Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 1 / 17J / c / Solution

Assume no deletion works. For each $x\in A_i$, some $A_j$ contains $A_i\setminus \{x\}$; otherwise deleting $x$ preserves the antichain. For different deleted elements these witnesses are distinct, since one witness containing both complements would contain $A_i$. Thus every element of $A_i$ belongs to at least $|A_i|$ sets.

Use the incidence graph with left class ${\bigcup }_i{A}_i$ and right class $\{A_i\}$. Along every incidence edge, the degree of the element is at least the degree $|A_i|$ of the set vertex. Part (b) gives a matching from the union into the $k$ sets, forcing $|{\bigcup }_i{A}_i|\le k$, a contradiction.

Solved by gpt-5.6-sol high.