Difference between revisions of "Supercompact"

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(Relation to strongly compact cardinals)
(Failure of GCH at a measurable)
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Gitik proved that the following statements are equiconsistent:
 
Gitik proved that the following statements are equiconsistent:
 
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* The generalized continuum hypothesis fails at a measurable cardinal $\kappa$, i.e. $2^\kappa > \kappa^+$
1. The generalized continuum hypothesis fails at a measurable cardinal $\kappa$, i.e. $2^\kappa > \kappa^+$
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* The singular cardinal hypothesis fails, i.e. there is a strong limit singular $\kappa$ such that $2^\kappa > \kappa^+$
 
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* There is a measurable cardinal of [[Mitchell rank | Mitchell order]] $\kappa^{++}$, i.e. $o(\kappa)=\kappa^{++}$
2. The singular cardinal hypothesis fails, i.e. there is a strong limit singular $\kappa$ such that $2^\kappa > \kappa^+$
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3. There is a measurable cardinal of [[Mitchell rank | Mitchell order]] $\kappa^{++}$, i.e. $o(\kappa)=\kappa^{++}$
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Thus violating GCH at a measurable (or violating the SCH at any strong limit cardinal) is strictly stronger consistency-wise than the existence of a measurable cardinal.
 
Thus violating GCH at a measurable (or violating the SCH at any strong limit cardinal) is strictly stronger consistency-wise than the existence of a measurable cardinal.

Revision as of 11:04, 11 September 2017


Supercompact cardinals are best motivated as a generalization of measurable cardinals, particularly the characterization of measurable cardinals in terms of elementary embeddings and strong closure properties. The notion of supercompactness and its consequences was initially developed by Solovay and Reinhardt and further elaborated on by Magidor and Gitik, among many others. Assuming the existence of a supercompact is a very strong assumption and the large cardinal strength of supercompact cardinals is seen in a wide (and bewildering) array of set-theoretic contexts, especially the development of strong forcing axioms and establishing regularity properties of sets of reals. The inner model program has yet to reach the level of a supercompact cardinal and this is considered a prominent open problem in the program itself. Curiously, by results of Woodin, should the inner program reach the level of a supercompact, there is a sense in which it will have reached all greater large cardinals, a startling contrast to previous advances in the program.

Formal Definition and Equivalent Characterizations

Generalizing the elementary embedding characterization of measurable cardinal, a cardinal $\kappa$ is $\theta$-supercompact if there is an elementary embedding $j:V\to M$ with $M$ a transitive class, such that $j$ has critical point $\kappa$ and $M^\theta\subset M$, i.e. $M$ is closed under arbitrary sequences of length $\theta$. Under AC, one may assume without loss of generality that $j(\kappa)\gt\theta$. $\kappa$ is then said to be supercompact if it is $\theta$-supercompact for all $\theta$. It is worth noting that, using this formulation, the collection of sets of Hereditary Cardinality less than $\theta^+$ must be contained in the transitive class $M$.

There is an alternative formulation that is expressible in ZFC using certain ultrafilters with somewhat technical properties: For $\kappa <\theta$, recall that $P_\kappa\theta$ denotes the collection of subsets of $\theta$ of size smaller than $\kappa$. Call an ultrafilter $U\subseteq P_\kappa\theta$ fine if for every $\alpha < \theta$ the set $\{a\in P_\kappa\theta: \alpha\in a\}$ is a member of $U$. Call an ultrafilter $U\subseteq P_\kappa\theta$ normal if $U$ is $\theta$-complete and for every "regressive" function $f:P_\kappa\theta\to\theta$ with $\{a\in P_\kappa\theta: f(a)\in a\}\in U$ there is some $\alpha <\theta$ where $\{a\in P_\kappa\theta: f(a)=\alpha\}\in U$. (An equivalent notion of normal using diagonal intersections can be found in Kanamori.)

Using these notions, one can define $\kappa$ as $\theta$-supercompact if there is a normal fine ultrafilter $U\subseteq P_\kappa\theta$. $\kappa$ is supercomact, then, if there is such an ultrafilter $U\subseteq P_\kappa\theta$ for every cardinal $\theta$ at least as big as $\kappa$.

One can see the equivalence of the two formulations by first considering the ultrafilter $U$ arising from the seed $j''\theta$, so that $X\in U\iff j''\theta\in j(X)$. It is easy to check that $U$ is a normal fine measure on $P_\kappa\theta$. Conversely, the ultrapower by a normal fine ultrafilter $U$ on $P_\kappa\theta$ gives rise to an embedding $j:V\to M$ (here $M$ is identified with the transitive collapse of the ultrapower by $U$). It is then straightforward to check that $\theta$ is the critical point of this embedding and that $M$ is sufficiently closed, thus witnessing $\theta$-supercompactness of $\kappa$.

A third characterization was given by Magidor in terms of elementary embeddings from initial segments of $V$ into other (larger) initial segments of $V$, but in this characterization, the supercompact cardinal $\kappa$ is the image of the critical point of this embedding, rather than the critical point itself. See...

Reflection Properties

Supercompact cardinals and forcing

Failure of GCH at a measurable

Gitik proved that the following statements are equiconsistent:

  • The generalized continuum hypothesis fails at a measurable cardinal $\kappa$, i.e. $2^\kappa > \kappa^+$
  • The singular cardinal hypothesis fails, i.e. there is a strong limit singular $\kappa$ such that $2^\kappa > \kappa^+$
  • There is a measurable cardinal of Mitchell order $\kappa^{++}$, i.e. $o(\kappa)=\kappa^{++}$

Thus violating GCH at a measurable (or violating the SCH at any strong limit cardinal) is strictly stronger consistency-wise than the existence of a measurable cardinal.

Laver preparation

Indestructibility, including the Laver diamond.

Proper forcing axiom

Baumgartner proved that if there is a supercompact cardinal, then the proper forcing axiom holds in a forcing extenion.


Martin's Maximum

Relation to strongly compact cardinals

If a cardinal $\theta$-supercompact then it also $\theta$-strongly compact. Consequently, every supercompact cardinal is also strongly compact. It is consistent with ZF(C) that every strongly compact cardinal is also supercompact, but it is not currently known whether the existence of a strongly compact cardinal is equiconsistent with the existence of a supercompact cardinal.


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