Another title for this question could be: where do inaccessible cardinals live? It may be that this question does not make any sense. So I will try to explain what I mean.

I think of the ZFC axioms as a recipe that allows us to structure the universe of sets in particularly nice way. Out of all the ways one could define a "collection of things" ZFC manages to pick out those "collections" which behave well. We call these collections sets. Using this recipe the collection of all sets can be bundled together into a proper class $V$ --- the von Neumann Universe.

Within $V$ we can pick out sets with particular properties. Of note there are ordinals and cardinals. Furthermore we can look at cardinals with particular properties; large cardinals and inaccessible cardinals in particular.

I guess I am adopting a Platonist view point when I ask the following questions:

(1) are the inaccessible cardinals in the class $V$?

(2) are these cardinals inaccessible in the sense that do not fit into the hierarchy of the von Neumann universe, $V$?

(3) are they inaccessible in the sense that the axioms of ZFC can't prove that they are in there?

(4) could it be that there are no cardinals with the added properties that make them inaccessible/large, but ZFC can't prove that this is the case?

Here is what I think: By definition they are particular types of cardinals i.e. particular types of sets. So they must be in $V$, right? I think inaccessible/large cardinals (if they exist) are sets (i.e. are in $V$) but ZFC fails to decide (i.e. is unable to access) whether or not there are any sets with these properties.

I am confused and would love to know what is really going on.

Thank you for persisting through my ramble :)

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    $\begingroup$ My suggestion is to take things in more guarded steps. You already asked a reasonably good question in the title, about the significance of "inaccessible" in inaccessible cardinals. Much of what wound up in the body of your Question is not germane to that terminology, $\endgroup$
    – hardmath
    May 10, 2018 at 4:43
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    $\begingroup$ Inaccessible cardinals are inaccessible in the same sense that $\omega$ is: you can't reach $\omega$ by taking finite sums, products, and powers of finite cardinals. The same goes for an inaccessible cardinal $\kappa,$ but with "finite" replaced by "smaller than $\kappa$". $\endgroup$
    – bof
    May 10, 2018 at 5:12

2 Answers 2


Yes, inaccessible cardinals are cardinals, making them by definition ordinals, and therefore sets. If you adopt the Platonist view, even for the sake of discussion, then there is a universe of sets $V$ which contains all the sets, and if there are inaccessible cardinals, then they are sets there.

Two examples

Let me digress and present to you two examples, which might help you grasp a little bit what inaccessibility really mean.

The case of $\omega$

There is a reason why large cardinal axioms are often called "strong infinity axioms". If you consider $\sf ZFC-Inf$, then $V_\omega$ is a model of that theory. And this theory is too weak to prove the existence of $\omega$. Because all you can do is finitary operations on finite sets, which can never lead you to the existence of an infinite set.

So in order to get $\omega$ as a set, you need to postulate its existence by an axiom. You cannot "access" $\omega$ by constructing it from below. And in that sense, inaccessible cardinals are inaccessible.

The case of $\omega_1$

Let's consider $\omega_1$, but replace the usual operations with ordinal arithmetic. So you're allowed to take infinite sums, products, exponentiations, whatever you like. But you are only allowed to do so over countable ordinals, both as indices and as summands/multiplicands/etc., the result is that you cannot write $\omega_1$ down. It is beyond your reach.

Ordinal arithmetic is cardinality preserving (when infinite ordinals are involved), so no matter what you apply to countable ordinals, the result is a countable ordinal. Never $\omega_1$.

So from an ordinal arithmetic point of view, $\omega_1$ is inaccessible.

So what does it mean to be inaccessible?

Really it means that there is no set of size less than $\kappa$ that can definably provide you with $\kappa$, or with a bigger ordinal. Either via exponentiation (so power sets are not enough), or summation (so unions of smaller sets remain smaller, if the index was smaller).

Inaccessible cardinals are those that no matter what we do, we cannot construct them from smaller sets. We must postulate their existence "by hand".

Here is a curious fact, though. The above implies that if $\kappa$ is inaccessible, then $V_\kappa$ is a model of $\sf ZFC$. One would be tempted to argue that the reverse implication holds. This is not true.

If $V_\kappa$ is a model of $\sf ZFC$, then all you can conclude is that there is no way to define inside $V_\kappa$ a recipe of getting $\kappa$ from smaller sets, but $V$ is larger, and in $V_{\kappa+1}$ we can find such object. So in order to get inaccessibility, you need to replace $\sf ZFC$ with its second-order counterpart.

Okay, what about their existence, then?

Finitism reject the idea of infinite sets. Some, like ultrafinitists even reject the idea of an infinite "class". But so far, the attempts to prove that infinite sets do not exist fail. And because this idea is so ingrained into modern mathematics, most of us sleep safely at night, believing that no one will outright (and seriously) disprove their existence.

Even if we cannot prove the existence of an infinite set without the help of that explicit axiom of infinity.

Large cardinal axioms, or strong infinity axioms, share the same fate. We cannot prove them, and so from a Platonist view, we cannot prove that inaccessible cardinals really exist if all we agree upon is $\sf ZFC$. But if you believe that inaccessible cardinals are consistent (and it seems that the majority of set theorists believe that way), then you sleep safe and sound at night knowing that no one will seriously disprove their existence tomorrow morning.

There is a catch, though, from a "non-set theorist point of view". Infinity is mysterious, but incompleteness is even worse and it can be misleading for non-logicians sometimes.

If we assume an inaccessible cardinal exists, then we can prove that $\sf ZFC$ is consistent. Just like when we assume that $\omega$ exists we can prove that $\sf ZFC-Inf$ is consistent (which is again a motivation for the term "strong infinity axioms").

But inaccessible cardinals don't usually come with this "very concrete description" like $\omega$ has as the smallest inductive set. An inaccessible cardinal cannot even be the smallest $\kappa$ such that $V_\kappa$ is a model of $\sf ZFC$. So you get this sort of ambiguous object, whose existence allegedly contradicts the incompleteness theorem.

When I was a masters student I remember feeling that something is very fishy with all those large cardinals, since they let you prove all sort of consistency results like that. But then Shimon Garti, who was finishing his Ph.D. at the time told me "yes, but you add axioms, your theory is stronger and you can prove more things". And that clicked everything together: you add axioms, then you can prove more.

But the fact remains, for a non-logician, inaccessible cardinals are also mentally inaccessible, since their existence is not given by some "nice construction". So they do not posses this ineffable understanding of what an inaccessible cardinal is in the hierarchy of sets. For them it's all just black magic anyway.

  • $\begingroup$ Thank you for taking the time to write such a detailed answer. I fall into the category "non-logician" so I will have to think about your answer, a lot. I think my main confusion is coming from the understanding of $V$ - I must not have a good enough understanding of the formation of each $V_\alpha$. $\endgroup$ May 11, 2018 at 0:36
  • $\begingroup$ I hope you don't mind with one follow up question to see if I have grasped something of the answers given here: Is the formation of $V$ done inside ZFC? Or "outside" ZFC? It seems to me that the formation of $V$ must be done from outside ZFC as it is done by taking a union indexed by a proper class (namely, the class of ordinals) --- In this case, one could ask which $V_\alpha$ are accessible from within ZFC? And the answer is that some are not accessible. $\endgroup$ May 11, 2018 at 0:48
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    $\begingroup$ I think that math.stackexchange.com/questions/1802655/… and math.stackexchange.com/questions/2495465/… might be helpful in clearing these issues. $\endgroup$
    – Asaf Karagila
    May 11, 2018 at 6:20
  • $\begingroup$ Thank you :) I think this "non-logician" will have to think about all of this a lot more than I already have. $\endgroup$ May 14, 2018 at 3:15
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    $\begingroup$ It's a long way to the top if you want to rock and roll. $\endgroup$
    – Asaf Karagila
    May 14, 2018 at 6:06
  1. Yes, they are sets, and like all sets, they are in $V$ by definition.
  2. No, they are ordinals and thus have Von-Neumann rank.
  3. It is true that their existence is not provable in ZFC. Not sure that's exactly the sense in which "inaccessible" is meant. I think of it as being inaccessible under iteration of set operations. An inaccessible cardinal is, roughly speaking, one that cannot be obtained from smaller ordinals by iterating union, replacement and power set. But this also gives a way of understanding why they can't be proved to exist in ZFC, since ZFC basically assumes that the way you get new sets is by iterating these operations. In fact, if $\kappa$ is inaccessible, $V_\kappa$ is a model of ZFC (cause of these closure properties), so by the incompleteness theorem, ZFC cannot prove $\kappa$ exists.
  4. Regardless if they exist or not (or if that is a meaningful question), we know that ZFC, if consistent, is not able to prove they exist. It is a possibility that ZFC is consistent and can also prove that there are no inaccessibles. (If we could non-circularly prove that ZFC can't prove this, it would be an argument that the consistency of ZFC implies the consistency of ZFC with inaccessibles... so the theory of ZFC plus inaccessibles could prove its own consistency, so would be inconsistent, which implies by the contrapositive of our argument the ZFC is inconsistent.) Most platonistically inclined set theorists believe in their existence. Basically, we don't assume all sets are finite just cause the finite sets are the only ones in reach of the closure of set operations starting from the empty set. We add the axiom of infinity. Why, after we've exhausted the closure of set-producing operations starting after the axiom of infinity, should we assume that's it and there is no more? Why not add another axiom of infinity?
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    $\begingroup$ "ZFC, if consistent, is not able to prove it one way or the other": That's not quite right. ZFC could be consistent and also prove that inaccessible cardinals do not exist. $\endgroup$ May 10, 2018 at 5:10
  • $\begingroup$ Axiom of moar infinity. $\endgroup$ May 10, 2018 at 5:12
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    $\begingroup$ @User0112358: Your universe is given. It is fixed. The von Neumann hierarchy does not construct the universe, as much as it gives you a nice filtration of the universe. $\endgroup$
    – Asaf Karagila
    May 10, 2018 at 6:22
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    $\begingroup$ @YuiToCheng I'm sure someone has, but nothing specific I can think of. But as to the more modest goal of proving larger large cardinals are inconsistent 1) The rank-into-rank hypotheses I0-I3 were named with "I" since it was originally believed they were inconsistent, but nobody has managed to prove that (presumably not for lack of trying). 2) As for further down in the hierarchy, I've heard that Jack Silver expended a lot of effort attempting to prove measurable cardinals were inconsistent, see here. $\endgroup$ Sep 20, 2021 at 19:21
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    $\begingroup$ @YuiToCheng (I might also mention a much more ambitious project: Edward Nelson's high-profile attempt to show PA is inconsistent, which culminated in a purported proof that was (fortunately) found to be flawed and retracted in short order. Of course this would have knocked out ZF and even a lot of weaker set theories, and, a fortiori, ZFC + inaccessibles.) $\endgroup$ Sep 21, 2021 at 4:56

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