Let $X\neq \emptyset$ be a set, and let ${\cal U}$ be a collection of subsets of $X$ such that

  1. $\bigcup {\cal U} = X$, and
  2. $U_1\neq U_2\in {\cal U}$ implies $|U_1\cap U_2| < \aleph_0$.

Is there ${\cal U}_0\subseteq {\cal U}$ such that

  1. $\bigcup {\cal U}_0 = X$, and
  2. if $U\in{\cal U}_0$ then $\bigcup \big({\cal U}_0 \setminus \{U\}\big) \ne X$ ?

No, consider the covering of naturals by initial segments.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.