FS iterated forcing preserves chain condition

Let $\kappa$ be a regular cardinal and let be a finite support iterated forcing where for every , ${⊩}_{P_{\beta }}{\widehat{Q}}_{\beta }$ has the $\mathrm{\kappa }$ chain condition.

By induction:

$P_0$ is the empty set.

If $P_{\alpha }$ satisfies the $\kappa$ chain condition then so does $P_{\alpha +1}$, since $P_{\alpha +1}$ is equivalent to $P_{\alpha }*Q_{\alpha }$ and composition preserves the $\kappa$ chain condition for regular $\kappa$.

Suppose $\alpha$ is a limit ordinal and $P_{\beta }$ satisfies the $\kappa$ chain condition for all . Let be a subset of $P_{\alpha }$ of size $\kappa$. The domains of the elements of $p_i$ form $\kappa$ finite subsets of $\alpha$, so if $\mathrm{cf}(\alpha )>\kappa$ then these are bounded, and by the inductive hypothesis, two of them are compatible.

Otherwise, if , let be an increasing sequence of ordinals cofinal in $\alpha$. Then for any there is some such that $\mathrm{dom}(p_i)\subseteq {\alpha }_{n(i)}$. Since $\kappa$ is regular and this is a partition of $\kappa$ into fewer than $\kappa$ pieces, one piece must have size $\kappa$, that is, there is some $j$ such that $j=n(i)$ for $\kappa$ values of $i$, and so $\{p_i\mid n(i)=j\}$ is a set of conditions of size $\kappa$ contained in $P_{{\alpha }_j}$, and therefore contains compatible members by the induction hypothesis.

Finally, if $\mathrm{cf}(\alpha )=\kappa$, let be a strictly increasing, continuous sequence cofinal in $\alpha$. Then for every there is some such that $\mathrm{dom}(p_i)\subseteq {\alpha }_{n(i)}$. When $n(i)$ is a limit ordinal, since $C$ is continuous, there is also (since $\mathrm{dom}(p_i)$ is finite) some such that $\mathrm{dom}(p_i)\cap [{\alpha }_{f(i)},{\alpha }_i)=\mathrm{\varnothing }$. Consider the set $E$ of elements $i$ such that $i$ is a limit ordinal and for any , . This is a club, so by Fodor’s lemma there is some $j$ such that $\{i\mid f(i)=j\}$ is stationary.

For each $p_i$ such that $f(i)=j$, consider $p_i^{\prime }=p_i\upharpoonright j$. There are $\kappa$ of these, all members of $P_j$, so two of them must be compatible, and hence those two are also compatible in $P$.