Let $\kappa$ be a regular, uncountable cardinal. We call $\kappa$
- $\dagger$ if for every sequence $(A_\alpha \colon \alpha < \kappa)$, $A_\alpha \subseteq \alpha$ and every $\xi_0 < \kappa$ there is some $\xi_0 < \xi_1 < \kappa$ such that $A_{\xi_0} = A_{\xi_1} \cap \xi_0$.
- subtle if for every sequence $(A_\alpha \colon \alpha < \kappa)$, $A_\alpha \subseteq \alpha$ and every club $C \subseteq \kappa$ there are $\xi_0 < \xi_1 \in C$ such that $A_{\xi_0} = A_{\xi_1} \cap \xi_0$.
$\dagger \Rightarrow \text{subtle}$, because we may fix $\xi_0 \in C$ and then look at the sequence $(A_\alpha' \colon \alpha < \kappa)$ where $A_\alpha' = A_\alpha$ for $\alpha \in C$ and for $\alpha \in \kappa \setminus C$ we make sure that $A_\alpha' \cap \xi_0 \neq A_{\xi_0}$ (e.g. $A_\alpha' = \emptyset$, if $A_{\xi_0} \neq \emptyset$ and $A_\alpha' = A_{\xi_0} \setminus \{\min A_{\xi_0} \}$ otherwise). Now the $\xi_0 < \xi_1 < \kappa$ witnesses $A_{\xi_0}' = A_{\xi_0} = A_{\xi_1}' \cap \xi_0$ and by the choice of $A_\alpha'$ we get that $A_{\xi_1}' = A_{\xi_1}$ and $\xi_1 \in C$.
Question
Is it also true that $\text{subtle} \Rightarrow \dagger$?