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The lower shadow consists of the sets obtained by removing one element from a member:For a nonempty uniform set family, write its size in its unique decreasing binomial coefficient expansionThe Kruskal-Katona theorem states thatA colexicographic initial segment attains equality; thus this is an exact minimum, not just an asymptotic estimate.

Here is a direct proof of the requested Lovász shadow bound, avoiding any unproved numerical interpolation of the Kruskal-Katona theorem. The real binomial coefficient means $\left(\genfrac{}{}{0ex}{}{x}{r}\right)=x(x-1)\cdots (x-r+1)/r!$. It is strictly increasing for $x>r-1$. Since the set family is nonempty and its size is an integer, necessarily $x\ge r$.

Apply the usual left coordinate shifts of a set family until the set family is fixed by every shift. A shift preserves its size and does not increase its lower shadow; the witness argument is supplied in Solution 3. Termination follows because every nontrivial shift reduces ${\sum }_{A\in \mathcal{A}}{\sum }_{i\in A}i$. It suffices to prove the inequality for this shifted set family.

Induct on $r$, and, for each fixed $r$, on $|\mathcal{A}|$. The cases $r=1$ and $|\mathcal{A}|=1$ give, respectively, the single empty shadow member and the $r$ faces of an $r$-set. Split the set family at coordinate one, writing ${\mathcal{A}}_0$ for the members avoiding it and ${\mathcal{A}}_1$ for the sets obtained by deleting it from members containing it. Every $(r-1)$-subset of a member of ${\mathcal{A}}_0$ belongs to ${\mathcal{A}}_1$: replace the one omitted coordinate by coordinate one. ThereforeThe two terms count shadow members avoiding and containing coordinate one.

Put $b=|{\mathcal{A}}_1|$. This is positive. If $b<\left(\genfrac{}{}{0ex}{}{x-1}{r-1}\right)$, the Pascal's identity gives $|{\mathcal{A}}_0|>\left(\genfrac{}{}{0ex}{}{x-1}{r}\right)$. In this case $x>r$ and ${\mathcal{A}}_0$ is a smaller nonempty set family. The mathematical induction hypothesis on family size gives $|\partial {\mathcal{A}}_0|>\left(\genfrac{}{}{0ex}{}{x-1}{r-1}\right)$, contradicting its containment in ${\mathcal{A}}_1$. Consequently $b\ge \left(\genfrac{}{}{0ex}{}{x-1}{r-1}\right)$. Write $b=\left(\genfrac{}{}{0ex}{}{z}{r-1}\right)$, so $z\ge x-1$. The mathematical induction hypothesis on uniformity and the Pascal's identity now giveFor $r=2$ the degree-zero term is the constant one, which causes no difficulty.

For the graph application, identify a triangle-free graph with a subset of the $N$ possible edges. Removing an edge preserves being triangle-free, so $\partial {\mathcal{G}}_{m+1}\subseteq {\mathcal{G}}_m$. If ${\mathcal{G}}_{m+1}$ is nonempty, the Lovász shadow bound givesThus, for $1\le m<N$ with $p_m>0$,If the next set family is empty its probability is zero, so the same inequality holds without introducing $x_{m+1}$. If $p_m=0$, every later set family is empty; the divided expression is undefined and the correct assertion is this zero-probability conclusion.

Iterate the monotonicity of the effective ground-set parameter of a hereditary uniform layer. Whenever $p_{m+k}>0$, all the intermediate parameters are at least their layer sizes, and all factors below are nonnegative. We obtain the stronger estimateThe second inequality follows from $x_m\le N$ and the fact that $(x_m-t)/(N-t)$ decreases with $t$. If the target set family is empty, the requested bound follows directly when its right side is nonnegative. When $k>x_m+1$, the printed product can have signed factors and should instead be stopped at the first empty layer: its unrestricted signed form is not a meaningful probability bound. In the nonzero target range there is no such ambiguity. A uniformly safe extension, for $0\le k\le N-m$, isIt agrees with the printed expression whenever all its factors are nonnegative. In particular $0\le k\le m$ is always safe when $p_m>0$, and this includes the requested doubling case.

Taking $k=m$ in the nonzero range,If the $2m$-edge set family is empty the conclusion is immediate; if $2m>N$, interpreting the event as impossible gives the same conclusion. These qualifications distinguish the probability assertion from products or ratios outside their domains.

Work over the finite field ${\mathbb{F}}_p$. For each set family member form the intersection polynomialAt the characteristic vector of a set $B$, its value is ${\prod }_i(|A\cap B|-{\ell }_i)$. The given restrictions make $f_A({\chi }_B)=0$ for $A\ne B$ and $f_A({\chi }_A)\ne 0$. Hence these functions are linearly independent: evaluate a purported relation at each set family member to eliminate its coefficient.

Use multilinear reduction on the Boolean cube, replacing every positive power of $x_i$ by $x_i$. This preserves all evaluations on $\{0,1{\}}^n$ and cannot increase degree. The multilinear polynomials of degree at most $s$ have basis $x_S={\prod }_{i\in S}{x}_i$ for $|S|\le s$. Their dimension is ${\sum }_{j=0}^{\min (s,n)}\left(\genfrac{}{}{0ex}{}{n}{j}\right)$, and thereforewhere $\left(\genfrac{}{}{0ex}{}{n}{j}\right)=0$ for $j>n$. The proof also covers a list whose distinct integers have repeated residues: no assumption of distinct residues is needed for this first bound.

To obtain the sharper conclusion, suppose the common size residue $r$ avoids $0,\dots ,s-1$. If $s>n$, no set on this ground set can have the required residue, so the set family is empty. Otherwise, add the modular-size auxiliary polynomialsagain understood as functions after multilinear reduction on the Boolean cube. They have degree at most $s$, and all vanish on the set family's characteristic vectors of sets.

The $q_S$ are linearly independent. Indeed, evaluate ${\sum }_S{c}_S{q}_S=0$ on ${\chi }_T$ with $|T|\le s-1$:The first factor is nonzero in the finite field. Starting with $T=\varnothing$ and proceeding in increasing cardinality shows $c_T=0$ for every such $T$.

Moreover, the combined list consisting of the $f_A$ and the $q_S$ is linearly independent. Evaluation at each ${\chi }_A$ first kills the coefficient of $f_A$; the preceding argument then kills all remaining coefficients. Comparing this list with the dimension of the degree-at-most-$s$ space yieldsThis is a nonuniform common-residue version of the Frankl-Wilson theorem; exact equality of all member sizes was never used.

The exclusion of the small residues is necessary. A concrete example with $s<n$ is $n=4$, $p=5$, $s=3$, the residue list $0,1,3$, and the set family of all two-element subsets. Every member has size residue $r=2$, outside that list, while distinct members intersect in zero or one element. ButHere the omitted condition fails at $j=2$. Thus the example satisfies all the other hypotheses, rather than relying on a vacuous $s>n$ case.

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.

The key is to widen the forbidden intersection while controlling a product of densities. We first derive the two ingredients, with constants explicit enough to yield the numerical base in the question.

For two nonempty set families $\mathcal{F},\mathcal{G}$ on $m$ coordinates, suppose every cross-intersection has size greater than an integer $w\ge 0$. Their complements are separated from $\mathcal{F}$ by Hamming distance greater than $w$. Let $a$ be the least integer with $|\mathcal{F}|\le S_m(a)$, where $S_m(a)={\sum }_{j=0}^a\left(\genfrac{}{}{0ex}{}{m}{j}\right)$. Iterating Harper inequality, and using $|\mathcal{F}|>S_m(a-1)$, shows that its radius-$w$ neighbourhood has at least $S_m(a+w-1)$ vertices. The complementary set family is disjoint from that neighbourhood, soFor $a=0$ the needed weaker neighbourhood bound follows simply by taking the ball around its single vertex; define $S_m(-1)=0$. If a displayed index forces a full neighbourhood, the other set family would be empty, already excluded.

The provided binary entropy function bound givesIndeed, for $t\le m/2$ this follows from $H_2(1/2-v)\le 1-2v^2/\log 2$, obtained by integrating $H_2^{\prime \prime }\le -4/\log 2$; for larger $t$ use $S_m(t)\le 2^m$. The two nonnegative deficits in the preceding product have sum at least $w$. Their squared sum is at least $w^2/2$. Thus the cross-intersection bound from cube separation is

Now suppose a pair of set families forbids every cross-intersection in $[a,b]\subseteq [0,m]$. Put $\rho ={\mu }_m(\mathcal{F}){\mu }_m(\mathcal{G})$, and fix $\delta =1/50$. Split both set families by the last coordinate. The following three replacements preserve the indicated forbidden intervals on $m-1$ coordinates:For example, in the third row the same second set has witnesses in both sections of $\mathcal{G}$, so its intersection with a first-section member must avoid both shifted intervals; their union is $[a-1,b]$.

If the first replacement increases the density product by a factor at least $1+\delta$, use it. Otherwise interchange the names of the two set families if needed so thatIf the second replacement gives a factor at least $1+\delta$, use it. If neither does, use the third. To bound its loss, putWe have $z\ge 1$, $(2-x)z<1+\delta$, and consequently $1-\delta <x\le \sqrt{1+\delta }\le 1+\delta /2$. The third replacement's product factor isSince $h^{\prime \prime }(x)=-4(1+\delta )/(2-x{)}^3<0$, its minimum on $[1-\delta ,1+\delta /2]$ is at an endpoint. ThereThus each step either gains a factor $1+\delta$ without increasing interval width, or widens the interval by one and loses at most a factor $1-\delta -2{\delta }^2$. This is the forbidden-intersection density increment, with its loss proved rather than hidden in an asymptotic error.

Start with $\mathcal{F}=\mathcal{G}=\mathcal{A}$ and $a=b=\ell$. Stop when $a=0$ or $b=m$. Dimension drops at every step. The positive lower product factors ensure that a nonempty starting pair stays nonempty; it therefore cannot reach a forbidden interval containing all $0,\dots ,m$. Write $v$ for the number of widening steps, $u$ for the number of gain steps, and putThe terminal density product ${\rho }_{\ast }$ satisfies ${\rho }_{\ast }\ge {\rho }_0{e}^{gu-hv}$. Always $v\le \ell$, since a widening step lowers $a$ by one.

If the procedure stops at $a=0$, the terminal forbidden interval is $[0,v]$. Every cross-intersection is therefore greater than $v$, and the cross-intersection bound from cube separation gives ${\rho }_{\ast }\le e^{-v^2/m}$. At least $\ell$ steps were needed to lower $a$ to zero, so $u\ge \ell -v$. Since $m\le n$,If instead the procedure stops at $b=m$, the number of steps is at least $n-\ell$, so $u\ge n-\ell -v$. Using ${\rho }_{\ast }\le 1$ and $v\le \ell$ givesThese estimates hold for every nonempty pair forbidding the original intersection, with ${\rho }_0=(|\mathcal{A}|/2^n{)}^2$ in our application.

For $\ell =n/4$, the two positive exponential decay rates are, respectively,Both exceed $c=-2\log (1.999/2)<0.001001$. Hence ${\rho }_0\le e^{-cn}$ andThe empty set family is immediate. This proves the numerical bound directly from Harper inequality and the supplied binary entropy function estimate.

For $\ell =\lfloor n/8\rfloor$, the identical forbidden-intersection density increment applies. For $n\ge 17$, use $\ell \ge n/8-7/8$ in the first stopping estimate. Its decay rate is at leastFor the second stopping estimate, $\ell \le n/8$ gives the still stronger rate $g-(2g+h)/8>0.01222$. At $n=16$ the floor is exact and the first rate is $g/8-D^2/4>0.002058>c$.

It remains to handle $1\le n\le 15$ without silently absorbing a constant into an exponential. Then $\ell$ is zero or one. Fix an $\ell$-set $T$. The $2^{n-\ell }$ sets containing $T$ are paired as $T\cup S$ and $T\cup ([n]\setminus (T\cup S))$, and each pair intersects in exactly $T$. Since $n>\ell$, these are distinct pairs, and at most one member of each belongs to $\mathcal{A}$. ThereforeFor $n=0$, the forbidden self-intersection is zero, so the set family is empty. We have proved the claimed base for all dimensions in the second case as well.

Finally, the same base is not forced by a forbidden intersection near $2n/3$. Take $n=2000$ and every subset of size at most $1000$. Every intersection has size at most $1000$, below $\lfloor 2n/3\rfloor =1333$, including self-intersections. By pairing complementary sets this set family has more than $2^{1999}$ members, whereassince $2000\log (0.9995)<-1<-\log 2$. Thus it is a large family avoiding a high intersection, and it contradicts the proposed bound.