Difference between revisions of "N-fold variants"

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($n$-fold variants which are simply the original large cardinal)
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{{DISPLAYTITLE: $n$-fold Variants of large cardinals}}
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{{DISPLAYTITLE: n-fold Variants of large cardinals}}
  
  
= $n$-fold variants and $M^{(n)}$ =
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= n-fold variants and $M^{(n)}$ =
  
The $n$-fold variants of large cardinals were given in a very large paper by Sato Kentaro. Most of the definitions involve giving large closure properties to the $M$ used in the original large cardinal in an [[elementary embedding]] $j:V\rightarrow M$. They are very large, but [[rank into rank | $I3$]] cardinals are stronger than most $n$-fold variants of large cardinals. The critical piece of information required in most $n$-fold variants was $M^{(n)}$, a modification of a transitive class $M$.
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The n-fold variants of large cardinals were given in a very large paper by Sato Kentaro. Most of the definitions involve giving large closure properties to the $M$ used in the original large cardinal in an [[elementary embedding]] $j:V\rightarrow M$. They are very large, but [[rank into rank | rank-into-rank]] cardinals are stronger than most n-fold variants of large cardinals. The critical piece of information required in most n-fold variants was $M^{(n)}$, a modification of a transitive class $M$.
  
 
Given a transitive class $M$ and an elementary embedding $j:V\rightarrow M$, $M^{(n)}$ and $j^{(n)}$ defined as follows:
 
Given a transitive class $M$ and an elementary embedding $j:V\rightarrow M$, $M^{(n)}$ and $j^{(n)}$ defined as follows:
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$$j^{(n)}=\bigcup_{\alpha\in\mathrm{Ord}}j^{n-1}(j\upharpoonright V_\alpha)$$
 
$$j^{(n)}=\bigcup_{\alpha\in\mathrm{Ord}}j^{n-1}(j\upharpoonright V_\alpha)$$
  
These can be thought of as a new version of $V$ using the $n$-th $j$-iteration on $V_\alpha$ rather than $V_\alpha$ itself; this of course makes $M^{(0)}=V$ for any transitive class $M$ and elementary embedding $j:V\rightarrow M$. Because of particularly this reason, the $0$-fold variants are particularly weak relative to $1$-fold variants. However, because most $n$-fold variants are more easily defined without $M^{(n)}$, although it is easier to prove things about them with $M^{(n)}$, the variants are defined in this page without them.
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These can be thought of as a new version of $V$ using the n-th $j$-iteration on $V_\alpha$ rather than $V_\alpha$ itself; this of course makes $M^{(0)}=V$ for any transitive class $M$ and elementary embedding $j:V\rightarrow M$. Because of particularly this reason, the 0-fold variants are particularly weak relative to 1-fold variants. However, because most n-fold variants are more easily defined without $M^{(n)}$, although it is easier to prove things about them with $M^{(n)}$, the variants are defined in this page without them.
  
== $n$-fold variants which are simply the original large cardinal ==
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== n-fold variants which are simply the original large cardinal ==
  
There were many $n$-fold variants which were simply renaming the original large cardinal. This was due to the fact that some $n$-fold variants, if only named $n$-variants instead, would be confusing to the reader (for example the $n$-fold extendibles rather than the [[extendible | $n$-extendibles]]). Here are a list of such cardinals:
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There were many n-fold variants which were simply renaming the original large cardinal. This was due to the fact that some n-fold variants, if only named n-variants instead, would be confusing to the reader (for example the n-fold extendibles rather than the [[extendible | n-extendibles]]). Here are a list of such cardinals:
  
*The '''$n$-fold superstrong''' cardinals are precisely the [[superstrong | $n$-superstrong]] cardinals
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*The '''n-fold superstrong''' cardinals are precisely the [[superstrong | n-superstrong]] cardinals
*The '''$n$-fold almost huge''' cardinals are precisely the [[huge | almost $n$-huge]] cardinals
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*The '''n-fold almost huge''' cardinals are precisely the [[huge | almost n-huge]] cardinals
*The '''$n$-fold huge''' cardinals are precisely the [[huge | $n$-huge]] cardinals
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*The '''n-fold huge''' cardinals are precisely the [[huge | n-huge]] cardinals
*The '''$n$-fold superhuge''' cardinals are precisely the [[huge | $n$-superhuge]] cardinals
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*The '''n-fold superhuge''' cardinals are precisely the [[huge | n-superhuge]] cardinals
  
== $n$-fold supercompact cardinals ==
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== n-fold supercompact cardinals ==
  
A cardinal $\kappa$ is '''$n$-fold $\lambda$-supercompact''' iff it is the critical point of some nontrivial elementary embedding $j:V\rightarrow M$ such that $\lambda<j(\kappa)$ and $M^{j^{n-1}(\lambda)}\subset M$ (i.e. $M$ is closed under all of its sequences of length $j^{n-1}(\lambda)$). This definition is very similar to that of the [[huge | $n$-huge]] cardinals.
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A cardinal $\kappa$ is '''n-fold $\lambda$-supercompact''' iff it is the critical point of some nontrivial elementary embedding $j:V\rightarrow M$ such that $\lambda<j(\kappa)$ and $M^{j^{n-1}(\lambda)}\subset M$ (i.e. $M$ is closed under all of its sequences of length $j^{n-1}(\lambda)$). This definition is very similar to that of the [[huge | n-huge]] cardinals.
  
A cardinal $\kappa$ is '''$n$-fold supercompact''' iff it is $n$-fold $\lambda$-supercompact for every $\lambda$. Consistency-wise, the $n$-fold supercompact cardinals are stronger than the [[superstrong | $n$-superstrong]] cardinals and weaker than the $n+1$-fold strong cardinals (see later in the page for such cardinals). In fact, if an $n$-fold supercompact cardinal exists, then it is consistent for there to be a proper class of $n$-superstrong cardinals.
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A cardinal $\kappa$ is '''n-fold supercompact''' iff it is n-fold $\lambda$-supercompact for every $\lambda$. Consistency-wise, the n-fold supercompact cardinals are stronger than the [[superstrong | n-superstrong]] cardinals and weaker than the (n+1)-fold strong cardinals (see later in the page for such cardinals). In fact, if an n-fold supercompact cardinal exists, then it is consistent for there to be a proper class of n-superstrong cardinals.
  
The $1$-fold supercompact cardinals are precisely the [[supercompact]] cardinals. The $0$-fold supercompact cardinals are precisely the [[measurable]] cardinals.
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The 1-fold supercompact cardinals are precisely the [[supercompact]] cardinals. The 0-fold supercompact cardinals are precisely the [[measurable]] cardinals.
  
== $n$-fold strong cardinals ==
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== n-fold strong cardinals ==
  
A cardinal $\kappa$ is '''$n$-fold $\lambda$-strong''' iff it is the critical point of some nontrivial elementary embedding $j:V\rightarrow M$ such that $\kappa+\lambda<j(\kappa)$ and:
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A cardinal $\kappa$ is '''n-fold $\lambda$-strong''' iff it is the critical point of some nontrivial elementary embedding $j:V\rightarrow M$ such that $\kappa+\lambda<j(\kappa)$ and:
  
 
$$V_{j^{n-1}(\kappa+\lambda)}\subset M$$
 
$$V_{j^{n-1}(\kappa+\lambda)}\subset M$$
  
A cardinal $\kappa$ is '''$n$-fold strong''' iff it is $n$-fold $\lambda$-strong for every $\lambda$. Consistency-wise, the $n+1$-fold strong cardinals are stronger than the $n$-fold supercompact cardinals, equivalent to the $n$-fold extendible cardinals, and weaker than the $n+1$-fold Woodin cardinals. More specifically, in the rank of an $n+1$-fold Woodin cardinal there is an $n+1$-fold strong cardinal.
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A cardinal $\kappa$ is '''n-fold strong''' iff it is n-fold $\lambda$-strong for every $\lambda$. Consistency-wise, the (n+1)-fold strong cardinals are stronger than the n-fold supercompact cardinals, equivalent to the n-fold extendible cardinals, and weaker than the (n+1)-fold Woodin cardinals. More specifically, in the rank of an (n+1)-fold Woodin cardinal there is an (n+1)-fold strong cardinal.
  
The $1$-fold Strong cardinals are precisely the [[strong]] cardinals. The $0$-fold strong cardinals are precisely the [[measurable]] cardinals.
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The 1-fold Strong cardinals are precisely the [[strong]] cardinals. The 0-fold strong cardinals are precisely the [[measurable]] cardinals.
  
== $n$-fold extendible cardinals ==
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== n-fold extendible cardinals ==
  
 
''(To be added)''
 
''(To be added)''
  
== $n$-fold Woodin cardinals ==
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== n-fold Woodin cardinals ==
  
A cardinal $\kappa$ is '''$n$-fold Woodin''' iff for every function $f:\kappa\rightarrow\kappa$ there is some ordinal $\alpha<\kappa$ such that $\{f(\beta):\beta<\alpha\}\subseteq\alpha$ and $V_{j^{n}(f)(j^{n-1}(\alpha))}\subset M$. Consistency-wise, the $n+1$-fold Woodin cardinals are stronger than the $n+1$-fold strong cardinals, and weaker than the $n+1$-fold Shelah cardinals. Specifically, in the rank of an $n+1$-fold Shelah cardinal there is an $n+1$-fold Woodin cardinal, and every $n+1$-fold Shelah cardinal is also an $n+1$-fold Woodin cardinal.
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A cardinal $\kappa$ is '''n-fold Woodin''' iff for every function $f:\kappa\rightarrow\kappa$ there is some ordinal $\alpha<\kappa$ such that $\{f(\beta):\beta<\alpha\}\subseteq\alpha$ and $V_{j^{n}(f)(j^{n-1}(\alpha))}\subset M$. Consistency-wise, the (n+1)-fold Woodin cardinals are stronger than the (n+1)-fold strong cardinals, and weaker than the (n+1)-fold Shelah cardinals. Specifically, in the rank of an (n+1)-fold Shelah cardinal there is an (n+1)-fold Woodin cardinal, and every (n+1)-fold Shelah cardinal is also an (n+1)-fold Woodin cardinal.
  
The $1$-fold Woodin cardinals are precisely the [[Woodin]] cardinals. The $0$-fold Woodin cardinals are precisely the [[measurable]] cardinals.
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The 1-fold Woodin cardinals are precisely the [[Woodin]] cardinals. The 0-fold Woodin cardinals are precisely the [[measurable]] cardinals.
  
 
''(More to be added)''
 
''(More to be added)''

Revision as of 11:21, 23 October 2017


n-fold variants and $M^{(n)}$

The n-fold variants of large cardinals were given in a very large paper by Sato Kentaro. Most of the definitions involve giving large closure properties to the $M$ used in the original large cardinal in an elementary embedding $j:V\rightarrow M$. They are very large, but rank-into-rank cardinals are stronger than most n-fold variants of large cardinals. The critical piece of information required in most n-fold variants was $M^{(n)}$, a modification of a transitive class $M$.

Given a transitive class $M$ and an elementary embedding $j:V\rightarrow M$, $M^{(n)}$ and $j^{(n)}$ defined as follows: $$M^{(n)}=\bigcup_{\alpha\in\mathrm{Ord}}j^n(V_\alpha)$$ $$j^{(n)}=\bigcup_{\alpha\in\mathrm{Ord}}j^{n-1}(j\upharpoonright V_\alpha)$$

These can be thought of as a new version of $V$ using the n-th $j$-iteration on $V_\alpha$ rather than $V_\alpha$ itself; this of course makes $M^{(0)}=V$ for any transitive class $M$ and elementary embedding $j:V\rightarrow M$. Because of particularly this reason, the 0-fold variants are particularly weak relative to 1-fold variants. However, because most n-fold variants are more easily defined without $M^{(n)}$, although it is easier to prove things about them with $M^{(n)}$, the variants are defined in this page without them.

n-fold variants which are simply the original large cardinal

There were many n-fold variants which were simply renaming the original large cardinal. This was due to the fact that some n-fold variants, if only named n-variants instead, would be confusing to the reader (for example the n-fold extendibles rather than the n-extendibles). Here are a list of such cardinals:

  • The n-fold superstrong cardinals are precisely the n-superstrong cardinals
  • The n-fold almost huge cardinals are precisely the almost n-huge cardinals
  • The n-fold huge cardinals are precisely the n-huge cardinals
  • The n-fold superhuge cardinals are precisely the n-superhuge cardinals

n-fold supercompact cardinals

A cardinal $\kappa$ is n-fold $\lambda$-supercompact iff it is the critical point of some nontrivial elementary embedding $j:V\rightarrow M$ such that $\lambda<j(\kappa)$ and $M^{j^{n-1}(\lambda)}\subset M$ (i.e. $M$ is closed under all of its sequences of length $j^{n-1}(\lambda)$). This definition is very similar to that of the n-huge cardinals.

A cardinal $\kappa$ is n-fold supercompact iff it is n-fold $\lambda$-supercompact for every $\lambda$. Consistency-wise, the n-fold supercompact cardinals are stronger than the n-superstrong cardinals and weaker than the (n+1)-fold strong cardinals (see later in the page for such cardinals). In fact, if an n-fold supercompact cardinal exists, then it is consistent for there to be a proper class of n-superstrong cardinals.

The 1-fold supercompact cardinals are precisely the supercompact cardinals. The 0-fold supercompact cardinals are precisely the measurable cardinals.

n-fold strong cardinals

A cardinal $\kappa$ is n-fold $\lambda$-strong iff it is the critical point of some nontrivial elementary embedding $j:V\rightarrow M$ such that $\kappa+\lambda<j(\kappa)$ and:

$$V_{j^{n-1}(\kappa+\lambda)}\subset M$$

A cardinal $\kappa$ is n-fold strong iff it is n-fold $\lambda$-strong for every $\lambda$. Consistency-wise, the (n+1)-fold strong cardinals are stronger than the n-fold supercompact cardinals, equivalent to the n-fold extendible cardinals, and weaker than the (n+1)-fold Woodin cardinals. More specifically, in the rank of an (n+1)-fold Woodin cardinal there is an (n+1)-fold strong cardinal.

The 1-fold Strong cardinals are precisely the strong cardinals. The 0-fold strong cardinals are precisely the measurable cardinals.

n-fold extendible cardinals

(To be added)

n-fold Woodin cardinals

A cardinal $\kappa$ is n-fold Woodin iff for every function $f:\kappa\rightarrow\kappa$ there is some ordinal $\alpha<\kappa$ such that $\{f(\beta):\beta<\alpha\}\subseteq\alpha$ and $V_{j^{n}(f)(j^{n-1}(\alpha))}\subset M$. Consistency-wise, the (n+1)-fold Woodin cardinals are stronger than the (n+1)-fold strong cardinals, and weaker than the (n+1)-fold Shelah cardinals. Specifically, in the rank of an (n+1)-fold Shelah cardinal there is an (n+1)-fold Woodin cardinal, and every (n+1)-fold Shelah cardinal is also an (n+1)-fold Woodin cardinal.

The 1-fold Woodin cardinals are precisely the Woodin cardinals. The 0-fold Woodin cardinals are precisely the measurable cardinals.

(More to be added)