# A question about supercompact cardinal numbers

Is it possible to tell if a cardinal number C is supercompact by looking only at the properties of subsets of C and paying no attention to sets whose cardinal numbers are greater than C? This can certainly be done with many smaller infinite cardinal numbers such as measurable cardinal numbers.

-
I added the large-cardinals tag. – Joel David Hamkins Apr 17 '10 at 17:13

No, this is not possible.

If κ is supercompact, then in particular it is measurable, and there is a normal measure μ on κ that concentrates on non-measurable cardinals. If j:V to M is the corresponding normal ultrapower embedding, then κ is not measurable in M. But M and V have the same power set of κ. Thus, any criterion that looks only at P(κ) cannot distinguish between V and M, but κ is supercompact in V and not in M.

(To find a measure μ concentrating on non-measurable cardinals, simply choose a normal measure μ for which the ultrapower jμ has the smallest possible value for jμ(κ). It follows that κ cannot be measurable in the corresponding Mμ, for if it were, we could make a better μ.)

A similar argument shows that the supercompactness of κ cannot be characterized by any property in Hδ for any particular δ above κ, since for every λ above δ there will be λ supercompactness ultrapowers j:V to M for which κ is not λ supercompact in M, but V and M have the same Hδ.

Finally, note that is it not really correct to say that the measurability of a cardinal κ can be characterized by looking only at P(κ), since my argument above shows two models M and V with the same P(κ), but κ is measurable in one of them and not the other. Rather, in order to tell if κ is measurable, you have to look at subsets of P(κ), that is, at P(P(κ)), since this is where the measure lives.

-
A similar argument shows that strong cardinals, strongly compact cardinals etc. also cannot be characterized inside P(kappa). On the other hand, superstrong, almost huge and huge cardinals are characterized by the existence of a single set object (the appropriate measure), and they do have Sigma_2 definitions in set theory. – Joel David Hamkins Apr 17 '10 at 16:35
I should have said, "No, this is not possible, assuming that supercompact cardinal are consistent." If the existence of a supercompact cardinal is inconsistent with ZFC, then they are very easy to characterize by a property of the desired type: just use any tautologically false property. – Joel David Hamkins Apr 17 '10 at 21:38