# iterated forcing and composition

There is a function satisfying forcings are equivalent if one is dense in the other $f:P_{\alpha}*Q_{\alpha}\rightarrow P_{\alpha+1}$.

## 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.

Title iterated forcing and composition IteratedForcingAndComposition 2013-03-22 12:54:51 2013-03-22 12:54:51 Henry (455) Henry (455) 6 Henry (455) Result msc 03E35 msc 03E40