# proof of complete partial orders do not add small subsets

Take any $x\in \U0001d510[G]$, $x\subseteq \kappa $. Let $\widehat{x}$ be a name for $x$. There is some $p\in G$ such that

$$ |

*Outline:*

For any $q\le p$, we construct by induction^{} a series of elements ${q}_{\alpha}$ stronger than $p$. Each ${q}_{\alpha}$ will determine whether or not $\alpha \in \widehat{x}$. Since we know the subset is bounded below $\kappa $, we can use the fact that $P$ is $\kappa $ complete^{} to find a single element stronger than $q$ which fixes the exact value of $\widehat{x}$. Since the series is definable in $\U0001d510$, so is $\widehat{x}$, so we can conclude that above any element $q\le p$ is an element which forces $\widehat{x}\in \U0001d510$. Then $p$ also forces $\widehat{x}\in \U0001d510$, completing the proof.

*Details:*

Since forcing^{} can be described within $\U0001d510$, $S=\{q\in P\mid q\u22a9\widehat{x}\in V\}$ is a set in $\U0001d510$. Then, given any $q\le p$, we can define ${q}_{0}=q$ and for any ${q}_{\alpha}$ ($$), ${q}_{\alpha +1}$ is an element of $P$ stronger than ${q}_{\alpha}$ such that either ${q}_{\alpha +1}\u22a9\alpha +1\in \widehat{x}$ or ${q}_{\alpha +1}\u22a9\alpha +1\notin \widehat{x}$. For limit $\alpha $, let ${q}_{\alpha}^{\prime}$ be any upper bound of ${q}_{\beta}$ for $$ (this exists since $P$ is $\kappa $-complete and $$), and let ${q}_{\alpha}$ be stronger than ${q}_{\alpha}^{\prime}$ and satisfy either ${q}_{\alpha +1}\u22a9\alpha \in \widehat{x}$ or ${q}_{\alpha +1}\u22a9\alpha \notin \widehat{x}$. Finally let ${q}^{*}$ be the upper bound of ${q}_{\alpha}$ for $$. ${q}^{*}\in P$ since $P$ is $\kappa $-complete.

Note that these elements all exist since for any $p\in P$ and any (first-order) sentence^{} $\varphi $ there is some $q\le p$ such that $q$ forces either $\varphi $ or $\mathrm{\neg}\varphi $.

${q}^{*}$ not only forces that $\widehat{x}$ is a bounded subset of $\kappa $, but for every ordinal^{} it forces whether or not that ordinal is contained in $\widehat{x}$. But the set $$ is defineable in $\U0001d510$, and is of course equal to $\widehat{x}[{G}^{*}]$ in any generic ${G}^{*}$ containing ${q}^{*}$. So ${q}^{*}\u22a9\widehat{x}\in \U0001d510$.

Since this holds for any element stronger than $p$, it follows that $p\u22a9\widehat{x}\in \U0001d510$, and therefore $\widehat{x}[G]\in \U0001d510$.

Title | proof of complete partial orders do not add small subsets |
---|---|

Canonical name | ProofOfCompletePartialOrdersDoNotAddSmallSubsets |

Date of creation | 2013-03-22 12:53:35 |

Last modified on | 2013-03-22 12:53:35 |

Owner | Henry (455) |

Last modified by | Henry (455) |

Numerical id | 4 |

Author | Henry (455) |

Entry type | Proof |

Classification | msc 03E40 |