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Let $X\neq \emptyset$ be a set. We say ${\cal C} \subseteq {\cal P}(X)\setminus\{\emptyset\}$ is a cover of $X$ if $\bigcup {\cal C} = X$. A subset $S\subseteq X$ is a choice set for ${\cal C}$ if $|S\cap K| = 1$ for all $K\in {\cal C}$.

Suppose $X$ is infinite and ${\cal C}$ is a cover of $X$ with the following properties:

  1. if $K\in {\cal C}$, then $|K| = |X|$, and
  2. for $K\neq L \in {\cal C}$ we have $|K\cap L| <|X|$.

Is there a choice set for ${\cal C}$?

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Not always. Say, we may achieve that for any $x\in X$ there exists a set $K\in \mathcal{C}$ such that $K\subset \cup_{L\in \mathcal{C};L\ni x} L\setminus\{x\}$. It means that $x$ can not belong to a choice set. This may be done as follows: let $X$ be a set of rationals, and each set in $\mathcal{C}$ be a convergent sequence of rationals, with all limits different. This is to be sure that all mutual intersections are finite. Enumerating the rationals as $x_1,x_2,\dots$, on $n$-th step we take a countable family of sets $L\ni x_n$ and a subset $$K\in \cup_{L\in \mathcal{C};L\ni x_n} L\setminus\{x_n\}$$ such that their limits are all different from previously chosen limits.

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