A question about supercompact cardinal numbers - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T22:38:15Z http://mathoverflow.net/feeds/question/21669 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/21669/a-question-about-supercompact-cardinal-numbers A question about supercompact cardinal numbers Garabed Gulbenkian 2010-04-17T15:46:11Z 2010-04-17T17:30:50Z <p>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.</p> http://mathoverflow.net/questions/21669/a-question-about-supercompact-cardinal-numbers/21672#21672 Answer by Joel David Hamkins for A question about supercompact cardinal numbers Joel David Hamkins 2010-04-17T16:13:35Z 2010-04-17T17:30:50Z <p>No, this is not possible. </p> <p>If &kappa; is supercompact, then in particular it is measurable, and there is a normal measure &mu; on &kappa; that concentrates on non-measurable cardinals. If j:V to M is the corresponding normal ultrapower embedding, then &kappa; is not measurable in M. But M and V have the same power set of &kappa;. Thus, any criterion that looks only at P(&kappa;) cannot distinguish between V and M, but &kappa; is supercompact in V and not in M.</p> <p>(To find a measure &mu; concentrating on non-measurable cardinals, simply choose a normal measure &mu; for which the ultrapower j<sub>&mu;</sub> has the smallest possible value for j<sub>&mu;</sub>(&kappa;). It follows that &kappa; cannot be measurable in the corresponding M<sub>&mu;</sub>, for if it were, we could make a better &mu;.)</p> <p>A similar argument shows that the supercompactness of &kappa; cannot be characterized by any property in H<sub>&delta;</sub> for any particular &delta; above &kappa;, since for every &lambda; above &delta; there will be &lambda; supercompactness ultrapowers j:V to M for which &kappa; is not &lambda; supercompact in M, but V and M have the same H<sub>&delta;</sub>.</p> <p>Finally, note that is it not really correct to say that the measurability of a cardinal &kappa; can be characterized by looking only at P(&kappa;), since my argument above shows two models M and V with the same P(&kappa;), but &kappa; is measurable in one of them and not the other. Rather, in order to tell if &kappa; is measurable, you have to look at <em>subsets</em> of P(&kappa;), that is, at P(P(&kappa;)), since this is where the measure lives.</p>