# composition of forcing notions

Suppose $P$ is a forcing^{} notion in $\U0001d510$ and $\widehat{Q}$ is some $P$-name such that ${\u22a9}_{P}\widehat{Q}$ is a forcing notion.

Then take a set of $P$-names $Q$ such that given a $P$ name $\stackrel{~}{Q}$ of $Q$, ${\u22a9}_{P}\stackrel{~}{Q}=\widehat{Q}$ (that is, no matter which generic^{} subset $G$ of $P$ we force with, the names in $Q$ correspond precisely to the elements of $\widehat{Q}[G]$). We can define

$$P*Q=\{\u27e8p,\widehat{q}\u27e9\mid p\in P,\widehat{q}\in Q\}$$ |

We can define a partial order^{} on $P*Q$ such that $\u27e8{p}_{1},{\widehat{q}}_{1}\u27e9\le \u27e8{p}_{2},{\widehat{q}}_{2}\u27e9$ iff ${p}_{1}{\le}_{P}{p}_{2}$ and ${p}_{1}\u22a9{\widehat{q}}_{1}{\le}_{\widehat{Q}}{\widehat{q}}_{2}$. (A note on interpretation^{}: ${q}_{1}$ and ${q}_{2}$ are $P$ names; this requires only that ${\widehat{q}}_{1}\le {\widehat{q}}_{2}$ in generic subsets contain ${p}_{1}$, so in other generic subsets that fact could fail.)

Then $P*\widehat{Q}$ is itself a forcing notion, and it can be shown that forcing by $P*\widehat{Q}$ is equivalent^{} to forcing first by $P$ and then by $\widehat{Q}[G]$.

Title | composition of forcing notions |
---|---|

Canonical name | CompositionOfForcingNotions |

Date of creation | 2013-03-22 12:54:20 |

Last modified on | 2013-03-22 12:54:20 |

Owner | Henry (455) |

Last modified by | Henry (455) |

Numerical id | 4 |

Author | Henry (455) |

Entry type | Definition |

Classification | msc 03E35 |

Classification | msc 03E40 |

Related topic | Forcing |