Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 10 (Combinatorics)

Write $[n{]}^{(r)}$ for the rank-$r$ uniform layer of the Boolean cube. For a uniform set family $\mathcal{F}\subseteq [n{]}^{(r)}$, with $1\le r\le n$, let $\partial \mathcal{F}$ be its lower shadow. The Local LYM inequality isFor its proof, count incidences $(B,A)$ with $A\in \mathcal{F}$, $B\subset A$, and $|B|=r-1$. Each member of $\mathcal{F}$ contributes $r$ incidences, while each member of the lower shadow contributes at most $n-r+1$. Thus $r|\mathcal{F}|\le (n-r+1)|\partial \mathcal{F}|$, which is the displayed bound by the ratio of consecutive binomial coefficients. Complementing every set gives the corresponding upper shadow bound, $|\nabla \mathcal{F}|/\left(\genfrac{}{}{0ex}{}{n}{r+1}\right)\ge |\mathcal{F}|/(\genfrac{}{}{0ex}{}{n}{r})$ for $r<n$.

For an antichain $\mathcal{A}$ in the Boolean lattice, the LYM inequality isHere is the proof using the Local LYM inequality. Denote the rank-$r$ part by ${\mathcal{A}}_r$, and let ${\mathcal{D}}_r$ consist of the $r$-sets containing some member of $\mathcal{A}$ of size at most $r$. These families satisfyThe union is disjoint because a member of the antichain cannot contain a strictly smaller member. Apply the upper shadow form of the Local LYM inequality to ${\mathcal{D}}_{r-1}$:Iterating gives a bound for the entire sum by $|{\mathcal{D}}_n|/(\genfrac{}{}{0ex}{}{n}{n})\le 1$. This also covers the empty antichain and an antichain containing the empty set.

For the second proof, each permutation of $[n]$ specifies a maximal chain in a Boolean lattice by taking its successive prefixes. There are $n!$ such maximal chains in a Boolean lattice, and a fixed $r$-set belongs to $r!(n-r)!$ of them. A maximal chain in a Boolean lattice meets an antichain at most once. Double counting these incidences therefore gives ${\sum }_r|{\mathcal{A}}_r|r!(n-r)!\le n!$, exactly the LYM inequality.

The Sperner theorem says that an antichain in $\mathcal{P}([n])$ has at most $\left(\genfrac{}{}{0ex}{}{n}{\lfloor n/2\rfloor }\right)$ members; a full middle rank attains this bound. Indeed, every binomial coefficient $\left(\genfrac{}{}{0ex}{}{n}{r}\right)$ is at most the central one, so the LYM inequality givesNow let $\mathcal{F}$ be an intersection-free uniform set family. If it is empty, there is nothing to prove. Fix $X\in \mathcal{F}$ and consider the tracesIf $A,B\ne X$ are distinct and $X\cap A\subseteq X\cap B$, then $X\cap A\subseteq B$. This contradicts the defining restriction on the three distinct members $X,A,B$. The intersection has size less than $r$, so it is a proper subset of $B$ even if the printed subset symbol is interpreted strictly. In particular, equal traces are impossible, and the traces form an antichain. Thus $|\mathcal{T}|=|\mathcal{F}|-1$. Applying the Sperner theorem to the $r$-element ground set $X$ proves the antichain trace bound for intersection-free familiesThe argument also handles $r=0$: a uniform set family then has at most one member.

The Kruskal-Katona theorem states that a colexicographic initial segment minimizes the lower shadow of a uniform set family of prescribed size. Its numerical form is as follows. For $m>0$, write the unique combinatorial number system expansionEvery $\mathcal{F}\subseteq [n{]}^{(r)}$ of size $m$ satisfiesThe empty family has empty lower shadow. Equality is attained by the first $m$ $r$-sets in colexicographic order, where the largest differing element belongs to the later set. Iterating the Kruskal-Katona theorem shows that the same colexicographic initial segment minimizes every iterated lower shadow.

For $1\le r<n/2$, the Erdős-Ko-Rado theorem givesfor an intersecting family $\mathcal{F}\subseteq [n{]}^{(r)}$. All $r$-sets containing one prescribed point show that the bound is sharp.

For the proof from the Kruskal-Katona theorem, put $k=n-r$ and form the complement family $\mathcal{G}=\{[n]\setminus A:A\in \mathcal{F}\}\subseteq [n{]}^{(k)}$. Let $\mathcal{S}$ be its rank-$r$ iterated lower shadow. No member $B\in \mathcal{F}$ belongs to $\mathcal{S}$: containment in $[n]\setminus A$ would give $A\cap B=\varnothing$, impossible for an intersecting family of nonempty sets. HenceSuppose $|\mathcal{F}|>\left(\genfrac{}{}{0ex}{}{n-1}{r-1}\right)=\left(\genfrac{}{}{0ex}{}{n-1}{k}\right)$. The first $\left(\genfrac{}{}{0ex}{}{n-1}{k}\right)$ members of the colexicographic order are all $k$-sets of $[n-1]$. Their rank-$r$ iterated lower shadow is all $r$-sets of $[n-1]$, since $r<k$. The next $k$-set contains $n$ and has an $r$-subset containing $n$, so taking even one more member strictly enlarges that iterated lower shadow. By the iterated Kruskal-Katona theorem, $|\mathcal{S}|>\left(\genfrac{}{}{0ex}{}{n-1}{r}\right)$. Together with $|\mathcal{F}|>\left(\genfrac{}{}{0ex}{}{n-1}{r-1}\right)$ this contradicts Pascal's identity and the preceding inequality. This proves the Erdős-Ko-Rado theorem.

For the Katona circle method proof, fix a cyclic ordering of $[n]$. A cyclic interval of length $r$ is specified by its final position. If no such interval belongs to $\mathcal{F}$, the bound below is automatic. Otherwise rotate one selected interval so that it ends at position $n$, and thus occupies positions $n-r+1,\dots ,n$. Intervals ending at positions $r,\dots ,n-r$ are disjoint from it and cannot be selected. Among the remaining endpoints, pair $j$ with $n-r+j$ for $1\le j\le r-1$. The corresponding intervals are disjoint when $n\ge 2r$, so at most one interval per pair is selected. Together with the interval ending at $n$, this gives the cyclic interval intersection bound of $r$ selected intervals.

Choose a uniformly random permutation and read its positions cyclically. A fixed $r$-set is a cyclic interval with probability $n/\left(\genfrac{}{}{0ex}{}{n}{r}\right)$: each cyclic position gives a uniformly distributed $r$-set, and for $r<n$ the $n$ intervals are distinct. Summing over $\mathcal{F}$, the expected value of the number of selected intervals is $n|\mathcal{F}|/(\genfrac{}{}{0ex}{}{n}{r})$. The cyclic interval intersection bound makes this at most $r$, giving $|\mathcal{F}|\le (r/n)\left(\genfrac{}{}{0ex}{}{n}{r}\right)=\left(\genfrac{}{}{0ex}{}{n-1}{r-1}\right)$.

Finally apply the Katona circle method to an arbitrary antichain $\mathcal{A}$. If it contains $\varnothing$ or $[n]$, it has exactly one member and the LYM inequality holds with equality. Otherwise all its members have sizes $1,\dots ,n-1$. In a fixed cyclic ordering, the cyclic intervals sharing a final position form a nested chain as their lengths increase. At most one of them can belong to the antichain. Summing over the $n$ final positions proves the cyclic interval antichain bound of $n$ intervals. Taking the expected value over a uniformly random permutation givesAfter division by $n$, this is the LYM inequality proved by cyclic intervals. The case $n=0$ consists only of the empty set and is immediate.

Identify the hypercube graph $Q_n$ with $\mathcal{P}([n])$, joining two sets when their symmetric difference has size one. Write $N[A]$ for the closed graph neighbourhood of $A$, and $N_t[A]$ for its radius-$t$ neighbourhood in Hamming distance. In the simplicial order on the discrete cube, smaller sets come first, with lexicographic order within each rank: the smallest differing coordinate belongs to the earlier set. Let $I_m$ be the first $m$ vertices in this order. Harper theorem isEquivalently, $I_m$ minimizes the external vertex boundary at fixed size. We will also obtain $|N_t[A]|\ge |N_t[I_{|A|}]|$ for every integer $t\ge 0$.

Here is a complete proof by simplicial section compression. First note two properties of the simplicial order on the discrete cube. Its restriction to either face obtained by fixing one coordinate is the corresponding order on the smaller hypercube graph. Also, the closed graph neighbourhood of an initial segment is an initial segment. To verify the latter, a nonempty initial segment contains every level below its last level $r$, and a lexicographic prefix $\mathcal{L}$ at level $r$. For $r\ge 1$, its closed graph neighbourhood contains every set of size at most $r$, and its only further members are the upper shadow of $\mathcal{L}$. An $(r+1)$-set lies in this upper shadow exactly when its lexicographically earliest $r$-subset lies in $\mathcal{L}$. That earliest subset is obtained by deleting the largest element. These earliest subsets vary monotonically in lexicographic order, so the upper shadow is another lexicographic prefix. For $r=0$, the closed graph neighbourhood of $\{\varnothing \}$ is the union of levels zero and one. The empty initial segment causes no exception.

Induct on $n$, with $n=0,1$ immediate. Fix a coordinate $i$ and write the sections of $A$ as $A_0,A_1\subseteq Q_{n-1}$, deleting $i$ from the latter section. Replace them by the initial segments $I_a,I_b$ with $a=|A_0|$, $b=|A_1|$. The sections of the old closed graph neighbourhood areBy induction their sizes are at least $\max \{|N[I_a]|,b\}$ and $\max \{|N[I_b]|,a\}$, respectively. These are exactly the sizes of the new sections, because both sets in each new union are initial segments and hence nested. Thus simplicial section compression preserves size and never enlarges the closed graph neighbourhood.

Repeatedly compress any section that is not already an initial segment. Every change strictly reduces the sum of the global simplicial positions of the included vertices. This nonnegative integer decreases only finitely often, so the process ends at a family $B$ whose sections in every coordinate are initial segments. If $B$ itself is an initial segment, the induction is complete. Otherwise choose the earliest absent vertex $X$ and the latest present vertex $Y$, with $X\prec Y$. They cannot agree in any coordinate: agreement would put them in one compressed section, where an earlier absent vertex cannot precede a later present one. Hence $Y=X^c$. Nor can a vertex $Z$ lie between them: if $Z$ is present, pairing it with $X$ would force $Z=X^c=Y$; if absent, pairing it with $Y$ would force $Z=Y^c=X$. Thus $X,Y$ are consecutive, and $B$ is obtained from $I_{|B|}$ by replacing its last vertex $X$ with the next vertex $Y$.

There are only two possibilities for these terminal families for simplicial section compression. If $|X|<|Y|$, consecutiveness makes $X$ the last $k$-set and $Y$ the first $(k+1)$-set. Complementarity forces $n=2k+1$, $X=\{k+2,\dots ,2k+1\}$ and $Y=\{1,\dots ,k+1\}$. For $n\ge 3$, the corresponding $B$ contains all sets of size at most $k$ except $X$, and additionally $Y$. Every $(k+1)$-set has at least two $k$-subsets, so at least one belongs to $B$. The missing $X$ has a $(k-1)$-subset in $B$. Consequently $N[B]$ contains every set of size at most $k+1$, which is exactly $N[I_{|B|}]$.

If $|X|=|Y|$, then $n=2k$. Since $X,Y$ are complementary, the smaller one contains coordinate $1$. Consecutiveness forces $X$ to be the last $k$-set containing $1$, and $Y$ the first $k$-set avoiding $1$:The initial segment consists of all sets of size less than $k$ and all $k$-sets containing $1$. Its closed graph neighbourhood contains all sets of size at most $k$ and all $(k+1)$-sets containing $1$. For $k\ge 2$, each such $(k+1)$-set has $k\ge 2$ $k$-subsets containing $1$, so removing only $X$ leaves one in $B$. All sets of size at most $k$ are still in $N[B]$, because all levels below $k$ belong to $B$. Hence again $N[I_{|B|}]\subseteq N[B]$. For $k=1$, the two families are related by exchanging coordinates and both have the whole $Q_2$ as their closed graph neighbourhood. These comparisons finish the induction and prove Harper theorem.

For the radius-$t$ version, repeatedly use the one-step result. At each step the comparison family remains an initial segment, and its closed graph neighbourhood is monotone in its size. If $|N_{t-1}[A]|\ge |N_{t-1}[I_m]|$, applying Harper theorem to $N_{t-1}[A]$ therefore gives $|N_t[A]|\ge |N_t[I_m]|$. Induction on $t$ proves the asserted extension.

A Lévy family of graphs has vanishing concentration of measure functions after distances have been scaled by the graph diameter. More explicitly, for connected finite graphs $G_n$ of diameter $D_n$, use normalized graph distance $d_n=d_{G_n}/D_n$ and uniform probability measure. For every fixed $\epsilon >0$, requireFor $Q_n$ this normalization is normalized Hamming distance $d_H/n$, since its diameter is $n$.

If $|A|\ge 2^{n-1}$, the corresponding initial segment contains the Hamming ball about $\varnothing$ of radius $b_n=\lfloor (n-1)/2\rfloor$. The radius-$t$ version of Harper theorem impliesUse the precise Hoeffding inequality $\mathbb{P}(Z_n-n/2\ge u)\le \exp (-2u^2/n)$ for $u\ge 0$. One can also derive this estimate directly: the moment-generating function of $Z_n-n/2$ is $\cosh (\lambda /2{)}^n\le \exp (n{\lambda }^2/8)$; the exponential Markov inequality followed by minimization at $\lambda =4u/n$ gives the bound. With $t=\lfloor \epsilon n\rfloor$, we have $b_n+t\ge n/2+\epsilon n-2$. Thus, whenever $\epsilon n>2$,This proves that the hypercube graphs form a Lévy family of graphs. The stronger fixed-positive-mass formulation follows as well. For any $a>0$, choose $c$ with $1/(4c^2)<a$. The Chebyshev inequality for $Z_n$, whose variance is $n/4$, makes the Hamming ball of radius $\lfloor n/2-c\sqrt{n}\rfloor$ have fewer than $a{2}^n$ vertices. An initial segment of size at least $a{2}^n$ contains that ball. Enlarging by $\lfloor \epsilon n\rfloor$ and using the same Hoeffding inequality gives a complement proportion tending to zero uniformly over all such $A$. The normalization matters: with unscaled unit Hamming distance, cubes do not satisfy the metric version at every fixed radius; for a radius below one, a half-cube has no enlargement.

Equality in Harper theorem does not determine a down-set up to isomorphism. In $Q_4$, takeThis is a down-set of size nine. Its closed graph neighbourhood contains every set of size at most two, because all singletons belong to $A$. Every triple contains an edge of the four-cycle among the listed pairs, so every triple also belongs to $N[A]$. The four-element set does not. Hence $|N[A]|=1+4+6+4=15$.

The size-nine simplicial order on the discrete cube initial segment is all sets of size at most one together with the pairs $12,13,14,23$. Every triple contains one of these pairs, and the four-element set is again absent from its closed graph neighbourhood. Its neighbourhood therefore also has size $15$, proving extremality of $A$. Nevertheless the induced graphs are not isomorphic: in $A$, the empty set has degree four, the four singletons have degree three, and the four pairs have degree two. The initial segment has two vertices of degree four, namely $\varnothing$ and $\{1\}$. Since a graph automorphism preserves induced degree of a vertex, no automorphism of the cube can carry one family to the other. This gives nonunique down-set extremizers for Harper theorem even under the stronger test of induced graph isomorphism.

The general modular form, the nonuniform Frankl-Wilson theorem, can be stated as follows. Let $p$ be a prime number, and let $L\subseteq {\mathbb{F}}_p$ have $s$ elements. Suppose $\mathcal{F}\subseteq \mathcal{P}([n])$ has $|A|\mathrm{mod}p\notin L$ for every member, while $|A\cap B|\mathrm{mod}p\in L$ for any distinct members. ThenThe uniform Frankl-Wilson theorem sharpens this to $|\mathcal{F}|\le \left(\genfrac{}{}{0ex}{}{n}{s}\right)$ when all members have size $k$ and $0\le s\le \min \{k,n-k\}$. Both forms require a prime modulus and exclusion of the self-intersection residue.

For the general proof, work over the finite field ${\mathbb{F}}_p$. Associate to each member $A$ the intersection polynomialAt the characteristic vectors of sets $1_B$, this is zero if $B\ne A$, and nonzero if $B=A$. Thus these polynomials, regarded as functions on the Boolean hypercube, are linearly independent: evaluating any relation at $1_A$ isolates its coefficient. Apply multilinear reduction on the Boolean cube, replacing every positive power of a variable by that variable. Values on the Boolean hypercube remain unchanged, and the resulting multilinear polynomials have degree at most $s$. Their ambient space has a monomial basis consisting of $x_S={\prod }_{i\in S}{x}_i$ for $|S|\le s$, with dimension ${\sum }_{j=0}^{\min (s,n)}\left(\genfrac{}{}{0ex}{}{n}{j}\right)$. This proves the general bound, including $s=0$.

For completeness, obtain the uniform sharpening without dividing by factorials in a finite field. For $s\ge 1$ put $a=k\mathrm{mod}p$ and $g(x)={\sum }_i{x}_i-a$. The functions $x_{S}g$ with $|S|<s$ are linearly independent. Indeed, a relation gives $gh=0$ with $\mathrm{deg}h<s$, so $h$ is supported only on weights in $H=\{j\in [0,n]:j\equiv a(\mathrm{mod}p)\}$. With boundary weights $-1,n+1$, this set has a gap of at least $s+1$: either consecutive allowed weights differ by $p\ge s+1$, or its terminal gap does, since $n\ge k+s\ge a+s$. The alternating sum of $h$ over any interval of $s$ free coordinates vanishes by its degree. Across that gap only one endpoint level can contribute, forcing $h$ to vanish there. Delete that level and repeat across the enlarged gap until $H$ is empty. Hence $h=0$, and independence of the square-free monomials gives the claim. Adjoin these functions to the $f_A$. Evaluation at each family vector eliminates the coefficients of $f_A$, since $g$ vanishes there; the claim eliminates all remaining coefficients. Counting dimensions yieldsThe case $s=0$ permits at most one member directly. The gap argument is an instance of the modular layer vanishing lemma.

For the final application, enumerate the distinct sets as $A_1,\dots ,A_m$ and let $v_i=1_{A_i}\in {\mathbb{R}}^n$ be their characteristic vectors of sets. If $m\le 1$, the desired bound is immediate for $n\ge 1$. Otherwise $|A_i|\ge k$, because $A_i$ intersects another member in $k$ points. At most one member can have size $k$: two distinct $k$-sets cannot have intersection size $k$. The Gram matrix of these vectors isFor real coefficients $c_i$,All terms are nonnegative, with $k>0$. If the expression vanishes, every coefficient corresponding to a set of size greater than $k$ is zero. At most one coefficient remains, and the first term forces it to be zero as well. Thus the characteristic vectors of sets are linearly independent in ${\mathbb{R}}^n$, proving the constant-intersection family boundThis proof explicitly covers the possible member of size exactly $k$; assuming all diagonal corrections were strictly positive would miss that case. Positivity of $k$ is essential: when $k=0$, the empty set and all $n$ singleton sets give $n+1$ members.