# Are those cardinals not weakly inaccessible?

Consider the set of aleph fixed points, $\{\gamma:\gamma=\aleph_\gamma\}$ and call $\daleth_\alpha$ the $\alpha$-th element of that set. Consider the function $f:\sf Ord\to\sf Ord$ defined by $\alpha\mapsto\daleth_\alpha$, this is the first Veblen function associated with the function $\alpha\mapsto\aleph_\alpha$ and as such it is normal, so it has a proper class of fixed points.

Exercise 1.13.17 of Kunen's "The Foundations of Mathematics" asks to prove that if $\kappa$ is weakly inaccessible then it is a fixed point of $f$ and clearly the assertion "$\kappa$ is a fixed point of $f$ $\implies$ $\kappa$ is inaccessible" is not provable in $\sf ZFC$.

Is its negation provable in $\sf ZFC$? More generally is it possible that $\sf ZFC$ proves the existence of a certain cardinal but can't show that it isn't inaccessible?

## 1 Answer

The answer to your specific question is: ZFC proves, for instance, that the first fixed point of the $\aleph$ function is not weakly inaccessible; this is because that first fixed point has cofinality $\omega$, and so is not regular. Indeed, the $\alpha$th fixed point (call it $\kappa_\alpha$) won't be inaccessible unless $\kappa_\alpha=\alpha$ (since otherwise it won't be regular).

And in fact $\kappa_\alpha=\alpha$ still doesn't guarantee inaccessibility; this is because the function $\alpha\rightarrow\kappa_\alpha$ is continuous, and so just as above its first fixed point has cofinality $\omega$, hence isn't regular, hence isn't weakly inaccessible.

The problem, as you can see, is cofinality: any construction of an ordinal in terms of fixed points or similar won't be able to get away from this problem on its own.

Meanwhile, your broader question - "is it possible that ZFC proves the existence of a certain cardinal but can't show that it isn't inaccessible?" - has a positive answer, but for silly reasons: consider the cardinal $\kappa$ defined to be $0$ if no inaccessibles exist, and the least inaccessible if there is an inaccessible at all. ZFC can't prove "$\kappa$ is not inaccessible" unless ZFC already proves "there are no inaccessible cardinals."

Now, presumably this isn't what you mean, but improving this question in a precise way is quite difficult. Broadly speaking, though, I would say that the following is a good heuristic: for the reasons above, no "concrete" construction which ZFC proves actually builds something can ever have the first thing it builds be inaccessible. And even more broadly, if there is a natural class of ordinals which ZFC proves is nonempty, then ZFC probably proves that the least element of this class isn't inaccessible.

EDIT: The above merely shows that in general, questions of this type are difficult to make precise. This specific one, however, has a natural positive solution: it is consistent that $2^{\aleph_0}$ is a regular limit cardinal, that is, weakly inaccessible. And of course it is consistent that $2^{\aleph_0}$ isn't weakly inaccessible.

The proof that $2^{\aleph_0}$ can be weakly inaccessible is quite hard - the relevant technique is forcing, which roughly speaking lets us enlarge a given model of ZFC by adding new sets in a controlled way. I can't try to explain this here (although you might be interested in this introduction by Timothy Chow), but let me say some words which, if you choose to dive into forcing, will help outline the path to this result:

In this case, the most natural approach is the following: we start with some beginning model $M$ with a weakly inaccessible cardinal $\kappa$ satisfying CH. Now we want, intuitively, to "add $\kappa$-many reals" to $M$. This much is easy to do via forcing; however, there are now two things that need to be checked in the bigger model:

• $2^{\aleph_0}=\vert\kappa\vert$ (that is, we didn't add more than $\kappa$-many reals, in the process of adding at least $\kappa$-many reals);

• $\kappa$ remains inaccessible (e.g. amongst the various things forcing can add are bijections between $\kappa$ and smaller ordinals, so after forcing $\kappa$ might not even be a cardinal at all!).

These requirements are addressed by studying the combinatorics of both $\kappa$ itself and of the forcing poset used: the key facts being the regularity of $\kappa$ and the fact that our poset has a combinatorial property called the "countable chain condition."

• Note that this is mirrored to a certain extent on a smaller scale when we try to build up to $\omega_1$ using countable ordinals; and some discussion of this topic can be found at this question. – Noah Schweber Aug 24 '17 at 20:59
• Thanks for your answer, so every fixed point of $\alpha\mapsto\kappa_\alpha$ which can be shown to be least fixed point above some ordinal can't be inaccessible because of the same cofinality problem, right? – Alessandro Codenotti Aug 24 '17 at 21:00
• Concerning my broader question I discovered after posting this question (thanks to one that was suggested as related) that it is consistent with $\sf ZFC$ that $2^{\aleph_0}$ is weakly inaccessible, although because of reasons I don't understand. Your example while being arguably sillier is definitely simpler – Alessandro Codenotti Aug 24 '17 at 21:04
• @AlessandroCodenotti Yes, your first comment is correct. And re: your second comment, that is indeed a better example - the point I wanted to make, though, is that questions of this form are inherently difficult to make precise. (That said, the example of $2^{\aleph_0}$ gives a natural positive answer, so this particular instance should be considered fully addressed. You mention that you don't understand how it can be weakly inaccessible; I've added a bit of explanation in my answer.) – Noah Schweber Aug 24 '17 at 21:08
• thanks for the added explanation, it makes a lot of sense intuitively! I'm currently self studying Kunen's book(s) with the objective of understanding forcing so hopefully I'll return to it in the future – Alessandro Codenotti Aug 24 '17 at 21:26