TITLE: A logic problem about set theory
QUESTION [10 upvotes]: In a group of n people, subgroups with common interest are formed (football,tennis,snooker). The number of subgroups equals $2^{n-1}$. Any 3 subgroups have a common member. Prove that there is a person who is a member of all the subgroups.
proof by contradiction seems like a good fit here, but I can't seem to get the details right.

REPLY [2 votes]: Denote the set by $S,$ and let $X$ be the collection of subsets of $S$ as given in the problem. First note that $A\in X$ iff $A^{C}\notin X$ since we have exactly $1/2$ of the total number of subsets. From this we deduce that if $A, B\in X,$ then $A\cap B \in X,$ since $(A\cap B)^{C} \cap A\cap B = \emptyset.$ Now, if $\{x\} \in X$ for some singleton, then we're immediately done, so suppose this isn't true. Then every set of the form $S-\{x\}$ is in $X.$ Taking the intersection of two of these gives a set of the form $S-\{x, y\},$ and we clearly get all such sets. Continuing in this manner gives that every set is in $X,$ a contradiction, so we're done.
Alternatively you can just do this by induction on the intersection of $k$ sets, with the base case being $k=2$ proven above.