Closure, distributivity, and absence of new short sequences

Suppose $P$ is $\kappa$-closed. Given $p$ and dense open $D_{\xi }$ for $\xi <\gamma <\kappa$, recursively choose $p_{\xi +1}\le p_{\xi }$ in $D_{\xi }$ and at each limit take a lower bound. Regularity keeps every stage below $\kappa$; a final lower bound belongs to every $D_{\xi }$. Thus $P$ is $\kappa$-distributive. AC is used for the recursive choices.

Suppose $P$ is $\kappa$-distributive and $p⊩\dot{f}:\check{\gamma }\to \check{V}$ for $\gamma <\kappa$. For each $\xi <\gamma$, the set of conditions deciding $\dot{f}(\xi )$ is dense open. A common extension $q$ decides every coordinate, say as $x_{\xi }\in V$. Replacement forms $f=⟨x_{\xi }:\xi <\gamma ⟩$ in the ground model and $q⊩\dot{f}=\check{f}$. Conversely, assume that separative $P$ adds no such sequence. Given maximal antichains $A_{\xi }$ for $\xi <\gamma$, let $\dot{f}(\xi )$ be the unique member of $A_{\xi }$ met by the generic. This is a name for a $\gamma$-sequence of ground-model conditions, hence conditions deciding its whole ground-model value are dense. If $q$ decides that value as $f_q$, then $q\le f_q(\xi )$ for every $\xi$: otherwise separativity gives $r\le q$ incompatible with $f_q(\xi )$, while a generic through $r$ must both realize the decision and meet $A_{\xi }$, a contradiction. A maximal antichain of such $q$ therefore refines every $A_{\xi }$. Replacing each dense open set by a maximal antichain contained in it proves that the intersection of the original family is dense. AC is used to choose the maximal antichains.

A subset of $\gamma <\kappa$ is a $2$-valued sequence of length $\gamma$, so closure adds none. A new cofinal map into a ground-model ordinal of cofinality at most $\kappa$, or a collapse of a cardinal at most $\kappa$, would yield after restricting to a cofinal domain a new sequence of ground-model ordinals of length below $\kappa$. Hence those cofinalities and cardinals are preserved.