It is well known that adding a subset of a regular cardinal $\kappa$ with partial functions of size $< \kappa$ forces $\Diamond\_\kappa$. One can also see that if $S \in V$ is a stationary subset of $\kappa$, then $Add(\kappa,1)$ also forces $\Diamond(S)$. Question: If $G \subseteq Add(\kappa,1)$ is generic, do we get $\Diamond(S)$ for all stationary $S \in V[G]$, $S \subseteq \kappa$?



Forcing Diamond
The answer is yes, and the construction is essentially the same as for $\diamondsuit$. Let $\dot S$ be a name for the stationary set, and define a diamond sequence $\langle A\_\alpha\mid\alpha\in S\rangle$ as follows: Let $f$ be the $Add(\kappa,1)$ function. Suppose $\alpha\in S$. Then let $\gamma$ be least such that $f\upharpoonright\gamma\Vdash\alpha\in\dot S$ and let $A\_\alpha=\{\nu<\alpha\mid f(\gamma+\nu)=1\}$.


In answer to the comment, let $\dot C$ be a name for a club set $C$ and $\dot x$ the name for a subset of $\kappa$. Let $C'$ be the set of $\nu\in C$ such that $f\upharpoonright\nu$ decides the value of $\dot x\cap\nu$ and forces $\nu$ is a limit point of $\dot C$ (and hence is in $\dot C$). Then it is forced that $\dot C'$ is club in $V[\dot f]$, so there is a dense set of $q$ such that, setting $\eta=\operatorname{domain}(q)$, $q\Vdash\eta\in\dot C'$ and $q$ does not force $\eta\notin\dot S$. Then let $q'\le q$ be of minimal length forcing $\eta\in\dot S$, and let $q''$ be $q'$ with the known value of $\dot x\cap \eta$ tacked onto the end. Then $q''$ forces that $\dot A\_\eta= \dot x\cap\eta$.


