Let $\mathcal A\subseteq[n]^{(r)}$ be an <intersecting family>, with $n\geq2r$. By the <Iterated local LYM inequality>, its upper shadow in level $n-r$ satisfies
$$
|\nabla^{,n-2r}\mathcal A|
\geq
\frac{\binom n{n-r}}{\binom nr}|\mathcal A|
=|\mathcal A|.
$$
The family of complements $\mathcal A^c=\{[n]\setminus A:A\in\mathcal A\}$ also lies in level $n-r$ and has cardinality $|\mathcal A|$. It is disjoint from the upper shadow: if $A\subseteq[n]\setminus B$ for $A,B\in\mathcal A$, then $A\cap B=\varnothing$, contradicting intersection. Both families fit inside the $(n-r)$th level, so
$$
2|\mathcal A|\leq\binom n{n-r}=\binom nr.
$$
Thus replacing the <Kruskal-Katona theorem> by Local LYM gives only
$$
\boxed{|\mathcal A|\leq\frac12\binom nr.}
$$
This agrees with the <Erdős-Ko-Rado theorem> when $n=2r$ but is weaker when $n>2r$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 109 / 1 / ii / Solution

{c}

Repeatedly applying the <Local LYM inequality> to upper shadows gives
$$
\frac{|\nabla^s\mathcal A|}{\binom n{r+s}}
\geq\frac{|\mathcal A|}{\binom nr}
$$
for $\mathcal A\subseteq[n]^{(r)}$ and $0\leq s\leq n-r$. The analogous statement for iterated lower shadows follows directly from the stated lower-shadow form.

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 Incoming links: Iterated local LYM inequality