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For cross-intersecting families $\mathcal{A},\mathcal{B}$ with member sizes at most $r,s$ respectively, there is a finite set $X$ with $|X|\le r{\sum }_{j=0}^s{r}^j$ such that $A\cap B\cap X\ne \varnothing$ for every $A\in \mathcal{A},B\in \mathcal{B}$. Take a bounded-size transversal kernel ${\mathcal{A}}_0$ preserving hitting sets of size at most $s$ and set $X={\bigcup }_{A\in {\mathcal{A}}_0}A$. For each $B$, the set $B\cap X$ is a hitting set for ${\mathcal{A}}_0$, hence for all of $\mathcal{A}$. This proves the assertion even for infinite set families; if either family is empty, take $X=\varnothing$.

The LYM inequality. For an antichain $\mathcal{F}$ in the Boolean lattice,To prove this by double counting, generate a maximal chain in a Boolean lattice from a permutation $\pi$ of $[n]$: its members are the first $j$ entries of $\pi$, for $0\le j\le n$. A particular $j$-set belongs to exactly $j!(n-j)!$ of the $n!$ resulting maximal chains in a Boolean lattice. Each such maximal chain in a Boolean lattice meets the antichain at most once. Counting pairs consisting of a member of $\mathcal{F}$ and a permutation whose associated maximal chain in a Boolean lattice contains it givesDivision by $n!$ proves the LYM inequality, including the empty antichain and the levels containing $\varnothing$ or $[n]$.

A symmetric chain decomposition of a Boolean lattice. A symmetric chain has one set at every size from $k$ to $n-k$, with consecutive members related by inclusion. Start with the single symmetric chain containing $\varnothing$ when $n=0$. Given a symmetric chainin a symmetric chain decomposition of a Boolean lattice on $[n]$, let $x=n+1$ and replace its two copies in the larger Boolean lattice byand, when it is nonempty, byThe endpoint sizes of the first symmetric chain sum to $n+1$; those of the second symmetric chain also sum to $n+1$. These symmetric chains partition the original symmetric chain together with its copy containing $x$. Thus this construction inductively partitions the entire Boolean lattice into symmetric chains. Each symmetric chain contains exactly one set of size $\lfloor n/2\rfloor$, so the number of symmetric chains is

The proposed sum bound for cross-intersecting families is false. Take $n=3$ andBoth are antichains, and any member of one meets any member of the other. Nevertheless,The hypothesis permits the two antichains to coincide. Even if distinct antichains were wanted, taking $\mathcal{B}=\{\{1,2\},\{1,3\}\}$ gives $5>3$.

A bounded-size transversal kernel gives a finite intersection witness for cross-intersecting families, even when the two set families are infinite. We first prove the kernel statement: for any set family $\mathcal{F}$ whose members have size at most $r$, and any integer $s\ge 0$, there is a finite subset ${\mathcal{F}}_0\subseteq \mathcal{F}$ withsuch that every set $H$ of size at most $s$ is a hitting set for ${\mathcal{F}}_0$ if and only if it is a hitting set for $\mathcal{F}$.

Use mathematical induction on $s$. If $\mathcal{F}$ is empty, take ${\mathcal{F}}_0=\varnothing$. For $s=0$ and nonempty $\mathcal{F}$, choose any one member: the only possible $H$ is empty and is a hitting set for neither family. For the induction step, choose $A_0\in \mathcal{F}$. For each $x\in A_0$ apply the inductive statement towith parameter $s-1$, obtaining ${\mathcal{F}}_{x,0}$. Put ${\mathcal{F}}_0=\{A_0\}\cup {\bigcup }_{x\in A_0}{\mathcal{F}}_{x,0}$. Its size is at most $1+rM(r,s-1)=M(r,s)$. If $H$ is a hitting set for ${\mathcal{F}}_0$, choose $x\in H\cap A_0$. Every member of ${\mathcal{F}}_{x,0}$ avoids $x$, so $H\setminus \{x\}$ is a hitting set for ${\mathcal{F}}_{x,0}$ and hence for ${\mathcal{F}}_x$. Members of $\mathcal{F}$ containing $x$ already meet $H$. Thus $H$ is a hitting set for all of $\mathcal{F}$. The reverse implication follows from inclusion. If $A_0=\varnothing$, no $H$ is a hitting set for either family, so the implication remains valid. This finite branching proof never assumes $\mathcal{F}$ is finite.

Apply the kernel statement with $\mathcal{F}=\mathcal{A}$ and define $X={\bigcup }_{A\in {\mathcal{F}}_0}A$. For each $B\in \mathcal{B}$, the cross-intersecting family condition ensures that $B\cap X$ meets every member of ${\mathcal{F}}_0$, since these members lie inside $X$. As $|B\cap X|\le s$, it therefore meets every $A\in \mathcal{A}$. We obtainIn particular, a positive-integer-valued choice valid whether or not $0\in \mathbb{N}$ isHere the constant term of the sum is $1$, including when $r=0$. If either set family is empty, $X=\varnothing$ suffices. If both are nonempty, the cross-intersecting family condition rules out $r=0$ or $s=0$.

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.