alpha=\aleph_{F(\alpha)}$. $\endgroup$ Jan 8, 2018 at 15:57

2 Answers 2


One can consistently generate new cardinals simply by combining the two methods you have mentioned.

For example, I claim that it is relatively consistent with ZF that the cardinal $\aleph_1+\beth_1$ is neither an $\aleph$ cardinal nor a $\beth$ cardinal.

To see this, consider a model of ZF in which there is no $\omega_1$ sequence of real numbers. This is true, for example, in any model of the axiom of determinacy AD. Let $X$ be a set of cardinality $\aleph_1+\beth_1$, formed by a disjoint union of a copy of the reals and a copy of $\omega_1$. This set is not an $\aleph$, since it is not well-orderable, as that would provide a well-ordering of the reals. I claim it is also not a $\beth$. For example, this set cannot be bijective with $\beth_1=2^\omega$, since then there would be an $\omega_1$-sequence of reals, contrary to assumption. And it cannot be as large as $\beth_2$, since $\beth_1$ surjects onto $X$, but $\beth_1$ cannot surject onto $\beth_2$ by Cantor's theorem.

Lastly, I noticed that you mentioned ZFC in your question, but of course, in ZFC, every infinite cardinal is an $\aleph$ cardinal, since every set is well-orderable in that theory. In fact, the assertion that every infinite cardinal is an $\aleph$ cardinal is equivalent to the axiom of choice, and indeed, the assertion that the cardinals are merely linearly ordered is equivalent to the axiom of choice.


I think the answer to your question lies in Scott's cardinals, the cardinality of a set x is defined as the equivalence class of all sets bijective to x that belongs to the minimal possible rank that a set bijective to x appears as a subset of. And along the same way ordinals also can be defined. If you work in ZF alone, then there are models of ZF in which some Scott's cardinals are not comparable to any $\aleph$ or $\beth$ number at all, like for example the Scott cardinals of Tarski Infinite Dedekind finite sets. Also I do think that there are many ways to define 'structure' even in $ZFC$ that can possess indecidability of cardinal comparisons similar to those between the $\aleph$s and the $\beth$s, like for example the cardinality of the set of all Hereditarily subnumerous sets to a set, you can build up stages of those in a similar manner to how you do with powers, call those as the $\daleth_i$ numbers (I don't know if this is used in other contexts), so there will be no clear comparisons between for example $\aleph_i$ and $\daleth_i$ numbers, nor even between the $\beth_i$s and the $\daleth_i$s.

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    $\begingroup$ You are mixing up two things here, which makes it difficult to understand what you are doing. What you call a Scott cardinal is simply a way of formalizing a cardinality as a set in the absence of choice. It is essentially noise, since this is not relevant. Then you mention that without choice some cardinalities are neither alephs nor beths. But this was already mentioned in Joel's answered. What is new here is that the cardinalities of infinite Dedekind finite sets are examples. This is not necessarily true though. But almost: Given any such set, either it is an example, or ... $\endgroup$ Jan 10, 2018 at 21:47
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    $\begingroup$ ... you obtain an example by removing (or adding) a single point. $\endgroup$ Jan 10, 2018 at 21:48
  • $\begingroup$ I believe I defined a structure that generates a sequence of cardinals that bears similar indecidability to what is present between the $\aleph$ and $\beth$ numbers, and even in $ZFC$ itself, that of the sequence of what I symbolized as the $\daleth$ numbers, so you can recursively define those by using the "set of all hereditarily subnumerous to a set" operator instead of the power set operator, see: cs.nyu.edu/pipermail/fom/2012-January/016121.html, so $\daleth$ numbers are the cardinalities of those iterative stages of the hierarchy mentioned in that link. $\endgroup$ Jan 11, 2018 at 20:40

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