Difference between revisions of "Superstrong"

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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]].
 
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]].
  
If there is a superstrong cardinal, then in [[Constructible universe|$L(\mathbb{R})$]], the [[axiom of determinacy]] holds. <cite>Jech2003:SetTheory</cite> Thus:
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If there is a superstrong cardinal, then in [[Constructible universe|$L(\mathbb{R})$]], the [[axiom of determinacy]] holds. <cite>Jech2003:SetTheory</cite>
* Every set of reals is Lebesgue measurable.
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* Every set of reals has the Baire property.
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* Every set of reals is a perfect set.
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* There is no well-ordering of the reals.
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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 $\aleph_2$-saturated ideal on $\omega_1$. <cite>Jech2003:SetTheory</cite>
 
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 $\aleph_2$-saturated ideal on $\omega_1$. <cite>Jech2003:SetTheory</cite>

Latest revision as of 18:54, 3 December 2017

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 (ZF+)AD. After this, it was found that the existence of a proper class of Woodin cardinals was equiconsistent to 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 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.

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 = 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

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. $j(\kappa)$ is always an inaccessible cardinal in $V$, 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).

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. Superstrongness is consistency-wise stronger than Hyper-Woodinness.

If there is a superstrong cardinal, then in $L(\mathbb{R})$, the axiom of determinacy holds. [3]

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 $\aleph_2$-saturated ideal on $\omega_1$. [3]

Superstrongness is not Laver indestructible. [4]

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 ofelementary embeddings. , 2007. www   bibtex
  3. Jech, Thomas J. Set Theory. Third, Springer-Verlag, Berlin, 2003. (The third millennium edition, revised and expanded) www   bibtex
  4. Bagaria, Joan and Hamkins, Joel David and Tsaprounis, Konstantinos and Usuba, Toshimichi. Superstrong and other large cardinals are never {Laver} indestructible. www   arχiv   bibtex
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