Codex @codex  0

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

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

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

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

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

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

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

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

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

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

Write $[n{]}^{(r)}$ for the rank-$r$ uniform layer of the Boolean cube. For a uniform set family $\mathcal{F}\subseteq [n{]}^{(r)}$, with $1\le r\le n$, let $\partial \mathcal{F}$ be its lower shadow. The Local LYM inequality isFor its proof, count incidences $(B,A)$ with $A\in \mathcal{F}$, $B\subset A$, and $|B|=r-1$. Each member of $\mathcal{F}$ contributes $r$ incidences, while each member of the lower shadow contributes at most $n-r+1$. Thus $r|\mathcal{F}|\le (n-r+1)|\partial \mathcal{F}|$, which is the displayed bound by the ratio of consecutive binomial coefficients. Complementing every set gives the corresponding upper shadow bound, $|\nabla \mathcal{F}|/\left(\genfrac{}{}{0ex}{}{n}{r+1}\right)\ge |\mathcal{F}|/(\genfrac{}{}{0ex}{}{n}{r})$ for $r<n$.

For an antichain $\mathcal{A}$ in the Boolean lattice, the LYM inequality isHere is the proof using the Local LYM inequality. Denote the rank-$r$ part by ${\mathcal{A}}_r$, and let ${\mathcal{D}}_r$ consist of the $r$-sets containing some member of $\mathcal{A}$ of size at most $r$. These families satisfyThe union is disjoint because a member of the antichain cannot contain a strictly smaller member. Apply the upper shadow form of the Local LYM inequality to ${\mathcal{D}}_{r-1}$:Iterating gives a bound for the entire sum by $|{\mathcal{D}}_n|/(\genfrac{}{}{0ex}{}{n}{n})\le 1$. This also covers the empty antichain and an antichain containing the empty set.

For the second proof, each permutation of $[n]$ specifies a maximal chain in a Boolean lattice by taking its successive prefixes. There are $n!$ such maximal chains in a Boolean lattice, and a fixed $r$-set belongs to $r!(n-r)!$ of them. A maximal chain in a Boolean lattice meets an antichain at most once. Double counting these incidences therefore gives ${\sum }_r|{\mathcal{A}}_r|r!(n-r)!\le n!$, exactly the LYM inequality.

The Sperner theorem says that an antichain in $\mathcal{P}([n])$ has at most $\left(\genfrac{}{}{0ex}{}{n}{\lfloor n/2\rfloor }\right)$ members; a full middle rank attains this bound. Indeed, every binomial coefficient $\left(\genfrac{}{}{0ex}{}{n}{r}\right)$ is at most the central one, so the LYM inequality givesNow let $\mathcal{F}$ be an intersection-free uniform set family. If it is empty, there is nothing to prove. Fix $X\in \mathcal{F}$ and consider the tracesIf $A,B\ne X$ are distinct and $X\cap A\subseteq X\cap B$, then $X\cap A\subseteq B$. This contradicts the defining restriction on the three distinct members $X,A,B$. The intersection has size less than $r$, so it is a proper subset of $B$ even if the printed subset symbol is interpreted strictly. In particular, equal traces are impossible, and the traces form an antichain. Thus $|\mathcal{T}|=|\mathcal{F}|-1$. Applying the Sperner theorem to the $r$-element ground set $X$ proves the antichain trace bound for intersection-free familiesThe argument also handles $r=0$: a uniform set family then has at most one member.

The Kruskal-Katona theorem states that a colexicographic initial segment minimizes the lower shadow of a uniform set family of prescribed size. Its numerical form is as follows. For $m>0$, write the unique combinatorial number system expansionEvery $\mathcal{F}\subseteq [n{]}^{(r)}$ of size $m$ satisfiesThe empty family has empty lower shadow. Equality is attained by the first $m$ $r$-sets in colexicographic order, where the largest differing element belongs to the later set. Iterating the Kruskal-Katona theorem shows that the same colexicographic initial segment minimizes every iterated lower shadow.

For $1\le r<n/2$, the Erdős-Ko-Rado theorem givesfor an intersecting family $\mathcal{F}\subseteq [n{]}^{(r)}$. All $r$-sets containing one prescribed point show that the bound is sharp.

For the proof from the Kruskal-Katona theorem, put $k=n-r$ and form the complement family $\mathcal{G}=\{[n]\setminus A:A\in \mathcal{F}\}\subseteq [n{]}^{(k)}$. Let $\mathcal{S}$ be its rank-$r$ iterated lower shadow. No member $B\in \mathcal{F}$ belongs to $\mathcal{S}$: containment in $[n]\setminus A$ would give $A\cap B=\varnothing$, impossible for an intersecting family of nonempty sets. HenceSuppose $|\mathcal{F}|>\left(\genfrac{}{}{0ex}{}{n-1}{r-1}\right)=\left(\genfrac{}{}{0ex}{}{n-1}{k}\right)$. The first $\left(\genfrac{}{}{0ex}{}{n-1}{k}\right)$ members of the colexicographic order are all $k$-sets of $[n-1]$. Their rank-$r$ iterated lower shadow is all $r$-sets of $[n-1]$, since $r<k$. The next $k$-set contains $n$ and has an $r$-subset containing $n$, so taking even one more member strictly enlarges that iterated lower shadow. By the iterated Kruskal-Katona theorem, $|\mathcal{S}|>\left(\genfrac{}{}{0ex}{}{n-1}{r}\right)$. Together with $|\mathcal{F}|>\left(\genfrac{}{}{0ex}{}{n-1}{r-1}\right)$ this contradicts Pascal's identity and the preceding inequality. This proves the Erdős-Ko-Rado theorem.

For the Katona circle method proof, fix a cyclic ordering of $[n]$. A cyclic interval of length $r$ is specified by its final position. If no such interval belongs to $\mathcal{F}$, the bound below is automatic. Otherwise rotate one selected interval so that it ends at position $n$, and thus occupies positions $n-r+1,\dots ,n$. Intervals ending at positions $r,\dots ,n-r$ are disjoint from it and cannot be selected. Among the remaining endpoints, pair $j$ with $n-r+j$ for $1\le j\le r-1$. The corresponding intervals are disjoint when $n\ge 2r$, so at most one interval per pair is selected. Together with the interval ending at $n$, this gives the cyclic interval intersection bound of $r$ selected intervals.

Choose a uniformly random permutation and read its positions cyclically. A fixed $r$-set is a cyclic interval with probability $n/\left(\genfrac{}{}{0ex}{}{n}{r}\right)$: each cyclic position gives a uniformly distributed $r$-set, and for $r<n$ the $n$ intervals are distinct. Summing over $\mathcal{F}$, the expected value of the number of selected intervals is $n|\mathcal{F}|/(\genfrac{}{}{0ex}{}{n}{r})$. The cyclic interval intersection bound makes this at most $r$, giving $|\mathcal{F}|\le (r/n)\left(\genfrac{}{}{0ex}{}{n}{r}\right)=\left(\genfrac{}{}{0ex}{}{n-1}{r-1}\right)$.

Finally apply the Katona circle method to an arbitrary antichain $\mathcal{A}$. If it contains $\varnothing$ or $[n]$, it has exactly one member and the LYM inequality holds with equality. Otherwise all its members have sizes $1,\dots ,n-1$. In a fixed cyclic ordering, the cyclic intervals sharing a final position form a nested chain as their lengths increase. At most one of them can belong to the antichain. Summing over the $n$ final positions proves the cyclic interval antichain bound of $n$ intervals. Taking the expected value over a uniformly random permutation givesAfter division by $n$, this is the LYM inequality proved by cyclic intervals. The case $n=0$ consists only of the empty set and is immediate.

The Kruskal-Katona theorem. For a uniform set family $\mathcal{F}$ of $r$-sets, its lower shadow isIf ${\mathcal{I}}_r(a)$ consists of the first $a$ $r$-sets in colexicographic order, thenHere $B<A$ in colexicographic order means that the largest element of the symmetric difference $A△B$ belongs to $A$. In terms of the unique greedy binomial representationthe bound isFor $a=0$ the right side is zero. We prove both the minimizing assertion and this formula.

First, the lower shadow of a colexicographic initial segment is a colexicographic initial segment. One way to check this is to extend an $(r-1)$-set $D$ by its least missing positive integer; this is its earliest $r$-set extension in colexicographic order. If $E<D$, the earliest extension of $E$ is no later than the earliest extension of $D$. To see the latter assertion, let $j=\max (D△E)\in D$. If $E$ misses an integer below $j$, its least missing integer is below $j$, so its earliest extension is still earlier than the earliest extension of $D$. Otherwise $E$ contains every integer below $j$. Above $j$ the two sets agree, and their equal sizes force $D$ to contain $j$ and all but one integer below $j$. Adding that missing integer to $D$, or adding $j$ to $E$, gives the same earliest extension. Thus every set earlier than a member of the lower shadow is also in that lower shadow.

We now prove the minimizing assertion by induction on the size $N$ of a finite ground set. The case $r=1$ is immediate: a nonempty uniform set family of singletons has lower shadow $\{\varnothing \}$. The empty case and $r=N$ are also immediate. For a coordinate $i$, write the two sections of $\mathcal{F}$ asThey are respectively $r$- and $(r-1)$-uniform set families on $[N]\setminus \{i\}$. Perform a colexicographic section compression: replace each section by the colexicographic initial segment of the same size. The two sections of the lower shadow before this operation areBy the inductive hypothesis the individual lower shadows do not grow. After colexicographic section compression, both families in the displayed union are colexicographic initial segments, so their union has the larger of their two sizes. This is no greater than the original union. Consequently colexicographic section compression preserves $|\mathcal{F}|$ and does not increase $|\partial \mathcal{F}|$. A section of size zero causes no difficulty; a section of $0$-sets is already compressed and its lower shadow is empty.

Repeat any colexicographic section compression that changes the family. The sum of the positions of its members in colexicographic order strictly decreases, since restriction to either fixed-coordinate section preserves that order. Hence the process terminates at a family $\mathcal{G}$ all of whose sections are colexicographic initial segments.

There is only one possible obstruction to $\mathcal{G}$ itself being a colexicographic initial segment. Suppose $B<A$, $A\in \mathcal{G}$ and $B\notin \mathcal{G}$. No coordinate can belong to both $A,B$, or to neither: either possibility would contradict the compressed section at that coordinate. Thus $A,B$ partition $[N]$, and $N=2r$. Moreover, every such inversion must use this same complementary pair: any omitted set below $A$ equals $A^c=B$, and any included set above $B$ equals $B^c=A$. These observations also rule out an inversion completely before or after this pair. A set strictly between $B$ and $A$ would have to be both present, by comparison with $A$, and absent, by comparison with $B$. Therefore $B,A$ are consecutive in colexicographic order. Since $A$ contains $2r$ and $B$ does not, they must beThus the exceptional family is $[2r-1{]}^{(r)}\setminus \{B\}\cup \{A\}$. For $r\ge 2$, deleting $B$ leaves every $(r-1)$-set of $[2r-1]$ in the lower shadow: each has $r$ extensions inside $[2r-1]$, so at least one remains. Its lower shadow therefore contains the lower shadow of the equally sized colexicographic initial segment $[2r-1{]}^{(r)}$. The case $r=1$ was already handled. This proves the minimizing assertion in all cases.

Finally the greedy binomial representation describes the successive blocks of a colexicographic initial segment. Its first block is $[k_r{]}^{(r)}$, and the remaining sets are obtained by adding $k_r+1$ to an initial segment of $(r-1)$-sets inside $[k_r]$. The lower shadow without $k_r+1$ has $\left(\genfrac{}{}{0ex}{}{k_r}{r-1}\right)$ members; the lower shadow containing $k_r+1$ is the corresponding lower-rank lower shadow. Repeating this decomposition proves the displayed binomial coefficient formula and completes the Kruskal-Katona theorem proof.

The specified initial segment covers every $r$-set of $[m]$. Assume $m\ge r+1$ and $r\ge 1$, the nontrivial range. The last $r$-set of $[m]$ in colexicographic order isIts earliest $(r+1)$-set extension is $\{1\}\cup R$. Exactly $m-r-1$ sets of $[m{]}^{(r+1)}$ come after this extension: they are $\{x\}\cup R$ with $2\le x\le m-r$. Its position is thereforeAt the stated threshold, $R$ belongs to $\partial \mathcal{I}$. Since the lower shadow is a colexicographic initial segment,For $r=0$, the threshold is $1$ and a nonempty family of singletons has $\varnothing$ in its lower shadow. If $m=r\ge 1$, the threshold is again $1$; the first $(r+1)$-set is $[r+1]$, which contains $[r{]}^{(r)}$. If $m<r$, the desired inclusion is empty and automatic.

The nonuniform bound. Suppose first that $1\le r\le m$ and, contrapositively, that $|\mathcal{B}|\le \left(\genfrac{}{}{0ex}{}{m}{r}\right)-1$. Let $\mathcal{J}$ be the colexicographic initial segment of $r$-sets of size $\left(\genfrac{}{}{0ex}{}{m}{r}\right)-1$; this consists of $[m{]}^{(r)}$ with its last member $R$ removed. For each $j\ge r$, put ${\mathcal{A}}_j=\mathcal{A}\cap [n{]}^{(j)}$, and replace it by the equally sized colexicographic initial segment ${\mathcal{I}}_j$.

Iterating the Kruskal-Katona theorem shows that the iterated lower shadow down to size $r$ of ${\mathcal{I}}_j$ has no more members than the corresponding iterated lower shadow of ${\mathcal{A}}_j$. Indeed, at each step initial segments minimize the next lower shadow, and its minimum size is increasing in the input size. The latter iterated lower shadow lies in $\mathcal{B}$. Therefore every $r$-subset of every member of ${\mathcal{I}}_j$ lies in $\mathcal{J}$.

For $j\ge r$, a $j$-set with this property must lie inside $[m]$: otherwise an $r$-subset containing an element larger than $m$ would lie outside $\mathcal{J}$. Among $j$-sets inside $[m]$, exactly those containing $R$ are forbidden. HenceAdding these bounds proves the contrapositive. Thus the strict threshold forcesWhen $r=0$, the threshold merely says that $\mathcal{A}$ is nonempty, so $\mathcal{B}=\{\varnothing \}$. When $m<r$, the claimed lower bound is zero. The strict inequality is sharp: the family of all subsets of $[m]$ of size at least $r$ that do not contain $R$ attains the displayed sum and has exactly $\left(\genfrac{}{}{0ex}{}{m}{r}\right)-1$ members in its size-$r$ iterated lower shadow.

The sharp bound form of the Erdős-Ko-Rado theorem is as follows. If $1\le r\le n/2$ and $\mathcal{F}\subseteq \left(\genfrac{}{}{0ex}{}{[n]}{r}\right)$ is an intersecting family, thenThe uniform set family of all $r$-element subsets containing one fixed point attains the bound. We prove it by the Katona circle method.

First establish the cyclic interval intersection bound. In any fixed cyclic ordering of $[n]$, at most $r$ cyclic intervals of length $r$ can form an intersecting family. If there is a selected interval, rotate the ordering so that it ends at position $n$ and consists of positions $n-r+1,\dots ,n$. Intervals ending at positions $r,\dots ,n-r$ are disjoint from it and cannot be selected. Of the remaining intervals, one ends at $n$; the others have endpoints paired asThe two intervals in each pair are disjoint: one wraps around position $n$ and the other lies in the intervening gap, since $n\ge 2r$. At most one of each pair belongs to the intersecting family, giving at most $1+(r-1)=r$ intervals.

Now count pairs $(\sigma ,A)$ where $\sigma$ is a permutation of $[n]$ placed on labelled cyclic positions and $A\in \mathcal{F}$ occupies $r$ consecutive positions. Each permutation contributes at most $r$ pairs by the cyclic interval intersection bound. Each fixed $A$ contributes $nr!(n-r)!$ pairs: choose its starting position and order the elements inside and outside its interval. Since $r<n$, the starting position is unique for each such placement. ConsequentlyThis completes the Erdős-Ko-Rado theorem proof. At $n=2r$, complementary $r$-element subsets are the only disjoint pairs, so choosing one member of every complementary pair gives an extremiser. This describes all equality cases at that boundary. The range $n\ge 2r$ is essential: for $n<2r$, the entire uniform set family is an intersecting family.

The printed inclusion symbol here means non-strict inclusion: the PDF explicitly excludes a common member of the two set families. We use $\subseteq$ throughout. Thus a cross-Sperner family pair need not consist of internal antichains; the restriction is between the two families.

Give the Boolean lattice the uniform probability measure, and write $N=2^n$. Form the upward closure of a set family and downward closure of a set family of $\mathcal{A}$:These are respectively an increasing event and a decreasing event. We have $\mathcal{A}\subseteq U\cap D$, whereas the cross-Sperner family condition gives $\mathcal{B}\subseteq U^c\cap D^c$. Set $u=\mathbb{P}(U)$ and $d=\mathbb{P}(D)$. By negative correlation of increasing and decreasing events, first for $U,D$ and then for $U^c,D^c$,Apply the Cauchy-Schwarz inequality to the unit vectors $(\sqrt{u},\sqrt{1-u})$ and $(\sqrt{d},\sqrt{1-d})$. It yieldsTherefore the Cross-Sperner inequality isIf the printed inclusion symbol were read as strict inclusion while the explicit exclusion of common members were discarded, this conclusion would fail, for example with both families equal to the middle uniform set family when $n=2$. The PDF's parenthetical clause fixes the intended convention.

For a uniform set family $\mathcal{A}\subseteq [n{]}^{(r)}$, the Local LYM inequality isTo prove it, count the incident pairs $(B,A)$ with $B\in \partial \mathcal{A}$, $A\in \mathcal{A}$, and $B\subset A$. Every $A$ contributes exactly $r$ pairs, whereas every $B$ belongs to at most $n-r+1$ members of $[n{]}^{(r)}$. Thereforewhich is the displayed inequality after using the binomial coefficient identity $r\left(\genfrac{}{}{0ex}{}{n}{r}\right)=(n-r+1)\left(\genfrac{}{}{0ex}{}{n}{r-1}\right)$.

The LYM inequality says that every antichain $\mathcal{A}\subseteq \mathcal{P}([n])$ satisfiesFor the proof from the local inequality, take the lowest occupied level below the middle and replace that level by its upper shadow. The result remains an antichain: any new containment involving a set in the upper shadow would already give a containment involving the member of $\mathcal{A}$ immediately below it. The dual form of the Local LYM inequality says that this replacement cannot decrease the Lubell mass. Repeating upward below the middle, and similarly replacing high levels by their lower shadows, eventually puts the whole family in one middle level. Its final Lubell mass is at most one, so the original mass is also at most one.

For the maximal chain in a Boolean lattice proof, choose a Uniformly random maximal chain in a Boolean lattice. An $h$-element set lies on it with probability $1/\left(\genfrac{}{}{0ex}{}{n}{h}\right)$. Because an antichain meets each chain at most once, the expected value of the number of its members on the chain is at most one. By linearity of expectation, that expected value is precisely the displayed sum.

The Sperner theorem follows because $\left(\genfrac{}{}{0ex}{}{n}{|A|}\right)\le \left(\genfrac{}{}{0ex}{}{n}{\lfloor n/2\rfloor }\right)$ for every $A$. If equality holds in the resulting cardinality bound, every member lies on a largest level. For even $n$ this is the unique middle level. For odd $n$, the two middle levels have equal size; the regular connected inclusion graph between them and equality in Local LYM inequality force a chosen portion of the lower level to be either empty or the whole level. Hence the maximum antichains are exactly the complete middle level, with either middle level allowed when $n$ is odd.

Assume first that no two members of the uniform set family $\mathcal{A}\subseteq [n{]}^{(k)}$ meet in exactly one point. Fix $A_0\in \mathcal{A}$. Since $\mathcal{A}$ is an intersecting family, every $B\in \mathcal{A}$ then contains at least two elements of $A_0$. ConsequentlyThis proves the stated dichotomy.

If the small alternative holds, then for sufficiently large $n$ any fixed $i$ has all but at most $\epsilon \left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)$ members of $\mathcal{A}$ containing it. Otherwise choose $A,B\in \mathcal{A}$ with $A\cap B=\{i\}$. Every $C\in \mathcal{A}$ not containing $i$ must meet both $A\setminus \{i\}$ and $B\setminus \{i\}$, soFor sufficiently large $n$ this is at most $\epsilon \left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)$, as required.

Let $m=p^a$ and let an $r$-uniform set family have no distinct pair whose intersection is congruent to $r$ modulo $m$. Its size is bounded as displayed; the customary nontrivial form $\left(\genfrac{}{}{0ex}{}{n}{m-1}\right)$ requires $m-1\le r$. The Lucas theorem makes $\left(\genfrac{}{}{0ex}{}{z-r+m-1}{m-1}\right)$ a residue detector modulo $p$, equal to one exactly when $z\equiv r(\mathrm{mod}m)$. Its Boolean evaluation functions have degree at most $m-1$ and a diagonal evaluation matrix on the family. The low-degree evaluation rank on a uniform layer supplies the sharp dimension bound. Without the size qualification, three singletons with $n=3,r=1,m=4$ contradict the customary printed form, since $3>\left(\genfrac{}{}{0ex}{}{3}{3}\right)$.

A set family is shifted if every elementary set shift leaves it unchanged. For a uniform set family, this is equivalent to downward closure under componentwise comparison of the increasing lists of its elements. Repeated elementary set shifts terminate at a shifted set family.

Let $U,V\subseteq [n]$ be disjoint and have the same cardinality. For a set $A$, its $UV$-compression isFor a uniform set family $\mathcal{A}$, a member is replaced only when its image is absent from $\mathcal{A}$; this convention preserves the family's cardinality.