The Erdős-Ko-Rado theorem says that if and is an intersecting family, then
The star of all -sets containing one fixed point attains equality.
For the shadow proof, let be the iterated upper shadow at level , and let
These two families are disjoint: if , then and are disjoint. The upper-shadow form of the Kruskal-Katona theorem says that if
then
But , so disjointness and Pascal's identity would give more than
members at level , a contradiction.
For the Katona circle method, place in a cyclic order. At most of its cyclic intervals of length can belong to an intersecting family. Indeed, after fixing one selected interval, every selected interval starts at one of the positions at cyclic distance below from its start; apart from the fixed interval, these positions form pairs whose corresponding intervals are disjoint. Double-count pairs consisting of and a cyclic order in which is consecutive. There are cyclic orders, at most selected intervals in each, and each is consecutive in cyclic orders. Therefore
which rearranges to the required bound.
Solved by gpt-5.6-sol high.
Suppose are cross-intersecting families. The iterated upper shadow is disjoint from
because would mean . Hence
If , the upper-shadow form of the Kruskal-Katona theorem gives
It follows from Pascal's identity that . Thus the two sizes cannot both exceed that number.
Solved by gpt-5.6-sol high.
No. Let , , take , and let
The families are nonempty and cross-intersecting. There are
members, so
Solved by gpt-5.6-sol high.
Fix a prime number . Distinct members of an intersecting -uniform family have intersection size in
whereas every member has size modulo . The Frankl-Wilson theorem with therefore gives
For fixed ,
This is the asserted asymptotic weakening of the Erdős-Ko-Rado theorem.
Solved by gpt-5.6-sol high.

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