Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 13 / 3

Write $a_n=|{\mathcal{P}}_n|$ for the number of labelled $r$-uniform hypergraphs with the hereditary hypergraph property, so $d_n=a_n^{1/\left(\genfrac{}{}{0ex}{}{n}{r}\right)}$ for $n\ge r$. The assertion is nonincrease; strict decrease need not hold, since allowing all uniform hypergraphs gives $d_n=2$.

If $a_n=0$, heredity forces $a_{n+1}=0$. If $a_{n+1}=0$, the desired inequality is immediate. Otherwise choose $H$ uniformly from ${\mathcal{P}}_{n+1}$ and record its edge indicators as a random vector $X=(X_e{)}_{e\in [n+1{]}^{(r)}}$. Its information entropy, with any fixed logarithm base, is $H(X)=\log a_{n+1}$. Deleting a vertex $v$ gives the restriction $X^{(v)}$ to the edges avoiding $v$. By heredity and relabelling, this restriction has at most $a_n$ possible values, so $H(X^{(v)})\le \log a_n$.

Each edge coordinate occurs in exactly $n+1-r$ of these restrictions. For completeness, order all the edge coordinates. The chain rule for information entropy expands a restriction entropy as a sum of conditional coordinate entropies. The predecessors retained in a restriction form a subset of all predecessors, and conditioning reduces entropy; hence each restriction term is at least the corresponding full-vector term. Summing over the restrictions counts each full-vector term $n+1-r$ times. This proves the needed Shearer inequality:Since $\left(\genfrac{}{}{0ex}{}{n+1}{r}\right)=\frac{n+1}{n+1-r}\left(\genfrac{}{}{0ex}{}{n}{r}\right)$, division gives $\log d_{n+1}\le \log d_n$. Therefore the normalized speed of a hereditary hypergraph property satisfies

Let $H$ be the $r$-uniform hypergraph and use strong saturation of a uniform hypergraph in its immediate-completion sense: each missing edge $e$ lies in an $(r+t)$-set $T$ whose other $r$-subsets are already edges of $H$. The argument allows existing hypergraph cliques. Take $n\ge r+t$, the usual parameter range; if $r\le n<r+t$, strong saturation forces every possible edge to be present, and the inequality is immediate whenever its binomial terms are defined. The case $t=0$ has the trivial lower bound zero.

For $t\ge 1$, choose distinct real numbers $z_1,\dots ,z_n$ and form the $t\times n$ Vandermonde matrix $M$ whose $i$th column is $(1,z_i,\dots ,z_i^{t-1}{)}^T$. Every $t$ columns are linearly independent by the Vandermonde determinant. Set $U=\ker M\subseteq {\mathbb{R}}^n$, so $\dim U=n-t$, and work in the quotient vector spaceHere the inclusion of ${\bigwedge }^{r}U$ is injective: extend a basis of $U$ to a basis of ${\mathbb{R}}^n$ and use the resulting wedge bases of the exterior powers. Write $w_e$ for the image in $W$ of the coordinate wedge indexed by an $r$-set $e$. All these images span $W$.

For any $(r+t)$-set $T$, let $V_T$ be its coordinate subspace. The restriction of $M$ has rank $t$, so $U\cap V_T$ has dimension $r$. Wedge a basis of this intersection to obtain a nonzero elementEvery coefficient $c_e$ is nonzero. Indeed, projection of $U\cap V_T$ onto the $e$ coordinates is an isomorphism: a vector in its kernel is supported on $T\setminus e$, whose $t$ columns of $M$ are independent. The coefficient is the determinant of that projection in the chosen bases. Passing to $W$ therefore gives a linear dependence

If $e$ is missing from $H$, take its strong-saturation witness $T$. The dependence expresses $w_e$ in the span of the $w_f$ for the other edges $f\subseteq T$, all of which are present. Thus the vectors indexed by the present edges span all of $W$. The exterior-power quotient bound for strong hypergraph saturation follows:The bound is sharp: include precisely the $r$-sets meeting one fixed $t$-set $S$. A missing edge $e$ is disjoint from $S$, and every other $r$-subset of $e\cup S$ meets $S$, so adding $e$ immediately completes a hypergraph clique of order $r+t$. The number of included edges is exactly the displayed bound.