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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.

We prove the vertex-isoperimetric inequality in the discrete cube by mathematical induction on $n$, with $n=0,1$ immediate. Fix a coordinate $i$ and write ${\mathcal{A}}_0,{\mathcal{A}}_1\subseteq Q_{n-1}$ for the sections without and with $i$. The two sections of its closed graph neighbourhood areA simplicial section compression replaces ${\mathcal{A}}_j$ by the initial segment $I_{a_j}$ of size $a_j=|{\mathcal{A}}_j|$ in the other coordinates. By the preceding observation, both $N(I_{a_j})$ and $I_{a_{1-j}}$ are initial segments, so their union has size $\max \{|N(I_{a_j})|,a_{1-j}\}$. The mathematical induction hypothesis givesAdding the two section bounds proves that simplicial section compression cannot increase the closed graph neighbourhood.

Apply changing simplicial section compressions until none remains. The sum of the global simplicial order on the discrete cube positions strictly decreases at each step: within a section the order is exactly the restriction of the global order. Thus the procedure terminates at a set family $\mathcal{B}$ every coordinate section of which is initial. We now classify the possible terminal families for simplicial section compression, rather than assuming this terminal set family must be globally initial.

If $X<Y$, $X\notin \mathcal{B}$ and $Y\in \mathcal{B}$, then $X,Y$ cannot agree in any coordinate, since that coordinate section would have an omitted earlier member and an included later one. Therefore $Y=X^c$. Let $X$ be the first omitted vertex and $Y$ the last included vertex. If $\mathcal{B}$ is noninitial, $X<Y$ and $Y=X^c$. Any vertex $Z$ strictly between them would have to be included, since an omitted $Z$ paired with $Y$ would have to equal $Y^c=X$. But it would also have to be omitted, since an included $Z$ paired with $X$ would have to equal $X^c=Y$. Thus there is no such $Z$: $X,Y$ are consecutive complementary vertices. Everything before $X$ is included and everything after $Y$ omitted. Hence $\mathcal{B}$ is obtained from the initial segment of its size by replacing $X$ with its immediate successor $X^c$.

There are only two forms of this exception. If $n=2r+1$, consecutive complementary vertices must straddle ranks $r,r+1$; otherwise an intermediate rank or a further vertex in one of their ranks separates them. They are consequentlywith $X$ the last $r$-subset and $Y$ the first $(r+1)$-subset. The corresponding initial segment is all ranks at most $r$. For $r\ge 1$, the omitted $X$ remains in $N(\mathcal{B})$ through a lower neighbour, and every $(r+1)$-subset has at least two $r$-subsets, only one of which was omitted. Thus $N(\mathcal{B})$ contains every rank at most $r+1$, namely $N(I_{|\mathcal{B}|})$. For $r=0$ both singleton closed graph neighbourhoods are the whole $Q_1$.

If $n=2r$, the two complementary vertices must both have rank $r$. The complement map reverses the lexicographic order in that rank, so a consecutive complementary pair must be its central pair. Exactly half the $r$-subsets contain $1$, and all of those precede those missing $1$. Therefore the pair isHere $I_{|\mathcal{B}|}$ comprises all ranks below $r$ and all $r$-subsets containing $1$. For $r\ge 2$, $N(\mathcal{B})$ contains every rank at most $r$, because all lower ranks were retained. Every $(r+1)$-subset containing $1$ has $r\ge 2$ different $r$-subsets containing $1$, so deletion of the one member $X$ removes no vertex of the upper shadow. Consequently $N(\mathcal{B})\supseteq N(I_{|\mathcal{B}|})$. For $r=1$, the two closed graph neighbourhoods both have size three in $Q_2$. These cases exhaust the exceptions. Since the successive simplicial section compressions never increased the closed graph neighbourhood, this proves the vertex-isoperimetric inequality in the discrete cube.

The same argument gives all integer radii. Write $N^t$ for $t$ repetitions of the closed graph neighbourhood. Its application to an initial segment is again initial; the sizes of these initial segments are monotone in the original size. Inducting on $t$ therefore gives

For the concentration question, use uniform probability measure on a finite connected graph and normalize its graph distance by its graph diameter. A sequence is a Lévy family of graphs if, for every fixed $\epsilon >0$ and $\delta >0$, the radius-$\epsilon$ neighbourhood of every set of measure at least $\delta$ has measure tending to one, uniformly over those sets. One may equivalently require this only for sets of measure at least one half. To see the reverse implication, first the radius-$\epsilon /2$ neighbourhood of every positive-density set must eventually have measure greater than one half. Otherwise its complement has measure at least one half, while the radius-$\epsilon /4$ neighbourhood of that complement misses the original positive-density set, contradicting the half-measure condition. Apply the half-measure condition once more to the radius-$\epsilon /2$ neighbourhood, enlarging it by another $\epsilon /2$, to obtain the full claim. The metric normalization is essential here.

For $Q_n$, the graph diameter is $n$. A uniformly chosen vertex has size $Z_n$ with binomial distribution $\mathrm{Bin}(n,1/2)$, expectation $n/2$ and variance $n/4$. Fix $\delta >0$ and choose $C>0$ so that $1/(4C^2)<\delta$. By Chebyshev's inequality,For sufficiently large $n$, put $k_n=\lfloor n/2-C\sqrt{n}\rfloor \ge 0$. Every initial segment of measure at least $\delta$ contains the complete ranks through $k_n$. Its radius-$t$ graph neighbourhood therefore contains every rank at most $k_n+t$. Taking $t=\lfloor \epsilon n\rfloor$, the all-radius vertex-isoperimetric inequality in the discrete cube gives, uniformly for $|\mathcal{A}|\ge \delta 2^n$,The last Chebyshev's inequality is used only when its denominator has positive square root. These are precisely the radius-$\epsilon$ neighbourhoods in the normalized graph distance. The discrete cubes form a Lévy family. In fact, taking $t=\lfloor n^{3/4}\rfloor$ in the same bound makes the omitted measure $O(n^{-1/2})$ for each fixed $\delta$: a radius that is $o(n)$ already suffices.

For fixed $d\ge 1$, the nearest-neighbour grid $[n{]}^d$ has graph distance ${\sum }_i|x_i-y_i|$ and graph diameter $d(n-1)$. Take the slab ${\mathcal{A}}_n=\{x:x_1\le \lceil n/2\rceil \}$, whose uniform measure is at least one half. Its integer radius-$t$ graph neighbourhood is exactly $\{x:x_1\le \min (n,\lceil n/2\rceil +t)\}$: changing the first coordinate alone attains the minimum distance to the slab. For $0<\epsilon <1/(2d)$ and $t=\lfloor \epsilon d(n-1)\rfloor$, the omitted proportion tends toThus there is failure of concentration in fixed-dimensional grids, and the fixed-dimensional grids do not form a Lévy family with normalized graph distance.

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

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

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

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

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

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

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

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

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

For an $r$-uniform set family $\mathcal{F}$, its lower shadow is the set family of all $(r-1)$-sets contained in a member. Write a positive integer $m$ in its unique binomial representationThe Kruskal-Katona theorem states thatFor $m=0$ the bound is zero. Equality is attained by the first $m$ $r$-sets in colexicographic order. In that order $A$ precedes $B$ when the largest element of their symmetric difference belongs to $B$.

Here is a complete UV-compression proof of the Kruskal-Katona theorem. For disjoint equal-sized sets $U,V$, put $C(A)=(A\setminus V)\cup U$ when $A$ contains every element of $V$ and no element of $U$, and put $C(A)=A$ otherwise. The UV-compression of a set family replaces an eligible $A$ only if its image is absent. It preserves cardinality: a retained collision keeps both the original set and its already present image.

We first prove the required Shadow lemma for UV-compressions. Suppose that, for each $x\in U$, a $y\in V$ can be chosen so the set family is fixed by $C_{U\setminus \{x\},V\setminus \{y\}}$. We claimConsider a lower shadow member obtained by deleting $z$ from a moved set $A^{\prime }=(A\setminus V)\cup U$. If $z\notin U$, it is the compressed image of $A\setminus \{z\}$, an old lower shadow member. If $z=x\in U$, the smaller compression of $A$ produces $A^{\prime }\setminus \{x\}\cup \{y\}$, which is in $\mathcal{F}$ by the assumed invariance. Thus $A^{\prime }\setminus \{x\}$ already belongs to the old lower shadow; it is not eligible to move since it lacks $x$.

It remains to consider a lower shadow member $B$ from a retained original member $A=B\cup \{z\}$. If $B$ is not eligible to move, it stays in the compressed lower shadow. If it is eligible, let $B^{\prime }=(B\setminus V)\cup U$. When $z\notin U$, retaining the eligible $A$ means its image already belongs to $\mathcal{F}$, and that image contains $B^{\prime }$. When $z=x\in U$, the smaller compression of $A$ belongs to $\mathcal{F}$ and equals $B^{\prime }\cup \{y\}$. In either case $B^{\prime }$ is in the old lower shadow too. Consequently the collision rule retains $B$ as well. This proves the claimed containment in every case. Since compression preserves cardinality, it cannot increase lower shadow size.

If $\mathcal{F}$ is not a colexicographic initial segment, an earlier absent set $A$ and a later present set $B$ give $U=A\setminus B$, $V=B\setminus A$ with $\max U<\max V$, and a nontrivial UV-compression. Choose such a compression with $|U|$ smallest. If $|U|>1$, choose $y\in V\setminus \{\max V\}$. For every $x\in U$, the smaller compression still has its largest differing element on the $V$ side and must fix $\mathcal{F}$ by minimality. If $|U|=1$, the smaller compression is the identity. Thus the lower shadow argument applies.

Each nontrivial step strictly decreases the positive integer weightbecause the largest differing coordinate belongs to $V$. Iteration therefore terminates. A terminal set family is a colexicographic initial segment, since an earlier gap and a later member would otherwise provide another nontrivial compression. The final set family has $m$ members and no larger lower shadow.

To compute its lower shadow, take $a_r$ maximal with $\left(\genfrac{}{}{0ex}{}{a_r}{r}\right)\le m$. All $r$-sets of $[a_r]$ occur first. The next block consists of sets containing $a_r+1$ whose remaining $(r-1)$-set runs through the initial segment of length $m-\left(\genfrac{}{}{0ex}{}{a_r}{r}\right)$. By Pascal's identity, the remainder is smaller than $\left(\genfrac{}{}{0ex}{}{a_r}{r-1}\right)$, so its next leading top index is strictly less than $a_r$. Repeating gives the stated unique binomial representation. The first block contributes all $\left(\genfrac{}{}{0ex}{}{a_r}{r-1}\right)$ $(r-1)$-sets in $[a_r]$ to the lower shadow. Deleting $a_r+1$ from the next block contributes only sets already counted; deleting another coordinate contributes $a_r+1$ joined to the lower shadow of the smaller initial segment. This recursive decomposition gives exactly the displayed sum. The base $r=1$ has lower shadow $\{\varnothing \}$ for every nonempty set family. This completes the proof and the sharpness assertion.

Now let $\mathcal{C}$ be the set family of vertex sets of the graph's cliques of order four. Its twice-iterated lower shadow consists of edges of the graph. If there were at least sixteen such copies, choose a sixteen-member subfamily. The relevant binomial representation isThe proved theorem gives at least $\left(\genfrac{}{}{0ex}{}{6}{3}\right)+\left(\genfrac{}{}{0ex}{}{3}{2}\right)=23$ triples in its first lower shadow. Apply it again to any twenty-three of those triples, using $23=\left(\genfrac{}{}{0ex}{}{6}{3}\right)+\left(\genfrac{}{}{0ex}{}{3}{2}\right)$. Their edge lower shadow has at leastmembers, contradicting the fifteen available edges. Conversely, the complete graph $K_6$ has $\left(\genfrac{}{}{0ex}{}{6}{2}\right)=15$ edges and $\left(\genfrac{}{}{0ex}{}{6}{4}\right)=15$ copies of $K_4$. Therefore the maximum number of copies is exactly 15.

Take the natural parameter range $1\le t\le r\le n$ and admissible indices $0\le i\le r-t$. Larger indices give empty set families. Write $q=r-t+1$ and set $n_0=\infty$ when discussing the first interval. For the nontrivial interval calculation, $t>1$ and $n>2r-t+1$; all prefixes used below then lie inside the ground set.

To compare consecutive complete-intersection candidate families, put $a=t+2i$ and $b=t+i$. A set in ${\mathcal{A}}_{i+1}\setminus {\mathcal{A}}_i$ has exactly $b-1$ points in $[a]$ and both of the next two points. A set in ${\mathcal{A}}_i\setminus {\mathcal{A}}_{i+1}$ has exactly $b$ points in $[a]$ and neither next point. ThereforeFor $0\le i<r-t$ these quantities are positive in the present range, and cancellation of binomial coefficients givesThe numerator minus the denominator isThus $|{\mathcal{A}}_{i+1}|$ is larger, equal or smaller than $|{\mathcal{A}}_i|$ according as $n$ is smaller, equal or larger than $n_{i+1}$. Since these thresholds decrease with the index, $n_{k+1}<n<n_k$ implies strict increase up to ${\mathcal{A}}_k$ and strict decrease thereafter:At an interior boundary $n=n_{k+1}$, the two consecutive maxima tie. The interval assertion concerns admissible $k$; taking $k>r-t$ would instead refer to an empty set family and is not a valid interpretation.

The Ahlswede-Khachatrian theorem, in its cardinality form, statesHere ${\mathcal{A}}_i$ is interpreted exactly by its intersection with $[t+2i]$, even if that prefix extends beyond $[n]$. Its size can always be written with $m_i=\min \{n,t+2i\}$ aswhere impossible binomial coefficients are zero. When $n\le 2r-t$, the whole level is $t$-intersecting and the formula gives $M=\left(\genfrac{}{}{0ex}{}{n}{r}\right)$. The nontrivial case has $n>2r-t$.

First, each candidate is a t-intersecting family: two members contain at least $2(t+i)-(t+2i)=t$ common points in the prefix. This proves the lower bound for $M$. We now prove the upper bound, using only the compression and generating-family facts expressly allowed in the question.

The first auxiliary fact is that the usual left coordinate shifts of a set family preserve cardinality and $t$-intersection. Iterating them produces a left-compressed set family of the same size. The shift $S_{ab}$ with $a<b$ replaces $b$ by $a$ only when $b$ is present, $a$ is absent, and the replacement is not already in the set family. A set family is left-compressed when it is fixed by every such shift. Right compression is the reflected notion.

For the second auxiliary fact, a generating family for a uniform set family $\mathcal{G}$ means a collection of sets of size at most $r$ withThe permitted small-support generating lemma for extremal intersecting families is this precise statement: if $\mathcal{F}$ is a maximum-cardinality left-compressed $t$-intersecting set family, $n>2r-t$, andthen it has a generating set family supported on $[t+2j]$. Reflection gives the same assertion for a right-compressed extremal set family, with the supporting interval at the right end. These two auxiliary results are being used as permitted lemmas; the extremal conclusion of the theorem is established in the remaining argument.

Choose a maximum left-compressed set family $\mathcal{F}$. Its complement set familyis right-compressed, has uniformity $r^{\prime }=n-r$ and intersection parameter $t^{\prime }=n-2r+t$, and is extremal for those parameters. Indeedso complementation gives a cardinality-preserving bijection between the two classes of intersecting set families. Notice that $r^{\prime }-t^{\prime }+1=q$ too.

We need a compatibility observation, which we prove. If $G$ generates $\mathcal{F}$ and $H$ generates ${\mathcal{F}}^{\ast }$, thenOtherwise extend $G\cup H$ to a set $T$ of size $n-r+t-1$. This size is at least both $r$ and $n-r$ because $n\ge 2r-t+1$ and $t\ge 1$. Choose an $r$-set $A\subseteq T$ containing $G$, and an $(n-r)$-set $B\subseteq T$ containing $H$. Their generating properties imply $A,[n]\setminus B\in \mathcal{F}$, buta contradiction. This compatibility bound for complementary generating families is the link between the two compression directions.

Suppose first that $t>1$ and $n>2r-t+1$ lies strictly between $n_{k+1}$ and $n_k$. Put $m=t+2k$ and $h=r-t-k$. The small-support lemma gives generators $\mathcal{G}$ for $\mathcal{F}$ supported on $[m]$. It also gives generators $\mathcal{H}$ for ${\mathcal{F}}^{\ast }$ supported on $[m+1,n]$. For completeness, the dual inequality required here isWriting $D=n-2q$, it is equivalent to $kD<q(t-1)$, exactly the upper interval inequality $n<n_k$; for $k=0$ it follows from $t>1$. The dual support length is $t^{\prime }+2h=n-m$.

Either every $G\in \mathcal{G}$ has $|G|\ge t+k$, or every $H\in \mathcal{H}$ has $|H|\ge n-r-k$. If neither assertion held, a pair of smaller generators would have union size at mostcontradicting the compatibility bound. In the first alternative, every member of $\mathcal{F}$ has at least $t+k$ points in $[m]$, so $\mathcal{F}\subseteq {\mathcal{A}}_k$. In the second alternative, each complement has at least $n-r-k$ points in $[m+1,n]$, leaving its original set with at most $r-t-k$ points there; again $\mathcal{F}\subseteq {\mathcal{A}}_k$. Therefore $M\le |{\mathcal{A}}_k|$, proving the theorem in every strict interval.

At an interior boundary $n=n_{k+1}$, with $0\le k\le r-t-1$, apply the small-support lemma at index $k+1$ to the original set family, since $n>n_{k+2}$. Its generators are supported on $[t+2k+2]$. The dual generators are still supported on $[t+2k+1,n]$, by the same dual calculation with $h=r-t-k$. Either all original generators have size at least $t+k+1$, or all dual generators have size at least $n-r-k$: failure of both would give union size at most $n-r+t-1$, again impossible. The first alternative places $\mathcal{F}$ inside ${\mathcal{A}}_{k+1}$, and the second places it inside ${\mathcal{A}}_k$. The comparison already proved gives $|{\mathcal{A}}_k|=|{\mathcal{A}}_{k+1}|$, so the common size is $M$.

It remains to cover the endpoint cases, not to assume them implicitly. For $t=1$ and $n>2r$, the small-support lemma with $j=0$ puts an extremal left-compressed set family's generators inside $\{1\}$. Its members must all contain one, giving $M\le \left(\genfrac{}{}{0ex}{}{n-1}{r-1}\right)$, attained by ${\mathcal{A}}_0$. For $t=1$, $n=2r$, at most one member of each complementary pair can occur, so $M\le \frac{1}{2}\left(\genfrac{}{}{0ex}{}{2r}{r}\right)=\left(\genfrac{}{}{0ex}{}{2r-1}{r-1}\right)$, again attained by ${\mathcal{A}}_0$. If $n<2r$, the whole level is intersecting.

For $t>1$ and $n=2r-t+1$, complementation changes the problem into a one-intersection problem with uniformity $q=r-t+1$. Since $n=2q+t-1>2q$, the preceding case givesattained by ${\mathcal{A}}_{r-t}$, the set family of $r$-sets avoiding the final point. For $n\le 2r-t$, every two $r$-sets meet in at least $2r-n\ge t$, so $M=\left(\genfrac{}{}{0ex}{}{n}{r}\right)$ and ${\mathcal{A}}_{r-t}$ is the whole level. Finally, $t=r$ gives $M=1$ directly. These cases, the strict intervals and their boundaries cover all $1\le t\le r\le n$, completing the proof of the stated maximum formula.