t-intersecting family ($|A\cap B|\ge t$)

A set family is t-intersecting if every pair of its members has at least $t$ common elements. For $r$-sets on an $n$-point ground set, the whole level is t-intersecting when $n\le 2r-t$. Ordinary intersecting families are the case $t=1$.

On eight points, the rank-four Frankl family consisting of sets containing at least three of $1,2,3,4$ has seventeen members and is two-intersecting. The first fifteen rank-four sets in lexicographic order contain $1,2$, while the sixteenth is $1345$. The earlier set $1278$ meets it only at one. This disproves preservation of two-intersection by lexicographic replacement even when the rank is half the ground-set size.

For two-intersecting families on four points, all supersets of a fixed pair form a maximal family of size four: any set missing part of that pair fails its intersection test with the pair itself. The maximum instead has size five, attained by all sets of size at least three. Thus maximality and maximum cardinality diverge for intersection parameter two.

If $|A\cap B|\ge t$, then $|A△B|=|A\cup B|-|A\cap B|\le n-t$. The Kleitman diametric theorem therefore gives the sharp nonuniform intersection bound. For $n-t=2q$, all sets of size at least $n-q$ attain it. For $n-t=2q+1$, take all sets of size at least $n-q$, together with all $(n-q-1)$-sets avoiding one fixed point.

The maximum size of a t-intersecting family of $r$-sets on $[n]$ is the largest complete-intersection candidate family. In strict nontrivial intervals $n_{k+1}<n<n_k$, the maximizing index is $k$, where $n_k=(r-t+1)(2+(t-1)/k)$ and $n_0=\infty$. At admissible interior boundaries the consecutive candidates tie. If $n\le 2r-t$, the whole level is optimal. A proof uses left compression, the small-support generating lemma for extremal intersecting families for the original and complement families, and the compatibility bound for complementary generating families. The latter forces one generator family to have the candidate threshold size, placing the original family inside a candidate.

Complementing all members preserves cardinality and gives $|A^c\cap B^c|=n-2r+|A\cap B|$. Thus complementation bijects t-intersecting families of $r$-sets with $(n-2r+t)$-intersecting families of $(n-r)$-sets. It swaps left and right coordinate shifts of a set family. When the transformed intersection parameter is positive, extremal families and their generating-support lemmas can therefore be compared in opposite compression directions.

Let $1\le t\le r\le n$, and choose an integer $i$ withThe Frankl family ${\mathcal{A}}_i$ consists of the $r$-element subsets of $[n]$ containing at least $t+i$ points of $[t+2i]$. Two members share at least $2(t+i)-(t+2i)=t$ points in this distinguished prefix. Hence it is a t-intersecting family. For $i=0$, it is the uniform set family of all $r$-element subsets containing a fixed $t$-element set.

For consecutive admissible indices $i,i+1$, write $N=n-t-2i-2$ and $a=r-t-i-1$. The members gained by enlarging the prefix contain both new points and exactly $t+i-1$ old prefix points; the lost members contain neither new point and exactly $t+i$ old prefix points. ThusWhen $n>2r-t$, the relevant binomial coefficients are positive and their ratio shows that the difference is positive, zero or negative according as $n$ is below, equal to or aboveThis comparison explains the consecutive-candidate transitions in the Ahlswede-Khachatrian theorem; the prefix range ensures both new coordinates actually exist.