# Difference between revisions of "Tall"

Zetapology (Talk | contribs) m (→Strongly Tall Cardinals: Consistency strength is known) |
Zetapology (Talk | contribs) m (→Strongly Tall Cardinals: completeness $\kappa^+$ -> $\kappa$-complete) |
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=== Ultrapower Characterization === | === Ultrapower Characterization === | ||

− | $\kappa$ is strongly $\theta$-tall iff $\kappa$ is uncountable and there is some set $S$ and | + | $\kappa$ is strongly $\theta$-tall iff $\kappa$ is uncountable and there is some set $S$ and a $\kappa$-complete [[filter|ultrafilter]] $U$ on $S$ such that, letting $j:V\prec M\cong Ult_U(V)$, $j(\kappa)>\theta$. |

=== Ultrafilter Characterization === | === Ultrafilter Characterization === | ||

− | $\kappa$ is strongly $\theta$-tall iff there is some set $S$, | + | $\kappa$ is strongly $\theta$-tall iff there is some set $S$, a $\kappa$-complete [[filter|ultrafilter]] $U$ on $S$, and a class $H$ of functions $H_\alpha:S\rightarrow\text{Ord}$ for each ordinal $\alpha$ such that: |

#$\kappa$ is uncountable. | #$\kappa$ is uncountable. | ||

− | |||

#$H_0(x)=0$ for each $x\in S$. | #$H_0(x)=0$ for each $x\in S$. | ||

#For each $\alpha$ and each $f:S\rightarrow\text{Ord}$, $\{x\in S:f(x)<H_\alpha(x)\}\in U$ iff there is some $\beta<\alpha$ such that $\{x\in S:f(x)=\beta\}\in U$. That is, $f(x)<H_\alpha(x)$ almost everywhere iff there is some $\beta<\alpha$ such that $f(x)=\beta$ almost everywhere. | #For each $\alpha$ and each $f:S\rightarrow\text{Ord}$, $\{x\in S:f(x)<H_\alpha(x)\}\in U$ iff there is some $\beta<\alpha$ such that $\{x\in S:f(x)=\beta\}\in U$. That is, $f(x)<H_\alpha(x)$ almost everywhere iff there is some $\beta<\alpha$ such that $f(x)=\beta$ almost everywhere. |

## Revision as of 19:23, 7 October 2018

## Contents

## Tall Cardinals

A cardinal $\kappa$ is **$\theta$-tall** iff there is an elementary embedding $j:V\to M$ into a transitive class $M$ with critical point $\kappa$ such that $j(\kappa)>\theta$ and $M^\kappa\subset M$. $\kappa$ is **tall** iff it is $\theta$-tall for every $\theta$; i.e. $j(\kappa)$ can be made arbitrarily large. Every strong cardinal is tall and every strongly compact cardinal is tall, but measurable cardinals are not necessarily tall. It is relatively consistent, however, that the least measurable cardinal is tall. Nevertheless, the existence of a tall cardinal is equiconsistent with the existence of a strong cardinal. Any tall cardinal $\kappa$ can be made indestructible by a variety of forcing notions,
including forcing that pumps up the value of $2^\kappa$ as high as desired. See [1].

### Extender Characterization

If $\theta$ is a cardinal, $\kappa$ is $\theta$-tall iff there exists some $(\kappa,\theta^+)$-extender $E$ such that, if $M\cong Ult_E$ is the ultrapower of $V$ by $E$, $M^\kappa\subset M$. Similarly, $\kappa$ is tall iff for any $\lambda$ there exists some $(\kappa,\lambda)$-extender such that $M^\kappa\subset M$ where $M$ is as above.

## Strongly Tall Cardinals

A cardinal $\kappa$ is **strongly $\theta$-tall** iff there is some measure $U$ on a set $S$ witnessing $\kappa$'s $\theta$-tallness in the ultrapower of $V$ by $U$. More precisely, the ultrapower embedding $j:V\prec M$ has critical point $\kappa$, $M^\kappa\subset M$, and $j(\kappa)>\theta$. $\kappa$ is **strongly tall** iff it is strongly $\theta$-tall for every $\theta$.

The existence of a strongly tall cardinal is equiconsistent to the existence of a strong cardinal with a proper class of measurables above it (below the consistency strength of a Woodin cardinal, above the consistency strength of a strong cardinal and therefore above a tall cardinal). Specifically, if $κ$ is strong and has a proper class of measurables above it and GCH holds, then in a forcing extension of $V$, $κ$ is strongly tall. On the other hand, if $κ$ is strongly tall and there is no inner model with two strong cardinals, then $κ$ is strong in $K$ and has a proper class of measurables above it in $K$ ($K$ being the core model).

### Ultrapower Characterization

$\kappa$ is strongly $\theta$-tall iff $\kappa$ is uncountable and there is some set $S$ and a $\kappa$-complete ultrafilter $U$ on $S$ such that, letting $j:V\prec M\cong Ult_U(V)$, $j(\kappa)>\theta$.

### Ultrafilter Characterization

$\kappa$ is strongly $\theta$-tall iff there is some set $S$, a $\kappa$-complete ultrafilter $U$ on $S$, and a class $H$ of functions $H_\alpha:S\rightarrow\text{Ord}$ for each ordinal $\alpha$ such that:

- $\kappa$ is uncountable.
- $H_0(x)=0$ for each $x\in S$.
- For each $\alpha$ and each $f:S\rightarrow\text{Ord}$, $\{x\in S:f(x)<H_\alpha(x)\}\in U$ iff there is some $\beta<\alpha$ such that $\{x\in S:f(x)=\beta\}\in U$. That is, $f(x)<H_\alpha(x)$ almost everywhere iff there is some $\beta<\alpha$ such that $f(x)=\beta$ almost everywhere.
- $\{x\in S:H_\theta(x)<\kappa\}\in U$. That is, $H_\theta(x)<\kappa$ almost everywhere.

## References

Main library