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Let $f_r(m)$ be the lower-shadow size prescribed by the binomial representation in the Kruskal-Katona theorem. ThenInduction using Pascal's identity proves the inequality and drives the standard ground-set induction for the theorem.

Let $c(A)=[n]\setminus A$ and reverse the order of the ground set by $j\mapsto n+1-j$. The resulting bijection sends lexicographic order on $[n{]}^{(r)}$ to colexicographic order on $[n{]}^{(n-r)}$. It also sends the iterated upper shadow of an $r$-uniform family at level $r+1$ to the lower shadow of the corresponding $(n-r)$-uniform family, up to the same harmless reversal of the ground set.

The Kruskal-Katona theorem says that this lower shadow is smallest for an initial colex segment. Undoing the complement and reversal therefore says that the upper shadow of a family of fixed size in $[n{]}^{(r)}$ is smallest for the initial lex segment.

Solved by gpt-5.6-sol high.

Every positive integer $m$ has a unique binomial representationDefineThe Kruskal-Katona theorem says that every $r$-uniform family $\mathcal{A}$ of size $m$ hasand the initial segment of length $m$ in colexicographic order attains equality.

We prove the inequality by simultaneous mathematical induction on the ground-set size and on $r$. We use the elementary binomial-shadow arithmetic lemmaTo verify the lemma, greedily remove the largest term $\left(\genfrac{}{}{0ex}{}{q}{r}\right)$ from the binomial representation of $a+b$. If $a$ reaches that term first, apply the induction hypothesis to the remainders; otherwise transfer the excess to $b$. In the crossing case Pascal's identitygives exactly the two terms on the right. This induction also proves uniqueness of the greedy representation.

Choose the largest ground-set element $x$ and splitThe members of the lower shadow that avoid $x$ form $\partial {\mathcal{A}}_0\cup {\mathcal{A}}_1$, while those containing $x$ are the sets $B\cup \{x\}$ with $B\in \partial {\mathcal{A}}_1$. Hence, writing $a=|{\mathcal{A}}_0|$ and $b=|{\mathcal{A}}_1|$,This completes the induction.

Finally, the initial colex segment first contains all $r$-subsets of $[a_r]$, followed recursively by sets containing $a_r+1$ whose remaining elements form the appropriate initial segment at level $r-1$. Its shadow has the same recursive decomposition, and therefore has exactly $f_r(m)$ members. This proves sharpness.

Solved by gpt-5.6-sol high.

Suppose $\mathcal{A},\mathcal{B}\subseteq [n{]}^{(r)}$ are cross-intersecting families. The iterated upper shadow ${\nabla }_{n-r}\mathcal{A}$ is disjoint frombecause $A\subseteq [n]\setminus B$ would mean $A\cap B=\varnothing$. HenceIf $|\mathcal{A}|>\left(\genfrac{}{}{0ex}{}{n-1}{r-1}\right)$, the upper-shadow form of the Kruskal-Katona theorem givesIt follows from Pascal's identity that $|\mathcal{B}|<\left(\genfrac{}{}{0ex}{}{n-1}{r-1}\right)$. Thus the two sizes cannot both exceed that number.

Solved by gpt-5.6-sol high.

The Erdős-Ko-Rado theorem says that if $n\ge 2r$ and $\mathcal{A}\subseteq [n{]}^{(r)}$ is an intersecting family, thenThe star of all $r$-sets containing one fixed point attains equality.

For the shadow proof, let ${\nabla }_{n-r}\mathcal{A}$ be the iterated upper shadow at level $n-r$, and letThese two families are disjoint: if $A\subseteq [n]\setminus B$, then $A$ and $B$ are disjoint. The upper-shadow form of the Kruskal-Katona theorem says that ifthenBut $|{\mathcal{A}}^c|=|\mathcal{A}|$, so disjointness and Pascal's identity would give more thanmembers at level $n-r$, a contradiction.

For the Katona circle method, place $[n]$ in a cyclic order. At most $r$ of its cyclic intervals of length $r$ can belong to an intersecting family. Indeed, after fixing one selected interval, every selected interval starts at one of the $2r-1$ positions at cyclic distance below $r$ from its start; apart from the fixed interval, these positions form $r-1$ pairs whose corresponding intervals are disjoint. Double-count pairs consisting of $A\in \mathcal{A}$ and a cyclic order in which $A$ is consecutive. There are $(n-1)!$ cyclic orders, at most $r$ selected intervals in each, and each $A$ is consecutive in $r!(n-r)!$ cyclic orders. Thereforewhich rearranges to the required bound.

Solved by gpt-5.6-sol high.