Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 14 / 2

We use the random alteration method, controlling large hypergraph independent sets and then removing the few forbidden overlaps. Fix $k=2005$, let $n$ tend to infinity, and independently include each three-element subset of an $n$-vertex set as an hypergraph edge with probability $p=n^{-7/4}$. Call the resulting three-uniform hypergraph $H_0$, and let $s=\lceil n/(2k)\rceil$.

For any prescribed $s$-vertex set, the probability that it is a hypergraph independent set is $(1-p{)}^{\left(\genfrac{}{}{0ex}{}{s}{3}\right)}$. Hence the union bound givesbecause $p\left(\genfrac{}{}{0ex}{}{s}{3}\right)$ has order $n^{5/4}$, with a positive constant depending only on $k$.

Let $Y$ count unordered pairs of distinct hypergraph edges meeting in two vertices. Each potential pair consists of its common pair and two distinct additional vertices, soThe Markov inequality implies $\mathbb{P}(Y\ge n/2)=O(n^{-1/2})$. We may therefore fix one realization for which $Y<n/2$ and no $s$-vertex hypergraph independent set exists. For each offending hypergraph edge pair choose one vertex from its union, and let $D$ be the set of all chosen vertices. Delete $D$, together with all incident hypergraph edges, and call the induced hypergraph $H$. Then $|D|\le Y<n/2$, so $N=|V(H)|>n/2$. The next two parts verify both desired properties of this same $H$.

Every hypergraph independent set in the induced hypergraph $H$ is also a hypergraph independent set in $H_0$. Thus its maximum independent-set size is at most $s-1<n/(2k)<N/k$. In any proper hypergraph colouring with at most $k$ colours, one colour class would have at least $N/k$ vertices, contradicting that bound. ThereforeThis is the weak hypergraph colouring convention: an hypergraph edge may use two colours, but it may not be monochromatic.

If two distinct hypergraph edges survived in $H$ and shared two vertices, they were an offending pair in $H_0$. The deletion set $D$ contains a chosen vertex from their union, so at least one of those hypergraph edges could not have survived. This contradiction proves that $H$ is a linear hypergraph. Deleting vertices leaves every surviving hypergraph edge of size three, so the construction meets the uniformity requirement as well as the chromatic-number requirement.