# Are certain collections of cardinals still sets?

I have been looking at certain collections of cardinals lately and am trying to figure out if they are sets or not.

Is it true, that if $P$ is some large cardinal property and there is some $\kappa_{\text{first}}$, for which this property holds, that the collection of cardinals $\mathcal{C} = \{\kappa \mid \kappa \text{ has not } P\}$ is a set? Can I argue that $\kappa_{\text{first}}$ must have greater cardinality than all elements of $\mathcal{C}$ and therefore contains $\mathcal{C}$ as a subset?

On the other hand, if I know that a collection of cardinals $\mathcal{D}$ contains cardinals of arbitrarily large cardinality, must $\mathcal{D}$ be a proper class? Is it sufficient to argue that if $\mathcal{D}$ was a set, then $\lambda = \sup{\mathcal{D}}^+$ would be a cardinal not in $\mathcal{D}$, but since $\mathcal{D}$ contains arbitrarily large cardinals, one of them would have cardinality greater than $\lambda$, which already yields a contradiction?

Tl;dr: A class of cardinals is a set iff it has an upper bound. The right-to-left direction follows from Separation applied to that upper bound, while the left-to-right direction is via the Burali-Forti paradox.

The answer to your second question is yes - any collection of cardinals without an upper bound is not a set. However, your argument isn't correct - in defining $\lambda$, you're assuming what you're trying to prove (that if $\mathcal{D}$ is a set, then it has a supremum). Instead, show that if there were an unbounded set of cardinals, then the class of ordinals would be a set - which violates the Burali-Forti argument.

Meanwhile, the answer to your first question is extremely negative: for instance, the successor of a cardinal with a large cardinal property $P$ tends not to have $P$ (e.g. the successor of an inaccessible is not inaccessible, the successor of a measurable is not measurable . . .). Similarly, most large cardinal properties imply regularity, that is, they fail for any singular cardinal - of which there are unboundedly many. So for any large cardinal property $P$ I'm aware of, ZFC proves that the collection of cardinals which don't have $P$ is unbounded, i.e. is a proper class.

The only exceptions I can think of are artificial - e.g. say that $\kappa$ is big if $\kappa$ is greater than an inaccessible. Then assuming there is an inaccessible, the collection of non-big cardinals is a set. But this is a very silly property. EDIT: See the comments below - shame on my failure of imagination!

• @JasonDeVito Thanks, fixed. Oct 19, 2016 at 18:15
• Yet, if I had a property $P$ and $\kappa_{\text{first}}$ was the first one to have it, and P holds for all cardinals $\geq \kappa_{\text{first}}$, this would work as an exception? (I'm thinking \$omega$-measurable cardinals as defined in Eklof-Mekler's Almost Free Modules ...) Oct 19, 2016 at 18:48
• @mars412 Yes, that would work. (I was unaware that $\omega$-measurability was ever considered in its own right.) However, this is by far not the norm. It's a good exercise to show that a class of cardinals is a set iff it is bounded. Oct 19, 2016 at 18:52
• @NoahSchweber It was sort of common, although much less so nowadays. For instance, there are several results in Federer's Geometric measure theory book that, in modern language, state that a result holds for all cardinals $\kappa$ below the first measurable. Federer call these $\kappa$ Ulam numbers. Oct 19, 2016 at 20:40
• @AndrésE.Caicedo Very cool, I did not know that! Oct 19, 2016 at 21:23

That really depends on what you mean by "large cardinal property".

For example, if you mean "$\kappa$ carries a $\sigma$-complete measure", then every cardinal above the first measurable has this property, and therefore in that case there is only a set of cardinals without such a measure.

On the other hand, if you mean "$\kappa$ carries a $\kappa$-complete measure", then we can show that this implies $\kappa$ is a measurable. So it could be there is only one measurable in the universe, in which case the class of cardinals which are not measurable is co-finite. But in any case, every successor cardinal is not measurable, so you provably get a proper class.

So this really depends on the definition of "a large cardinal property". But more often than not, the largeness property is not "automatically upwards", namely a cardinal greater than a certain large cardinal need not be large in itself. The elevator pitch to remember here, is that being a large cardinal does not mean that your cardinality is very large.

• Unfortunately, this answer fell victim to someone being disgruntled with the way tags work on this website, and decided to reward me with two downvotes as a "thank you" message for removing irrelevant tags from their question. Oct 21, 2016 at 22:09