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Repeatedly take the lower shadow of a uniform set family. After $j$ steps the result consists of all subsets of its members whose size is $j$ smaller. Iterating the Kruskal-Katona theorem shows that colexicographic initial segments minimize every iterated lower shadow.

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.