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Is there a set $X\neq\emptyset$ and a collection ${\cal F}\subseteq {\cal P}(X)\setminus\{\emptyset\}$ of non-empty subsets of $X$ with the following properties?

  1. $a\in {\cal F} \implies |a|\geq 2$,
  2. $a\neq b\in {\cal F} \implies |a\cap b| \leq 1$, and
  3. there is no function $f: {\cal F} \to X$ such that
    • $f(a) \in a$ for all $a\in {\cal F}$, and
    • if $a, b\in {\cal F}$ with $a\cap b\neq \emptyset$ then $f(a)\neq f(b)$?
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    $\begingroup$ Of course $fa)\ne f(b)$ if $a\cap b=\emptyset$ so you could have just said $f$ is injective. Your selector can't exist if $|\mathcal F|\gt|X|$ so $\mathcal F=\binom X1+\binom X2$ is a counterexample if $|X|\ge2.$ $\endgroup$
    – bof
    Jan 11, 2016 at 14:25
  • $\begingroup$ Sorry forgot criterion 1 $\endgroup$ Jan 11, 2016 at 16:20

2 Answers 2

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Satisfying the new Condition 1, If $|X|\geqslant4$, let ${\cal F}$ consist of all subsets of $X$ with two elements.

Condition 1 can be weakened. For example, if $X=\{a_{i,j}:1\leqslant i,j\leqslant 3\}$, then the following 12 subsets, each of size 3, satisfy the conditions. $$ \{a_{1,1},a_{1,2},a_{1,3}\},\{a_{2,1},a_{2,2},a_{2,3}\},\{a_{3,1},a_{3,2},a_{3,3}\} $$ $$\{a_{1,1},a_{2,1},a_{3,1}\},\{a_{1,2},a_{2,2},a_{3,2}\},\{a_{1,3},a_{2,3},a_{3,3}\} $$ $$ \{a_{1,1},a_{2,2},a_{3,3}\},\{a_{1,3},a_{2,2},a_{3,1}\} $$ $$ \{a_{1,1},a_{2,3},a_{3,2}\},\{a_{1,3},a_{2,1},a_{3,2}\},\{a_{1,2},a_{2,1},a_{3,3}\},\{a_{1,2},a_{2,3},a_{3,1}\} $$

It seems likely that we can permit any minimum size for sets in ${\cal F}$.

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  • $\begingroup$ Provided $|X|\gt1.$ $\endgroup$
    – bof
    Jan 11, 2016 at 21:36
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Let ${\cal F}$ consist of all subsets with one element and, in addition, the whole set $X$. We must have $f(\{a\})=a$, which leaves no possibility for $f(X)$.

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