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The LYM inequality. For an antichain $\mathcal{F}$ in the Boolean lattice,To prove this by double counting, generate a maximal chain in a Boolean lattice from a permutation $\pi$ of $[n]$: its members are the first $j$ entries of $\pi$, for $0\le j\le n$. A particular $j$-set belongs to exactly $j!(n-j)!$ of the $n!$ resulting maximal chains in a Boolean lattice. Each such maximal chain in a Boolean lattice meets the antichain at most once. Counting pairs consisting of a member of $\mathcal{F}$ and a permutation whose associated maximal chain in a Boolean lattice contains it givesDivision by $n!$ proves the LYM inequality, including the empty antichain and the levels containing $\varnothing$ or $[n]$.

A symmetric chain decomposition of a Boolean lattice. A symmetric chain has one set at every size from $k$ to $n-k$, with consecutive members related by inclusion. Start with the single symmetric chain containing $\varnothing$ when $n=0$. Given a symmetric chainin a symmetric chain decomposition of a Boolean lattice on $[n]$, let $x=n+1$ and replace its two copies in the larger Boolean lattice byand, when it is nonempty, byThe endpoint sizes of the first symmetric chain sum to $n+1$; those of the second symmetric chain also sum to $n+1$. These symmetric chains partition the original symmetric chain together with its copy containing $x$. Thus this construction inductively partitions the entire Boolean lattice into symmetric chains. Each symmetric chain contains exactly one set of size $\lfloor n/2\rfloor$, so the number of symmetric chains is

The proposed sum bound for cross-intersecting families is false. Take $n=3$ andBoth are antichains, and any member of one meets any member of the other. Nevertheless,The hypothesis permits the two antichains to coincide. Even if distinct antichains were wanted, taking $\mathcal{B}=\{\{1,2\},\{1,3\}\}$ gives $5>3$.

For a family $\mathcal{F}\subseteq \left(\genfrac{}{}{0ex}{}{[n]}{r}\right)$, let its lower shadow $\partial \mathcal{F}$ consist of all $(r-1)$-sets contained in members of $\mathcal{F}$. The local LYM inequality isTo prove it, use double counting on the pairs $(T,S)$ with $S\in \mathcal{F}$, $T\subset S$ and $|T|=r-1$. Each $S$ contributes $r$ pairs; each $T\in \partial \mathcal{F}$ contributes at most $n-r+1$. Hence $r|\mathcal{F}|\le (n-r+1)|\partial \mathcal{F}|$, exactly the displayed Local LYM inequality after using $r\left(\genfrac{}{}{0ex}{}{n}{r}\right)=(n-r+1)\left(\genfrac{}{}{0ex}{}{n}{r-1}\right)$.

To deduce the LYM inequality, start with an antichain $\mathcal{A}$ in the Boolean lattice and assign it weight ${\sum }_{S\in \mathcal{A}}{\left(\genfrac{}{}{0ex}{}{n}{|S|}\right)}^{-1}$. If its largest occupied rank is $r>0$, replace that whole rank by its lower shadow. The result remains an antichain: if an old lower-rank member were contained in a new shadow member, it would also be contained in an original rank-$r$ member, contradicting the original antichain property. Nor can a new member be contained in an old one, since every old member has size at most $r-1$ and equality would give the same contradiction. The Local LYM inequality shows that this replacement does not decrease the weight. Repeating eventually leaves an antichain on rank zero, whose weight is at most one. ThereforeFor the empty family the conclusion is immediate; the same argument includes the singleton family containing the empty set.

The antichains of maximum size are exactly the full middle levels. More precisely, for even $n$ only $\left(\genfrac{}{}{0ex}{}{[n]}{n/2}\right)$ attains the bound in Sperner theorem; for odd $n$ either $\left(\genfrac{}{}{0ex}{}{[n]}{(n-1)/2}\right)$ or $\left(\genfrac{}{}{0ex}{}{[n]}{(n+1)/2}\right)$ does. To prove that there are no others, put $M=\left(\genfrac{}{}{0ex}{}{n}{\lfloor n/2\rfloor }\right)$. If $|\mathcal{A}|=M$, the LYM inequality and $\left(\genfrac{}{}{0ex}{}{n}{r}\right)\le M$ imply equality in every term's bound $1/\left(\genfrac{}{}{0ex}{}{n}{r}\right)\ge 1/M$. Thus every member lies on a middle level. For even $n$, there is only one such level, so it must be full.

For odd $n=2r+1$, let $X$ be the rank-$r$ members and $Y$ the rank-$(r+1)$ members. Consider the bipartite graph of inclusions between those levels. Both parts have size $M$, and the degree of a vertex is $r+1$. If $N(X)$ is the neighbourhood of $X$, double counting gives $|N(X)|\ge |X|$. Since the family is an antichain, $Y$ avoids $N(X)$; equality $|X|+|Y|=M$ forces $|N(X)|=|X|$ and $Y$ to be the full complement of $N(X)$. Equality in the degree count means that every edge into $N(X)$ comes from $X$, so there are no edges between $X\cup N(X)$ and its complement.

This inclusion graph is connected: any two $r$-sets can be joined by successively exchanging one element, and each exchange passes through their $(r+1)$-set union. Every upper vertex has a lower neighbour. Thus $X$ must be empty or the whole lower level, proving the classification. For $n=1$ the graph is just one edge, and for $n=0$ the unique maximum family is $\{\varnothing \}$. This establishes equality in Sperner theorem without assuming that equality in the LYM inequality alone always forces a middle level.

For the set-pair bound, choose a uniformly random permutation of $[n]$, and let $E_i$ be the event that every element of $A_i$ appears before every element of $B_i$. Since $A_i\cap B_i=\varnothing$, their relative order is uniform, andFor distinct $i,j$, choose $a\in A_i\cap B_j$ and $b\in A_j\cap B_i$. These are distinct because $A_i\cap B_i$ is empty. On $E_i$, $a$ precedes $b$; on $E_j$, $b$ precedes $a$. Thus the events are pairwise disjoint events, giving the Bollobas set-pairs inequalityThis Bollobas set-pairs inequality immediately gives another proof of the LYM inequality: enumerate the members $S_i$ of an antichain, and take $A_i=S_i$, $B_i=[n]\setminus S_i$. For $i\ne j$, incomparability means $S_i\setminus S_j\ne \varnothing$, while $A_i\cap B_i=\varnothing$. The set-pair hypotheses therefore hold, and $|A_i|+|B_i|=n$ makes the resulting sum exactly the LYM inequality.

The printed inclusion symbol here means non-strict inclusion: the PDF explicitly excludes a common member of the two set families. We use $\subseteq$ throughout. Thus a cross-Sperner family pair need not consist of internal antichains; the restriction is between the two families.

Give the Boolean lattice the uniform probability measure, and write $N=2^n$. Form the upward closure of a set family and downward closure of a set family of $\mathcal{A}$:These are respectively an increasing event and a decreasing event. We have $\mathcal{A}\subseteq U\cap D$, whereas the cross-Sperner family condition gives $\mathcal{B}\subseteq U^c\cap D^c$. Set $u=\mathbb{P}(U)$ and $d=\mathbb{P}(D)$. By negative correlation of increasing and decreasing events, first for $U,D$ and then for $U^c,D^c$,Apply the Cauchy-Schwarz inequality to the unit vectors $(\sqrt{u},\sqrt{1-u})$ and $(\sqrt{d},\sqrt{1-d})$. It yieldsTherefore the Cross-Sperner inequality isIf the printed inclusion symbol were read as strict inclusion while the explicit exclusion of common members were discarded, this conclusion would fail, for example with both families equal to the middle uniform set family when $n=2$. The PDF's parenthetical clause fixes the intended convention.

Harris' inequality states that, for a product measure on $\{0,1{\}}^n$ with independent Bernoulli random variables of parameters $p_i\in [0,1]$, increasing events $E,F$ satisfyIn particular, the uniform probability measure on the Boolean lattice gives $|E\cap F|\ge |E||F|/2^n$. We give a direct mathematical induction proof, so no stronger correlation theorem is assumed.

Prove more generally that $\mathbb{E}[fg]\ge \mathbb{E}[f]\mathbb{E}[g]$ for real coordinatewise increasing functions $f,g$ on the finite cube. For $n=0$ both are constant and equality holds. For the inductive step let $p=p_n$, and let $f_j,g_j$ be the restrictions obtained by setting the last coordinate equal to $j\in \{0,1\}$. Write $a_j=\mathbb{E}[f_j]$, $b_j=\mathbb{E}[g_j]$, using the product measure on the other coordinates. Monotonicity gives $a_0\le a_1$, $b_0\le b_1$. By the inductive hypothesis and conditional expectation,The nonnegative correction term also handles $p=0$ and $p=1$. Taking indicator functions $f=1_E$, $g=1_F$ proves Harris' inequality.

For later use, an increasing event $E$ and a decreasing event $G$ have negative correlation of increasing and decreasing events. The complement $G^c$ is an increasing event, so

The Hall marriage theorem states that a finite bipartite graph with vertex classes $U$ and $V$ has a matching in a graph saturating $U$ if and only ifNecessity is immediate: distinct vertices of $A$ must be matched to distinct vertices of $N(A)$.

For sufficiency, induct on $|U|$. If every nonempty proper $A\subset U$ satisfies the strict bound $|N(A)|\ge |A|+1$, choose any edge $uv$ and delete its endpoints. Every $A\subseteq U\setminus \{u\}$ loses at most the neighbour $v$, so Hall's condition remains true; induction supplies a matching of the smaller graph, to which $uv$ is added.

Otherwise there is a nonempty proper tight set $A\subset U$ with $|N(A)|=|A|$. Hall's condition holds on the induced graph with parts $A,N(A)$, so induction matches $A$. For $B\subseteq U\setminus A$, Hall's condition on $A\cup B$ givesand hence $|N(B)\setminus N(A)|\ge |B|$. Induction therefore matches $U\setminus A$ into $V\setminus N(A)$. The two matchings are disjoint and together saturate $U$, completing the proof through Hall induction through a tight set.

Identify a subset of $[n]$ with its binary characteristic vector. Then $Q$ is the hypercube graph, and every edge changes set size by exactly one. The induced subgraph on $X_{i-1}\cup X_i$ is therefore bipartite with those two levels as its vertex classes.

Let $\mathcal{A}\subseteq X_{i-1}$. Every member of $\mathcal{A}$ has $n-i+1$ neighbours in $X_i$, obtained by adding one element, while every member of $N(\mathcal{A})$ is incident with at most $i$ of these edges. Double-counting the edges between the two families givesSince $i\le n/2$, we have $n-i+1\ge i$, and hence $|N(\mathcal{A})|\ge |\mathcal{A}|$. Hall's theorem yields the requested adjacent-level matching in a Boolean lattice, saturating $X_{i-1}$.

Finally, every chain in a partially ordered set in the Boolean lattice contains at most one member of the middle level $X_{n/2}$. Any chain partition therefore uses at leastchains.

For every $i\le n/2$, choose one of the matchings just constructed and orient its edges upward. These edges partition all levels at or below $n/2$ into chains ending at distinct middle-level sets. Taking complements gives, for every $i\ge n/2$, a matching from $X_{i+1}$ into $X_i$; orienting its edges upward extends those chains through all upper levels. Every subset belongs to exactly one resulting chain and every chain meets $X_{n/2}$ exactly once. Thus this symmetric chain decomposition of a Boolean lattice has one chain per middle-level set, and the minimum is

The power set $\mathcal{P}(X)$ is the set of all subsets of $X$, including the empty set and $X$ itself. It is useful whenever an object is a selection rather than an individual member: a possible collection of active vertices in a graph, for example, belongs to the power set of the vertex set.

For $X=\{a,b,c\}$, the power set has eight members:Each of the three set elements is either included or excluded. More generally, if $|X|=n<\infty$, these $n$ independent binary choices give $|\mathcal{P}(X)|=2^n$. This is a bijection between subsets and binary strings of length $n$, not a probabilistic independence assumption.

Ordering the subsets by inclusion gives a Boolean lattice. In the picture an upward edge adds one member. Complementation reverses inclusion, and union and intersection combine selections. The empty-input case is worth checking: $\mathcal{P}(\varnothing )=\{\varnothing \}$ has one member, not zero, consistently with $2^0=1$.

A set family on a finite set separates ordered pairs if each ordered pair of distinct points has a member containing the first point and omitting the second. If its members are $A_1,\dots ,A_m$, assign each point the characteristic vector of its incidences, or equivalently $T_i=\{a:i\in A_a\}$. Separation in both directions says precisely that the $T_i$ form an antichain in the Boolean lattice on $m$ coordinates. Thus the Sperner theorem gives $n\le \left(\genfrac{}{}{0ex}{}{m}{\lfloor m/2\rfloor }\right)$. Conversely, any antichain of $n$ distinct subsets of $[m]$ supplies such a set family by transposing its incidence table.

For a family in the $n$-dimensional Boolean lattice, the loss of trace cardinality on deleting coordinate $i$ equals the number of pairs $S,S\cup \{i\}$ both in the family. If every direction has at least two such edges, downward coordinate compression preserves that property. The resulting down-set contains the empty set and all $n$ singletons. Each coordinate must also occur in a two-element member. Those pairs cover all $n$ coordinates and number at least $\lceil n/2\rceil$, giving the bound. For $n\ge 2$, the empty set, all singletons and a minimum collection of pairs covering every coordinate attain it.

The uniform layer consists of the characteristic vectors of sets of size $r$ in $[n]$. It has size $\left(\genfrac{}{}{0ex}{}{n}{r}\right)$, a binomial coefficient. Restricting polynomials to a uniform layer introduces identities absent on the entire Boolean lattice, since the coordinate sum is fixed. In particular, homogenisation on a uniform layer spans all restricted multilinear polynomials of degree at most $s\le r$ using just the degree-$s$ monomials.

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.

For odd $n$ and $w\ge 1$, an intersecting family contains at most one set from each complementary pair in the Boolean lattice. Choosing the larger member in every pair gives the displayed bound and is feasible, because all sets of size greater than $n/2$ intersect. For $w>1$ it is the unique maximizer; for $w=1$ other maximizing families can exist.