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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 $\bigcup \big({\cal U}_0 \setminus \{U\}\big) \ne X$ ?
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No, consider the covering of naturals by initial segments.

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