Do inaccessible-sized sets have any properties that smaller sets don't?

When going from finite to infinite sets, properties arise that are impossible for finite sets. For example, an infinite set has bijections to proper subsets, and to the Cartesian product of the set with itself.

Now I wonder if for sets whose cardinality is $$\ge$$ an inaccessible cardinal, there are also new properties arising that are true for all such sets, but are not possible for smaller sets.

Now obviously there are new properties like “has a cardinality larger than an inaccessible cardinal” but I'm interested in things that can be formulated without reference to large cardinals (just as my two examples of infinite set properties nowhere reference infinity).

Clarification: The properties I'm after are set properties that under the assumption of ZFC + large cardinal, are equivalent to the property “the set's cardinality is greater or equal to an inaccessible cardinal”.

• Clearly if $\kappa$ is above an inaccessible $\mathfrak c=\kappa$ cannot happen, so I guess you can cook up a property of cardinals above an inaccessible in terms of what can be forced through Cohen forcing, I don't know if this is the kind of answer you're looking for though – Alessandro Codenotti Jun 23 at 9:18
• @AlessandroCodenotti: If I understand correctly, anything you cook up by forcing is not true in all models of ZFC (or, in this case, in all models of ZFC + inaccessoble cardinal). I'm interested in things that automatically come with large size. – celtschk Jun 23 at 9:21
• What is true in every model of $\mathsf{ZFC}$ (regardless of whether this model actually has an inaccessible cardinal) then is "if $\kappa$ is above an inaccessible cardinal, then $\kappa\neq\mathfrak c$" – Alessandro Codenotti Jun 23 at 9:25
• @AlessandroCodenotti: But the property “does not have the cardinality of the continuum” also is true for the vast majority of smaller sets, and therefore fails the “not possible for smaller sets” condition. – celtschk Jun 23 at 9:27
• Sure, but it is in some sense a property of inaccessible cardinals, for example the slight weakening "if $\kappa$ is above a weakly inaccessible cardinal then $\kappa\neq\mathfrak c$" can fail very badly, since $\mathfrak c$ can be weakly Mahlo – Alessandro Codenotti Jun 23 at 9:30

In general all the properties will somehow end up being somehow equivalent to "There is an inaccessible of size $$\leq|X|$$".

Sometimes, this can be boring. For example, suppose that there is only a single inaccessible cardinal $$\kappa$$, and there are no transitive models of $$\sf ZFC$$ which contain $$V_\kappa$$ (i.e. all transitive models of $$\sf ZFC$$ have height $$\leq\kappa$$), then a set has size $$\geq\kappa$$ if and only if it is not equipotent to a set which lies in a transitive model of set theory.

Yes, that's not a particularly attractive property.

When you go up in the food chain of large cardinals these can be nicer. For example, if $$\kappa$$ is a measurable cardinal, then $$|X|\geq\kappa$$ if and only if there is a $$\kappa$$-complete free ultrafilter on $$X$$. And if $$\kappa$$ is strongly compact, then you can even require that this ultrafilter is uniform.

But this is ultimately just a fancy restatement of $$|X|\geq\kappa$$ and $$\kappa$$ is a large cardinal of type $$\varphi$$. Maybe that's what you're looking for.

• Well, it's not exactly what I'm looking for (it's all "ZFC+large cardinal * additional assumptions). The part about measureable cardinals is, however, interesting and in the spirit of the question (thus +1). BTW, looking again at my examples for infinite sets, I notice that they fail without choice, so maybe the interesting inaccessible-size properties would be those that fail without choice. – celtschk Jun 23 at 10:43
• Well, how do you define inaccessible cardinals without choice? There are several non-equivalent definitions. – Asaf Karagila Jun 23 at 10:45
• It may fail with any of the possible definitions. After all, it's a heuristic. – celtschk Jun 23 at 10:58
• Asaf, can't your statement about measurables be made in a more "attractive" way, i.e. "being larger than a measurable" is equivalent to there being a $\sigma$-complete nonprincipal ultrafilter on you, right ? This seems to be more in the spirit of the question (since your property refers to $\kappa$) – Max Jun 23 at 13:19
• @AsafKaragila: On the question how much more attractive: Very much more attractive. I managed to somehow overlook that instance of $\kappa$ when reading the answer; a $\kappa$-free formulation is definitely what I want. – celtschk Jun 23 at 13:34

There are a few examples that you can find in these slides that somewhat hide the underlying large cardinal property.

Here are two:

1. $$\vert X\vert$$ is larger than a measurable iff for every $$\omega$$-model $$(M, E)$$ with $$X\subseteq M$$ there is a proper elementary extension which is still an $$\omega$$-model.

Here $$(M, E)$$ is an $$\omega$$-model means that $$\omega$$ is in the transitive collapse of the wellfounded-part of $$(M, E)$$.

1. $$\vert X\vert$$ is larger than an extendible cardinal iff $$\vert X\vert$$ is a strong reflection cardinal for all invariant $$\Sigma_3$$ properties of structures.

Here we call $$\varphi(x)$$ with only $$x$$ free an invariant property of structures iff whenever $$\mathcal A$$ and $$\mathcal B$$ are isomorphic structures then $$\varphi(\mathcal A)\Leftrightarrow\varphi(\mathcal B)$$. Furthermore, "$$\lambda$$ is a strong reflection cardinal for $$\varphi$$" means that whenever $$\mathcal A$$ is a structure in a countable language with $$\varphi(\mathcal A)$$ there is a substructure $$\mathcal B$$ of $$\mathcal A$$ of cardinality less than $$\lambda$$ with $$\varphi(\mathcal B)$$.

Unfortunatley I do not know any interesting example for just an inaccessible cardinal.

• The first one can't be right. Assume that there are no $\omega$-models... – Asaf Karagila Jun 23 at 19:07
• Oh, I do not make any assumption on what theory the $\omega$ model satisfies. The backwards direction follows from making use of a proper elementary extension of $(V_{\vert X\vert+1}, \in)$ which is still an $\omega$ model. (This maybe a somewhat nonstandard definition of $\omega$ model...) – Andreas Lietz Jun 23 at 19:20
• Thank you. Maybe the answer to my question is simply "no", at least as far as interesting properties go. Given that I now independently got two properties for measurable cardinals (not counting any that may still be in the slides), maybe those are the more “structurally relevant” ones, not the “merely” inaccessible ones. – celtschk Jun 23 at 20:32