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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 $s$.

Use the uniform Frankl-Wilson theorem: if $p$ is a prime number, $L\subseteq {\mathbb{F}}_p$ has $s$ distinct residues, $s\le \min \{r,n-r\}$, and an $r$-uniform set family $\mathcal{F}$ has distinct-member intersection sizes in $L$ modulo $p$ while $r\mathrm{mod}p\notin L$, thenHere is a polynomial method in combinatorics proof. For each $A\in \mathcal{F}$, form the intersection polynomialover ${\mathbb{F}}_p$, and use multilinear reduction on the Boolean cube so its degree is at most $s$. At the characteristic vector of $B\in \mathcal{F}$, this evaluates to zero for $A\ne B$ and is nonzero for $A=B$. Consequently these evaluation functions are linearly independent.

The low-degree evaluation rank on a uniform layer bounds their space by $\left(\genfrac{}{}{0ex}{}{n}{s}\right)$. To justify that step without division by a possibly zero residue, form the integer incidence matrix with rows $T\subseteq [n]$, $|T|\le s$, columns $r$-sets $B$, and entries $1_{T\subseteq B}$. Over the rational numbers, a row with $|T|=j$ equals ${\left(\genfrac{}{}{0ex}{}{r-j}{s-j}\right)}^{-1}$ times the sum of the rows of all $s$-sets containing $T$. Thus its rational matrix rank is at most $\left(\genfrac{}{}{0ex}{}{n}{s}\right)$. Every larger minor is an integer determinant equal to zero and stays zero modulo $p$, so the same upper bound holds over ${\mathbb{F}}_p$. This proves the theorem.

For the forbidden midpoint problem, split $\mathcal{A}$ according to whether its members contain coordinate 1. Leave the containing half unchanged; complement every member of the avoiding half. Both resulting families consist of $2p$-sets containing 1, and complementation preserves their internal intersection sizes becauseWithin either transformed family, distinct intersections lie between 1 and $2p-1$ and are not $p$. Their residues are therefore in $L=\{1,\dots ,p-1\}$, whereas the set-size residue is $2p\equiv 0$. Apply the Frankl-Wilson theorem to each half and add:The factor 2 comes from splitting into two families. The unsplit family can contain disjoint sets, whose intersection residue zero equals the set-size residue and prevents this direct application. Removing the shared coordinate after splitting also gives the stronger complement splitting for forbidden midpoint intersections bound $2\left(\genfrac{}{}{0ex}{}{4p-1}{p-1}\right)$.

For the construction, fix a $(p+1)$-set $B$ and let $R=[4p]\setminus B$, of size $3p-1$. The fixed-core construction avoiding midpoint intersections isTwo members of $\mathcal{F}$ intersect in at least $p+1$ points; the same is true for two complements. An intersection across the two halves has size $|X\setminus Y|\le p-1$. No intersection is $p$. The halves are disjoint because one contains all of $B$ and the other none, so

The uniform modular Frankl-Wilson theorem states the following. Let $p$ be prime, let $L\subseteq {\mathbb{F}}_p$ have size $s$, and let $\mathcal{F}$ be a rank-$k$ family on $[N]$, with $0\le s\le \min \{k,N-k\}$. Assume $k\mathrm{mod}p\notin L$ and $|A\cap B|\mathrm{mod}p\in L$ for every distinct $A,B\in \mathcal{F}$. ThenThe restrictions concern distinct pairs; the diagonal size residue is explicitly excluded from $L$.

For each member $A$, over ${\mathbb{F}}_p$ form the intersection polynomialAt the characteristic vector $1_B$, its value vanishes when $B\ne A$ and is nonzero when $B=A$. Thus these functions are linearly independent on the whole rank-$k$ layer: evaluate any linear relation at each family member in turn. Replacing every positive power of $z_i$ by $z_i$ preserves all Boolean evaluations, so the functions are restrictions of multilinear polynomials of degree at most $s$.

We need the sharp low-degree evaluation rank on a uniform layer, not the larger whole-cube dimension ${\sum }_{j\le s}\left(\genfrac{}{}{0ex}{}{N}{j}\right)$. Let $M$ be the integer incidence matrix with rows indexed by subsets $S$ of size at most $s$, columns by rank-$k$ sets $B$, and entries $1_{S\subseteq B}$. Over the rationals, if $|S|=j\le s$, thenThis coefficient counts the possible $T$ inside each containing column set and is positive because $s\le k$. Consequently the degree-$s$ rows span all rows over $\mathbb{Q}$, giving rank at most $\left(\genfrac{}{}{0ex}{}{N}{s}\right)$. Every larger square minor is therefore the integer zero. Reducing those minors modulo $p$ shows the rank over ${\mathbb{F}}_p$ is also at most $\left(\genfrac{}{}{0ex}{}{N}{s}\right)$. This transfer avoids dividing by a binomial coefficient that might vanish modulo $p$. The independent functions $f_A$ lie in this row space, so their number is at most its rank. This proves the theorem in full.

For the requested forbidden-intersection application, split $\mathcal{A}$ into sets containing coordinate one and sets avoiding it. In the first part, delete one. Distinct original sets have intersection size between one and $2p-1$, excluding $p$. Their reduced intersections therefore lie between zero and $2p-2$, excluding $p-1$, so their residues belong to $L=\{0,\dots ,p-2\}$. The reduced set size is $2p-1$, whose residue is $p-1\notin L$. Frankl-Wilson on the ground set of size $4p-1$ bounds this part by $\left(\genfrac{}{}{0ex}{}{4p-1}{p-1}\right)$.

For the avoiding part, first complement every set. The complements contain one, retain size $2p$, and satisfyDelete one and apply the same argument. The complement splitting for forbidden midpoint intersections yields the stronger boundThe split is essential: a family may contain complementary pairs with intersection zero, preventing direct use of $L=\{1,\dots ,p-1\}$ on the original layer.

The Borsuk conjecture asserted that a bounded subset of ${\mathbb{R}}^d$ of positive diameter can be partitioned into $d+1$ subsets of strictly smaller diameter. Here is the Kahn-Kalai counterexample to the Borsuk conjecture, implemented by balanced cut vectors. For each $X\in [4p{]}^{(2p)}$, define $v_X\in {\mathbb{R}}^{\left(\genfrac{}{}{0ex}{}{4p}{2}\right)}$ whose coordinate indexed by an unordered pair $\{i,j\}$ is one when exactly one endpoint lies in $X$, and zero otherwise. Complementary sets give the same cut. Choose the representative with $1\in X$, obtaining exactlydistinct points: a cut determines its two parts, and fixing the side containing one removes its only ambiguity. Every cut has $(2p{)}^2=4p^2$ occupied coordinates, so all points lie in the affine hyperplane with coordinate sum $4p^2$. Its dimension isWe can identify this hyperplane isometrically with ${\mathbb{R}}^d$; no claim of full affine span is needed.

For two side sets with $t=|X\cap Y|$, their symmetric difference has $4p-2t$ points. The two cuts disagree exactly on pairs with one endpoint in that symmetric difference, giving the balanced cut-vector distance formulaThe diameter is $2p$, attained exactly when $t=p$; such pairs exist even among representatives containing one. Thus a smaller-diameter part contains no pair with side intersection $p$. Since all representative sets already contain one, deleting that common coordinate and applying the preceding Frankl-Wilson argument bounds each part by $\left(\genfrac{}{}{0ex}{}{4p-1}{p-1}\right)$. There is no extra factor two at this step. This is the cut construction described in the original Kahn-Kalai paper; the distance and part-size calculations above give the required justification directly.

An explicit choice is $p=13$. The point set lies in dimensionThe exact counts areTheir ratio requires at least $\lceil M/B\rceil =1562$ smaller-diameter parts, exceeding $d+1=1326$. In particular,Rescaling the point set to diameter one changes none of these partition counts. Therefore the Borsuk counterexample in dimension 1325 is fully explicit. This is a concrete dimension, not a claim that it is the smallest possible counterexample dimension.

The Frankl-Wilson theorem in its uniform form says: let $p$ be a prime number, let $L\subseteq {\mathbb{F}}_p$ have size $s\le \min (k,n-k)$, and let $\mathcal{F}\subseteq \left(\genfrac{}{}{0ex}{}{[n]}{k}\right)$ have $k\mathrm{mod}p\notin L$, while $|A\cap B|\mathrm{mod}p\in L$ for every distinct pair of members. Then

The modular-intersection, nonuniform form of the Frankl-Wilson theorem is as follows. Let $p$ be a prime number, let $L\subseteq {\mathbb{F}}_p$ have $s$ elements, and let $\mathcal{F}\subseteq \mathcal{P}([n])$ satisfy $|A|\mathrm{mod}p\notin L$ for each $A\in \mathcal{F}$, whereas $|A\cap B|\mathrm{mod}p\in L$ for distinct $A,B\in \mathcal{F}$. ThenThis nonuniform Frankl-Wilson theorem allows different sizes and different diagonal residues; in particular it applies when all sizes have one common residue outside $L$.

To prove it by the polynomial method in combinatorics, work over the finite field ${\mathbb{F}}_p$ and associate to each member the intersection polynomialReplace each positive power of $x_i$ in its expansion by $x_i$ to obtain its multilinear reduction on the Boolean cube ${\tilde{P}}_A$. This preserves evaluation on every characteristic vector, whose coordinates are zero or one, and leaves degree at most $s$. At the characteristic vector $1_B$,It is zero for distinct $A,B$, and nonzero for $B=A$ by the diagonal hypothesis. If ${\sum }_{A\in \mathcal{F}}{c}_A{\tilde{P}}_A=0$, evaluating at each $1_B$ forces $c_B=0$. Thus the multilinear polynomials are linearly independent. They belong to the vector space with basis the square-free monomials ${\prod }_{i\in S}{x}_i$ for $|S|\le s$, whose dimension is the displayed sum. This proves the nonuniform form, including $s=0$, where distinct members are impossible and the empty product is one.

For the sharp uniform form, we need the low-degree evaluation rank on a uniform layer. Form the integer matrix $M$ whose rows are indexed by $S\subseteq [n]$ with $|S|\le s$, columns by $k$-subsets $T$, and entries $M_{S,T}=1_{S\subseteq T}$. These are evaluations of the square-free monomials. Over the rational numbers, the row of a $j$-subset $S$ satisfiesIndeed, when $S\subseteq T$ the left side counts the $s$-subsets between $S$ and $T$, and otherwise both sides vanish. Since $s\le k$, the coefficient is a positive integer. Thus the degree-$s$ rows span all the rows over the rational numbers, giving ${\mathrm{rank}}_{\mathbb{Q}}M\le \left(\genfrac{}{}{0ex}{}{n}{s}\right)$. Every square submatrix of larger size has integer determinant zero, so its determinant remains zero modulo $p$. HenceThis reduction of matrix rank is the reason one must not try to divide by $\left(\genfrac{}{}{0ex}{}{k-j}{s-j}\right)$ directly in ${\mathbb{F}}_p$.

The evaluations of the ${\tilde{P}}_A$ on the $k$th uniform layer of the Boolean cube lie in the row linear span of $M$ over ${\mathbb{F}}_p$. They are linearly independent as functions there, because restriction to the columns corresponding to $\mathcal{F}$ still gives the nonzero diagonal evaluation matrix. Their number is consequently at most $\left(\genfrac{}{}{0ex}{}{n}{s}\right)$. This completes the proof of the uniform Frankl-Wilson theorem as well.

For the odd-cardinality problem the sharper Oddtown theorem follows directly from linear algebra. Let $v_A=1_A\in {\mathbb{F}}_2^n$ be the characteristic vector. The parity conditions sayFor any linear combination ${\sum }_A{c}_A{v}_A=0$, take its dot product with $v_B$. All off-diagonal terms vanish, leaving $c_B=0$. Hence these vectors are linearly independent in an $n$-dimensional vector space, givingThe singleton sets show that the bound can be attained. Notice why the constant term in the general nonuniform Frankl-Wilson theorem is unnecessary here: these dot products already give a diagonal evaluation matrix using only the $n$ coordinate functions.

For even $n$, partition $[n]$ into the pairs $\{1,2\},\{3,4\},\dots ,\{n-1,n\}$ and take every union of a selection of these pairs. This gives exactly $2^{n/2}$ distinct sets, including the empty set. Each has even size; every pairwise intersection is again a union of entire pairs and therefore has even size. This attains the proposed bound.

For any set family satisfying the even-cardinality conditions, let $W$ be the linear span of its characteristic vectors over ${\mathbb{F}}_2$. Now the self-dot products vanish as well as the off-diagonal ones. By bilinearity, $u\cdot v=0$ for all $u,v\in W$, so $W\subseteq W^{\perp }$: it is a self-orthogonal binary subspace.

For completeness the standard bilinear form on ${\mathbb{F}}_2^n$ is nondegenerate, because a vector orthogonal to every coordinate vector has every coordinate zero. Thus the map $v\mapsto (w\mapsto v\cdot w)$ from ${\mathbb{F}}_2^n$ onto the dual space $W^{\ast }$ is surjective: every linear functional on $W$ extends to the whole vector space by extending a basis, and every such functional on ${\mathbb{F}}_2^n$ is a dot product with a vector. Its kernel is $W^{\perp }$. The rank-nullity theorem consequently gives $\dim W^{\perp }=n-\dim W$. The inclusion $W\subseteq W^{\perp }$ implies $\dim W\le n/2$. Finally, distinct members have distinct characteristic vectors, all in $W$, and a $k$-dimensional vector space over ${\mathbb{F}}_2$ has exactly $2^k$ elements. ThereforeThis proves the Eventown theorem and also explains the suggested dimension contradiction if the set family had more than $2^{n/2}$ members.

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.

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.

The Frankl-Wilson theorem. Let $p$ be a prime number, let $L\subseteq {\mathbb{F}}_p$ have $s$ elements, and let $\mathcal{F}\subseteq \mathcal{P}([n])$. Suppose that $|A|\mathrm{mod}p\notin L$ for every $A\in \mathcal{F}$, whereas $|A\cap B|\mathrm{mod}p\in L$ for all distinct members. Thenwith terms beyond $n$ understood as zero.

Work over the finite field ${\mathbb{F}}_p$. For each member define its intersection polynomialReplace every positive power $x_i^d$ by $x_i$ to obtain a multilinear polynomial ${\tilde{f}}_A$ of polynomial degree at most $s$. This Boolean multilinearization preserves its values at every characteristic vector of a set. At the characteristic vector of a set $B$,It is zero if $A\ne B$ and nonzero if $A=B$. Evaluating a proposed linear relation at each $1_B$ therefore establishes linear independence of the ${\tilde{f}}_A$. The vector space of multilinear polynomials of polynomial degree at most $s$ has the basis of monomials ${\prod }_{i\in T}{x}_i$, $|T|\le s$, of size ${\sum }_{j=0}^s\left(\genfrac{}{}{0ex}{}{n}{j}\right)$. This proves the Frankl-Wilson theorem by the polynomial method in combinatorics. The empty product for $s=0$ is $1$, and the argument still applies.

The constant-intersection family bound, directly. Write $\mathcal{F}=\{F_1,\dots ,F_q\}$ and assume $n\ge 1$. The cases $q\le 1$ are immediate. If $\lambda =0$, the nonempty members are pairwise disjoint and there can also be the empty set, givingFor $\lambda >0$, every member has size at least $\lambda$. If every member has size strictly greater than $\lambda$, form the real matrix $X$ whose rows are their characteristic vectors of sets. Its Gram matrix satisfiesFor any nonzero real vector $z$,Thus the rows are linearly independent and $q\le n$. If a member $F_1$ has size $\lambda$, it lies inside every other member. The sets $F_i\setminus F_1$, $i>1$, are nonempty and pairwise disjoint, so $q-1\le n-\lambda$. This also givesThese bounds are sharp: the singletons together with $\varnothing$ attain $n+1$ at $\lambda =0$, while all complements of singletons attain $n$ at $\lambda =n-2>0$ when $n\ge 3$.

The constant t-wise intersection dichotomy. In its substantive form the last argument requires $m\ge t$. The printed statement does not explicitly impose this. Without it the $t$-fold condition can be vacuous and the claimed dichotomy is false: take $t=3$, $m=2$, $n=1$, $A_1=\varnothing$, $A_2=\{1\}$ and $\lambda =1$. There is no $3$-tuple to test, no common $1$-set, and $k=0$, so $m=2>k+t-2=1$. We therefore prove the intended assertion for $m\ge t$ and record this necessary qualification.

If $\lambda =0$, the first alternative always holds with $S=\varnothing$. Suppose $\lambda >0$. Choose $t-2$ indices attaining the minimum intersection size $k$, and writeFor each of the $q=m-t+2\ge 2$ remaining indices put $B_j=K\cap A_j$. The hypothesis gives $|B_j\cap B_{\ell }|=\lambda$ for distinct remaining indices. This reduces the constant t-wise intersection dichotomy to a constant-intersection family bound on the $k$-element set $K$. The two possibilities are treated below.

Use the polynomial method in combinatorics to prove the Ray-Chaudhuri–Wilson theorem. For each $A_i$ define an intersection polynomial over $\mathbb{R}$,At the characteristic vector of a set $A_j$, the inner sum is $|A_i\cap A_j|$. ThereforeIt follows that the restrictions of $p_1,\dots ,p_m$ to the uniform layer of the Boolean cubehave linear independence: evaluating any vanishing linear combination at each ${\chi }_{A_j}$ forces its $j$th coefficient to be zero.

Apply multilinear reduction on the Boolean cube by replacing each positive power $x_a^t$ in a monomial with $x_a$. This preserves evaluations on $\{0,1{\}}^n$ and gives multilinear polynomials of polynomial degree at most $s$. Counting all monomials of degrees up to $s$ would give only ${\sum }_{j=0}^s\left(\genfrac{}{}{0ex}{}{n}{j}\right)$, which is too weak. Instead use homogenisation on a uniform layer.

Because $L$ has $s$ distinct integers in $\{0,\dots ,r-1\}$, we have $s\le r$. For $T\subseteq [n]$ with $|T|=j\le s$, write $x_T={\prod }_{a\in T}{x}_a$, with $x_{\varnothing }=1$. On ${\Omega }_r$,Indeed, at ${\chi }_A$ both sides vanish if $T⊈A$; otherwise exactly $\left(\genfrac{}{}{0ex}{}{r-j}{s-j}\right)$ summands equal $1$. The denominator is positive since $j\le s\le r$. Thus every monomial of degree at most $s$, restricted to ${\Omega }_r$, lies in the span of the $\left(\genfrac{}{}{0ex}{}{n}{s}\right)$ degree-$s$ monomials. This vector space consequently has dimension at most $\left(\genfrac{}{}{0ex}{}{n}{s}\right)$. The linear independence already proved now givesNo additional condition such as $r+s\le n$ is needed: only the spanning upper bound is used, not independence of all degree-$s$ monomials. The argument also covers $s=0$ if empty $L$ is allowed, with the empty product equal to $1$ and at most one member in the set family.