IteratedForcingAndComposition 2002-08-04 00:44:02 2003-01-12 21:47:48 Result iterated forcing % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here %\PMlinkescapeword{theory} There is a function satisfying forcings are equivalent if one is dense in the other $f:P_{\alpha}*Q_\alpha\rightarrow P_{\alpha+1}$. \section*{Proof} Let $f(\langle g,\hat{q}\rangle)=g\cup\{\langle \alpha,\hat{q}\rangle\}$. This is obviously a member of $P_{\alpha+1}$, since it is a partial function from $\alpha+1$ (and if the domain of $g$ is less than $\alpha$ then so is the domain of $f(\langle g,\hat{q}\rangle)$), if $i<\alpha$ then obviously $f(\langle g,\hat{q}\rangle)$ applied to $i$ satisfies the definition of iterated forcing (since $g$ does), and if $i=\alpha$ then the definition is satisfied since $\hat{q}$ is a name in $P_i$ for a member of $Q_i$. $f$ is order preserving, since if $\langle g_1,\hat{q}_1\rangle\leq \langle g_2,\hat{q}_2\rangle$, all the appropriate characteristics of a function carry over to the image, and $g_1\upharpoonright\alpha\Vdash_{P_i} \hat{q}_1\leq \hat{q}_2$ (by the definition of $\leq$ in $*$). If $\langle g_1,\hat{q}_1\rangle$ and $\langle g_2,\hat{q}_2\rangle$ are incomparable then either $g_1$ and $g_2$ are incomparable, in which case whatever prevents them from being compared applies to their images as well, or $\hat{q}_1$ and $\hat{q}_2$ aren't compared appropriately, in which case again this prevents the images from being compared. Finally, let $g$ be any element of $P_{\alpha+1}$. Then $g\upharpoonright\alpha\in P_\alpha$. If $\alpha\notin \operatorname{dom}(g)$ then this is just $g$, and $f(\langle g,\hat{q}\rangle)\leq g$ for any $\hat{q}$. If $\alpha\in\operatorname{dom}(g)$ then $f(\langle g\upharpoonright\alpha,g(\alpha)\rangle)=g$. Hence $f[P_\alpha*Q_\alpha]$ is dense in $P_{\alpha+1}$, and so these are equivalent.