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An augmenting path in a matching $M$ is a nonempty alternating path in a matching with distinct endpoints that are both unmatched vertices in a matching. Its first and last edges lie outside $M$. Thus it has $k+1$ unselected edges and $k$ selected edges, for some $k\ge 0$. Taking the symmetric differenceremoves the selected edges on $P$ and inserts its unselected edges. Each internal vertex exchanges one selected edge for another; each endpoint gains its first selected edge. All choices away from $P$ stay in place, so $M^{\prime }$ is a matching in a graph with $|M^{\prime }|=|M|+1$.^[ref]

For an explicit repair, let the path graph $P_6$ have vertices $1,2,3,4,5,6$ in that order and start withThe endpoints $1$ and $6$ are initially unmatched vertices in a matching. Along $P$ the edges are outside, inside, outside, inside, outside $M$. Remove $(2,3)$ and $(4,5)$, then add $(1,2),(3,4),(5,6)$ together:This raises the cardinality from two to three and gives a perfect matching. Six vertices allow at most three disjoint pairs, so it is also a maximum cardinality matching. The initial $M$ is already a maximal matching: every unused edge touches a matched vertex. Augmentation succeeds because it can replace earlier choices instead of only adding to them.

One unused edge joining two unmatched vertices in a matching is the shortest possible augmenting path in a matching. Longer augmenting paths in a matching reassign several partners, explaining why a stalled greedy matching need not be optimal. For edge benefits $w_e$, the change in total benefit iswhich need not be positive even though the cardinality increases. A maximum-weight matching therefore needs a different optimality test. In an assignment equality graph an augmenting path in a matching moves toward a perfect matching while preserving zero reduced costs on matched edges; this is the setting used by the Hungarian algorithm.

Represent a balanced partition of $[4p]$ by the zero-one vector recording which unordered pairs lie across its cut. Complementary subsets represent the same cut; choose the representative containing coordinate one. Two such cuts differ on the pairs crossing the symmetric difference of their sides, giving the displayed distance. Its maximum is $4p^2$, attained exactly when $t=p$. Every cut has $4p^2$ occupied coordinates, so the vectors lie in an affine hyperplane of dimension at most $\left(\genfrac{}{}{0ex}{}{4p}{2}\right)-1$.

Events turn Boolean operations into operations on sets. For a die with sample space $U=\{1,2,3,4,5,6\}$, let $A=\{2,4,6\}$ be the event “even outcome” and $B=\{4,5,6\}$ the event “outcome at least four”. At outcome $u$, use the indicator functions $P=1_A(u)$ and $Q=1_B(u)$. ThenThus logical conjunction becomes set intersection, logical disjunction becomes set union, exclusive or becomes symmetric difference, and logical negation becomes the complement of a set within $U$. Outcome four satisfies both events, so inclusive logical disjunction includes it and exclusive or excludes it. If the die is fair, these four events have probabilities $1/3$, $2/3$, $1/3$, and $1/2$, respectively. No assumption of independent events is needed for the set identities; indeed $A$ and $B$ here are not independent, since $1/3\ne (1/2)(1/2)$.

Factor subsets multiply by symmetric difference; their sign columns multiply pointwise. The subgroup generated by independent defining words has $2^k$ elements. On a selected fractional factorial design, every word in this subgroup is constant, with sign determined by the chosen generator signs. The cosets classify identical or opposite contrast columns.

Two functions $f:A\to X$ and $g:B\to X$, whose domains lie in one ambient set, are equivalent modulo finite changes when $A△B$ is a finite set and the functions disagree at only finitely many points of $A\cap B$. The symmetric difference inclusion $A△C\subseteq (A△B)\cup (B△C)$ handles domains in a transitive chain. An exceptional set for values must also include points of $A\cap C$ absent from the middle domain $B$; these lie in the finite difference $A\setminus B$.

In a two-level factorial design, a factor subset defines a sign column. The coefficient of this column is its response contrast divided by the number of runs when the columns are orthogonal. A singleton describes a main effect and a pair describes a two-factor interaction. On the full cube, distinct columns have zero inner product: summing over any coordinate in their symmetric difference cancels the terms.

Choose one representative $R$ for each equivalence class of infinite sets under finite symmetric difference. The colour $|A△R|\mathrm{mod}2$ changes on removing one point. Every cone $[M{]}^{\omega }$ therefore contains opposite-coloured sets $M$ and $M\setminus \{\min M\}$. Either colour family is a non-Ramsey set of infinite subsets. Choice of representatives is essential to this construction, which does not claim a definable regular family.

A partition relation whose colored subsets are infinite. For countably infinite arity, choosing representatives modulo finite symmetric difference and coloring its parity shows $\kappa \nrightarrow (\omega {)}_2^{\omega }$ for every infinite cardinal $\kappa$.

Identify a subset $A\subseteq \Omega$ with its indicator vector $a=(a_1,\dots ,a_n)\in {\mathbb{F}}_2^n$. Symmetric difference becomes coordinatewise addition modulo two, so the subsets form a vector space of dimension $n$, with the singleton subsets as a basis. The proposed form becomes the dot productDistributivity proves bilinearity, and commutativity proves symmetry. Its Gram matrix in the singleton basis is $I_n$, so it is nondegenerate. A permutation merely permutes coordinates and preserves the dot product, proving invariance under $S_n$.

Write $u=(1,\dots ,1)$ for the full subset. Its bilinear orthogonal complement $E=u^{\perp }$ consists exactly of even-cardinality subsets. If $n\ge 2$ is even, $B(u,u)=n=0$ in ${\mathbb{F}}_2$, so $⟨u⟩\le E$. For any $v\in E$, $B(v,v)={\sum }_i{v}_i^2={\sum }_i{v}_i=0$, so the restricted form is alternating.

Nondegeneracy of the full form gives $E^{\perp }=⟨u⟩$; one can also see this directly by testing against all vectors $e_i+e_j$, which forces all coordinates of an element of $E^{\perp }$ to be equal. Hence the radical of a bilinear form of the restricted form isDefine the quotient form by $\overline{B}(v+⟨u⟩,w+⟨u⟩)=B(v,w)$. Adding $u$ to either representative does not change the pairing with $E$, so it is well-defined. It is alternating, and a class pairing to zero with all classes has a representative in $\mathrm{rad}(B{|}_E)=⟨u⟩$, hence is zero. ThereforeBoth spaces are $S_n$-invariant, so the coordinate action descends and preserves the quotient form. This is the symplectic quotient of the binary subset module. The evenness assumption is essential because otherwise $u$ does not lie in its own bilinear orthogonal complement.

Code each two-level factor as a sign $x_j\in \{-1,1\}$. A factorial contrast indexed by a factor subset $A$ has sign column ${\chi }_A(x)={\prod }_{j\in A}{x}_j$, with the empty subset giving the intercept. Construct a regular fractional factorial design by choosing $k$ independent words $G_1,\dots ,G_k$ and retaining the settings satisfying ${\chi }_{G_i}(x)=s_i$, where each $s_i\in \{-1,1\}$. In binary exponent notation these are $k$ independent linear equations over ${\mathbb{F}}_2$, leaving $m-k$ freely chosen coordinates and exactly $2^{m-k}$ settings. Dependence of the chosen words would fail to give the stated fraction size.

The words generated by multiplying the $G_i$ form the defining contrast subgroup $\mathcal{H}$, of order $2^k$. Multiplication cancels a repeated factor, so word multiplication is symmetric difference of subsets. On the fraction, ${\chi }_H=s_H$ is constant for every $H\in \mathcal{H}$, and henceTo see that these are all the aliases, average ${\chi }_A{\chi }_B={\chi }_{AB}$ over the fraction. If $AB\notin \mathcal{H}$, this is a nonconstant character of its freely varying binary coordinates; pairing settings that reverse that character makes the average zero. If $AB\in \mathcal{H}$, the product is constant. Thus the aliasing in a fractional factorial design classes are exactly the cosets of $\mathcal{H}$. For a half fraction, $\mathcal{H}=\{I,G\}$ with $G\ne I$, and every contrast column is aliased with exactly the other word $AG$, with a plus or minus sign according to the chosen fraction. This includes the defining word paired with the intercept.

For the six-factor experiment, take the 32-run half fractionIts resolution of a fractional factorial design is VI. Every main effect is aliased with a five-factor interaction, and every two-factor interaction with a four-factor interaction. In particular, no two required low-order effects are aliased with one another. Under the stated negligible-higher-interaction assumption, all six main effects and the fourteen required two-factor interactions are estimable before introducing blocks.

To form four equal days, classify runs by the two signsTheir product is $x_B{x}_C{x}_F$. The three nonconstant block contrasts therefore coincide with the treatment-column alias classesOnly $AF$ among the main effects and two-factor interactions is lost to block confounding in a factorial design; the other two block directions involve only negligible three-factor interactions.

Here is an explicit construction of every run without an ambiguous generator convention. For each day sign pair $(u,v)\in \{-1,1{\}}^2$, choose $(s,t,w)\in \{-1,1{\}}^3$ freely and setThere are eight distinct settings per day and 32 in total. Direct multiplication gives $ABCDEF=1$, $AF=u$ and $ABC=v$. Conversely those three equations solve $C,E,F$ uniquely from $A,B,D$, so there are no missing or repeated settings. Assign the four sign pairs randomly to the four days, and randomize the eight runs within each day.

Distinct nonaliased factorial contrasts are orthogonal over the fraction. None of the 20 required low-order contrast columns belongs to a block class, so all are also orthogonal to the three block-effect columns. Their joint design matrix, including an intercept, therefore has rank $1+3+6+14=24$. The analysis of variance partition isThe eight residual directions are the remaining unconfounded three-factor alias classes. They estimate error only under the negligible-higher-interaction assumption; they are not a separate replicated pure error stratum. This provides the requested comparisons within exactly four days of eight runs.

Write $[\mathbb{N}{]}^{\omega }$ for the space of infinite subsets of the natural numbers, with each subset identified with its increasing enumeration. In the homogeneous-cone convention, a family $\mathcal{A}\subseteq [\mathbb{N}{]}^{\omega }$ is a Ramsey family in the homogeneous-cone sense when some infinite $M$ satisfiesThe stronger Ramsey set of infinite subsets convention asks for such an infinite $M\subseteq A$ inside every given infinite ground set $A$. We will prove the open-set assertion in this stronger form, and our Ramsey-but-not-completely-Ramsey example will satisfy it as well. Thus the distinction in conventions does not affect any of the conclusions below.

For a non-Ramsey example, declare $X\sim Y$ if their symmetric difference is finite, and choose a representative $R$ from each equivalence class using the axiom of choice. Setfor the representative of the class of $X$. Removing one point reverses this colour, because its symmetric difference with $R$ changes by exactly one point. Consequently every cone $[M{]}^{\omega }$ contains the oppositely coloured sets $M$ and $M\setminus \{\min M\}$. Either colour class of this finite-symmetric-difference parity colouring is not a Ramsey family in the homogeneous-cone sense. No definability is claimed for the representative selection.

The ordinary topology on infinite subsets, denoted $\tau$, has basic cylinders $[s]=\{X:s⊏X\}$, where $s$ is a finite stem of the increasing enumeration. Let $O$ be a $\tau$-open family and fix an arbitrary infinite ground set $A$. We give a complete fusion proof of a homogeneous cone inside $A$.

Use the convention $\max \varnothing =0$. For finite $s$ and an infinite reservoir $B$ above $\max s$, say that $B$ accepts $s$ if $[s,B]\subseteq O$, whereSay that $B$ rejects $s$ if no infinite subset of $B$ accepts $s$. There is always a refinement deciding $s$: choose an accepting subreservoir if one exists, and otherwise retain $B$, which rejects. Both decisions persist under further thinning.

A first fusion produces an infinite $H\subseteq A$ deciding every finite $s\subseteq H$ on its tail $H/s=\{h\in H:h>\max s\}$. First decide the empty stem. Choose the first point, and thin the remaining reservoir successively to decide every subset of the selected prefix. Repeat after choosing each new point. There are finitely many subsets to decide at each stage. The diagonal sequence of chosen points has the required property, since its tail after the largest point of any stem lies in the reservoir where that stem was decided. The empty-stem decision persists too. This is deciding all finite stems by fusion.

If $H$ accepts the empty stem, $[H{]}^{\omega }\subseteq O$ and we are done. Suppose it rejects. For a rejected finite $s\subseteq H$, only finitely many $h\in H/s$ can have $H/h$ accepting $s\cup \{h\}$. Otherwise let $D$ be their infinite set. Every infinite $X\subseteq D$ starts with some such $h$, and its remaining tail lies in $H/h$, so $s\cup X\in O$. Then $D$ accepts $s$, contradicting rejection. This proves finitely many accepting extensions of a rejected stem.

A second fusion constructs $K\subseteq H$ whose every finite subset is rejected. If a finite prefix of chosen points has been selected, all its subsets are rejected by induction. Avoid the union of their finitely many accepting successor sets when choosing the next point. The first fusion already decides every successor, so every new subset is rejected. An infinite reservoir remains after each finite exclusion.

If some $X\in [K{]}^{\omega }$ belonged to $O$, openness would give an initial segment $s⊏X$ with $[s]\subseteq O$. Its reservoir $H/s$ would then accept $s$, contradicting the second fusion. Therefore $[K{]}^{\omega }\cap O=\varnothing$. In either case we have a homogeneous cone inside the arbitrary ground set $A$. Every $\tau$-open family is a Ramsey set of infinite subsets, under both conventions.

A completely Ramsey set requires more: for every finite $s$ and infinite reservoir $A$ above $\max s$, there is an infinite $B\subseteq A$ such thatThe finite stem must be kept fixed. These $[s,A]$ are the basic neighbourhoods of the Ellentuck topology, also called the star topology.

Let $E$ be the even positive integers, and construct the finite-symmetric-difference parity colouring $\epsilon$ on $[E{]}^{\omega }$. DefineFor any infinite ground set $A$, the set $A\setminus \{1\}$ is still infinite and its whole cone misses $\mathcal{R}$. Thus $\mathcal{R}$ is a Ramsey set of infinite subsets even in the stronger convention. But inside $[\{1\},E]$, every infinite refinement $B\subseteq E$ gives the two sets $\{1\}\cup B$ and $\{1\}\cup (B\setminus \{\min B\})$ of opposite colours. Hence no such refinement is homogeneous. The family $\mathcal{R}$ is Ramsey but not completely Ramsey. The stem-supported Ramsey family need not be completely Ramsey construction explains this example: forgetting the stem $\{1\}$ is precisely what loses the obstruction.

Finally take $D=[E{]}^{\omega }$. It is nonempty and star-open, since it equals $[\varnothing ,E]$. It is $\tau$-closed: if an infinite set contains an odd integer, an initial segment witnessing that integer has a cylinder disjoint from $D$. Its $\tau$-interior is empty: after any finite stem of even integers, append an arbitrarily large odd integer and then infinitely many further points. ThusThis is the cone on an infinite coinfinite ground set. Choose its subset $Y=\{X\in [E{]}^{\omega }:\epsilon (X)=0\}$. Since ${\overline{Y}}^{\tau }\subseteq D$, $Y$ is also $\tau$-nowhere dense. Nevertheless every $[\varnothing ,B]$ with infinite $B\subseteq E$ contains both $B$ and $B\setminus \{\min B\}$, of opposite colours. Hence $Y$ is not completely Ramsey. We have proved explicitly that ordinarily nowhere-dense sets need not be completely Ramsey; the topology in the nowhere-dense hypothesis cannot be silently changed to the star topology.

Let $P$ be the transition matrix of a finite irreducible Markov chain with at least two states, reversible with stationary distribution $\pi$. Its Dirichlet form of a Markov chain isThe Discrete Poincare inequality states that ${\mathrm{Var}}_{\pi }f\le C\mathcal{E}(f,f)$ for every real $f$. Its optimal constant is $C=(1-{\lambda }_2{)}^{-1}$, where ${\lambda }_2$ is the second largest eigenvalue of $P$. Indeed, detailed balance makes $P$ self-adjoint on $L^2(\pi )$. Expand $f-\pi (f)={\sum }_{j\ge 2}{a}_j{v}_j$ in an orthonormal eigenbasis. Then ${\mathrm{Var}}_{\pi }f=\sum a_j^2$ and $\mathcal{E}(f,f)=\sum (1-{\lambda }_j)a_j^2$, proving the inequality and its sharpness.

The useful constructive version is the canonical paths Poincare bound. Choose transition graph paths ${\gamma }_{xy}$ between each ordered pair, write $Q(u,v)=\pi (u)P(u,v)$, and definewhere $N_{uv}$ counts occurrences of the directed transition. The Cauchy-Schwarz inequality along each path in a graph gives $(f(x)-f(y){)}^2\le |{\gamma }_{xy}|{\sum }_{{\gamma }_{xy}}(f(u)-f(v){)}^2$. Multiply by $\pi (x)\pi (y)/2$ and sum. The identity ${\mathrm{Var}}_{\pi }f=\frac{1}{2}{\sum }_{x,y}\pi (x)\pi (y)(f(x)-f(y){)}^2$ proves ${\mathrm{Var}}_{\pi }f\le \rho \mathcal{E}(f,f)$.

For a lazy Markov chain this also proves the mixing time estimateTo see this, the centred initial density has $L^2(\pi )$ norm at most $\pi (x{)}^{-1/2}$; laziness puts all nonconstant eigenvalues between zero and $1-{\rho }^{-1}$. Their powers contract that norm, and the Cauchy-Schwarz inequality bounds total variation distance by half the density's $L^2(\pi )$ norm.

Here is a precise sufficient density condition for the permanent application: every row and every column has more than $n/2$ ones. Interpret the permanent of a matrix as the number $m_n$ of perfect matchings in its bipartite graph, with $n$ vertices per side. Write $m_k$ for the number of $k$-edge matchings in a graph.

First prove the short augmenting paths in a dense bipartite graph bound. Given a nonperfect matching in a graph, choose unmatched $u,v$ on opposite sides. If either has an unmatched neighbour, augment with one edge. Otherwise the matched partners of $N(u)$ and the neighbours $N(v)$ are subsets of the same $n$-element side whose total size exceeds $n$. Their intersection gives an alternating path in a graph $u,b,a,v$ of length three, with $ab$ a matching edge. Repeating proves $m_n>0$. Choose augmentations by a fixed deterministic rule. A reverse augmentation is specified by at most four vertices, so a next-size matching in a graph has at most $C_0=2n^4$ preimages. Therefore

Use the matching generating function $Z(\lambda )={\sum }_{k=0}^n{m}_k{\lambda }^k$ and the weighted matching Markov chain on all matchings in a graph, with stationary weights ${\pi }_{\lambda }(M)={\lambda }^{|M|}/Z(\lambda )$. It stays put with probability $1/2$; otherwise it chooses a graph edge uniformly and proposes its deletion, its addition if both endpoints are unmatched, or its exchange for the unique incident matching edge if precisely one endpoint is matched. Other proposals stay put. Accept with probability $\min (1,{\lambda }^{|M^{\prime }|-|M|})$. Reverse proposals use the opposite exchanged edge, so detailed balance holds. Deletions connect every state to the empty matching in a graph.

For completeness, the canonical paths for the weighted matching chain give a polynomial bound without assuming a mixing theorem. The symmetric difference $I△F$ consists of alternating graph paths and graph cycles. Order its components deterministically, orient open components from their smaller endpoint, and convert them from $I$ to $F$ by moving an unmatched endpoint with exchanges. If the first edge belongs to $I$, delete it first. For a graph cycle, delete one $I$-edge, move the unmatched endpoint around the opened cycle, then insert the last $F$-edge. Each intermediate state is a matching in a graph, and each entire transition path in a graph has length at most $4n$.

Fix a directed transition with initial state $M$. The complementary edge set $K=I△F△M$ agrees with one of the two matchings in a graph outside the active component and has at most two degree-two vertices on that component. Remove at most two offending edges to obtain a matching in a graph $K^{\prime }$. Store those edges and one bit specifying the initial alternating colour of the active component. This encoding is injective for the fixed transition: it restores $K$ and $I△F=M△K$, identifies the active component from the changed edges, and recovers the two colours using the fixed component order. Completed components have $F$'s colour in $M$, pending components have $I$'s colour, and the stored bit resolves the active component. Common edges are $M\cap K$.

There are at most $2(m+1{)}^2$ auxiliary records per $K^{\prime }$, where $m$ is the number of graph edges. If $d=|K|-|K^{\prime }|\le 2$, thenAlso $Q(M,M^{\prime })\ge {\pi }_{\lambda }(M)/(2m\max (\lambda ,{\lambda }^{-1}))$. Sum over the encodings and use the path length bound to obtainThis is polynomial whenever $\lambda$ and ${\lambda }^{-1}$ are polynomially bounded. Starting at the empty matching in a graph, $\log (1/{\pi }_{\lambda }(\varnothing ))=\log Z(\lambda )\le m\log (1+\lambda )$, so the Discrete Poincare inequality gives polynomial-time sampling to any prescribed total variation distance. The addition/deletion/exchange construction is also described in Section 12.4 of Jerrum and Sinclair's survey.

It remains to turn sampling into a relative count; merely sampling perfect matchings would not do that. For $0<\epsilon \le 1/2$, choose ${\lambda }_0=\epsilon /(8m)$ and ${\lambda }_{\ast }=8C_0/\epsilon$. The elementary bounds above giveChoose a schedule ${\lambda }_0<{\lambda }_1<\cdots <{\lambda }_J={\lambda }_{\ast }$ whose successive ratios are at most $1+1/n$. It has $J=O(n\log (n/\epsilon ))$ stages. The annealing ratios for a matching generating function satisfyEach observable lies in $[1,e]$, so the Hoeffding inequality gives relative accuracy $\eta =\epsilon /(10J)$ with $O({\eta }^{-2}\log (J/\delta ))$ independent samples per stage. Run independent copies of the chain from the empty matching in a graph; make the total variation distance per sample at most $\delta /(3JK)$, where $K$ is that sample count. The coupling characterization of total variation distance and the union bound make the entire sample collection coincide with exact stationary samples except with probability at most $\delta /3$. Choose the sample-count constant so all stationary sample means are accurate except with probability at most $\delta /3$.

Let $\hat{R}$ be the product of the sample means, and output $\hat{p}=\hat{R}/{\lambda }_{\ast }^n$. On the successful event its multiplicative error relative to $Z({\lambda }_{\ast })/Z({\lambda }_0)$ lies between $e^{-\epsilon /5}$ and $e^{\epsilon /5}$. Combining the two endpoint bounds provesActivities, sample counts, simulation lengths and rational bit lengths are polynomial in $n,{\epsilon }^{-1},\log ({\delta }^{-1})$. Thus this is a fully polynomial randomized approximation scheme. Small dimensions can be evaluated directly.

Represent a subset by the coordinate vector of its indicator function in ${\mathbb{F}}_2^n$. Symmetric difference then becomes coordinatewise addition, and the proposed pairing is the dot productIt is a symmetric bilinear form, and it is nondegenerate because its Gram matrix in the singleton basis is the identity. Coordinate permutations preserve the sum, so the natural $S_n$ action preserves the bilinear form.

Let $j=(1,\dots ,1)$, the vector representing the full set. Thenis the even-weight subspace. For even $n$, $B(j,j)=n=0$ in ${\mathbb{F}}_2$, so $j\in W$. Moreover $B(x,x)={\sum }_i{x}_i^2={\sum }_i{x}_i=0$ for $x\in W$, making the restricted bilinear form alternating. Nondegeneracy on the ambient space implies $W^{\perp }=⟨j⟩$, so the radical of $B{|}_W$ is exactlyThus it descends to an alternating bilinear form on $W/⟨j⟩$, independent of the representatives, and that quotient form has zero radical. The induced form is nondegenerate, and its dimension is $n-2$. This is the symplectic quotient of the binary subset module; at $n=2$ it is the zero-dimensional nondegenerate space.

Identify the hypercube graph $Q_n$ with subsets of $[n]$, adjacent when they differ by one element. Write $N(\mathcal{F})$ for a family's closed vertex neighbourhood. The simplicial order on the discrete cube orders first by size and then, within a rank, by lexicographic order: the least element of a symmetric difference belongs to the earlier set. The Harper theorem says that if $I$ is the initial simplicial segment with $|I|=|\mathcal{F}|$, thenEquivalently this minimizes the external vertex boundary at fixed cardinality. This is the vertex-isoperimetric assertion, distinct from the edge version in Question 3.

Here is the deduction of Kruskal-Katona theorem. For a family $\mathcal{B}\subseteq [n{]}^{(k)}$, adjoin every lower rank and set $\mathcal{D}=[n{]}^{(<k)}\cup \mathcal{B}$. For $k\ge 1$,The corresponding simplicial initial segment is $[n{]}^{(<k)}\cup L$, where $L$ is the rank-$k$ lexicographic initial segment of size $|\mathcal{B}|$. Canceling the common lower-rank count in Harper's inequality gives $|\nabla \mathcal{B}|\ge |\nabla L|$. The rank-zero case is immediate by considering its two possible families.

To turn upper shadows into lower shadows, let $\rho (i)=n+1-i$ and $T(B)=\rho ([n]\setminus B)$. This bijection reverses inclusion and changes rank $k$ to $n-k$. If $B$ precedes $C$ in lex order, the largest element of $T(B)△T(C)$ belongs to $T(C)$, so $T(B)$ precedes $T(C)$ in colexicographic order. Also $T(\nabla \mathcal{B})=\partial (T\mathcal{B})$. Hence applying the upper-shadow conclusion to the transformed family proveswhere $C$ is the colex initial segment of the same size and rank as $\mathcal{A}$. This is the Kruskal-Katona lower-shadow theorem. A colex segment's shadow is itself a colex segment: split its sets by their largest element, giving a full initial block of the preceding ground set followed by an initial block with the new largest element. The same description holds one rank lower. Iteration therefore shows that colex segments minimize shadows at every lower rank. This proves the Harper theorem implies the Kruskal-Katona theorem deduction, including the necessary complement and coordinate reversal.

The Erdős-Ko-Rado theorem states that for $1\le r\le n/2$, every intersecting family $\mathcal{A}\subseteq [n{]}^{(r)}$ satisfiesThe family of all $r$-sets containing a fixed point attains the bound. We use the standard convention of positive ranks here.

For the first proof, let $\mathcal{B}$ be the complements of members of $\mathcal{A}$, at rank $k=n-r\ge r$, and let $\mathcal{S}$ be its rank-$r$ lower shadow. It is disjoint from $\mathcal{A}$, since $A\subseteq [n]\setminus A^{\prime }$ would make two family members disjoint. Suppose $|\mathcal{A}|>\left(\genfrac{}{}{0ex}{}{n-1}{r-1}\right)=\left(\genfrac{}{}{0ex}{}{n-1}{k}\right)$. The colex rank-$k$ segment of that size contains every $k$-set on $[n-1]$ and at least one set containing $n$. Its rank-$r$ shadow therefore contains all $\left(\genfrac{}{}{0ex}{}{n-1}{r}\right)$ old $r$-sets and at least one new $r$-set containing $n$. The iterated Kruskal-Katona bound givesThen $|\mathcal{A}|+|\mathcal{S}|>\left(\genfrac{}{}{0ex}{}{n-1}{r-1}\right)+\left(\genfrac{}{}{0ex}{}{n-1}{r}\right)=\left(\genfrac{}{}{0ex}{}{n}{r}\right)$, impossible for disjoint subfamilies of rank $r$. This proves the Erdős-Ko-Rado theorem from shadows.

For the second proof use the Katona circle method. In any cyclic ordering of the $n$ points, an intersecting collection of length-$r$ intervals has at most $r$ members. To prove this, rotate one selected interval to end at position $n$, so it occupies $n-r+1,\dots ,n$. Intervals ending at $r,\dots ,n-r$ miss it and cannot be selected. Among the remaining endpoints other than $n$, pair $j$ with $j+n-r$ for $1\le j\le r-1$. The two intervals in each pair are disjoint because $n\ge 2r$, so at most one can be selected. Including the fixed interval gives at most $1+(r-1)=r$.

Now count pairs consisting of a family member and a permutation in which it is a cyclic interval. A fixed $r$-set has $nr!(n-r)!$ such permutations: choose its final position, its internal order, and the order of the remaining points. Each of the $n!$ permutations contributes at most $r$ family intervals by the proved cyclic interval intersection bound. ThusThis is the required cyclic-ordering proof, with its interval lemma established explicitly.

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.

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 $n=2$.

Write $[M{]}^{\omega }$ for the set of all infinite subsets of $M$. A family $Y\subseteq [\mathbb{N}{]}^{\omega }$ is a Ramsey family in the homogeneous-cone sense if there is an infinite $M$ such that either $[M{]}^{\omega }\subseteq Y$ or $[M{]}^{\omega }\cap Y=\varnothing$. Thus membership in $Y$ is constant on an infinite homogeneous cone. We will prove the stronger version for open families: such a cone exists inside every prescribed infinite ground set, as in the definition of a Ramsey set of infinite subsets.

The symbols in the question refer to the ordinary topology on infinite subsets $\tau$ and the Ellentuck topology $\ast$. For a finite set $s$, letThese sets form a topological basis for $\tau$, equivalently the subspace topology inherited from the product topology on $\{0,1{\}}^{\mathbb{N}}$. For an infinite $A$ with $\max s<\min A$, the setsform a topological basis for $\ast$; take $\max \varnothing =0$. In particular $[s]=[s,\{n:n>\max s\}]$, so $\ast$ is finer than $\tau$.

For a non-Ramsey example use the finite-symmetric-difference parity colouring. On $[\mathbb{N}{]}^{\omega }$, define the equivalence relation $X\sim Z$ if their symmetric difference $X△Z$ is finite. By the axiom of choice, select a representative $R$ in every equivalence class. Give $X$ the colourwhere $R_X$ is its class representative. Deleting one point stays in the same class and toggles the parity. Thus for every infinite $M$, the sets $M$ and $M\setminus \{\min M\}$ are both in $[M{]}^{\omega }$ but have opposite colours. Either colour class, regarded as a family $Y$, is therefore not Ramsey.

Now let $U$ be $\tau$-open. We prove that for every infinite $A_0$ there is an infinite $C\subseteq A_0$ with $[C{]}^{\omega }\subseteq U$ or $[C{]}^{\omega }\cap U=\varnothing$. The proof uses acceptance and rejection of finite stems, which we define explicitly. For a finite stem $s$ and an infinite tail $A$ above $\max s$, say that $A$ accepts $s$ if $[s,A]\subseteq U$, and rejects $s$ if no infinite subset of $A$ accepts $s$. Both properties persist on taking infinite subsets. Every tail has an infinite subset that decides $s$: take an accepting subset if one exists, and otherwise it already rejects $s$.

First thin $A_0$ to decide the empty stem. Then construct $b_1<b_2<\cdots$ and nested infinite tails. At step $n$, choose $b_n$ from the current tail, discard its points at most $b_n$, and successively thin what remains to decide every stem $s\subseteq \{b_1,\dots ,b_n\}$. There are only finitely many such stems, so an infinite tail remains. Let $B=\{b_1,b_2,\dots \}$ and write $B/s=\{b\in B:b>\max s\}$. For every finite $s\subseteq B$, the tail $B/s$ decides $s$, since it lies in the tail obtained at the stage when the largest point of $s$ was selected. The empty stem is also decided. This is deciding all finite stems by fusion.

If $B$ accepts the empty stem, then $[B{]}^{\omega }\subseteq U$, as required. Otherwise $B$ rejects it. We need the finitely many accepting extensions of a rejected stem observation. Suppose a tail $B/s$ rejects $s$. Only finitely many $a\in B/s$ can have $B/a$ accepting $s\cup \{a\}$. Indeed, if infinitely many such $a$ formed a set $D$, every member of $[s,D]$ would have some first new point $a$ and all later points in $B/a$, hence would lie in $U$. This would make $D$ an accepting subset of $B/s$, contradicting rejection.

Choose $c_1<c_2<\cdots$ in $B$ recursively so that for every finite $s\subseteq \{c_1,\dots ,c_n\}$, the tail $B/s$ rejects $s$. The empty stem satisfies this initially. At the next step, each of the finitely many previous stems has only finitely many accepting one-point extensions by the observation. Choose $c_{n+1}$ larger than all previous points and outside all those finite exceptional sets. The first fusion ensured that the tail for every new stem decides it; since it does not accept it, it rejects it. This maintains the recursion.

Put $C=\{c_1,c_2,\dots \}$. If some $X\in [C{]}^{\omega }$ belonged to $U$, ordinary openness would give a finite initial segment $s$ of $X$ with $[s]\subseteq U$. In particular $[s,C/s]\subseteq U$. But $C/s$ is an infinite subset of $B/s$, which rejects $s$, a contradiction. Hence $[C{]}^{\omega }\cap U=\varnothing$. We have proved

The Erdős-Ko-Rado theorem asserts that, for $1\le r<n/2$, every intersecting family $\mathcal{A}\subseteq \left(\genfrac{}{}{0ex}{}{[n]}{r}\right)$ satisfiesThe bound is attained by the set family of all $r$-subsets containing one prescribed point. Here lexicographic order means that $X<Y$ when the smallest element of their symmetric difference belongs to $X$.

The first $M=\left(\genfrac{}{}{0ex}{}{n-1}{r-1}\right)$ members in lexicographic order are exactly those containing $1$. An initial segment longer than $M$ contains all of them and also the first member avoiding $1$, namely $B=\{2,\dots ,r+1\}$. There is an $r$-subset containing $1$ and disjoint from $B$: choose its other $r-1$ elements from $\{r+2,\dots ,n\}$, which has $n-r-1\ge r-1$ elements. Consequently any longer lexicographic initial segment fails to be an intersecting family. Thus preservation of the intersection property upon replacing a set family by the equally large lexicographic initial segment proves the displayed bound.

Here is a direct UV-compression proof of that preservation statement. For disjoint nonempty sets $U,V$ with $|U|=|V|$ and $\min U<\min V$, an eligible member $A$ contains $V$ and misses $U$; its proposed image is $A^{\prime }=(A\setminus V)\cup U$. The UV-compression moves $A$ to $A^{\prime }$ only if $A^{\prime }$ is not already present. Each eligible pair is either kept as a pair or moved from its later to its earlier member, so this operation preserves cardinality and uniformity and strictly decreases the sum of lexicographic order positions whenever it changes the set family.

The essential intersection-preserving lexicographic UV-compression lemma is that this operation preserves an intersecting family whenever all smaller lexicographically improving UV-compressions already leave that set family unchanged. Suppose instead that the compressed set family has disjoint members. At least one, say $A^{\prime }$, must be newly moved. Both newly moved members would contain $U$, so the other, $B$, is a retained old member. Write $A=(A^{\prime }\setminus U)\cup V$ for the old preimage. Since $A^{\prime }\cap B=\varnothing$, we have $B\cap U=\varnothing$ and $(A\setminus V)\cap B=\varnothing$. Old intersectingness forcesIf $V^{\prime }=V$, then $B$ was eligible for the same UV-compression. Since it was retained, $B^{\prime }=(B\setminus V)\cup U$ was already present. But $A\cap B^{\prime }=\varnothing$, contradicting old intersectingness.

Otherwise $0<|V^{\prime }|<|V|$. Choose $U^{\prime }\subset U$ with $|U^{\prime }|=|V^{\prime }|$ and $\min U\in U^{\prime }$. This is a smaller improving pair, since $\min U^{\prime }=\min U<\min V\le \min V^{\prime }$. The old member $A$ is eligible for its UV-compression. The assumed stability therefore forces $A^{\prime \prime }=(A\setminus V^{\prime })\cup U^{\prime }$ to be present already. Yet its portion outside $V$ misses $B$, its remaining portion $V\setminus V^{\prime }$ misses $B$, and $U^{\prime }$ misses $B$. Thus $A^{\prime \prime }\cap B=\varnothing$, the required contradiction. This proves the lemma, including the size-one case, where only the first alternative can arise.

Starting with $\mathcal{A}$, repeatedly choose a changing improving UV-compression with the smallest possible $|U|$. The lemma applies at each step because every smaller improving UV-compression is then unchanged. Thus every intermediate set family is intersecting. The strictly decreasing nonnegative integer sum of lexicographic order positions ensures termination. If the terminal set family were not an initial segment, there would be a missing $X$ earlier than a present $Y$. Taking $U=X\setminus Y$ and $V=Y\setminus X$ gives equally sized nonempty disjoint sets with $\min U<\min V$, and moves $Y$ to the missing $X$. That contradicts termination. The terminal set family is therefore exactly the equally large lexicographic initial segment, and it is intersecting. This proves the required lexicographic statement directly, and hence the Erdős-Ko-Rado theorem.

Identify the vertices of the Boolean hypercube $Q_n$ with subsets of $[n]$, adjacent when their symmetric difference has size one. Write $N(\mathcal{A})$ for the closed graph neighbourhood, including $\mathcal{A}$ itself. In simplicial order on the discrete cube, smaller sets come first, and equal-sized sets are ordered by lexicographic order: the least element of the symmetric difference belongs to the earlier member. Let $I_m$ denote the first $m$ vertices. The vertex-isoperimetric inequality in the discrete cube statesSubtracting $|\mathcal{A}|$ gives the equivalent assertion for the external vertex boundary.

We first check that $N(I_m)$ is also an initial segment in simplicial order on the discrete cube. Apart from the empty and full cases, $I_m$ consists of complete lower layers and a lexicographic initial segment $\mathcal{F}$ in its top rank $r$. Its closed graph neighbourhood contains all ranks at most $r$ and the upper shadow $\nabla \mathcal{F}$ in rank $r+1$. This upper shadow is lexicographic initial: an $(r+1)$-subset $C$ belongs to it precisely when its first $r$-subset in lexicographic order, obtained by deleting its largest element, belongs to $\mathcal{F}$. Taking the first $r$ elements preserves weak lexicographic order, proving the assertion. This also handles $r=0$, where $N(\{\varnothing \})$ consists of ranks zero and one.