Difference between revisions of "Superstrong"
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*Superstrongness is consistency-wise stronger than [[Woodin|Hyper-Woodinness]]. <cite>Kanamori2009:HigherInfinite</cite> | *Superstrongness is consistency-wise stronger than [[Woodin|Hyper-Woodinness]]. <cite>Kanamori2009:HigherInfinite</cite> | ||
*Superstrongness is not Laver indestructible. <cite>BagariaHamkinsTsaprounisUsuba:SuperstrongAndOtherLargeCardinalsAreNeverLaverIndestructible</cite> | *Superstrongness is not Laver indestructible. <cite>BagariaHamkinsTsaprounisUsuba:SuperstrongAndOtherLargeCardinalsAreNeverLaverIndestructible</cite> | ||
− | *If there is a superstrong cardinal, then in [[Constructible universe|$L[\mathbb{R}]$]] <cite>Jech2003:SetTheory</cite> | + | *If there is a superstrong cardinal, then in [[Constructible universe|$L[\mathbb{R}]$]], the [[axiom of determinacy]] holds. Thus: <cite>Jech2003:SetTheory</cite> |
**Every set of reals is Lebesgue measurable. | **Every set of reals is Lebesgue measurable. | ||
**Every set of reals has the Baire property. | **Every set of reals has the Baire property. | ||
**Every set of reals is a perfect set. | **Every set of reals is a perfect set. | ||
− | **There is no | + | **There is no well-ordering of the reals. |
*Although it may seem like superstrong cardinals are quite large, letting $\kappa$ be superstrong, $\kappa$ can be [[forcing|forced]] to $\aleph_2$ with an $\omega$-distributive, $\kappa$-c.c. notion of forcing, and in fact, in this new universe there is a normal $\aleph_2$-saturated ideal on $\omega_1$. More intuitively, one can easily create force universes such that $\kappa$ is no longer superstrong and is in fact quite small. <cite>Jech2003:SetTheory</cite> | *Although it may seem like superstrong cardinals are quite large, letting $\kappa$ be superstrong, $\kappa$ can be [[forcing|forced]] to $\aleph_2$ with an $\omega$-distributive, $\kappa$-c.c. notion of forcing, and in fact, in this new universe there is a normal $\aleph_2$-saturated ideal on $\omega_1$. More intuitively, one can easily create force universes such that $\kappa$ is no longer superstrong and is in fact quite small. <cite>Jech2003:SetTheory</cite> | ||
{{References}} | {{References}} |
Revision as of 10:25, 8 November 2017
Superstrong cardinals were first utilized by Hugh Woodin in 1981 as an upper bound of consistency strength for AD. However, Shelah had then discovered that Shelah cardinals were a weaker bound that still sufficed to imply the consistency strength for AD. After this, it was found that the existence of a proper class of Woodin cardinals was equiconsistent to AD. This is a significant weakening of superstrongness.
Contents
Definitions
There are, like most critical point variations on measurable cardinals, multiple equivalent definitions of superstrongness. In particular, there is an elementary embedding definition and an extender definition.
Elementary Embedding Definition
A cardinal $\kappa$ is $n$-superstrong (or $n$-fold superstrong when referring to the $n$-fold variants) iff it is the critical point of some elementary embedding $j:V\rightarrow M$ such that $M$ is a transitive class and $V_{j^n(\kappa)}\subset M$ (in this case, $j^{n+1}(\kappa):=j(j^n(\kappa))$ and $j^0(\kappa):=\kappa$).
A cardinal is superstrong iff it is $1$-superstrong.
The definition quite clearly shows that $\kappa$ is $j^n(\kappa)$-strong. However, the least superstrong cardinal is never strong. [1]
Extender Definition
A cardinal $\kappa$ is $n$-superstrong (or $n$-fold superstrong when referring to the $n$-fold variants) iff there is a $(\kappa,\beta)$-extender $\mathcal{E}$ for a $\beta>\kappa$ with $V_{j^n_{\mathcal{E}}(\kappa)}\subseteq$ $Ult_{\mathcal{E}}(V)$ (where $j_{\mathcal{E}}$ is the canonical ultrapower embedding from $V$ into $Ult_{\mathcal{E}}(V)$). [1]
A cardinal is superstrong iff it is $1$-superstrong.
Relation to other large cardinal notions
- The consistency strength of $n$-superstrongness follows the double helix pattern. [2] Specifically:
- measurable = $0$-superstrong = almost $0$-huge = super almost $0$-huge = $0$-huge = super $0$-huge
- $n$-superstrong
- $n$-fold supercompact
- $(n+1)$-fold strong, $n$-fold extendible
- $(n+1)$-fold Woodin, $n$-fold Vopěnka
- $(n+1)$-fold Shelah
- almost $n$-huge
- super almost $n$-huge
- $n$-huge
- super $n$-huge
- $(n+1)$-superstrong
- The consistency strength of the existence of a superstrong cardinal is less than that of the existence of a $\kappa$ which is $2^\kappa$-supercompact. [1]
- If $\kappa$ is superstrong, then the necessary transitive class $M$ has $M\cap V_{j(\kappa)}$ satisfying $\kappa$'s strongness (although $\kappa$ may not be strong in $V$).
- There are many measurable cardinals above a superstrong cardinal. [1]
- Every $n$-huge cardinal is $n$-superstrong. [1]
- If $\kappa$ is strong but there is a superstrong cardinal at least the size of $\kappa$, then there are also $\kappa$ superstrong cardinals less than $\kappa$. [1]
- Every $1$-extendible cardinal is superstrong and has a normal measure containing all of the superstrongs less than said $1$-extendible. [1]
- Every $\kappa$ which is $2^\kappa$-supercompact is larger than a superstrong cardinal and has a normal measure containing all of the superstrongs less than it. [1]
- Every superstrong cardinal is Woodin and has a normal measure containing all of the Woodin cardinals less than it. [1]
- Superstrongness is consistency-wise stronger than Hyper-Woodinness. [1]
- Superstrongness is not Laver indestructible. [3]
- If there is a superstrong cardinal, then in $L[\mathbb{R}]$, the axiom of determinacy holds. Thus: [4]
- Every set of reals is Lebesgue measurable.
- Every set of reals has the Baire property.
- Every set of reals is a perfect set.
- There is no well-ordering of the reals.
- Although it may seem like superstrong cardinals are quite large, letting $\kappa$ be superstrong, $\kappa$ can be forced to $\aleph_2$ with an $\omega$-distributive, $\kappa$-c.c. notion of forcing, and in fact, in this new universe there is a normal $\aleph_2$-saturated ideal on $\omega_1$. More intuitively, one can easily create force universes such that $\kappa$ is no longer superstrong and is in fact quite small. [4]
References
- Kanamori, Akihiro. The higher infinite. Second, Springer-Verlag, Berlin, 2009. (Large cardinals in set theory from their beginnings, Paperback reprint of the 2003 edition) www bibtex
- Kentaro, Sato. Double helix in large large cardinals and iteration of elementary embeddings. Annals of Pure and Applied Logic 146(2-3):199-236, May, 2007. www DOI bibtex
- Bagaria, Joan and Hamkins, Joel David and Tsaprounis, Konstantinos and Usuba, Toshimichi. Superstrong and other large cardinals are never Laver indestructible. Archive for Mathematical Logic 55(1-2):19--35, 2013. www arχiv DOI bibtex
- Jech, Thomas J. Set Theory. Third, Springer-Verlag, Berlin, 2003. (The third millennium edition, revised and expanded) www bibtex