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For a set family $\mathcal{A}\subseteq \mathcal{P}([n])$, its downward closure is $\downarrow \mathcal{A}=\{S:\text{some }A\in \mathcal{A}\text{ has }S\subseteq A\}$. It is the smallest down-set containing $\mathcal{A}$, just as the upward closure of a set family is the smallest up-set containing it. The two closures allow Harris' inequality to control sizes of a cross-Sperner family pair.

A maximal intersecting family is an up-set. It cannot contain both a set and its complement. If neither were present, maximality would supply a member disjoint from the first set, and upward closure would then include its complement. Thus precisely one member of each complementary pair occurs, so the family has $2^{n-1}$ members for $n\ge 1$.

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.

Form the family $\mathcal{F}=\{A\subseteq [n]:{\sum }_{i\in A}{c}_i>1/2\}$. The nonnegative weights make it an up-set. It is an intersecting family, since two disjoint members would have combined weight exceeding one. The no-tie hypothesis makes it a self-dual set family: exactly one of $A,A^c$ belongs.

Let $a_j=|\mathcal{F}\cap (\genfrac{}{}{0ex}{}{[n]}{j})|/(\genfrac{}{}{0ex}{}{n}{j})$. Self-duality gives $a_{n-j}=1-a_j$. The Erdős-Ko-Rado theorem gives $a_j\le j/n$ for $j<n/2$; at $j=0$ this is immediate. If $n$ is even, $a_{n/2}=1/2$.

The biased measure of a set family is ${\mu }_p(\mathcal{F})={\sum }_j\left(\genfrac{}{}{0ex}{}{n}{j}\right)a_j{p}^j(1-p{)}^{n-j}$. By independence of the Bernoulli random variables, it equals $\mathbb{P}(Z\ge 1/2)$, since equality is excluded. Subtract $p={\sum }_j\left(\genfrac{}{}{0ex}{}{n}{j}\right)(j/n)p^j(1-p{)}^{n-j}$ and pair complementary levels. The complementary-layer bound for biased measure givesBoth factors are nonnegative for $p\ge 1/2$. The endpoints $p=1/2,1$ also follow directly, or by continuity. Thus the weighted Bernoulli majority bound is

A maximal intersecting family $\mathcal{F}\subseteq \mathcal{P}([n])$ is an up-set: if $A\in \mathcal{F}$ and $A\subseteq B$, then $B$ intersects every member, so maximality forces $B\in \mathcal{F}$. For every complementary pair $\{A,A^c\}$, at most one member lies in $\mathcal{F}$. If neither did, maximality would give $B\in \mathcal{F}$ disjoint from $A$; then $B\subseteq A^c$, and the up-set property would put $A^c$ in $\mathcal{F}$, a contradiction. Exactly one set from each complementary pair occurs, so

For $n\ge 4$, three pairwise nonisomorphic examples are:

Their smallest members have different sizes except at $n=4$, where the inclusion graphs of the minimal two-sets are respectively a triangle and a three-edge star. Thus they are nonisomorphic.

The Erdős-Ko-Rado theorem says that if $n\ge 2r$ and $\mathcal{F}\subseteq [n{]}^{(r)}$ is intersecting, thenFor the Kruskal-Katona theorem proof, the family of complements ${\mathcal{F}}^c\subseteq [n{]}^{(n-r)}$ is disjoint from the $(n-r)$th upper shadow of $\mathcal{F}$: otherwise one member of $\mathcal{F}$ would be contained in the complement of another. Kruskal–Katona says that when $|\mathcal{F}|=\left(\genfrac{}{}{0ex}{}{n-1}{r-1}\right)$ this upper shadow has at least $\left(\genfrac{}{}{0ex}{}{n-1}{r}\right)$ members, with a strict corresponding inequality above that threshold. Since these two disjoint families lie in one level of size $\left(\genfrac{}{}{0ex}{}{n}{r}\right)$, the EKR bound follows.

For the averaging proof, place $[n]$ in a cyclic order. Among the $n$ cyclic intervals of length $r$, an intersecting family contains at most $r$: after fixing one interval, the possible intersecting intervals can be paired by their first separating endpoint. Double-counting pairs consisting of a cyclic order and a member of $\mathcal{F}$ that appears as an interval giveswhich is the same bound. This is the Katona circle method.

The minimum size $f_r(n)$ does not tend to infinity. Fix a core $S\subseteq [n]$ of size $2r-1$ and takeAny two members intersect. If an $r$-set $A$ is not contained in $S$, then $|A\cap S|\le r-1$, so the remaining at least $r$ points of $S$ contain a member of $\mathcal{F}$ disjoint from $A$. Thus $\mathcal{F}$ is maximal intersecting in $[n{]}^{(r)}$, andindependently of $n$.

In a partially ordered set $P$, an upper set $U$ satisfies $x\in U$ and $x\le y\Rightarrow y\in U$; a lower set $L$ satisfies $x\in L$ and $y\le x\Rightarrow y\in L$. The complement of a set exchanges the two conditions: if $x\notin U$ and $y\le x$, then $y\in U$ would force $x\in U$, so $P\setminus U$ is lower. In a Boolean lattice ordered by inclusion, these are the up-set and down-set conditions.