# 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 ProofOfCompletePartialOrdersDoNotAddSmallSubsets 2013-03-22 12:53:35 2013-03-22 12:53:35 Henry (455) Henry (455) 4 Henry (455) Proof msc 03E40