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For $i<j$, replace a member $A$ containing $j$ but not $i$ by $(A\setminus \{j\})\cup \{i\}$ only when the replacement is absent. An elementary set shift preserves family size and the intersecting family property. The inclusion $\partial (S_{ij}\mathcal{F})\subseteq S_{ij}(\partial \mathcal{F})$ shows that it cannot increase the lower shadow.

For an intersecting family of $r$-sets with $r\ge 1$, its lower shadow has at least as many members as the family. Apply elementary set shifts and then the ballot reflection injection to obtain this bound.

A family in the Boolean lattice whose pairwise Hamming distances are at most $d<n$ has size at most $D(n,d)$. Downward compression followed by elementary set shifts preserves the distance bound. In the resulting down-set, the section containing the last coordinate has diameter at most $d-2$: two members leave a coordinate free, and shifting the last coordinate to it creates a pair with two additional disagreements. This gives the recurrence $D(n,d)=D(n-1,d)+D(n-1,d-2)$. The boundary case $d=n-1$ follows by pairing complementary sets. Hamming balls attain the even case; the union of two radius-$q$ balls with adjacent centers attains the odd case.

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.

We use the standard convention that a two-intersecting family satisfies $|A\cap B|\ge 2$ for every pair of members, including $A=B$. In particular every member has at least two elements.

An intersecting shadow lemma. If $\mathcal{F}$ is an intersecting family of $r$-sets, $r\ge 1$, thenHere is a proof, so no additional shadow theorem is being assumed. For $i<j$, an elementary set shift replaces a member $A$ containing $j$ but not $i$ by $(A\setminus \{j\})\cup \{i\}$ exactly when that replacement is not already in the family. This preserves cardinality and the intersecting family property. The only potentially troublesome pair consists of a moved member and an unmoved member $B$ containing $j$ but not $i$. But then $B$'s replacement was already present, and the two original members obtained by using that replacement would have been disjoint if the new pair were disjoint.

Alsoso an elementary set shift cannot increase the lower shadow. For completeness, a shadow set containing neither $i,j$ or both has an unchanged witness in the original lower shadow. A shadow set containing $i$ but not $j$ is either originally present or the replacement of an originally present shadow set. If a shadow set contains $j$ but not $i$, its witnessing member survived the elementary set shift; either that member contains both coordinates, or its replacement was already present. In either case both the shadow set and its replacement were originally present, so the former survives in the shifted lower shadow.

Repeated elementary set shifts terminate because the sum of the elements of all members decreases whenever the family changes. In the resulting shifted set family $\mathcal{H}$, if $a_1<\cdots <a_r$ is a member and $b_1<\cdots <b_r$ satisfies $b_i\le a_i$ for every $i$, then $\{b_1,\dots ,b_r\}$ is also a member. Every $A\in \mathcal{H}$ has a prefix $[2q-1]$ containing $q$ of its elements. Otherwise its sorted elements satisfy $a_i\ge 2i$, and the first $r$ elements outside $A$ form a disjoint member componentwise no larger than $A$, contradicting the intersecting family property.

Choose the first such prefix and reflect membership inside it:This has size $r-1$. The set $D\cup \{2q-1\}$ is componentwise no larger than $A$: before the first excess, the prefix counts of $A$ are at most half the prefix length, and reflection only moves selected positions to the left. Thus $D$ belongs to $\partial \mathcal{H}$. This ballot reflection injection is injective: in $D$, the reflected prefix ends at the first position where unselected positions outnumber selected ones, and reflecting it again recovers $A$. Consequently $|\partial \mathcal{H}|\ge |\mathcal{H}|=|\mathcal{F}|$. Since elementary set shifts did not increase the lower shadow, this proves the intersecting shadow lemma.

The nonuniform two-intersecting family bound. Write ${\mathcal{A}}_i=\mathcal{A}\cap [n{]}^{(i)}$. No member of $\partial {\mathcal{A}}_i$ can be the complement of a member of ${\mathcal{A}}_{n-i+1}$: such an inclusion would make the intersection of the two original members have at most one element. The intersecting shadow lemma therefore givesFor $n=2h$, sum this over $1\le i\le h$. These pairs cover all nonempty levels, soEquality is attained by all subsets of size at least $h+1$: any two intersect in at least two elements.

For $n=2h+1$, the same pairs cover every level except $h+1$. Complements of the members of ${\mathcal{A}}_{h+1}$ form an intersecting family of $h$-sets, sinceFor $h\ge 1$, the permitted Erdős-Ko-Rado theorem bounds this middle level by $\left(\genfrac{}{}{0ex}{}{2h}{h-1}\right)=\left(\genfrac{}{}{0ex}{}{2h}{h+1}\right)$. HenceEquality is attained by all $(h+1)$-sets inside $[2h]$, together with all subsets of $[2h+1]$ of size at least $h+2$. Two members of the first group intersect in at least two elements; a member from each group intersects in at least two; two members of the second group intersect in at least three. For $n=1$ the only two-intersecting family is empty, which also gives equality. The empty-ground-set case is likewise immediate.

The pair-cover bound for a two-intersecting uniform family. If $\mathcal{A}$ is nonempty, fix $A_0\in \mathcal{A}$, with $|A_0|=r$. Every member contains one of the $\left(\genfrac{}{}{0ex}{}{r}{2}\right)$ pairs inside $A_0$. Each fixed pair belongs to exactly $\left(\genfrac{}{}{0ex}{}{n-2}{r-2}\right)$ $r$-sets, so counting their union givesThe empty family is immediate; a nonempty two-intersecting family requires $r\ge 2$.

Witness-pair concentration for an intersecting uniform family. Let $\mathcal{A}$ now be an intersecting family of $r$-sets. If it is two-intersecting, the preceding bound, together with $\left(\genfrac{}{}{0ex}{}{r}{2}\right)\le (r-1{)}^2$ for $r\ge 2$, shows that any chosen $x\in [n]$ has the required property. Otherwise there are members $A,B$ with $A\cap B=\{x\}$. Every member $C$ not containing $x$ must contain some $a\in A\setminus \{x\}$ and some $b\in B\setminus \{x\}$. These two sets of possible witnesses are disjoint. There are $(r-1{)}^2$ witness pairs, and each is contained in $\left(\genfrac{}{}{0ex}{}{n-2}{r-2}\right)$ $r$-sets. ThusFor $r=1$, a nonempty intersecting family consists of one singleton; choose its element and the number avoiding it is zero. This is the natural separate interpretation of the degenerate case, where the printed binomial coefficient has lower index $-1$. For an empty family any $x\in [n]$ works, provided $n\ge 1$ as usual.

A set family is shifted if every elementary set shift leaves it unchanged. For a uniform set family, this is equivalent to downward closure under componentwise comparison of the increasing lists of its elements. Repeated elementary set shifts terminate at a shifted set family.