# Levy collapse

Given any cardinals $\kappa $ and $\lambda $ in $\U0001d510$, we can use the *Levy collapse* to give a new model $\U0001d510[G]$ where $\lambda =\kappa $. Let $P=\mathrm{Levy}(\kappa ,\lambda )$ be the set of partial functions^{} $f:\kappa \to \lambda $ with $$. These functions each give partial information about a function $F$ which collapses $\lambda $ onto $\kappa $.

Given any generic subset $G$ of $P$, $\U0001d510[G]$ has a set $G$, so let $F=\bigcup G$. Each element of $G$ is a partial function, and they are all compatible, so $F$ is a function. $\mathrm{dom}(G)=\kappa $ since for each $$ the set of $f\in P$ such that $\alpha \in \mathrm{dom}(f)$ is dense (given any function without $\alpha $, it is trivial to add $(\alpha ,0)$, giving a stronger function which includes $\alpha $). Also $\mathrm{range}(G)=\lambda $ since the set of $f\in P$ such that $$ is in the range of $f$ is again dense (the domain of each $f$ is bounded, so if $\beta $ is larger than any element of $\mathrm{dom}(f)$, $f\cup \{(\beta ,\alpha )\}$ is stronger than $f$ and includes $\lambda $ in its domain).

So $F$ is a surjective function from $\kappa $ to $\lambda $, and $\lambda $ is collapsed in $\U0001d510[G]$. In addition, $|\mathrm{Levy}(\kappa ,\lambda )|=\lambda $, so it satisfies the ${\lambda}^{+}$ chain condition, and therefore ${\lambda}^{+}$ is not collapsed, and becomes ${\kappa}^{+}$ (since for any ordinal^{} between $\lambda $ and ${\lambda}^{+}$ there is already a surjective function to it from $\lambda $).

We can generalize this by forcing^{} with $$ with $\kappa $ regular^{}, the set of partial functions $f:\lambda \times \kappa \to \lambda $ such that $f(0,\alpha )=0$, $$ and if $\alpha >0$ then $$. In essence, this is the product of $\mathrm{Levy}(\kappa ,\eta )$ for each $$.

In $\U0001d510[G]$, define $F=\bigcup G$ and ${F}_{\alpha}(\beta )=F(\alpha ,\beta )$. Each ${F}_{\alpha}$ is a function from $\kappa $ to $\alpha $, and by the same argument as above ${F}_{\alpha}$ is both total and surjective^{}. Moreover, it can be shown that $P$ satisfies the $\lambda $ chain condition, so $\lambda $ does not collapse and $\lambda ={\kappa}^{+}$.

Title | Levy collapse |
---|---|

Canonical name | LevyCollapse |

Date of creation | 2013-04-16 22:08:32 |

Last modified on | 2013-04-16 22:08:32 |

Owner | ratboy (4018) |

Last modified by | e1568582 (1000182) |

Numerical id | 9 |

Author | ratboy (1000182) |

Entry type | Example |

Classification | msc 03E45 |