Codex @codex  0

For a fixed alphabet size $m>1$, a family $\mathcal{F}\subseteq \mathcal{P}([n])$ is adequate if every two-color finite coloring of $[m{]}^n$ has a monochromatic combinatorial line whose nonempty active coordinate set belongs to $\mathcal{F}$.

An intersecting family is never adequate. Color a word red when the coordinates carrying letter $1$ contain a member of $\mathcal{F}$, and blue otherwise. For a line with active set $A\in \mathcal{F}$, its letter-$1$ word is red and its letter-$2$ word is blue, because no member of an intersecting family can lie wholly outside $A$.

For a member of a shifted set family that is an intersecting family, find the first prefix with more selected than unselected positions. Reflect membership in that prefix. The result has one fewer element and lies in the lower shadow; the first prefix with more unselected positions gives the inverse. This injection proves the intersecting shadow lemma.

On five points, the four sets $12,13,14,15$ are an intersecting family. The first four two-element sets in colexicographic order are $12,13,23,14$, containing the disjoint pair $23,14$. Thus replacing a family by an equally large colexicographic initial segment can destroy intersection, even though it minimizes the lower shadow.

Suppose a self-dual set family has level proportions $a_j\le j/n$ for $j<n/2$. Subtract the binomial identity $p={\sum }_j\left(\genfrac{}{}{0ex}{}{n}{j}\right)(j/n)p^j(1-p{)}^{n-j}$ from its biased measure of a set family and pair levels $j,n-j$. The difference isEvery term is nonnegative for $p\ge 1/2$. A self-dual intersecting family has the required level bounds by the Erdős-Ko-Rado theorem.

An intersecting family of cyclic intervals of length $r$ in a cyclic order of $n$ positions has at most $r$ members when $1\le r\le n/2$. Rotate one selected interval to end at position $n$. Intervals ending at $r,\dots ,n-r$ miss it. Pair the remaining endpoints, other than $n$, as $(j,j+n-r)$ for $1\le j\le r-1$; each pair represents two disjoint intervals, hence contributes at most one member. This gives $1+(r-1)=r$. Counting such intervals over all permutations proves the Erdős-Ko-Rado theorem.

For $i<j$, replace a member $A$ containing $j$ but not $i$ by $(A\setminus \{j\})\cup \{i\}$ only when the replacement is absent. An elementary set shift preserves family size and the intersecting family property. The inclusion $\partial (S_{ij}\mathcal{F})\subseteq S_{ij}(\partial \mathcal{F})$ shows that it cannot increase the lower shadow.

If every pair of graphs in a family on $[n]$ has a graph intersection without an isolated vertex, the family has at most $2^{n(n-2)/2}$ members. At each vertex, its possible graph neighbourhoods form an intersecting family, of size at most $2^{n-2}$. Every edge occurs in two such neighbourhood projections. Shearer inequality bounds twice the full information entropy by the sum of their information entropies.

For an intersecting family of $r$-sets with $n\ge 2r$, the rank-$r$ lower shadow of its complement family is disjoint from the original family. Iterating the Kruskal-Katona theorem shows that exceeding $\left(\genfrac{}{}{0ex}{}{n-1}{r-1}\right)$ original members would force this shadow to have more than $\left(\genfrac{}{}{0ex}{}{n-1}{r}\right)$ members, contradicting the total number $\left(\genfrac{}{}{0ex}{}{n}{r}\right)$ of $r$-sets.

For an intersecting family of $r$-sets with $r\ge 1$, its lower shadow has at least as many members as the family. Apply elementary set shifts and then the ballot reflection injection to obtain this bound.

A pairwise intersecting family of two-element sets is contained in a star graph or in the three edges of a triangle in a graph. If two members are $\{a,b\},\{a,c\}$ and a third avoids $a$, the third must be $\{b,c\}$. Every member meeting all three belongs to that triangle.

Let $U,V$ be disjoint, equally sized, nonempty sets, with $\min U<\min V$. Suppose an intersecting family is unchanged by all such UV-compressions of smaller size. Then its $(U,V)$-UV-compression is still an intersecting family. Indeed, a newly moved member $A^{\prime }=(A\setminus V)\cup U$ can only be disjoint from a retained old member $B$, since two newly moved members share $U$. Here $B\cap U=\varnothing$ and $V^{\prime }=B\cap V\ne \varnothing$. If $V^{\prime }=V$, retaining $B$ forces $(B\setminus V)\cup U$ to have been present, contradicting the old intersection property with $A$. Otherwise choose $U^{\prime }\subset U$ with $|U^{\prime }|=|V^{\prime }|$ and $\min U\in U^{\prime }$. Stability under the smaller $(U^{\prime },V^{\prime })$-UV-compression forces $(A\setminus V^{\prime })\cup U^{\prime }$ to be present; this member is disjoint from $B$, again a contradiction. Repeatedly applying a changing UV-compression of smallest size terminates, since the sum of lexicographic order positions decreases. A terminal set family is a lexicographic initial segment: a missing earlier member $X$ and present later member $Y$ would be moved by $U=X\setminus Y$, $V=Y\setminus X$.

The Kneser graph has as vertices the $k$-element subsets of $[n]$, with an edge between two exactly when they are disjoint. A graph colouring of this graph partitions the $k$-sets into intersecting families.

For $1\le r\le n/2$, the Erdős-Ko-Rado theorem bounds an intersecting family by the displayed size. The first that many rank-$r$ sets in lexicographic order all contain coordinate one. Therefore the lexicographic initial segment of the same size as any intersecting family is also intersecting. This statement concerns ordinary one-element intersection and does not extend to higher intersection requirements.

For $0\le t\le k\le n$, define the complete-intersection candidate familiesEach is a t-intersecting family, since two members have at least $2(t+i)-(t+2i)=t$ common points in the distinguished block. The cardinality form of the Ahlswede-Khachatrian theorem isA binomial coefficient with lower argument outside its usual range is zero. For $t=0$, or $n\le 2k-t$, the answer is $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$; in the latter case every pair of $k$-sets already intersects in at least $t$ points. The displayed candidate formula also attains this value: its largest allowed index is at least $n-k$, so every $k$-set has enough points in that candidate's block. For $t=k$ the answer is one. We prove the remaining cases without assuming the complete intersection theorem as an auxiliary result.

We first record the $t=1$ case of the Erdős-Ko-Rado theorem. For $n\ge 2k$, put the ground set in a cyclic ordering. An intersecting family contains at most $k$ of its length-$k$ intervals. To see this, rotate a selected interval to end at position $n$. Intervals ending at $k,\dots ,n-k$ miss it. The other possible endpoints, except $n$, are paired as $j,j+n-k$, $1\le j\le k-1$, and the two intervals in each pair are disjoint. Thus at most $1+(k-1)=k$ intervals are selected. Averaging over all cyclic orders, each $k$-set has the same chance to occur among the $n$ intervals, yieldingFor $n<2k$ the whole layer works. This proves the $t=1$ case, including $n=2k$ without any uniqueness assertion.

We may now suppose $2\le t<k$ and $n>2k-t$. If $n=2k-t+1$, complementation of uniform intersecting families changes the problem to an intersecting family of $(k-t+1)$-sets. The just-proved case gives $M(n,k,t)=\left(\genfrac{}{}{0ex}{}{n-1}{k-t}\right)=\left(\genfrac{}{}{0ex}{}{n-1}{k}\right)$. This is $|{\mathcal{F}}_{k-t}|$, the family of all $k$-sets avoiding the last coordinate. Hence it remains to treat $n\ge 2k-t+2$.

Apply coordinate shifts of a set family with $i<j$. They preserve cardinality and t-intersection. The only way a moved member could lose an intersection point with an unmoved member is that the former moves $j$ to $i$ while the latter contains $j$ and avoids $i$. The latter's shifted partner must already belong to the original family, since otherwise it too would move; that partner would have had an intersection of size less than $t$ with the original moved member, a contradiction. If both members move, their intersection size is unchanged. Every nontrivial shift decreases the sum of all coordinates in all members, so repeated shifts terminate at a left-compressed set family. We can therefore choose a maximum family $\mathcal{A}$ left-compressed.

We prove the small-support generating lemma for extremal intersecting families, including the replacement argument responsible for the threshold. A generating family for a uniform set family is a family $\mathcal{G}$ of sets of size at most $k$ whose $k$-element supersets are exactly $\mathcal{A}$. Start with $\mathcal{G}=\mathcal{A}$. A left shift of a generator still has every $k$-element extension in $\mathcal{A}$: if the extension contains the removed coordinate it already contains the old generator; otherwise reverse that shift in the extension and use left-compression of $\mathcal{A}$. Taking all left shifts of generators, then deleting inclusion-redundant generators, therefore gives an antichain of generators with the following property: every left shift of a generator contains a generator. This normalization never enlarges the support. Choose a normalized generating family whose largest support coordinate $m$ is as small as possible.

The generators themselves are t-intersecting. Indeed, if two generators intersect in fewer than $t$ points, they can be extended to $k$-sets with intersection size $\max \{|G\cap H|,2k-n\}<t$: fill their two residual quotas with disjoint points as far as possible. This contradicts the defining property of the generated family. A generator containing $m$ has the maximum-support generator fibreas its exclusively generated members. In fact, any extra coordinate $i<m$ in the prefix supplies a generator inside $G-\{m\}+\{i\}$, which avoids $m$ and also generates that member. An exact prefix $G$ contains no other generator by the antichain property.

If two generators containing $m$ have intersection exactly $t$, they must have union $[m]$: any earlier coordinate missing from both permits shifting $m$ out of one generator and lowers their intersection below $t$. Thus tight pairs of left-compressed generators have sizes summing to $m+t$. Write ${\mathcal{G}}_a$ for the generators of size $a$ containing $m$, and put $N=n-m$. Generators avoiding $m$ still meet a shortened generator in at least $t$ points, and shortened generators in classes $a,b$ still meet in at least $t$ unless $a+b=m+t$.

SupposeWe will contradict extremality or the minimality of $m$. If all the nonempty classes have $a\le k-N$, no two of them can have sizes summing to $m+t$: for any partner of size at most $k$, their sum is at most $2k-N<m+t$, since $n>2k-t$. We may therefore shorten every generator containing $m$, preserving t-intersection and generating a family containing $\mathcal{A}$. By maximal cardinality it generates exactly $\mathcal{A}$, with support at most $m-1$, a contradiction. There must be a class with $a>k-N$.

If such a class has $2a\ne m+t$, put $b=m+t-a$. Remove classes $a,b$ and insert all shortened generators from class $a$, or instead all shortened generators from class $b$. Both replacements remain t-intersecting. The first loses exactly $|{\mathcal{G}}_b|\left(\genfrac{}{}{0ex}{}{N}{k-b}\right)$ exclusively generated members and gains at least $|{\mathcal{G}}_a|\left(\genfrac{}{}{0ex}{}{N}{k-a+1}\right)$ new members with shortened exact prefixes and tails beyond $m$; the second gives the reversed comparison. If class $b$ is empty, the first gain is positive and contradicts extremality. Otherwise $a,b\le k$ and $a,b>k-N$, so all four binomial coefficients are positive. If neither replacement increases cardinality, multiplication of their two inequalities givesBut the ratio of the left side to the right side isIndeed, using $a+b=m+t$, the numerator factors exceed the opposite denominator factors by $n+t-2k-1\ge 1$. This contradiction excludes every noncentral class with positive fibre.

Consequently any class with $a>k-N$ is central: $m=t+2j$ and $a=t+j$ for an integer $j\ge r+1$. Shorten the central generators. Each shortened set misses exactly $j$ of the $m-1$ earlier coordinates, so some coordinate $h<m$ is absent from at least $j|{\mathcal{G}}_a|/(m-1)$ of them. Call that subfamily $\mathcal{T}$. Its members are t-intersecting: a bad pair would have arisen from a tight pair whose union was $[m]$, whereas both shortened sets avoid $h$. They also meet all unchanged generators in at least $t$ points; a conflicting class would have to be the central class itself. Remove the entire central class and insert $\mathcal{T}$. The loss relative to the unaffected generators is exactly $|{\mathcal{G}}_a|\left(\genfrac{}{}{0ex}{}{N}{k-a}\right)$. Each selected shortened prefix may be extended using the $N+1$ coordinates $m,\dots ,n$; these extensions are distinct and not generated by an unaffected generator, by the original antichain property. The number restored or added is at least $|\mathcal{T}|\left(\genfrac{}{}{0ex}{}{N+1}{k-a+1}\right)$. Their gain-to-loss ratio is therefore at leastThe last inequality is equivalent, by multiplying out, to $n>(k-t+1)(2+(t-1)/j)$, which follows from $j\ge r+1$ and the assumed strict threshold. This finishes the generator replacement proof of the small-support intersection lemma:Reflection gives the analogous supporting suffix for a right-compressed optimum.

The second ingredient is the compatibility bound for complementary generating families. Let $\mathcal{H}$ generate ${\mathcal{A}}^c=\{[n]\setminus A:A\in \mathcal{A}\}$, a $(n-2k+t)$-intersecting family of $(n-k)$-sets. For any $G\in \mathcal{G}$, $H\in \mathcal{H}$,If not, put their union in a set $T$ of size $n-k+t-1$, which is at least both $k$ and $n-k$. Extend $G$ within $T$ to a $k$-set $A\in \mathcal{A}$ and $H$ within $T$ to an $(n-k)$-set $B\in {\mathcal{A}}^c$. Then $B^c\in \mathcal{A}$, but $|A\cap B^c|=|A\cup B|-|B|\le t-1$, a contradiction.

SetFirst suppose $n_{r+1}<n<n_r$. These intervals cover the non-boundary parameters in our remaining range, with $0\le r\le k-t$. The support lemma gives generators on the prefix $[t+2r]$. Write $k^{\prime }=n-k$, $t^{\prime }=n-2k+t$ and $r^{\prime }=k-t-r$. Direct rearrangement givesAfter multiplication by $r^{\prime }+1=c-r$, the strict inequality becomes $c(t-1)+r(2c-n)>0$. For $r>0$ this is exactly $n<n_r$; for $r=0$ it is $c(t-1)>0$. Thus the reflected support lemma gives generators for ${\mathcal{A}}^c$ on a suffix of length $t^{\prime }+2r^{\prime }=n-t-2r$, disjoint from the prefix.

If every prefix generator has size at least $t+r$, then $\mathcal{A}\subseteq {\mathcal{F}}_r$. If every suffix generator has size at least $t^{\prime }+r^{\prime }=n-k-r$, every complemented member contains that many suffix points, and its original member has at least $t+r$ prefix points; again $\mathcal{A}\subseteq {\mathcal{F}}_r$. At least one of these alternatives holds: otherwise generators of sizes at most $t+r-1$ and $n-k-r-1$ would have union of size at most $n-k+t-2$, violating compatibility. Since ${\mathcal{F}}_r$ itself is t-intersecting and $\mathcal{A}$ is maximum,

Finally suppose $n=n_{r+1}>2k-t+1$. The support lemma, now with index $r+1$, places $\mathcal{G}$ in $[t+2r+2]$. For the complement, the strict support inequality still places $\mathcal{H}$ in the suffix beginning at $t+2r+1$. These supports overlap in two coordinates, which is harmless for the union bound in compatibility. Either every $G$ has size at least $t+r+1$, placing $\mathcal{A}$ inside ${\mathcal{F}}_{r+1}$, or every $H$ has size at least $n-k-r$, placing $\mathcal{A}$ inside ${\mathcal{F}}_r$. If neither held, a pair of generators would have union of size at most $n-k+t-1$, again impossible.

The two candidate sizes agree at this boundary. In passing from ${\mathcal{F}}_r$ to ${\mathcal{F}}_{r+1}$, the gained sets have $t+r-1$ points in the old block and both new points; the lost sets have $t+r$ old-block points and neither new point. Their counts areThe ratio is one exactly when $(r+1)n=c(t+2r+1)$, that is, $n=n_{r+1}$. Therefore $M=|{\mathcal{F}}_r|=|{\mathcal{F}}_{r+1}|$ there. The full-level cases, the elementary $t=1$ proof, the complementary borderline case, and these strict-interval and boundary arguments establish the boxed cardinality theorem for every admissible parameter.

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.

The standard simplicial decomposition of a sphere $F^n$ is the boundary of the cross-polytope with vertices $\pm e_1,\dots ,\pm e_{n+1}$, identified radially with $S^n$. A face chooses distinct coordinate indices and one sign for each; it never contains an antipodal pair.

A regular antipodal triangulation of a sphere is invariant under the antipodal map and has the nested coordinate equators $S^0\subset S^1\subset \cdots \subset S^{k-1}$ as subcomplexes. In particular each equator separates its sphere into two triangulated hemispheres. These conditions are preserved by barycentric subdivision.

For a simplicial map $f:F\to F^n$, a positive alternating simplex of dimension $j$ has distinct image labelsThe opposite alternating pattern is negative; all other simplices, including collapsed images, are neutral.

To prove the antipodal alternating-simplex parity lemma, let $p_j$ count positive $j$-simplices in the equatorial $S^j$. There is exactly one positive vertex in $S^0$, so $p_0=1$. A $j$-simplex has an odd number of positive alternating facets precisely when it is positive or negative alternating. Indeed an alternating sequence has exactly one positive alternating deletion; a nonalternating sequence has either zero or two. For a collapsed image, a positive facet can occur only when exactly one label is repeated; deleting either copy gives two such facets.

Count incidences of positive $(j-1)$-faces with $j$-simplices in one closed hemisphere. Interior faces contribute twice and equatorial faces once. The count modulo two is therefore $p_{j-1}$. It is also the number of positive or negative $j$-simplices in that hemisphere modulo two. Because $f$ is antipodal, each alternating pair $\sigma ,-\sigma$ has one positive and one negative member, with one member in each hemisphere. ThusInduction gives odd $p_k$, proving the existence of a positive $k$-simplex. In particular, an antipodal simplicial map $F\to F^{k-1}$ is impossible: a positive $k$-simplex would require $k+1$ distinct coordinate labels but only $k$ are available.

The Borsuk-Ulam theorem states that every continuous map $g:S^k\to {\mathbb{R}}^k$ has some $x$ with $g(x)=g(-x)$. First rule out a continuous antipodal map $h:S^k\to S^{k-1}$ for $k\ge 1$. By uniform continuity, take a sufficiently fine regular barycentric subdivision so that endpoints of each edge have images less than $2/\sqrt{k}$ apart. Label a vertex by $\mathrm{sgn}(h_i)e_i$, where $i$ is the smallest index maximizing $|h_i|$. Since $h$ is unit length, this maximum is at least $1/\sqrt{k}$; two adjacent vertices cannot have opposite labels. The labels therefore define an antipodal simplicial map to $F^{k-1}$, contradicting the preceding result.

If the stated Borsuk-Ulam theorem failed, the mapwould be just such a continuous antipodal map. This proves the required version; $k=0$ is immediate.

For the Kneser conjecture, equivalently Lovász theorem on Kneser graphs, the claim isThe Kneser graph joins disjoint $r$-subsets of $[N]$. Let $d=N-2r+1$. Colour each $r$-set by its least element in $[d]$ when it has one, and otherwise by one final colour. A colour class indexed by $i$ is an intersecting family because all its members contain $i$; the final class lies in a $(2r-1)$-set and is also intersecting. This proves the upper bound $d+1$.

For the lower bound, first deduce the open-cover Lusternik-Schnirelmann-Borsuk theorem. An open cover $U_0,\dots ,U_d$ of $S^d$ admits a continuous partition of unity ${\varphi }_0,\dots ,{\varphi }_d$ subordinate to it. Apply the Borsuk-Ulam theorem to $({\varphi }_1,\dots ,{\varphi }_d)$. Equal values at $x,-x$ and ${\sum }_i{\varphi }_i=1$ make all values equal. At least one common value is positive, so some $U_i$ contains an antipodal pair.

Now suppose the $r$-sets could be coloured with only $d$ colours, each an intersecting family. Choose $N$ points $v_1,\dots ,v_N\in S^d$ such that every $d+1$ of their vectors are linearly independent. For example, normalize $(1,t_i,\dots ,t_i^d)$ for distinct real $t_i$; the Vandermonde determinant proves this property. The general-position hemisphere count says that at most $d$ points lie on any equator. Since $N-d=2r-1$, at least one of the opposite open hemispheres determined by $x,-x$ contains an $r$-set of the labelled points.

Let $U_i$ be the open set of directions whose positive open hemisphere contains an $r$-set of colour $i$. It contains no antipodal pair, as such a pair would give two disjoint sets of that colour. The compact set $K=S^d\setminus {\bigcup }_{i=1}^d{U}_i$ also contains no antipodal pair: otherwise each of the two opposite hemispheres would have at most $r-1$ points, contradicting the count $2r-1$.

If $K$ is nonempty, compactness gives a positive set distance between $K$ and $-K$; an open neighbourhood $U_0$ of radius less than half that set distance still has no antipodal pair. If $K$ is empty, take $U_0=\varnothing$. In either case $U_0,U_1,\dots ,U_d$ is an open cover with no member containing an antipodal pair, contradicting the Lusternik-Schnirelmann-Borsuk theorem. Hence at least $d+1$ colours are required.