5
$\begingroup$

Definition: Call a set of sets $\mathcal S$ cooperative when for every $\mathcal S'\subseteq\mathcal S$,

  • if every two elements in $\mathcal S'$ have nonempty intersection (so $\forall X,Y\in\mathcal S'. X\cap Y\neq\varnothing$)
  • then $\mathcal S'$ has nonempty intersection (so $\bigcap \mathcal S'=\varnothing$).

Intuitively, think of $\mathcal S$ as a set of agents (so an agent is an element $X\in\mathcal S$), and imagine that a set of agents can cooperate precisely when its intersection is nonempty. Then $\mathcal S$ is cooperative when a group of agents $\mathcal S'\subseteq \mathcal S$ can cooperate, if and only if its members can cooperate pairwise.

Note that is not required here that $\bigcap\mathcal S\neq\varnothing$, since it need not be the case that every two elements in $\mathcal S$ has nonempty intersection.

Examples:

  1. $\mathcal S=\bigl\{\{0\},\{1\},\{2\}\bigr\}$ is trivially cooperative, because no distinct elements intersect.
  2. $\mathcal S=\bigl\{\{0\},\{0,1\},\{0,2\}\bigr\}$ is trivially cooperative, because $\bigcap\mathcal S=\{0\}$.
  3. $\mathcal S=\bigl\{\{0\},\{0,1\},\{0,2\},\{4\}\bigr\}$ is cooperative (even though $\bigcap\mathcal S=\varnothing$).
  4. $\mathcal S=\bigl\{ \{0,1\}, \{1,2\}, \{2,0\} \}$ is not cooperative, because if we take $\mathcal S'=\mathcal S$ then every pair of elements in $\mathcal S'$ intersects, but $\bigcap\mathcal S'=\varnothing$.

Has a theory of cooperative sets been explored, and if so where?

Thank you.

(This question is not identical to, but seems related to, Prove that the intersection of all the sets is nonempty. )

$\endgroup$
7
  • $\begingroup$ The definition seems a bit odd. First of all, in your second condition, I expect you meant to write $ \bigcap \,S'\neq \emptyset$, right? Secondly, if $S$ is such that no two distinct elements in it intersect, then the conditions are vacuously satisfied. Is that what you intended? In any case, I have not seen a definition like this before. Seems like something Economists might look at. $\endgroup$
    – lulu
    Aug 27, 2023 at 16:08
  • $\begingroup$ @lulu Your second remark was already addressed in the fourth-last paragraph of the post. $\endgroup$ Aug 27, 2023 at 20:44
  • $\begingroup$ @legionwhale I don't see how. I am remarking that the definition vacuously implies that if, in $S$, $S_1\neq S_2\implies S_1\cap S_2=\emptyset$, then $S$ is "cooperative." That strikes me as odd, but of course there's nothing contradictory about it. If the OP is ok with that consequence of the definition, so be it. $\endgroup$
    – lulu
    Aug 27, 2023 at 20:49
  • $\begingroup$ @lulu But isn't the fourth-last paragraph by the OP acknowledging that may be a consequence, and thereby implying they are OK with it? There's no error in what you said, so my reply was not important, but I just thought I'd point this out. $\endgroup$ Aug 27, 2023 at 20:50
  • 1
    $\begingroup$ This is reminiscent of the result that if sub-trees $(T_n)$ of a tree $T$ intersect pairwise, then $\bigcap T_n \neq \varnothing$. I think the converse is also true, where I take the converse to be "Let $G$ be a connected graph such that whenever a family of connected sub-graphs intersect pairwise, the intersection over all of them is non-empty. Then, $G$ is a tree." Thus, a graph-theoretic rephrasing may be fruitful. But considering your post history, I expect you had already thought of this. $\endgroup$ Aug 28, 2023 at 0:55

0

You must log in to answer this question.

Browse other questions tagged .