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Assume first $1\le t\le n$, and write $d=n-t$. The sharp nonuniform t-intersecting family bound isWe prove the bound by deriving the needed Kleitman diametric theorem. Identify subsets with binary words, with Hamming distance $|A△B|$. Let $D(n,d)$ be the two expressions above. A set family of Hamming diameter at most $d<n$ has size at most $D(n,d)$, as follows.

First apply downward coordinate compression: delete coordinate $i$ from a member only when its deletion is not already present. This preserves size and cannot increase the Hamming diameter. The only possible increased distance would be between a newly lowered member $A\setminus \{i\}$ and a retained member $B$ containing $i$. The deletion $B\setminus \{i\}$ must have belonged to the old set family, or $B$ would have been lowered too. The old pair $A,B\setminus \{i\}$ has precisely that increased distance, proving it was already allowed. Repeated downward compressions terminate in a down-set.

Next apply all elementary set shifts $S_{ij}$ with $i<j$. They preserve size and Hamming diameter. If shifting one member but retaining the other increased their distance, the retained member must contain $j$ but not $i$; its shifted partner was already present, and the old pair with that partner has the same distance. These shifts preserve the down-set property: a subset of a shifted member either was already present or is obtained by shifting its corresponding old subset. For a retained member, a subset cannot disappear, because the shifted partner of that subset is also a subset of the member or of its retained shifted partner. Repeated shifts terminate, since the sum of the selected coordinate labels decreases whenever a change occurs. We now have a shifted down-set $\mathcal{F}$ of the same size and Hamming diameter.

Split it into sections on $[n-1]$:The first section has Hamming diameter at most $d$. We claim the second has Hamming diameter at most $d-2$. For $X,Y\in {\mathcal{F}}_1$, downward closure gives $X\setminus Y,Y\in {\mathcal{F}}_1$. Compare $(X\setminus Y)\cup \{n\}$ with $Y\in {\mathcal{F}}_0$. Their distance is $|X\cup Y|+1$, so $|X\cup Y|\le d-1<n-1$. Choose $i<n$ outside their union. Shiftedness puts $X\cup \{i\}$ in ${\mathcal{F}}_0$, while $Y\cup \{n\}$ is a member of $\mathcal{F}$. Their distance is $|X△Y|+2$, proving the claim. For $d<2$ this argument forces ${\mathcal{F}}_1$ to be empty.

The induction now giveswhere negative-diameter set families are empty. The equality of the numerical expressions is Pascal's identity. The base case $d=0$ has at most one member. At the boundary $d=n-1$, at most one member of each complementary pair is allowed, giving $2^{n-1}$, which equals $D(n,n-1)$ by binomial symmetry. For the remaining cases $1\le d\le n-2$, both section bounds belong to smaller-dimensional induction cases. This completes the diametric proof.

For a $t$-intersecting set family,so the theorem applies. For $d=2q$, equality is attained by all sets of size at least $n-q$, since two such sets intersect in at least $n-2q=t$ elements. For $d=2q+1$, take all sets of size at least $n-q$, and also all $(n-q-1)$-sets avoiding one fixed point. Two of these smaller sets intersect in at least $2(n-q-1)-(n-1)=t$ points; a smaller and a larger set intersect in at least $t$; two larger sets intersect in at least $t+1$. The size of this set family isso it attains the odd bound. If $t=0$, the full power set is admissible; if $t>n$, the only admissible set family is empty.

Every maximal 1-intersecting set family attains $2^{n-1}$, for $n\ge 1$. Such a set family is an up-set, since adding a superset preserves set intersections. It contains $[n]$ and omits the empty set. It contains at most one member of any complementary pair. If neither of two nonempty complements $A,A^c$ were present, maximality would give a member $B$ disjoint from $A$; upward closure would then include $A^c$, a contradiction. Thus it chooses precisely one member of each complementary pair, as in maximal intersecting families choose one member of every complementary pair.

Not every maximal 2-intersecting set family attains the bound. On four points, let $\mathcal{A}=\{A:\{1,2\}\subseteq A\}$. This set family has four members and is maximal: every missing set meets its member $\{1,2\}$ in at most one point. The sharp bound is $1+4=5$, attained by all sets of size at least three. This proves maximal intersection does not imply maximum size.