0
$\begingroup$

I'm currently working on union-closed sets and I couldn't understand what a separating set is. I found that if A is a separating (union-closed) set it means that for any two elements of A, there exists a member-set containing exactly one of them. I am not sure of what I have understood. If we randomly choose any two elements of a set, there is a subset (contained in the set) that has exactly one of those two elements and not both of them at once. Is that correct?

Example:

Let S be a set defined as follows: S={{1},{2},{1,2},{1,3},{1,2,3}}

I'd assume that S is separating considering the fact if I choose 2 and 3 (the two elements) I find 2 in {1,2} and 3 in {1,3}. If {1,3} did not exist, S wouldn't be considered as a separating set.

  • 1
    In your example $2$ and $3$ are not elements of $S$. The sets $\{2\}$ and $\{3\}$ are. – 2017-01-03
  • 0
    @JSchoone I thought the elements of a set include the elements of the sets it contains. – 2017-01-03

0 Answers 0