"Theorem": There is no $\omega$th inaccessible cardinal.

"Proof": Assume ZFC. Let $\kappa_n$ be the $n$-th inaccessible cardinal; since $V_\kappa$ is a model of ZFC for inaccessible $\kappa$, $V_{\kappa_n}$ is a model of ZFC + "there are $n-1$ inaccessible cardinals."

Now suppose there are $\aleph_0$ many inaccessible cardinals. Then in particular we can take the $\omega$-th inaccessible $\kappa_\omega$, and there will still be $\aleph_0$ many inaccessibles below it. Thus $V_{\kappa_\omega}$ is a model of ZFC+"there are $\aleph_0$ many inaccessible cardinals", and so by the (second) incompleteness theorem that theory is inconsistent. $\square$

But this "result" is literally too good to be true. So where am I wrong here?

Edit: Yep, I realize now what the fallacy is. Daniel Fischer and GME are correct - the existence of $\aleph_0$ many inaccessibles does not imply the existence of an $\omega$th inaccessible, just as the fact that there are $\aleph_0$ many natural numbers does not imply the existence of the ordinal $\omega$.

In other words, there can be models of ZFC + "there are $\aleph_0$ many inaccessibles" such that every inaccessible has finitely many inaccessibles below it, so the argument fails. In fact $V_{\kappa_\omega}$ is just such a model: it knows of $\kappa_n$ for all finite $n$, but it sees $\kappa_\omega$ as a proper class.

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    $\begingroup$ It does not follow from there being $\omega$ inaccessibles that there is an $\omega^{th}$ inaccessible. $\endgroup$
    – user104955
    Jun 7, 2015 at 19:42
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    $\begingroup$ If there are $\aleph_0$ many inaccessible cardinals, then it may be that for every single one of them, there are only finitely many smaller inaccessible cardinals. After all, there are $\aleph_0$ many natural numbers. $\endgroup$ Jun 7, 2015 at 19:42
  • $\begingroup$ Sorry, I deleted my answer because I'm feeling tired and not entirely coherent. I'll take a second look tomorrow morning and see what needs to be changed. $\endgroup$
    – Asaf Karagila
    Jun 7, 2015 at 21:33
  • $\begingroup$ @Daniel: There are also only $\aleph_0$ ordinals smaller than $\omega+\omega$. And yet infinitely many of them have $\aleph_0$ ordinals smaller than themselves. $\endgroup$
    – Asaf Karagila
    Jun 7, 2015 at 21:34
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    $\begingroup$ @Asaf Sure. The point is that $\aleph_0$ many doesn't guarantee that there is one among them with infinitely many predecessors, not that it rules that out. $\endgroup$ Jun 7, 2015 at 21:37

2 Answers 2


In many of these fake proofs, if you look closely you'll see that the infinity part plays a red herring.

The same argument, supposedly, would have been that if there are two inaccessible cardinals, $\kappa<\lambda$, then $V_\lambda$ would satisfy "There exists an inaccessible cardinal", thus the theory "$\sf ZFC$+There exists an inaccessible cardinal" proves its own consistency.

But all that we did was to show that there is one model of $\sf ZFC$ with an inaccessible; and in fact we just showed that $\sf ZFC$ with two inaccessible cardinals prove the consistency of $\sf ZFC$ with one inaccessible; while $V_\kappa$ is a model of $\sf ZFC$.

And of course the mistake is easily noted here, one inaccessible is not the same as two inaccessible cardinals. And we used the fact that there are two inaccessible cardinals and not the one.

The same mistake, as noted in the comments and the edit, was made here. You have assumed that $\omega$ inaccessibles mean $\omega+1$ inaccessibles. But there is a very large gap between $\sup\kappa_n$ and $\kappa_\omega$.

More to the point, you should measure order type and "discernible properties". So you shouldn't say "There are $\aleph_0$ inaccessible cardinals" but rather "There are $\omega$ inaccessible cardinals" vs. "There are $\omega+1$ inaccessible cardinals".

This can blow up into undefinable order types, or into proper classes, and indeed at some point you will get some large cardinal $\lambda$ such that $V_\kappa$ is an elementary model of $V_\lambda$ for an inaccessible cardinal $\kappa$. In particular all the first-order properties of $\lambda$ will be reflected by $\kappa$. But this just means that we are using properties of $\lambda$ which are first-order expressible in $V$, but not in $V_\lambda$ (as statement about the class of ordinals there).


I don't see the problem with your proof, but I look at it like this. Let $\theta_0$ be the first (I assume always strongly) inaccessible cardinal. Let $\theta_1$ be the smallest inaccessible larger than $\theta_0$, $\theta_2$ be the smallest inaccessible cardinal larger than $\theta_1$ and so forth. So what is $\theta_\omega$? Method 1 says it is the sup or limit of all of the $\theta_i$ as $i$ ranges over the finite ordinals. Method 2 says it is the smallest inaccessible cardinal larger than all the $\theta_i$. These two are NOT the same. The cardinal defined by Method 1 is the limit of a countable set, namely $\theta_i$, so it is accessible; in fact it is singular. The next inaccessible cardinal is therefore $\theta_{\omega+1}$. So it appears that a skip appears every time we hit a subscript that is a limit ordinal; in particular, the set of all inaccessibles less than a large cardinal $\kappa$ is not a club set.


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