Difference between revisions of "Superstrong"

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[[Category:Large cardinal axioms]]
 
[[Category:Large cardinal axioms]]
 
[[Category:Critical points]]
 
[[Category:Critical points]]
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.
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Superstrong cardinals were first utilized by Hugh Woodin in 1981 as an upper bound of consistency strength for the [[axiom of determinacy]]. However, Shelah had then discovered that [[Shelah]] cardinals were a weaker bound that still sufficed to imply the consistency strength of $\text{(ZF+)AD}$. After this, it was found that the existence of infinitely many [[Woodin]] cardinals was equiconsistent to $\text{AD}$. Woodin-ness is a significant weakening of superstrongness.
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''Most results in this article can be found in <cite>Kanamori2009:HigherInfinite</cite> unless indicated otherwise.''
  
 
== Definitions ==
 
== 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.
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There are, like all 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 ===
 
=== Elementary Embedding Definition ===
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A cardinal is '''superstrong''' iff it is $1$-superstrong.
 
A cardinal is '''superstrong''' iff it is $1$-superstrong.
  
The definition quite clearly shows that $\kappa$ is [[strong|$j^n(\kappa)$-strong]]. However, the least superstrong cardinal is never [[strong]].
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The definition quite clearly shows that $\kappa$ is $j^n(\kappa)$-[[strong]]. However, the least superstrong cardinal is never strong.
<cite>Kanamori2009:HigherInfinite</cite>
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$j^k(\kappa)$ for $1 \le k \le n$ are called targets.<cite>Kentaro2007:DoubleHelix</cite>
  
 
=== Extender Definition ===
 
=== Extender Definition ===
  
A cardinal $\kappa$ is '''$n$-superstrong''' (or $n$-fold superstrong when referring to the [[n-fold variants|$n$-fold variants]]) iff there is a [[extender|$(\kappa,\beta)$-extender]] $\mathcal{E}$ for a $\beta>\kappa$ with $V_{j^n_{\mathcal{E}}(\kappa)}\subseteq$ [[ultrapower|$Ult_{\mathcal{E}}(V)$]] (where $j_{\mathcal{E}}$ is the canonical ultrapower embedding from $V$ into $Ult_{\mathcal{E}}(V)$). <cite>Kanamori2009:HigherInfinite</cite>
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A cardinal $\kappa$ is '''$n$-superstrong''' (or $n$-fold superstrong) iff there is a [[extender|$(\kappa,\beta)$-extender]] $\mathcal{E}$ for a $\beta>\kappa$ with $V_{j^n_{\mathcal{E}}(\kappa)}\subseteq$ [[ultrapower|$Ult_{\mathcal{E}}(V)$]] (where $j_{\mathcal{E}}$ is the canonical ultrapower embedding from $V$ into $Ult_{\mathcal{E}}(V)$).
  
 
A cardinal is '''superstrong''' iff it is $1$-superstrong.
 
A cardinal is '''superstrong''' iff it is $1$-superstrong.
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== Relation to other large cardinal notions ==
 
== Relation to other large cardinal notions ==
  
*The consistency strength of $n$-superstrongness follows the [[n-fold variants|double helix pattern]]. <cite>Kentaro2007:DoubleHelix</cite> Specifically:
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The consistency strength of $n$-superstrongness follows the [[n-fold variants|double helix pattern]] <cite>Kentaro2007:DoubleHelix</cite>. Specifically:
**[[measurable]] = $0$-superstrong = [[huge|almost $0$-huge]] = super almost $0$-huge = $0$-huge = super $0$-huge  
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*[[measurable]] = $0$-superstrong = [[huge|almost $0$-huge]] = $0$-huge
**$n$-superstrong
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* $n$-superstrong
**[[n-fold variants|$n$-fold supercompact]]
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* $n$-fold supercompact
**[[n-fold variants|$(n+1)$-fold strong]], [[n-fold variants|$n$-fold extendible]]
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* $&#40;n+1)$-fold strong, $n$-fold extendible
**[[n-fold variants|$(n+1)$-fold Woodin]], [[n-fold variants|$n$-fold Vopěnka]]
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* $&#40;n+1)$-fold Woodin, $n$-fold Vopěnka
**[[n-fold variants|$(n+1)$-fold Shelah]]
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* $&#40;n+1)$-fold Shelah
**almost $n$-huge
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* almost $n$-huge
**super almost $n$-huge
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* super almost $n$-huge
**$n$-huge
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* $n$-huge
**super $n$-huge
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* super $n$-huge
**$(n+1)$-superstrong
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* $&#40;n+1)$-superstrong
*The consistency strength of the existence of a superstrong cardinal is less than that of the existence of a $\kappa$ which is [[supercompact|$2^\kappa$-supercompact]]. <cite>Kanamori2009:HigherInfinite</cite>
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*If $\kappa$ is superstrong, then the necessary transitive class $M$ has $M\cap V_{j(\kappa)}$ satisfying $\kappa$'s [[strong|strongness]] (although $\kappa$ may not be strong in $V$).
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Let $M$ be a transitive class $M$ such that there exists an elementary embedding $j:V\to M$ with $V_{j&#40;\kappa)}\subseteq M$, and let $\kappa$ be its superstrong critical point. While $j&#40;\kappa)$ need not be an inaccessible cardinal in $V$, it is always [[worldly]] and the rank model $V_{j&#40;\kappa)}$ satisfies $\text{ZFC+}$"$\kappa$ is strong" &#40;although $\kappa$ may not be strong in $V$).
*There are many [[measurable]] cardinals above a superstrong cardinal. <cite>Kanamori2009:HigherInfinite</cite>
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*Every $n$-huge cardinal is $n$-superstrong. <cite>Kanamori2009:HigherInfinite</cite>  
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Superstrong cardinals have strong upward reflection properties, in particular there are many [[measurable]] cardinals ''above'' a superstrong cardinal. Every $n$-huge cardinal is $n$-superstrong, and so $n$-huge cardinals also have strong reflection properties. Remark however that if $\kappa$ is [[strong]] or [[supercompact]], then it is consistent that there is no inaccessible cardinals larger than $\kappa$: this is because if $\lambda>\kappa$ is inaccessible, then $V_\lambda$ satisfies $\kappa$'s strongness/supercompactness. Thus it is clear that supercompact cardinals need not be superstrong, even though they have higher consistency strength. In fact, because of the downward reflection properties of strong/supercompact cardinals, if there is a superstrong above a strong/supercompact $\kappa$, then there are $\kappa$-many superstrong cardinals below $\kappa$; same with hugeness instead of superstrongness. In particular, the least superstrong is strictly smaller than the least strong &#40;and thus smaller than the least supercompact).
*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 $\kappa$. <cite>Kanamori2009:HigherInfinite</cite>  
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*Every [[extendible|$1$-extendible]] cardinal is superstrong and has a [[filter|normal measure]] containing all of the superstrongs less than said $1$-extendible. <cite>Kanamori2009:HigherInfinite</cite>
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* $n$-fold supercompactness implies $n$-fold “super”-superstrongness &#40;$n$-superstrongness with arbitrary large first target).<cite>Kentaro2007:DoubleHelix</cite>
*Every $\kappa$ which is [[supercompact|$2^\kappa$-supercompact]] is larger than a superstrong cardinal and has a normal measure containing all of the superstrongs less than it. <cite>Kanamori2009:HigherInfinite</cite>
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* Every [[extendible|$1$-extendible]] cardinal is superstrong and has a [[filter|normal measure]] containing all of the superstrongs less than said $1$-extendible. This means that the set of all superstronges less than it is [[stationary]]. Similarly, every cardinal $\kappa$ which is [[supercompact|$2^\kappa$-supercompact]] is larger than the least superstrong cardinal and has a normal measure containing all of the superstrongs less than it.
*Every superstrong cardinal is [[Woodin]] and has a normal measure containing all of the Woodin cardinals less than it. <cite>Kanamori2009:HigherInfinite</cite>  
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* Every superstrong cardinal is [[Woodin]] and has a normal measure containing all of the Woodin cardinals less than it. Thus the set of all Woodin cardinals below it is stationary, and so is the set of all measurables smaller than it.
*Superstrongness is consistency-wise stronger than [[Woodin|Hyper-Woodinness]]. <cite>Kanamori2009:HigherInfinite</cite>
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* If $κ$ is superstrong, then it is [[Shelah]] and there are $κ$ Shelah cardinals below it.<cite>Golshani2017:EastonLikeInPresenceShelah</cite>
*Superstrongness is not Laver indestructible. <cite>BagariaHamkinsTsaprounisUsuba:SuperstrongAndOtherLargeCardinalsAreNeverLaverIndestructible</cite>  
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* Superstrongness is consistency-wise stronger than [[Woodin|hyper-Woodinness]].<cite>Schimmerling2002:WoodinShelahAndCoreModel</cite>
*If there is a superstrong cardinal, then in [[Constructible universe|$L[\mathbb{R}]$]] <cite>Jech2003:SetTheory</cite>:
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* If there is a superstrong cardinal, then in [[Constructible universe|$L&#40;\mathbb{R})$]], the [[axiom of determinacy]] holds. <cite>Jech2003:SetTheory</cite>
**Every set of reals is Lebesgue measurable.
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* Letting $\kappa$ be superstrong, $\kappa$ can be [[forcing|forced]] to $\aleph_2$ with an $\omega$-distributive, $\kappa$-c.c. notion of forcing, and in this forcing extension there is a normal $\omega_2$-saturated ideal on $\omega_1$. <cite>Jech2003:SetTheory</cite>
**Every set of reals has the Baire property.
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* Superstrongness is not Laver [[indestructible]]. <cite>BagariaHamkinsTsaprounisUsuba:SuperstrongAndOtherLargeCardinalsAreNeverLaverIndestructible</cite>
**Every set of reals is a perfect set.
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* If [[I4|$\mathrm{I}_4^1&#40;\kappa)$]], then $\kappa$ is superstrong and $\{\alpha\lt\kappa|\alpha\text{ is superstrong}\}$ has measure 1.<cite>Corazza2003:GapBetweenI3andWA</cite>
**There is no [[projective]] well-ordering of the reals.
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*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>
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A cardinal $κ$ is '''[[correct|$C^{&#40;n)}$-superstrong]]''' iff there exists an elementary embedding $j : V → M$ for transitive $M$, with $crit&#40;j) = κ$, $V_{j&#40;κ)} ⊆ M$ and $j&#40;κ) ∈ C^{&#40;n)}$.<cite>Bagaria2012:CnCardinals</cite>
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* Every $C^{&#40;n)}$-superstrong cardinal belongs to $C^{&#40;n)}$.
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* Every superstrong cardinal is $C^{&#40;1)}$-superstrong.
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* For every $n ≥ 1$, if $κ$ is $C^{&#40;n+1)}$-superstrong, then there is a $κ$-complete normal [[ultrafilter]] $U$ over $κ$ such that $\{α < κ : α$ is $C^{&#40;n)}$-superstrong$\} ∈ U$. Hence, the first $C^{&#40;n)}$-superstrong cardinal, if it exists, is not $C^{&#40;n+1)}$-superstrong.
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* If $κ$ is $2^κ$-[[supercompact]] and belongs to $C^{&#40;n)}$, then there is a $κ$-complete normal ultrafilter $U$ over $κ$ such that the set of $C^{&#40;n)}$-superstrong cardinals smaller than $κ$ belongs to $U$.
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* If $κ$ is $κ+1$-$C^{&#40;n)}$-[[extendible]] and belongs to $C^{&#40;n)}$, then $κ$ is $C^{&#40;n)}$-superstrong and there is a $κ$-complete normal ultrafilter $U$ over $κ$ such that the set of $C^{&#40;n)}$-superstrong cardinals smaller than $κ$ belongs to $U$.
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* Every $C^{&#40;n)}$-[[huge|almost-huge]] cardinal is $C^{&#40;n)}$-superstrong.
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* Assuming [[rank into rank|$\mathrm{I3}&#40;κ, δ)$]], if $δ$ is a limit cardinal &#40;instead of a successor of a limit cardinal – Kunen’s Theorem excludes other cases), it is equal to $sup\{j^m&#40;κ) : m ∈ ω\}$ where $j$ is the elementary embedding. Then $κ$ and $j^m&#40;κ)$ are $C^{&#40;n)}$-superstrong &#40;inter alia) in $V_δ$, for all $n$ and $m$.
  
 
{{References}}
 
{{References}}

Latest revision as of 06:32, 29 May 2022

Superstrong cardinals were first utilized by Hugh Woodin in 1981 as an upper bound of consistency strength for the axiom of determinacy. However, Shelah had then discovered that Shelah cardinals were a weaker bound that still sufficed to imply the consistency strength of $\text{(ZF+)AD}$. After this, it was found that the existence of infinitely many Woodin cardinals was equiconsistent to $\text{AD}$. Woodin-ness is a significant weakening of superstrongness.

Most results in this article can be found in [1] unless indicated otherwise.

Definitions

There are, like all 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.

$j^k(\kappa)$ for $1 \le k \le n$ are called targets.[2]

Extender Definition

A cardinal $\kappa$ is $n$-superstrong (or $n$-fold superstrong) 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 $Ult_{\mathcal{E}}(V)$).

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 = $0$-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

Let $M$ be a transitive class $M$ such that there exists an elementary embedding $j:V\to M$ with $V_{j(\kappa)}\subseteq M$, and let $\kappa$ be its superstrong critical point. While $j(\kappa)$ need not be an inaccessible cardinal in $V$, it is always worldly and the rank model $V_{j(\kappa)}$ satisfies $\text{ZFC+}$"$\kappa$ is strong" (although $\kappa$ may not be strong in $V$).

Superstrong cardinals have strong upward reflection properties, in particular there are many measurable cardinals above a superstrong cardinal. Every $n$-huge cardinal is $n$-superstrong, and so $n$-huge cardinals also have strong reflection properties. Remark however that if $\kappa$ is strong or supercompact, then it is consistent that there is no inaccessible cardinals larger than $\kappa$: this is because if $\lambda>\kappa$ is inaccessible, then $V_\lambda$ satisfies $\kappa$'s strongness/supercompactness. Thus it is clear that supercompact cardinals need not be superstrong, even though they have higher consistency strength. In fact, because of the downward reflection properties of strong/supercompact cardinals, if there is a superstrong above a strong/supercompact $\kappa$, then there are $\kappa$-many superstrong cardinals below $\kappa$; same with hugeness instead of superstrongness. In particular, the least superstrong is strictly smaller than the least strong (and thus smaller than the least supercompact).

  • $n$-fold supercompactness implies $n$-fold “super”-superstrongness ($n$-superstrongness with arbitrary large first target).[2]
  • Every $1$-extendible cardinal is superstrong and has a normal measure containing all of the superstrongs less than said $1$-extendible. This means that the set of all superstronges less than it is stationary. Similarly, every cardinal $\kappa$ which is $2^\kappa$-supercompact is larger than the least superstrong cardinal and has a normal measure containing all of the superstrongs less than it.
  • Every superstrong cardinal is Woodin and has a normal measure containing all of the Woodin cardinals less than it. Thus the set of all Woodin cardinals below it is stationary, and so is the set of all measurables smaller than it.
  • If $κ$ is superstrong, then it is Shelah and there are $κ$ Shelah cardinals below it.[3]
  • Superstrongness is consistency-wise stronger than hyper-Woodinness.[4]
  • If there is a superstrong cardinal, then in $L(\mathbb{R})$, the axiom of determinacy holds. [5]
  • Letting $\kappa$ be superstrong, $\kappa$ can be forced to $\aleph_2$ with an $\omega$-distributive, $\kappa$-c.c. notion of forcing, and in this forcing extension there is a normal $\omega_2$-saturated ideal on $\omega_1$. [5]
  • Superstrongness is not Laver indestructible. [6]
  • If $\mathrm{I}_4^1(\kappa)$, then $\kappa$ is superstrong and $\{\alpha\lt\kappa|\alpha\text{ is superstrong}\}$ has measure 1.[7]

A cardinal $κ$ is $C^{(n)}$-superstrong iff there exists an elementary embedding $j : V → M$ for transitive $M$, with $crit(j) = κ$, $V_{j(κ)} ⊆ M$ and $j(κ) ∈ C^{(n)}$.[8]

  • Every $C^{(n)}$-superstrong cardinal belongs to $C^{(n)}$.
  • Every superstrong cardinal is $C^{(1)}$-superstrong.
  • For every $n ≥ 1$, if $κ$ is $C^{(n+1)}$-superstrong, then there is a $κ$-complete normal ultrafilter $U$ over $κ$ such that $\{α < κ : α$ is $C^{(n)}$-superstrong$\} ∈ U$. Hence, the first $C^{(n)}$-superstrong cardinal, if it exists, is not $C^{(n+1)}$-superstrong.
  • If $κ$ is $2^κ$-supercompact and belongs to $C^{(n)}$, then there is a $κ$-complete normal ultrafilter $U$ over $κ$ such that the set of $C^{(n)}$-superstrong cardinals smaller than $κ$ belongs to $U$.
  • If $κ$ is $κ+1$-$C^{(n)}$-extendible and belongs to $C^{(n)}$, then $κ$ is $C^{(n)}$-superstrong and there is a $κ$-complete normal ultrafilter $U$ over $κ$ such that the set of $C^{(n)}$-superstrong cardinals smaller than $κ$ belongs to $U$.
  • Every $C^{(n)}$-almost-huge cardinal is $C^{(n)}$-superstrong.
  • Assuming $\mathrm{I3}(κ, δ)$, if $δ$ is a limit cardinal (instead of a successor of a limit cardinal – Kunen’s Theorem excludes other cases), it is equal to $sup\{j^m(κ) : m ∈ ω\}$ where $j$ is the elementary embedding. Then $κ$ and $j^m(κ)$ are $C^{(n)}$-superstrong (inter alia) in $V_δ$, for all $n$ and $m$.

References

  1. 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
  2. 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
  3. Golshani, Mohammad. An Easton like theorem in the presence of Shelah cardinals. M Arch Math Logic 56(3-4):273-287, May, 2017. www   DOI   bibtex
  4. Schimmerling, Ernest. Woodin cardinals, Shelah cardinals, and the Mitchell-Steel core model. Proc Amer Math Soc 130(11):3385-3391, 2002. DOI   bibtex
  5. Jech, Thomas J. Set Theory. Third, Springer-Verlag, Berlin, 2003. (The third millennium edition, revised and expanded) www   bibtex
  6. 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
  7. Corazza, Paul. The gap between $\mathrm{I}_3$ and the wholeness axiom. Fund Math 179(1):43--60, 2003. www   DOI   MR   bibtex
  8. Bagaria, Joan. $C^{(n)}$-cardinals. Archive for Mathematical Logic 51(3--4):213--240, 2012. www   arχiv   DOI   bibtex
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