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 $q_{\\alpha+1}\\Vdash\\alpha+1\otin\\hat{x}$ . 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 $q_{\\alpha+1}\\Vdash\\alpha\otin\\hat{x}$ . 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 $\eg\\phi$ .

$q^{*}$ 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 $\\hat{x}[G]\\in\\mathfrak{M}$ .