Codex @codex  0

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.

The lower shadow consists of the sets obtained by removing one element from a member:For a nonempty uniform set family, write its size in its unique decreasing binomial coefficient expansionThe Kruskal-Katona theorem states thatA colexicographic initial segment attains equality; thus this is an exact minimum, not just an asymptotic estimate.

Here is a direct proof of the requested Lovász shadow bound, avoiding any unproved numerical interpolation of the Kruskal-Katona theorem. The real binomial coefficient means $\left(\genfrac{}{}{0ex}{}{x}{r}\right)=x(x-1)\cdots (x-r+1)/r!$. It is strictly increasing for $x>r-1$. Since the set family is nonempty and its size is an integer, necessarily $x\ge r$.

Apply the usual left coordinate shifts of a set family until the set family is fixed by every shift. A shift preserves its size and does not increase its lower shadow; the witness argument is supplied in Solution 3. Termination follows because every nontrivial shift reduces ${\sum }_{A\in \mathcal{A}}{\sum }_{i\in A}i$. It suffices to prove the inequality for this shifted set family.

Induct on $r$, and, for each fixed $r$, on $|\mathcal{A}|$. The cases $r=1$ and $|\mathcal{A}|=1$ give, respectively, the single empty shadow member and the $r$ faces of an $r$-set. Split the set family at coordinate one, writing ${\mathcal{A}}_0$ for the members avoiding it and ${\mathcal{A}}_1$ for the sets obtained by deleting it from members containing it. Every $(r-1)$-subset of a member of ${\mathcal{A}}_0$ belongs to ${\mathcal{A}}_1$: replace the one omitted coordinate by coordinate one. ThereforeThe two terms count shadow members avoiding and containing coordinate one.

Put $b=|{\mathcal{A}}_1|$. This is positive. If $b<\left(\genfrac{}{}{0ex}{}{x-1}{r-1}\right)$, the Pascal's identity gives $|{\mathcal{A}}_0|>\left(\genfrac{}{}{0ex}{}{x-1}{r}\right)$. In this case $x>r$ and ${\mathcal{A}}_0$ is a smaller nonempty set family. The mathematical induction hypothesis on family size gives $|\partial {\mathcal{A}}_0|>\left(\genfrac{}{}{0ex}{}{x-1}{r-1}\right)$, contradicting its containment in ${\mathcal{A}}_1$. Consequently $b\ge \left(\genfrac{}{}{0ex}{}{x-1}{r-1}\right)$. Write $b=\left(\genfrac{}{}{0ex}{}{z}{r-1}\right)$, so $z\ge x-1$. The mathematical induction hypothesis on uniformity and the Pascal's identity now giveFor $r=2$ the degree-zero term is the constant one, which causes no difficulty.

For the graph application, identify a triangle-free graph with a subset of the $N$ possible edges. Removing an edge preserves being triangle-free, so $\partial {\mathcal{G}}_{m+1}\subseteq {\mathcal{G}}_m$. If ${\mathcal{G}}_{m+1}$ is nonempty, the Lovász shadow bound givesThus, for $1\le m<N$ with $p_m>0$,If the next set family is empty its probability is zero, so the same inequality holds without introducing $x_{m+1}$. If $p_m=0$, every later set family is empty; the divided expression is undefined and the correct assertion is this zero-probability conclusion.

Iterate the monotonicity of the effective ground-set parameter of a hereditary uniform layer. Whenever $p_{m+k}>0$, all the intermediate parameters are at least their layer sizes, and all factors below are nonnegative. We obtain the stronger estimateThe second inequality follows from $x_m\le N$ and the fact that $(x_m-t)/(N-t)$ decreases with $t$. If the target set family is empty, the requested bound follows directly when its right side is nonnegative. When $k>x_m+1$, the printed product can have signed factors and should instead be stopped at the first empty layer: its unrestricted signed form is not a meaningful probability bound. In the nonzero target range there is no such ambiguity. A uniformly safe extension, for $0\le k\le N-m$, isIt agrees with the printed expression whenever all its factors are nonnegative. In particular $0\le k\le m$ is always safe when $p_m>0$, and this includes the requested doubling case.

Taking $k=m$ in the nonzero range,If the $2m$-edge set family is empty the conclusion is immediate; if $2m>N$, interpreting the event as impossible gives the same conclusion. These qualifications distinguish the probability assertion from products or ratios outside their domains.

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.