Given a set $S$, one can easily find a set with greater cardinality -- just take the power set of $S$. In this way, one can construct a sequence of sets, each with greater cardinality than the last. Hence there are at least countably infinite many orders of infinity.

But do there exist uncountably infinite orders of infinity?

To be precise, does there exist an uncountable set of sets whose elements all have distinct cardinalities?

The first answer to Types of infinity suggests the answer is "yes", but only establishes a countable number of cardinalities (which, to be fair, was what the question was asking about).

I've been exposed to enough mathematical logic to realize that I'm walking in a minefield; let me know if I've already mis-stepped.

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    $\begingroup$ possibly related: math.stackexchange.com/questions/1467/… $\endgroup$
    – Isaac
    Oct 27, 2010 at 7:30
  • $\begingroup$ @Isaac: I had the distinct feeling that the question, as well as my answer, was familiar. Thanks for checking up on this. $\endgroup$ Oct 27, 2010 at 7:33
  • $\begingroup$ you may like to see this: mathoverflow.net/questions/29624/… $\endgroup$
    – anonymous
    Oct 27, 2010 at 7:54
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    $\begingroup$ Note: "orders of infinity" is not the correct phrase to use here since it is normally used to to describe growth rates of functions. Instead, use "cardinal numbers". $\endgroup$ Oct 27, 2010 at 18:29

2 Answers 2


Pete's excellent notes have correctly explained that there is no set containing sets of unboundedly large size in the infinite cardinalities, because from any proposed such family, we can produce a set of strictly larger size than any in that family.

This observation by itself, however, doesn't actually prove that there are uncountably many infinities. For example, Pete's argument can be carried out in the classical Zermelo set theory (known as Z, or ZC, if you add the axiom of choice), but to prove that there are uncountably many infinities requires the axiom of Replacement. In particular, it is actually consistent with ZC that there are only countably many infinities, although this is not consistent with ZFC, and this fact was the historical reason for the switch from ZC to ZFC.

The way it happened was this. Zermelo had produced sets of size $\aleph_0$, $\aleph_1,\ldots,\aleph_n,\ldots$ for each natural number $n$, and wanted to say that therefore he had produced a set of size $\aleph_\omega=\text{sup}_n\aleph_n$. Fraenkel objected that none of the Zermelo axioms actually ensured that $\{\aleph_n\mid n\in\omega\}$ forms a set, and indeed, it is now known that in the least Zermelo universe, this class does not form a set, and there are in fact only countably many infinite cardinalities in that universe; they cannot be collected together there into a single set and thereby avoid contradicting Pete's observation. One can see something like this by considering the universe $V_{\omega+\omega}$, a rank initial segment of the von Neumann hierarchy, which satisfies all the Zermelo axioms but not ZFC, and in which no set has size $\beth_\omega$.

By adding the Replacement axiom, however, the Zermelo axioms are extended to the ZFC axioms, from which one can prove that $\{\aleph_n\mid n\in\omega\}$ does indeed form a set as we want, and everything works out great. In particular, in ZFC using the Replacement axiom in the form of transfinite recursion, there are huge uncountable sets of different infinite cardinalities.

The infinities $\aleph_\alpha$, for example, are defined by transfinite recursion:

  • $\aleph_0$ is the first infinite cardinality, or $\omega$.
  • $\aleph_{\alpha+1}$ is the next (well-ordered) cardinal after $\aleph_\alpha$. (This exists by Hartog's theorem.)
  • $\aleph_\lambda$, for limit ordinals $\lambda$, is the supremum of the $\aleph_\beta$ for $\beta\lt\lambda$.

Now, for any ordinal $\beta$, the set $\{\aleph_\alpha\mid\alpha\lt\beta\}$ exists by the axiom of Replacement, and this is a set containing $\beta$ many infinite cardinals. In particular, for any cardinal $\beta$, including uncountable cardinals, there are at least $\beta$ many infinite cardinals, and indeed, strictly more.

The cardinal $\aleph_{\omega_1}$ is the smallest cardinal having uncountably many infinite cardinals below it.

  • $\begingroup$ Just a comment: this first set of lecture notes treats set theory from a "cheerfully naive" standpoint, which I maintain (although it would be obnoxious to insist on it when talking to an eminent professional set theorist such as yourself) is a good viewpoint for most students of mathematics. The reader who wants axiomatic set theory should certainly look elsewhere. $\endgroup$ Oct 27, 2010 at 14:31
  • $\begingroup$ Your notes are great, Pete; I meant only to add some gloss to your answer. Indeed, my view is that the naive set theoretic viewpoint includes perhaps sub-conscious uses of the Replacement principle, as the historical situation with Zermelo shows. $\endgroup$
    – JDH
    Oct 27, 2010 at 15:47
  • $\begingroup$ No worries, I'm not offended in the least (I hope you can say the same). And I completely agree that naive set theory includes unwitting use of certain axioms beyond the minimal set of axioms that set theorists have agreed upon. (E.g., I also proved that every infinite set has a countable subset...) Still, it's a funny situation that the better answer is "No, in order to prove that you need to use the following axiom." It seems that the rest of mathematics is a little more platonistic than set theory: maybe I even believe that Replacement is -- dare I say it -- true? $\endgroup$ Oct 27, 2010 at 17:37
  • $\begingroup$ I had to end the comment there, so it ended a little flippantly. More seriously, I have no doubt that if I studied sets in more depth, my "mathematical-religious beliefs" regarding them would become more nuanced and formalist. It's obviously required for doing work in the area. So it's amusing to me that I can happily maintain (the illusion of) faith that there are sets out there which work in ways which seem intuitively clear. $\endgroup$ Oct 27, 2010 at 17:42
  • $\begingroup$ Why do you say that there are only countably many infinite cardinalities in that least Zermelo model? You said yourself (if I understood you right) that the cardinals $\aleph_n$ ($n\in\omega$) are not a countable set in that universe, they are even a proper class. They form a countable set "seen from outside" but doesn't the same go for the cardinalities in the least model of ZFC set theory? $\endgroup$
    – bof
    Sep 5, 2022 at 3:56

The answer is yes. See "Fact 20" at the end of the following handout:


Edit: Fact 20 states

For no set $I$ does there exists a family of sets $\{S_i\}_{i∈I}$ such that every set $S$ is equivalent to $S_i$ for at least one $i$.

Proof: Take $S_\text{bigger} := 2^{\cup_{i\in I}S_i}$. There is no surjection from $\cup_{i\in I}S_i$ onto $S_\text{bigger}$, so for sure there is no surjection from any $S_i$ onto $S_\text{bigger}$.$\quad\square$

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    $\begingroup$ 404! If only you had been thinking 11 years ahead and embedded Fact 20 into your answer :'( $\endgroup$
    – Matt
    May 14, 2021 at 14:52
  • $\begingroup$ web.archive.org/web/20201205105659/https://math.uga.edu/~pete/… $\endgroup$
    – Quantum7
    May 17, 2021 at 9:11
  • $\begingroup$ @Matt: Please see the note about how to fix broken links on my profile page. (Sorry! It annoys me at least as much as it annoys you.) $\endgroup$ May 21, 2021 at 21:21
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    $\begingroup$ (+1) The proof of Fact 20 is very short and could well be included with this answer. $\endgroup$ Aug 30, 2022 at 9:58

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