A question about supercompact cardinal numbers - MathOverflow most recent 30 from http://mathoverflow.net2013-05-23T22:38:15Zhttp://mathoverflow.net/feeds/question/21669http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/21669/a-question-about-supercompact-cardinal-numbersA question about supercompact cardinal numbersGarabed Gulbenkian2010-04-17T15:46:11Z2010-04-17T17:30:50Z
<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#21672Answer by Joel David Hamkins for A question about supercompact cardinal numbersJoel David Hamkins2010-04-17T16:13:35Z2010-04-17T17:30:50Z<p>No, this is not possible. </p>
<p>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.</p>
<p>(To find a measure μ concentrating on non-measurable cardinals, simply choose a normal measure μ for which the ultrapower j<sub>μ</sub> has the smallest possible value for j<sub>μ</sub>(κ). It follows that κ cannot be measurable in the corresponding M<sub>μ</sub>, for if it were, we could make a better μ.)</p>
<p>A similar argument shows that the supercompactness of κ cannot be characterized by any property in H<sub>δ</sub> 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 H<sub>δ</sub>.</p>
<p>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 <em>subsets</em> of P(κ), that is, at P(P(κ)), since this is where the measure lives.</p>