Generating family for a uniform set family ($\mathcal{F}=\{A\in [n{]}^{(r)}:\exists G\in \mathcal{G},G\subseteq A\}$)

A generating family consists of sets of size at most $r$ whose $r$-element supersets are exactly the prescribed uniform set family. Small-support generators express dependence on a bounded prefix rather than on the full ground set. Taking the entire family itself as generators always works, but extremal intersection problems admit much smaller supporting intervals.

For a t-intersecting family with left-compressed generators supported on $[m]$, a tight pair containing $m$ cannot both miss an earlier coordinate $i$. Replacing $m$ by $i$ in one generator gives a shifted set containing a generator, whose intersection with the other has at most $t-1$ points. Thus the union is $[m]$, and their sizes sum to $m+t$. This identifies precisely which size classes conflict when the largest support coordinate is removed.

Suppose a left-compressed set family of $k$-sets has an antichain of generators on $[m]$, closed under left shifts followed by taking inclusion-minimal members. For a generator $G$ containing $m$, the members generated only by $G$ have intersection with $[m]$ exactly $G$. If another prefix coordinate $i$ occurred, the left shift $G-\{m\}+\{i\}$ would supply another generator. Conversely, an exact prefix $G$ contains no other generator by the antichain condition. The free tail gives the displayed binomial coefficient.

Let $n>2r-t$ and let $G,H$ generate respectively a t-intersecting family of $r$-sets and its complement family. If their union had size at most $n-r+t-1$, contain it in a set $T$ of that size. Since $|T|\ge r,n-r$, extend $G,H$ within $T$ to an $r$-set $A$ and an $(n-r)$-set $B$. Both $A$ and $B^c$ belong to the original family, but $|A\cap B^c|=|A\cup B|-|B|\le t-1$. This contradiction proves the bound.

Assume $1\le t\le r\le n$, $n>2r-t$, and $j\ge 0$. A maximum-cardinality left-compressed set family of $r$-sets that is t-intersecting has a generating family for a uniform set family supported on $[t+2j]$ under the displayed strict inequality. Reflection gives a right-end supporting interval for a right-compressed optimum. The strict threshold is important: at a tie between consecutive candidate families, use the next larger support bound. This auxiliary generating result, combined with the compatibility bound for complementary generating families, supplies the upper bound in the Ahlswede-Khachatrian theorem.

Let $m$ be the smallest supporting prefix for a maximum left-compressed set family of $k$-sets that is t-intersecting, and set $N=n-m$. The maximum-support generator fibre gives exclusive count $\left(\genfrac{}{}{0ex}{}{N}{k-a}\right)$ for a size-$a$ generator containing $m$. By tight pairs of left-compressed generators, only classes $a,b$ with $a+b=m+t$ can conflict after deleting $m$. For unequal such sizes, replacing both classes by the shortened generators of either class would implyWhen $n\ge 2k-t+2$ and the classes have positive fibres, the reverse inequality is strict. For the remaining central class $a=t+j$, $m=t+2j$, some earlier coordinate is absent from at least $j/(m-1)$ of its shortened generators. Those generators are mutually t-intersecting. Replacing the central class by that subfamily has gain-to-loss ratio at leastwhich exceeds one exactly when $n>(k-t+1)(2+(t-1)/j)$. This forces the support bound in the small-support generating lemma for extremal intersecting families.