Let's consider $\kappa$ an uncountable regular cardinal and $\lambda<\kappa$. Given any decreasing family $\{A_\alpha\}_{\alpha<\lambda}$ of sets with $\sharp A_\alpha=\kappa$, does it true that $\bigcap_{\alpha<\lambda} A_\alpha\neq \emptyset$? I think, though I'm not sure, that this should be true by regularity of $\kappa$.
Does anyone have any idea to prove or refute this? Thank you in advance.
I’m afraid not.
Let $\kappa=\omega_1$ and $\lambda=\omega$. Each $\xi\in\omega_1$ can be written uniquely in the form $\eta_\xi+n_\xi$, where $\eta_\xi$ is $0$ or a limit ordinal, and $n_\xi\in\omega$. For $n\in\omega$ let $A_n=\{\xi\in\omega_1:n_\xi\ge n\}$. Then $\langle A_n:n\in\omega\rangle$ is a strictly decreasing family of uncountable subsets of $\omega_1$ whose intersection is empty.