 for every $\langle {{\mathcal{U}}}_\alpha\rangle_{\alpha<\kappa}\in {{\mathcal{A}}}^\kappa$ such that ${{\mathcal{U}}}_\beta$ is a refinement of ${{\mathcal{U}}}_\alpha$, for all $\alpha<\beta<\kappa$, there exists $\langle U_\alpha\rangle_{\alpha<\kappa}$ with $\{U_\alpha\colon \alpha<\kappa\}\in{{\mathcal{B}}}$, and $U_\alpha\in{{\mathcal{U}}}_\alpha$ for all $\alpha<\kappa$.  