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A set $H$ is a hitting set for a set family $\mathcal{F}$ if $H\cap A\ne \varnothing$ for every $A\in \mathcal{F}$. Every set is a hitting set for an empty set family; none is a hitting set for a family containing the empty set. Bounded hitting sets permit finite certificates for some infinite set families, as in the bounded-size transversal kernel.

For a finite ground set and an indexed set family, this zero-one matrix records membership of points in members. The diagonal entries of $B^{T}B$ are member sizes, and its off-diagonal entries count pairwise intersections. For the Fano plane, either point-line orientation gives the identity $B^{T}B=2I+J$, since every point lies on three lines and each distinct point pair lies on one line.

A family of sets is intersecting when every two of its members have nonempty intersection.

This is a property of a set family, distinct from extremal theorems bounding its size.

Let $U,V$ be disjoint, equally sized, nonempty sets, with $\min U<\min V$. Suppose an intersecting family is unchanged by all such UV-compressions of smaller size. Then its $(U,V)$-UV-compression is still an intersecting family. Indeed, a newly moved member $A^{\prime }=(A\setminus V)\cup U$ can only be disjoint from a retained old member $B$, since two newly moved members share $U$. Here $B\cap U=\varnothing$ and $V^{\prime }=B\cap V\ne \varnothing$. If $V^{\prime }=V$, retaining $B$ forces $(B\setminus V)\cup U$ to have been present, contradicting the old intersection property with $A$. Otherwise choose $U^{\prime }\subset U$ with $|U^{\prime }|=|V^{\prime }|$ and $\min U\in U^{\prime }$. Stability under the smaller $(U^{\prime },V^{\prime })$-UV-compression forces $(A\setminus V^{\prime })\cup U^{\prime }$ to be present; this member is disjoint from $B$, again a contradiction. Repeatedly applying a changing UV-compression of smallest size terminates, since the sum of lexicographic order positions decreases. A terminal set family is a lexicographic initial segment: a missing earlier member $X$ and present later member $Y$ would be moved by $U=X\setminus Y$, $V=Y\setminus X$.

The join of two elements of a lattice is their least upper bound. In a lattice of closed set families, it is the closure of their union.

A set family is laminar if any two members are disjoint or one contains the other. Its distinct inclusion-maximal members are therefore disjoint. Applied to the true sets in a hierarchical family of intersection hypotheses, this gives the disjoint-block bound used for familywise error control for a laminar hypothesis family.

If a forbidden intersection size exceeds $n/2$, every subset of size at most $\lfloor n/2\rfloor$ avoids it, including on self-intersection. This set family contains at least half the Boolean lattice. Consequently a fixed exponential bound $c^n$ with $c<2$ cannot hold uniformly as $n$ grows. Forbidden-intersection estimates below the middle rank therefore cannot simply be extrapolated to high intersections.

For a nonempty $r$-uniform set family, define $x>r-1$ by $|\mathcal{F}|=\left(\genfrac{}{}{0ex}{}{x}{r}\right)$ using real binomial coefficients. Then its lower shadow has at least $\left(\genfrac{}{}{0ex}{}{x}{r-1}\right)$ members. This continuous estimate is often easier to use than the exact integer expansion in the Kruskal-Katona theorem.

A short proof uses coordinate shifts of a set family. In a fully left-shifted set family split at coordinate one into ${\mathcal{F}}_0$ and ${\mathcal{F}}_1$, deleting one in the present section. The shifting property implies $\partial {\mathcal{F}}_0\subseteq {\mathcal{F}}_1$, and the total lower shadow has size $|{\mathcal{F}}_1|+|\partial {\mathcal{F}}_1|$. Induct on uniformity and, at fixed uniformity, on set family size. If $|{\mathcal{F}}_1|<\left(\genfrac{}{}{0ex}{}{x-1}{r-1}\right)$, the Pascal's identity and the size mathematical induction force $|\partial {\mathcal{F}}_0|>\left(\genfrac{}{}{0ex}{}{x-1}{r-1}\right)$, a contradiction. The uniformity mathematical induction applied to ${\mathcal{F}}_1$, followed by the Pascal's identity, proves the estimate. Integer $x$ is sharp for the full $r$-level on $x$ coordinates.

The meet of two elements of a lattice is their greatest lower bound. For set families ordered by inclusion, it is their intersection whenever that intersection lies in the specified lattice.

In the polynomial method in combinatorics, a set family of sets with common size residue $r$ modulo a prime number $p$ annihilates the displayed auxiliary functions. After multilinear reduction on the Boolean cube, functions with $|S|\le s-1$ have degree at most $s$. If $r$ avoids every $j=0,\dots ,s-1$ modulo $p$, these functions are linearly independent: evaluation at a set $T$ gives $(|T|-r){\sum }_{S\subseteq T}{c}_S$, and increasing $|T|$ eliminates coefficients. Combining them with diagonal intersection polynomials removes the lower-degree monomial count and strengthens the dimension bound from ${\sum }_{j\le s}\left(\genfrac{}{}{0ex}{}{n}{j}\right)$ to $\left(\genfrac{}{}{0ex}{}{n}{s}\right)$.

Let $p$ be a prime number and $L\subseteq {\mathbb{F}}_p$ have $s$ elements. A set family whose member sizes modulo $p$ avoid $L$, and whose distinct pairwise intersection sizes modulo $p$ belong to $L$, has at most ${\sum }_{j=0}^{\min (s,n)}\left(\genfrac{}{}{0ex}{}{n}{j}\right)$ members. The intersection polynomials ${\prod }_{\ell \in L}({\sum }_{i\in A}{x}_i-\ell )$ have a diagonal, nonzero evaluation matrix on the family's characteristic vectors of sets. Multilinear reduction on the Boolean cube places these linearly independent functions in the space spanned by square-free monomials of degree at most $s$.

If all members of a set family on $[n]$ have odd size and all distinct pairwise intersections have even size, its characteristic vectors over the finite field ${\mathbb{F}}_2$ are linearly independent. Their pairwise dot products are zero, and each vector has dot product one with itself. Taking the dot product of a linear combination with each vector makes every coefficient vanish, so there are at most $n$ vectors. The set family of all singleton sets attains this bound.

The four functions theorem concerns nonnegative functions $\alpha ,\beta ,\gamma ,\delta$ on the Boolean lattice. Its pointwise hypothesis and conclusion areWe prove it by eliminating coordinates.

The elementary step is the two-point four-functions inequality. Suppose $a_i{b}_j\le c_{\max (i,j)}{d}_{\min (i,j)}$ for $i,j\in \{0,1\}$. Put $x=a_0{b}_1$, $y=a_1{b}_0$, $M=c_1{d}_0$, $R=c_0{d}_1$. Then $x,y\le M$ and, using the two diagonal bounds,If $M>0$, $(M-x)(M-y)\ge 0$ yields $x+y\le M+xy/M\le M+R$. If $M=0$, both cross terms vanish and the same conclusion holds. Adding the diagonal bounds gives

For subsets $S,T\subseteq [n-1]$, take $a_i=\alpha (S\cup E_i)$, $b_j=\beta (T\cup E_j)$, $c_i=\gamma ((S\cup T)\cup E_i)$ and $d_i=\delta ((S\cap T)\cup E_i)$, where $E_0=\varnothing$, $E_1=\{n\}$. The original hypothesis supplies precisely the four inequalities of the elementary step. Therefore the marginal functionsand similarly ${\beta }^{\prime },{\gamma }^{\prime },{\delta }^{\prime }$, satisfy the same pointwise hypothesis on $[n-1]$. Induction, starting with the one-element Boolean lattice on the empty ground set, proves the four functions theorem. This proof includes zero-valued functions without division by zero.

For the Harris-Kleitman inequality, let $\mathcal{F},\mathcal{G}$ be increasing set families, meaning they contain every superset of each member. In the theorem take $\alpha =1_{\mathcal{F}}$, $\beta =1_{\mathcal{G}}$, $\gamma =1_{\mathcal{F}\cap \mathcal{G}}$ and $\delta =1$. If $A\in \mathcal{F}$ and $B\in \mathcal{G}$, their union belongs to both increasing families, so the hypothesis holds. Thus, for the uniform probability measure $\mu$ on subsets,Mapping every subset to its complement proves the same Harris-Kleitman inequality for two decreasing families. An increasing and a decreasing family have the reverse inequality: apply positive correlation to the increasing family and the complement, as a family, of the decreasing one. More generally, increasing nonnegative functions obey the corresponding positive-correlation bound by the same four-functions substitution, since both function values at the union dominate the original values. For an independent-coordinate product measure, multiply all four functions by its weight $w$; the identity $w(A)w(B)=w(A\cup B)w(A\cap B)$ preserves the hypothesis and proves the weighted Harris inequality too.

Finally, replace each intersecting family ${\mathcal{A}}_i$ by its upward closure of a set family ${\mathcal{U}}_i$. It is still intersecting: two supersets of intersecting members have nonempty intersection. It contains at most one of each pair $S,S^c$, and hence $|{\mathcal{U}}_i|\le 2^{n-1}$. Let ${\mathcal{D}}_i$ be the complement of ${\mathcal{U}}_i$ as a family. These families are decreasing and have measure at least one half. Their intersections are decreasing too, so repeated Harris-Kleitman inequality givesTaking complements and using ${\mathcal{A}}_i\subseteq {\mathcal{U}}_i$ proves the union bound for intersecting families:For $k\le n$, equality is attained by the families of sets containing coordinate $i$, $1\le i\le k$. For $k>n$, the displayed real bound implies the integer bound $2^n-1$, which is attained by using the $n$ coordinate families and repeating families as necessary.

Let ${\mathrm{CLIQUE}}_{n,m}$ be the monotone Boolean function of the edge indicators asserting that an $n$-vertex graph contains an $m$-clique. We prove a superpolynomial monotone clique lower bound using a finite lattice of approximations. The only circuit theorem assumed is the Razborov gate-by-gate approximation lemma.

Fix integers $\ell ,r\ge 2$. Consider upward-closed set families $A$ of subsets of $[n]$ of size at most $\ell$. Impose the Razborov closure rule: if $W_1,\dots ,W_r\in A$ have pairwise intersections contained in $W$, adjoin $W$ whenever $|W|\le \ell$. Also normalize a family containing a set of size zero or one to the full family, since its clique predicate is identically true. Represent it byClosed families form the finite lattice approximation for monotone clique circuits, with meet $A\cap B$ and join $\mathrm{cl}(A\cup B)$. An input edge is represented by the family of its supersets up to size $\ell$. The constant predicates have the empty and full families. The assumed approximation theorem says that a size-$S$ circuit's output disagreement is covered by at most $S$ local gate errors:

We need the Erdős–Rado sunflower lemma: an $s$-uniform set family with more than $s!(r-1{)}^s$ members contains $r$ members with a common pairwise intersection. Its short induction proves the bound. A maximal disjoint subfamily either has $r$ members, already the required empty-core delta-system, or its union has at most $s(r-1)$ points and meets every member. Some point then occurs in more than $(s-1)!(r-1{)}^{s-1}$ members. Delete it, apply the induction hypothesis, and restore it to the core. Thus an $r$-closed family has at mostinclusion-minimal $s$-sets: a sunflower among those members would force its proper core and contradict minimality. This is the sunflower bound for minimal members of a Razborov-closed family.

For positive inputs choose a uniform $m$-subset $Z$ and the graph $K_Z$ having exactly the edges of its clique. Put $q=\ell rm/n$ and suppose $q\le 1/2$. A nonuniversal family has no minimal member of size at most one, so it accepts at mostof positive inputs. Here $\mathbb{P}(W\subseteq Z)=(m{)}_{|W|}/(n{)}_{|W|}\le (m/n{)}^{|W|}$.

If $K_Z$ is in a meet-error set, it contains minimal witnesses $X\in A,Y\in B$. Their union must have size greater than $\ell$, since otherwise upward closure would put it in $A\cap B$. At least one witness therefore has size greater than $\ell /2$. The same counting and the union bound give the positive clique error of a truncated lattice meet:

For negative inputs independently assign one of $m-1$ colours to each vertex, joining differently coloured vertices. Every resulting complete multipartite graph is $m$-clique-free. Consider one forcing step adjoining $W$. If its clique is newly accepted, $W$ is rainbow while all witnesses $W_i$ are not. Condition on the colours of $W$. The sets $W_i\setminus W$ are disjoint, so their non-rainbow events are conditionally independent. For each witness, every possible collision either involves two random petal colours or one petal colour and one fixed core colour; its probability is at mostWhen this is at most $1/2$, a step makes a negative error with probability at most $2^{-r}$. Upward closure itself introduces no extra accepted graphs, and normalization introduces no extra error beyond the forcing step that produced a set of size at most one. At most ${\sum }_{j=0}^{\ell }\left(\genfrac{}{}{0ex}{}{n}{j}\right)\le n^{\ell +1}$ sets can be forced. The negative colouring error of a forced-set closure is therefore

Set $m=r=\lfloor n^{1/4}\rfloor$ and $\ell =\lfloor {\log }_{2}n\rfloor$. For sufficiently large $n$, both probability conditions hold and $2q^2<1/2$. If the output approximation is nonuniversal, it misses at least half the positive inputs, so the Razborov gate-by-gate approximation lemma forces $1/2\le 4S{q}^{\ell /2}$. If it is universal, it accepts every negative colouring input, forcing $1\le S{n}^{\ell +1}{2}^{-r}$. ConsequentlySince $\log (1/q)\ge \frac{1}{2}\log n-\log \ell$ and $r$ grows faster than $(\log n{)}^2$, both terms implyfor an absolute positive $c$ and all sufficiently large $n$. This is superpolynomial in the $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ input bits. The argument concerns AND/OR monotone circuits; it gives no such bound for circuits allowed negation.

The Frankl-Wilson theorem is a striking example of the polynomial method in combinatorics: an intersection restriction creates a family of linearly independent low-degree polynomials, whose dimension bounds the set family. Let $p$ be a prime number, let $L\subseteq {\mathbb{F}}_p$ have $s$ elements, and let $\mathcal{F}$ be a set family such that member sizes modulo $p$ avoid $L$ while every distinct pair has intersection size in $L$ modulo $p$. Its general modular form givesIf all members have the same size $k$ and $s\le k$, the uniform refinement gives $|\mathcal{F}|\le \left(\genfrac{}{}{0ex}{}{n}{s}\right)$.

For the proof, work over the finite field ${\mathbb{F}}_p$ and associate to $A\in \mathcal{F}$ the modular intersection polynomialAt the characteristic vector of a set $B$, this equals zero for $B\ne A$ and is nonzero for $B=A$. A linear relation among the evaluation functions therefore has every coefficient zero. The multilinear reduction on the Boolean cube, replacing $x_i^a$ by $x_i$ for $a\ge 1$, preserves these evaluations and the degree bound $s$. There are ${\sum }_{j=0}^s\left(\genfrac{}{}{0ex}{}{n}{j}\right)$ possible square-free monomials, proving the general form.

For uniform members, the low-degree evaluation rank on a uniform layer supplies the sharper dimension bound. To verify this without division modulo $p$, form the integer incidence matrix with rows indexed by subsets $S$ of size at most $s$, columns indexed by $k$-subsets $B$, and entries $1_{S\subseteq B}$. Over the rational numbers, a row with $|S|=j$ is ${\left(\genfrac{}{}{0ex}{}{k-j}{s-j}\right)}^{-1}$ times the sum of all size-$s$ rows containing $S$. Its matrix rank is at most $\left(\genfrac{}{}{0ex}{}{n}{s}\right)$. Every larger integer minor is consequently zero and remains zero modulo $p$, giving the same bound over ${\mathbb{F}}_p$. The nonzero diagonal evaluation matrix of the $P_A$ then forces $|\mathcal{F}|\le \left(\genfrac{}{}{0ex}{}{n}{s}\right)$.

The theorem, proved by Frankl and Wilson in 1981, extends the theme of the Ray-Chaudhuri–Wilson theorem from ordinary intersection sizes to residues. Modular restrictions can be weaker than exact intersection restrictions yet still force unexpectedly small families. Here are three consequences from one explicit modular intersection graph.

Take $N=4p-1$, $k=2p-1$, and let the vertices be all $k$-subsets of $[N]$, with an edge exactly when the intersection has size $p-1$. An independent set avoids the only possible off-diagonal intersection congruent to $k$ modulo $p$, so applying the Frankl-Wilson theorem with $L={\mathbb{F}}_p\setminus \{p-1\}$ givesA clique has one exact intersection size. Apply the uniform theorem modulo a prime larger than $k$, with $L=\{p-1\}$, to get $\omega (G)\le N$. Since there are $v=\left(\genfrac{}{}{0ex}{}{N}{k}\right)$ vertices, this gives the explicit asymmetric Ramsey number bound $R(N+1,B+1)>v$. It illustrates how algebraic constructions produce graphs with simultaneously restricted cliques and independent sets.

For a geometric application, map each vertex to $x_A=1_A/\sqrt{2p}$ in ${\mathbb{R}}^N$. Then $|x_A-x_B{|}^2=(k-|A\cap B|)/p$, so edges have unit Euclidean distance. Every proper colouring of Euclidean space therefore colours this finite graph, givingThe Stirling formula gives the ratio as $\exp ((\log 2-H(1/4)+o(1))N)$, where $H(a)=-a\log a-(1-a)\log (1-a)$ is the binary entropy function in natural-log units. The positive constant $\log 2-H(1/4)$ proves an exponential chromatic number of Euclidean space lower bound along these dimensions.

Finally, a quadratic embedding converts this same forbidden intersection into a forbidden diameter pair. Set $u_A=21_A-1$ and $Q_A=u_A{u}_A^T$. The quadratic sign-vector diameter formula follows from the Frobenius inner product:For distinct members the smallest possible absolute inner product is one, attained exactly when the intersection is $p-1$. Thus maximum-distance pairs are exactly edges of $G$. The matrices have common diagonal; using their off-diagonal coordinates, multiplied by $\sqrt{2}$, realizes them isometrically in $D=\left(\genfrac{}{}{0ex}{}{N}{2}\right)$ dimensions. They are distinct because equal outer products require $u_A=\pm u_B$, and complements have size $2p$, not $k$.

Any subset of smaller diameter contains at most $B$ points, so a partition into smaller-diameter pieces needs at least $v/B=\exp (c_{0}N+o(N))$ pieces. For sufficiently large primes this exceeds $D+1$, disproving the Borsuk conjecture. This is the mechanism behind the Kahn-Kalai counterexample to the Borsuk conjecture: the number of required pieces grows exponentially in the original set-system parameter while dimension grows quadratically. Their 1993 paper establishes the resulting high-dimensional counterexample.

Identify the discrete cube $Q_n$ with the subsets of $[n]$, with hypercube graph adjacency given by changing one element. Write $N[A]$ for the closed graph neighbourhood and $\partial A=N[A]\setminus A$ for the external vertex boundary. The simplicial order on the discrete cube first orders by set size and then by lexicographic order, with the smallest differing coordinate belonging to the earlier set.

Harper theorem states that, for every set family $A\subseteq Q_n$,Here is the simplicial section compression proof. Induct on $n$, the one-dimensional case being immediate. Split $A$ by a coordinate into sections $A_0,A_1\subseteq Q_{n-1}$. Its neighbourhood sections areReplace $A_0,A_1$ by equally large simplicial initial segments $I_0,I_1$. By induction $|N[I_j]|\le |N[A_j]|$. Also the neighbourhood of a simplicial initial segment is itself a simplicial initial segment: at its last occupied level the extra vertices form the upper shadow of a lexicographic initial segment, again lexicographically initial. Consequently the compressed unions have sizes $\max \{|N[I_0]|,|I_1|\}$ and $\max \{|N[I_1]|,|I_0|\}$, no larger than the original unions. The compression preserves $|A|$ and cannot increase $|N[A]|$.

Repeat in every coordinate until all sections are initial. Every nontrivial compression strictly decreases the sum of the global simplicial positions, so the process terminates. To describe the terminal families for simplicial section compression, suppose an earlier absent vertex $S$ precedes a later present vertex $T$. They cannot share a coordinate value: otherwise they lie in the same compressed section, contradicting its initiality. Thus $T=[n]\setminus S$. No vertex can lie strictly between them in the order, since it would make another absent/present pair that is not complementary. Hence the terminal family is either $C(|A|)$ or that segment with its last vertex $S$ exchanged for its immediate complementary successor $T$.

For odd $n=2m+1$, that exceptional exchange crosses the middle levels: $S=\{m+2,\dots ,2m+1\}$ and $T=\{1,\dots ,m+1\}$. For $m\ge 1$, all sets of size at most $m+1$ remain in the neighbourhood after the exchange, since an $(m+1)$-set has at least two lower neighbours and only one was removed. The new $T$ can only add vertices of size $m+2$.

For even $n=2m$, the only consecutive complementary pair in one level is $S=\{1,m+2,\dots ,2m\}$ and $T=\{2,\dots ,m+1\}$. The original segment contains all sets below size $m$ and all $m$-sets containing 1. For $m\ge 2$, its neighbourhood contains all sets of size at most $m$ and all $(m+1)$-sets containing 1; every such upper set still has an included lower neighbour after the single removal. The new $T$ can only add further neighbours. The cases $n=1,2$ follow directly, with the exceptional families related to the initial ones by a graph automorphism. Thus neither exception improves the neighbourhood, completing the induction.

For the half-cube vertex-boundary extrema, if $n=2m+1$, the initial half cube is all sets of size at most $m$; its boundary is the next level, of size $\left(\genfrac{}{}{0ex}{}{2m+1}{m+1}\right)$. If $n=2m$, it contains all levels below $m$ and the $m$-sets containing 1. The boundary consists of the $m$-sets avoiding 1 and the $(m+1)$-sets containing 1, giving $2\left(\genfrac{}{}{0ex}{}{2m-1}{m}\right)=\left(\genfrac{}{}{0ex}{}{2m}{m}\right)$.

No boundary can exceed the $2^{n-1}$ vertices outside a half cube. Taking all vertices of even Hamming weight attains that bound, because every odd-weight vertex is adjacent to one of them. Therefore, for $n\ge 1$,