proof of complete partial orders do not add small subsets

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

p\Vdash\hat{x}\text{ is a subset of }\kappa\text{ bounded by }\lambda<\kappa

Outline:

For any $q\leq p$ , we construct by induction a series of elements $q_{\alpha}$ stronger than $p$ . Each $q_{\alpha}$ will determine whether or not $\alpha\in\hat{x}$ . 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 $\hat{x}$ . Since the series is definable in $\mathfrak{M}$ , so is $\hat{x}$ , so we can conclude that above any element $q\leq p$ is an element which forces $\hat{x}\in\mathfrak{M}$ . Then $p$ also forces $\hat{x}\in\mathfrak{M}$ , completing the proof.

Details:

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

Note that these elements all exist since for any $p\in P$ and any (first-order) sentence $\phi$ there is some $q\leq p$ such that $q$ forces either $\phi$ or $\neg\phi$ .

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

Since this holds for any element stronger than $p$ , it follows that $p\Vdash\hat{x}\in\mathfrak{M}$ , and therefore $\hat{x}[G]\in\mathfrak{M}$ .

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