Extremal problem
Named by 3 essays across 2 fields — each of them below, with the objects they name alongside it.
Everybody's share of the chains
There are twenty-four ways to build a four-element set one element at a time. Every subset lies on some of them, and no two incomparable subsets share one — so an antichain is a set of disjoint shares of a single whole.
The largest family that always meets
Change the question from "no two comparable" to "every two share an element" and the answer changes shape. The best antichain is a whole layer; the best intersecting family is a star, and the proof is a circle.
The least area a width can hold
Barbier's theorem says every curve of constant width has the same perimeter, which removes perimeter as a way of telling the family apart. Area is not like that — the circle holds the most and the Reuleaux triangle the least — and the reason the minimiser has corners is a constraint rather than a preference.
Named alongside it
The objects these essays reach for when they reach for this one.
Binomial coefficientCounting argumentCounting two waysPosetAntichainAreaChainConstant widthConvexityCyclic orderErdos ko radoIntersecting family