# Difference between revisions of "Supercompact"

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Every cardinal $\kappa$ that is $2^\kappa$-supercompact is a stationary limit of [[superstrong]] cardinals, but need not be superstrong itself. In fact $2^\kappa$-supercompact are stationary limits of quasicompacts, themselves stationary limits of 1-[[extendible|extendibles]]. | Every cardinal $\kappa$ that is $2^\kappa$-supercompact is a stationary limit of [[superstrong]] cardinals, but need not be superstrong itself. In fact $2^\kappa$-supercompact are stationary limits of quasicompacts, themselves stationary limits of 1-[[extendible|extendibles]]. | ||

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+ | If $\theta=\beth_\theta$ then every $\theta$-supercompact cardinal is [[strong|$\theta$-strong]]. This is because $H_{\theta^+}\in M$ so $H_{\theta^+}\subset M$ by transitivity and $V_\theta\subset H_\theta\in M$ so $V_\theta\subset M$, as desired. | ||

If a cardinal $\theta$-supercompact then it also $\theta$-[[strongly compact]]. Consequently, every supercompact cardinal is also strongly compact. It is consistent with $\text{ZFC}$ 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. The [[ultrapower axiom]] gives a positive answer to this, but itself isn't known to be consistent with the existence of a supercompact in the first place. If $\kappa$ is supercompact, then there is a forcing extension in which $\kappa$ remains supercompact and is also the least strongly compact cardinal. If there exists a measurable cardinal that is a limit of strongly compact cardinals, then the least such cardinal is strongly compact but not supercompact, in fact not even $2^\kappa$-supercompact. | If a cardinal $\theta$-supercompact then it also $\theta$-[[strongly compact]]. Consequently, every supercompact cardinal is also strongly compact. It is consistent with $\text{ZFC}$ 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. The [[ultrapower axiom]] gives a positive answer to this, but itself isn't known to be consistent with the existence of a supercompact in the first place. If $\kappa$ is supercompact, then there is a forcing extension in which $\kappa$ remains supercompact and is also the least strongly compact cardinal. If there exists a measurable cardinal that is a limit of strongly compact cardinals, then the least such cardinal is strongly compact but not supercompact, in fact not even $2^\kappa$-supercompact. |

## Latest revision as of 19:36, 7 October 2018

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.

## Contents

## 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 the axiom of choice, 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, $H_{\theta^+}$ must be contained in the transitive class $M$.

There is an alternative formulation that is expressible in $\text{ZFC}$ using certain ultrafilters with somewhat technical properties: for $\theta\geq\kappa$, $\kappa$ if $\theta$-supercompact if there is a normal fine measure on $\mathcal{P}_\kappa(\theta)$. $\kappa$ is supercompact if for every set $A$ with $|A|\geq\kappa$, there is a normal fine measure on $\mathcal{P}_\kappa(A)$.

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 $\mathcal{P}_\kappa(\theta)$. Conversely, the ultrapower by a normal fine measure $U$ on $\mathcal{P}_\kappa(\theta)$ 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: $\kappa$ is supercompact if and only if for every $\eta>\kappa$ there is $\alpha<\kappa$ such that there exists a nontrivial elementary embedding $j:V_\alpha\to V_\eta$ such that $j(\mathrm{crit}(j))=\kappa$.

## Properties

If $\kappa$ is supercompact, then there are $2^{2^\kappa}$ normal fine measures on $\kappa$, also for every $\lambda\geq\kappa$ there are $2^{2^{\lambda^{<\kappa}}}$ normal fine measures on $\mathcal{P}_\kappa(\lambda)$.

Every supercompact has Mitchell order $(2^\kappa)^+\geq\kappa^{++}$.

If $\kappa$ is $\lambda$-supercompact then it is also $\mu$-supercompact for every $\mu<\lambda$. If $\lambda\geq\kappa$ is regular, $\kappa$ is $\lambda$-supercompact, then every $\alpha<\kappa$ that is $\gamma$-supercompact for all $\gamma<\kappa$ (if any exists) is also $\lambda$-supercompact. In the same vein, for every cardinals $\kappa<\lambda$, if $\lambda$ is supercompact and $\kappa$ is $\gamma$-supercompact for all $\gamma<\lambda$, then $\kappa$ is also supercompact.

*Laver's theorem* asserts that if $\kappa$ is supercompact, there exists a function $f:\kappa\to V_\kappa$ such that for every $x$ and $\lambda\geq\kappa$ with $|tc(x)|\leq\lambda$ there exists a normal fine measure $U$ on $\mathcal{P}_\kappa(\lambda)$ such that $j_U(f)(\kappa)=x$, where $j_U$ is the elementary embedding generated from $U$. Here $tc(x)$ is the *transitive closure* of $x$ (i.e. the smallest transitive set containing $x$), and $f$ is called a *Laver function*.

## Supercompact cardinals and forcing

### The continuum hypothesis and supercompact cardinals

If $\kappa$ is $\lambda$-supercompact and $2^\alpha=\alpha^{+}$ for every $\alpha<\kappa$, then $2^\alpha=\alpha^{+}$ for every $\alpha\leq\lambda$. Consequently, if the generalized continuum hypothesis holds below a supercompact cardinal, then it holds everywhere.

The existence of a supercompact implies the consistency of the failure of the *singular cardinal hypothesis*, i.e. it is consistent that the generalized continuum hypothesis fails at a strong limit singular cardinal. It also implies the consistency of the failure of the $\text{GCH}$ at a measurable cardinal.

By combining results of Magidor, Shelah and Gitik, one can show that the existence of a supercompact also implies the existence of a generic extension in which $2^{\aleph_\alpha}<\aleph_{\omega_1}$ for all $\alpha<\omega_1$, but also $2^{\aleph_{\omega_1}}>\aleph_{\omega_1+\alpha+1}$ for any prescribed $\alpha<\omega_2$. Similarly, one can have a generic extension in which the $\text{GCH}$ holds below $\aleph_\omega$ but $2^{\aleph_\omega}>\aleph_{\omega+\alpha+1}$ for any prescribed $\alpha<\omega_1$.

Woodin and Cummings furthermore showed that if there exists a supercompact, then there is a generic extension in which $2^\kappa=\kappa^{++}$ for *every* cardinal $\kappa$, i.e. the $\text{GCH}$ fails *everywhere*(!).

The ultrapower axiom, if consistent with a supercompact, implies that the $\text{GCH}$ holds above the least supercompact.

### 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 extension. $\text{PFA}$'s strengthening, $\text{PFA}^{+}$, is also consistent relative to the existence of a supercompact cardinal.

### Martin's Maximum

## Relation to other large cardinals

Every cardinal $\kappa$ that is $2^\kappa$-supercompact is a stationary limit of superstrong cardinals, but need not be superstrong itself. In fact $2^\kappa$-supercompact are stationary limits of quasicompacts, themselves stationary limits of 1-extendibles.

If $\theta=\beth_\theta$ then every $\theta$-supercompact cardinal is $\theta$-strong. This is because $H_{\theta^+}\in M$ so $H_{\theta^+}\subset M$ by transitivity and $V_\theta\subset H_\theta\in M$ so $V_\theta\subset M$, as desired.

If a cardinal $\theta$-supercompact then it also $\theta$-strongly compact. Consequently, every supercompact cardinal is also strongly compact. It is consistent with $\text{ZFC}$ 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. The ultrapower axiom gives a positive answer to this, but itself isn't known to be consistent with the existence of a supercompact in the first place. If $\kappa$ is supercompact, then there is a forcing extension in which $\kappa$ remains supercompact and is also the least strongly compact cardinal. If there exists a measurable cardinal that is a limit of strongly compact cardinals, then the least such cardinal is strongly compact but not supercompact, in fact not even $2^\kappa$-supercompact.

Under the axiom of determinacy, $\omega_1$ is <$\Theta$-supercompact, where $\Theta$ is at least an aleph fixed point, and under $V=L(\mathbb{R})$ is even weakly hyper-Mahlo. The existence of a supercompact cardinals also implies the axiom $\text{AD}^{L(\mathbb{R})}$.

If $\kappa$ is $|V_{\kappa+\eta}|$-supercompact with $\eta<\kappa$ then it is preceeded by a stationary set of $\eta$-extendible cardinals. If $\kappa$ is $(\eta+2)$-extendible then it is $|V_{\kappa+\eta}|$-supercompact. The least supercompact is not 1-extendible, in fact any cardinal that is both supercompact and 1-extendible is preceeded by a stationary set of cardinals that are both supercompact and limits of supercompact cardinals.

The least supercompact is larger than the least huge cardinal (if such a cardinal exists). It is also larger than the least n-huge cardinal, for all n. If $\kappa$ is supercompact and there is an n-huge cardinal above $\kappa$, then there are $\kappa$-many n-huge cardinals below $\kappa$.

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