Consider the set $S$, $\{1,2,3\}$. The set I want to find, $P$, is the set of all subsets of $S$ which contain a majority of elements of $S$ - $\{\{1,2\},\{1,3\},\{2,3\},\{1,2,3\}\}$. What are the sufficient conditions for defining $P$? Stated another way: given some subset $Q$ of the powerset of $S$, how can we tell whether $Q$ is the set of all subsets of $S$ containing the majority of elements in $S$?


All I can think of so far is $\forall a, b \in P : a \cap b \neq \emptyset$. A necessary condition, but not sufficient.


I'd like to avoid using set cardinality and arithmetic if possible.



What are the sufficient conditions for a set P to be the set of all majority subsets of a set S?
I think this is equivalent (for an odd number of elements of $S$) to:


1. $Q$ is closed under supersets;
2. For any set $A\subseteq S$, exactly one of $A\in Q$, $A^c\in Q$ holds;
3. For any permutation $\pi$ of $S$, $\pi[Q]=Q$.


Because by (1) and (3), $Q$ has to be some collection of all "sufficiently large" subsets of $S$, and then by (2) it must be the set of all majority subsets of $S$.


The combination of (1) and (2) was discussed in [another post](https://mathoverflow.net/a/261849/4600) and called "ultra upset" there.


