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For nonnegative functions on a finite Boolean lattice, the condition $\alpha (A)\beta (B)\le \gamma (A\cup B)\delta (A\cap B)$ for all pairs implies $(\sum \alpha )(\sum \beta )\le (\sum \gamma )(\sum \delta )$. Coordinate elimination and the two-point four-functions inequality prove it by induction. Indicators give inequalities for the union and intersection of set families. The Boolean-lattice statement appears in Theorem 3.1 of Nicholas Ruozzi's paper.

The bipartite inclusion graph between consecutive nonempty ranks of the Boolean lattice is connected. Two rank-$r$ sets differing by one exchanged element share their rank-$(r+1)$ union as a neighbour; successive exchanges connect any two rank-$r$ sets. Every upper vertex has a lower neighbour. For the two middle ranks of an odd-dimensional Boolean lattice the graph is regular, so equality in the Local LYM inequality for a nonempty upper family forces the whole upper rank: equality admits no edges from its lower shadow to the remaining upper vertices.

Two set families $\mathcal{A},\mathcal{B}$ are cross-Sperner when no member of either is a subset of a member of the other, including equality. Thus $\mathcal{A}\cap \mathcal{B}=\varnothing$. Neither family need itself be an antichain. On the Boolean lattice of subsets of $[n]$, the Cross-Sperner inequality bounds $\sqrt{|\mathcal{A}|}+\sqrt{|\mathcal{B}|}$ by $2^{n/2}$.

For a down-set $D$ in the Boolean lattice, every member $A$ has all its $|A|$ immediate lower neighbors in $D$. Counting an internal edge at its upper endpoint gives $e(D)={\sum }_{A\in D}|A|$. The degree identity for the edge boundary of the hypercube graph then gives the displayed formula.

Let $\mathcal{F}$ be a down-set in an $N$-element Boolean lattice, and let its layer ${\mathcal{F}}_m$ be nonempty, with $1\le m\le N$. Define $x_m$ byThe real binomial coefficient is continuous and strictly increasing for $x_m\ge m$, starting at one, so this parameter exists uniquely. It measures an effective number of coordinates rather than the actual size of the support.

The Lovász shadow bound and $\partial {\mathcal{F}}_{m+1}\subseteq {\mathcal{F}}_m$ give $\left(\genfrac{}{}{0ex}{}{x_{m+1}}{m}\right)\le \left(\genfrac{}{}{0ex}{}{x_m}{m}\right)$, hence $x_{m+1}\le x_m$ whenever the later layer is nonempty. For all subsets of a fixed $q$-element set, embedded in the larger ground set, one has $x_m=q$ for every $1\le m\le q$.

The positive-rank condition is necessary: a nonempty zero layer has size one, while $\left(\genfrac{}{}{0ex}{}{x}{0}\right)=1$ for every $x$, so it determines no unique parameter. Empty layers likewise have no parameter in the displayed range.

The maximum antichains in the Boolean lattice are exactly its full middle levels. For even $n$ there is one such level, and for odd $n$ either middle level works. The LYM inequality first confines a maximum antichain to those levels; connectedness of the regular inclusion bipartite graph rules out a mixture of the two odd middle levels.

An antichain in the Boolean lattice has Lubell mass one if and only if it is a full rank level. Equality means every maximal chain in a Boolean lattice meets the family exactly once. Swapping two adjacent entries in the defining permutation then forces the family to contain every set obtained from one member by exchanging an included and an excluded element, hence every set of that rank.

For a strictly positive law $\mu$ satisfying the FKG lattice condition on a finite Boolean lattice, and an increasing function $g$, tilt to ${\mu }_t(\omega )\propto e^{tg(\omega )}\mu (\omega )$. For $t\ge 0$, the lattice inequality and $g(\omega \vee \eta )\ge g(\omega )$ give the Holley condition for ${\mu }_t{\ge }_{\mathrm{st}}\mu$. For any increasing $f$, differentiate ${\mathbb{E}}_{{\mu }_t}f\ge {\mathbb{E}}_{\mu }f$ at $t=0$ from the right. The derivative is ${\mathrm{Cov}}_{\mu }(f,g)$, proving positive association of random variables.

In extremal set theory, a set family $\mathcal{F}\subseteq \mathcal{P}([n])$ is constrained by its intersections, inclusions or traces, and one asks for the largest possible cardinality of $\mathcal{F}$. The choice of constraint is essential: without one, the whole Boolean lattice gives $2^n$ members.

For example, let $n\ge 1$ and require an intersecting family. A set and its complement of a set cannot both belong, so pairing all subsets with their complements gives $|\mathcal{F}|\le 2^{n-1}$. All subsets containing a fixed point attain equality. Adding a uniform-size requirement leads to the Erdős-Ko-Rado theorem.

An antichain instead forbids containment between distinct members. Sperner's theorem gives the sharp central-binomial bound, attained by a full middle level. Coordinate shifts of a set family simplify extremizers while preserving size; the polynomial method in combinatorics turns other intersection restrictions into linear independence. These two methods explain why a problem about subsets can become either an order argument or a vector space dimension bound.

The FKG inequality turns the FKG lattice condition into positive association of random variables. Let $\mu$ be a strictly positive probability measure on a finite distributive lattice and supposefor every $x,y$. Then for increasing real-valued functions $f,g$,Taking indicator functions gives $\mu (A\cap B)\ge \mu (A)\mu (B)$ for increasing events. The theorem applies to suitable dependent laws as well as product measures.^[ref]

For example, on the Boolean lattice $\{0,1{\}}^2$, give the configurations $00,01,10,11$ unnormalized masses $2,1,1,2$. The only nontrivial lattice check is $2\cdot 2\ge 1\cdot 1$. Each coordinate-one event has probability $1/2$, while their intersection has probability $1/3>1/4$. Thus the coordinates are dependent but positively associated. For a product measure, the FKG lattice condition holds with equality and recovers the Harris-FKG inequality.

This lattice condition on configuration masses is the hypothesis of the FKG inequality. A product measure on a finite Boolean lattice satisfies it with equality, since in each coordinate the minimum and maximum retain the original two coordinate values. It implies positive association of random variables for increasing functions.

The cross-lattice sufficient condition for stochastic comparison of two measures:The join and meet are coordinatewise maximum and minimum on a Boolean configuration cube. Normalization factors cancel, so the condition can be checked using unnormalized positive weights.

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.

Replace each positive power $x_i^a$ in a monomial by $x_i$ and combine equal monomials. The result is a multilinear polynomial agreeing with the original polynomial at every point of $\{0,1{\}}^n$, because $x_i^a=x_i$ there for $a\ge 1$. Its polynomial degree does not increase. This reduction identifies polynomial functions on the Boolean lattice with their unique square-free monomial representations: uniqueness follows by successively evaluating at characteristic vectors of sets in increasing order of size.

Under a product measure on a finite Boolean lattice, an increasing event $E$ and a decreasing event $G$ obey the displayed inequality. The complement $G^c$ is an increasing event, so Harris' inequality gives $\mathbb{P}(E\cap G^c)\ge \mathbb{P}(E)\mathbb{P}(G^c)$. Subtract from $\mathbb{P}(E)$ to obtain the result. This implication includes degenerate Bernoulli distribution parameters $0$ and $1$.

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.

Proceed by induction on the ground-set size, starting with the one-element chain $\{\varnothing \}$ in the Boolean lattice on no coordinates. Suppose a symmetric chain decomposition of a Boolean lattice on $[n]$ has been constructed. Write one of its chains asOn adjoining the new coordinate $v=n+1$, replace its two copies byand, when nonempty,The first chain has endpoint ranks $r,n-r+1$, summing to $n+1$. The second has endpoint ranks $r+1,n-r$, also summing to $n+1$. Both are saturated chains, increasing rank by one at every step. If the original chain was a singleton, the second chain is omitted.

These new chains are disjoint: the first contains every old set without $v$ and just the last old set with $v$; the second contains precisely the remaining old sets with $v$. Thus together they cover both lifted copies of the old chain. Different old chains have disjoint copies, so performing this construction for each yields a partition of all subsets of $[n+1]$ into symmetric chains. This completes the induction and proves a symmetric chain partition exists for every $n$.

The law is a ferromagnetic Ising model with unit couplings and no field. Its masses are strictly positive. Put $H(\sigma )={\sum }_{\{x,y\}\in E}{\sigma }_x{\sigma }_y$, so $\pi (\sigma )=Z^{-1}{e}^{H(\sigma )}$. The order is coordinatewise with $-1<1$, which identifies the configuration space with a Boolean lattice.

Use the permitted reduction to pairs that disagree at no more than two graph vertices. If such a pair is comparable, its join and meet are the two original configurations, so the lattice inequality is equality. In the incomparable case, let the two differing graph vertices be $x,y$, with one configuration taking $(1,-1)$ there and the other $(-1,1)$. Their join takes $(1,1)$ and their meet $(-1,-1)$.

Every edge not joining $x$ to $y$ cancels ina fixed edge has the same contribution throughout, while an edge from a changed graph vertex to a fixed graph vertex has the same sum of its two endpoint-spin products before and after. An edge between $x$ and $y$ contributes $1+1-(-1)-(-1)=4$. Thus the difference is nonnegative, whether or not the two changed graph vertices are adjacent. Exponentiation, with the normalizing constants cancelling, proves the FKG lattice condition for these pairs, and hence for all pairs by the given reduction.

Equivalently one can see the full ferromagnetic Ising lattice inequality directly. Let $U=\{x:{\sigma }_x=1,{\tau }_x=-1\}$ and $W=\{x:{\sigma }_x=-1,{\tau }_x=1\}$. Exactly the edges between $U$ and $W$ contribute, each by four. ThereforeThe preceding part now givesso the measure is positively associated. The sign of the coupling is essential to this argument: positive couplings favor agreement and give the nonnegative lattice difference.

View a set family as a set of vertices of the Boolean lattice. The operation $D_i$ acts independently on each pair $\{S,S\cup \{i\}\}$, where $i\notin S$: it keeps a full or empty pair and moves a sole upper member down. Thus downward coordinate compression preserves cardinality.

Fix a tracing set $Y$. If $i\notin Y$, compression leaves the trace of a set family on $Y$ unchanged. If $i\in Y$, write $B_0,B_1\subseteq \mathcal{P}(Y\setminus \{i\})$ for the two sections of the old trace, according to absence or presence of $i$. The absent-$i$ section of the new trace is $B_0\cup B_1$. Its present-$i$ section is contained in $B_0\cap B_1$: an upper member survives only when its lower partner was also present in the original family. ConsequentlyIn particular, downward coordinate compression cannot create new set shattering.

Repeatedly apply any changing downward coordinate compression. Each change strictly decreases ${\sum }_{A\in \mathcal{F}}|A|$, so the process terminates at a down-set $\mathcal{D}$ with the same cardinality and with no larger trace on any $Y$. The Sauer-Shelah lemma says that a family shattering no $k$-element set satisfiesIndeed, the compressed down-set also shatters no $k$-element set. If it contained any member of size at least $k$, all subsets of a $k$-subset of that member would belong to the down-set, contradicting this restriction. Every member therefore has size at most $k-1$, giving the bound. Equivalently, VC dimension at most $d$ gives size at most ${\sum }_{j=0}^d\left(\genfrac{}{}{0ex}{}{n}{j}\right)$. The family of all sets of size at most $d$ attains this bound.

There is a different extremal example when $n\ge 3$. Toggle a fixed coordinate in the family of sets of size at most two:The symmetric difference is a bijection, so this family has $1+n+\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ members. On each tracing set $Y$, it simply toggles coordinate $1$ if $1\in Y$; thus it preserves trace cardinality and set shattering. A triple has seven trace patterns, not eight. The new family differs from the original one because it contains $\{1,2,3\}$. For $n\le 2$ the specified cardinality is $2^n$, so the entire power set is the unique family of that size; the different example requires $n\ge 3$.

For the final bound, let $e_i(\mathcal{F})$ count pairs $S,S\cup \{i\}$ both in the family. Deleting coordinate $i$ identifies precisely these pairs, soSuppose every deletion lost at least two members, so $e_i(\mathcal{F})\ge 2$ for every $i$. The trace inequality and preservation of cardinality show that $e_i(\mathcal{D})\ge e_i(\mathcal{F})$ after downward coordinate compression. The down-set $\mathcal{D}$ contains the empty set and every singleton. For each $i$, its second edge in direction $i$ has an upper endpoint of size at least two. Downward closure then supplies a pair $\{i,j\}\in \mathcal{D}$. The pairs in $\mathcal{D}$ therefore cover all $n$ coordinates, so at least $\lceil n/2\rceil$ pairs are needed. This proves two cube edges in every direction force many vertices:Taking the contrapositive gives the required answer:The threshold is sharp for every $n\ge 2$. Take the empty set, all singletons and $\lceil n/2\rceil$ pairs covering every coordinate. For even $n$, use the disjoint pairs $12,34,\dots ,(n-1)n$; for odd $n$, pair the first $n-1$ coordinates and add $(n-1)n$. This down-set has exactly $\lceil 3n/2\rceil +1$ members and at least two edges in every direction, so every $(n-1)$-coordinate trace has size at most $|\mathcal{F}|-2$. Adding further members preserves those pairs and therefore preserves the failure whenever a larger cardinality is feasible. For $n=1$, the putative counterexample cardinality is three, exceeding the size of the Boolean lattice; the sharpness construction naturally starts at $n=2$.

Use the external external vertex boundary conventionEquivalently, its closed graph neighbourhood is $N[\mathcal{A}]=\mathcal{A}\cup B(\mathcal{A})$. Some boundary conventions use this closed graph neighbourhood itself; because the compared set families have equal size, either convention gives the same inequality.

Order the Boolean lattice first by increasing size, and within a size by lexicographic order, with the smallest differing coordinate in the earlier set. If $I$ is the initial segment in this simplicial order on the discrete cube with $|I|=|\mathcal{A}|$, Harper inequality isIn particular, writing $S_n(t)={\sum }_{j=0}^t\left(\genfrac{}{}{0ex}{}{n}{j}\right)$, if $|\mathcal{A}|\ge S_n(t)$ then $|N[\mathcal{A}]|\ge S_n(t+1)$.

Here is the full simplicial section compression proof. The cases $n=0,1$ are immediate. A simplicial initial segment has an initial-segment closed graph neighbourhood: it contains every smaller level, and its remaining upper level is the upper shadow of a lexicographic initial segment. That upper shadow is lexicographic initial because a set has a face in the segment exactly when its earliest face does, and earliest faces preserve lexicographic order.

Assume the result in dimension $n-1$. For a coordinate $i$, delete it from the present section and write the two sections as ${\mathcal{A}}_0,{\mathcal{A}}_1$. Their neighbourhood sections are exactlyReplace both sections by equally large simplicial initial segments $I_0,I_1$. The mathematical induction hypothesis controls each internal neighbourhood. Both sets in each new union are initial segments, so that union has size equal to the larger of their sizes. The corresponding old union has at least that size. Therefore this simplicial section compression does not enlarge $N[\mathcal{A}]$.

Repeatedly compress any noninitial section. Every change strictly reduces the sum of the global simplicial positions of the set family members, so the process terminates at a set family $\mathcal{C}$ compressed in every section. Suppose an earlier vertex $X$ is missing while a later vertex $Y$ is present. If they agree in any coordinate, they contradict compression of that coordinate's section. Thus $Y=[n]\setminus X$. They must also be consecutive in simplicial order: an intermediate vertex, whether present or absent, would create another inversion and would have to equal one of these two complementary vertices. No other inversion is possible for the same reason. Hence $\mathcal{C}$ is either the required initial segment or that segment with its last vertex $X$ exchanged for the next vertex $Y$, with $X,Y$ complementary.

For completeness, these terminal families for simplicial section compression can be checked explicitly. If $n=2k+1$, the exceptional consecutive vertices are the last $k$-set $X=\{k+2,\dots ,2k+1\}$ and first $(k+1)$-set $Y=\{1,\dots ,k+1\}$. For $k\ge 1$, the initial segment is all levels through $k$. After the exchange every vertex of size at most $k+1$ is still in its closed graph neighbourhood: a $(k+1)$-set has at least two $k$-faces, so deleting the one face $X$ cannot remove it, and $X$ itself neighbours a retained smaller set. Thus the new neighbourhood contains the old one. For $k=0$, the two single-vertex set families have the same neighbourhood.

If $n=2k$, the exceptional pair is $X=\{1,k+2,\dots ,2k\}$ and $Y=\{2,\dots ,k+1\}$. The initial segment consists of every level below $k$ and all $k$-sets containing coordinate one. Its closed graph neighbourhood consists of every level through $k$ and the $(k+1)$-sets containing coordinate one. For $k\ge 2$, each of the latter has $k\ge 2$ faces containing coordinate one, so deleting $X$ removes no such neighbour. Every lower level remains covered. For $k=1$, the added vertex $Y$ itself keeps the sole two-element neighbour covered. Again the exchange does not decrease the neighbourhood. These are the only complementary consecutive pairs, as is seen by checking the central ranks and the lexicographic transition between middle-rank sets containing and avoiding coordinate one.

We have reached a set family whose neighbourhood is at least the initial segment's, without increasing the original neighbourhood. This proves Harper inequality by mathematical induction, including the exceptional compressed set families often omitted from the argument.

For the final request, the usual two-coordinate shift does not increase the boundary of a uniform set family. To see this carefully, let $i\ne j$ and let $C_{ij}$ move a set containing $j$ but not $i$ to its partner with $j$ replaced by $i$, only if that partner is absent. For any uniform set family $\mathcal{F}$,Here is the local witness check for the lower shadow inclusion. A shadow member containing neither or both of $i,j$ has an old witness after possibly exchanging $i,j$ in that witness. For a paired shadow vertex containing $i$ only, either its old version or its $j$-partner has an old witness, which suffices for its membership in the shifted shadow. For a vertex containing $j$ only that survives in the new shadow, its new witness either contains both coordinates, supplying both old shadow partners directly, or is an unshifted $j$-only witness whose $i$-partner was already present. In either event both old shadow partners existed, which is exactly the condition for the $j$-vertex to survive the shadow shift. This exhausts the coordinate patterns.

Taking complements converts an upper shadow into a lower shadow and converts $C_{ij}$ into $C_{ji}$, proving the second inclusion. Each shift preserves cardinality on the shadow level. The external boundary of an $r$-uniform set family is the disjoint union of its lower and upper shadows, soThe endpoint levels $r=0,n$ have an empty shadow on one side and obey the same conclusion.