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

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

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

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

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

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

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

Identify the Boolean hypercube $Q_n$ with the subsets of $[n]$. For a family $\mathcal{H}$, write $N(\mathcal{H})$ for its closed vertex neighbourhood, consisting of the family and all vertices at Hamming distance one from it. Order subsets first by increasing size and, within a fixed size, use lexicographic order: at the smallest coordinate on which they differ, the set containing that coordinate comes first. This is the simplicial order on the discrete cube. Harper's theorem states that if $I$ is its initial segment of size $|\mathcal{H}|$, thenThus simplicial initial segments minimize the external vertex boundary for each prescribed family size. The closed-neighbourhood and external-boundary formulations are equivalent because $|N(\mathcal{H})\setminus \mathcal{H}|=|N(\mathcal{H})|-|\mathcal{H}|$. No edge boundary is involved in this assertion.

We now give the deduction Harper theorem implies the Kruskal-Katona theorem, keeping track of the two different orders. For a family $\mathcal{F}$ of $r$-sets with $r\ge 1$, adjoin every smaller set:Its closed vertex neighbourhood is exactlywhere $\nabla \mathcal{F}$ is the upper shadow. Applying Harper theorem to $\mathcal{H}$ shows that, among $r$-uniform families of fixed size, the lexicographic initial segment minimizes the upper shadow: all the lower-level contributions to the two neighbourhood sizes are identical.

To convert this into a statement about the lower shadow of a $k$-uniform family $\mathcal{G}$, take $r=n-k$ and map each member $S$ to $R([n]\setminus S)$, where $R(i)=n+1-i$ reverses the ground-set coordinates. Complementation changes lower shadows to upper shadows, and the coordinate reversal preserves their sizes. Moreover, the resulting lexicographic order corresponds exactly to colexicographic order on the original $k$-sets: $S$ precedes $T$ in colexicographic order precisely when the largest element of $S△T$ lies in $T$. After complementation and reversal, the smallest differing coordinate lies in the image of $S$, as required for the stated lexicographic order. Hence a colexicographic initial segment minimizes the lower shadow, which is the Kruskal-Katona theorem. The cases $k=n$ and $k=0$ are immediate and need no adjoining construction.

For its usual numerical form, write the unique greedy binomial representationThen the Kruskal-Katona theorem saysHere is why the shadow of the colexicographic initial segment has exactly that size. Its first block consists of all $k$-sets in $[a_k]$. If any members remain, they have the form $\{a_k+1\}\cup T$, where $T$ runs through a colexicographic initial segment of $(k-1)$-sets. The lower shadow consists of all $(k-1)$-sets in $[a_k]$, together with $\{a_k+1\}$ adjoined to the lower shadow of that remainder. These two blocks are disjoint. Recursing gives precisely the displayed sum. For $m=0$, both the initial segment and its shadow are empty.

The Erdős-Ko-Rado theorem states that an intersecting family $\mathcal{A}\subseteq \left(\genfrac{}{}{0ex}{}{[n]}{r}\right)$, with $1\le r\le n/2$, satisfiesThe bound is attained by all $r$-sets containing one fixed element. For the proof Erdős-Ko-Rado theorem from shadows, take complements to obtain an $(n-r)$-uniform family $\mathcal{B}$ of the same size, and let $D$ be its rank-$r$ lower shadow. Every member of $D$ is disjoint from some member of $\mathcal{A}$, so $D\cap \mathcal{A}=\varnothing$. Iterating the Kruskal-Katona theorem bounds $|D|$ below by the rank-$r$ shadow of the colexicographic initial segment of size $|\mathcal{B}|$. This iteration is valid because lower shadows of colexicographic initial segments are again colexicographic initial segments, as the preceding block description shows.

If $|\mathcal{A}|>\left(\genfrac{}{}{0ex}{}{n-1}{r-1}\right)=\left(\genfrac{}{}{0ex}{}{n-1}{n-r}\right)$, that initial segment contains all $(n-r)$-sets in $[n-1]$ and at least one further set containing $n$. Since $r\le n-r\le n-1$, its rank-$r$ shadow contains every $r$-set in $[n-1]$ and at least one $r$-set containing $n$. Thus $|D|>\left(\genfrac{}{}{0ex}{}{n-1}{r}\right)$, implyingcontrary to their disjointness. This proves the bound. If $r>n/2$, all $r$-sets intersect and the trivial maximum is instead $\left(\genfrac{}{}{0ex}{}{n}{r}\right)$.

For an intersecting family in a product alphabet, label the alphabet by $\mathbb{Z}/k\mathbb{Z}$. Partition the words into the classesEach class has $k$ words, and any two distinct words in it disagree in every coordinate. An intersecting family therefore contains at most one word per class. There are $k^{n-1}$ classes, and fixing the first coordinate gives an intersecting family of that size. Consequently the exact maximum isFor $k=1$ this just says that the single word may be included.

Finally, for a weakly intersecting family in a product alphabet, associate to each word its support $S(x)=\{i:x_i>1\}$. The occurring supports form an intersecting family of nonempty subsets, and the fibre over support $S$ has $(k-1{)}^{|S|}$ words. Put $w=k-1$. When $k\ge 2$, $w\ge 1$, and a support family can contain at most one of each complementary pair $S,[n]\setminus S$. For odd $n$, the larger support in each pair has size greater than $n/2$, soAll supports of size greater than $n/2$ are mutually intersecting, so taking every word over those supports achieves this bound. Thus the required extremal family is the strict-majority support family, with sizeThis is the weighted intersecting family bound on an odd Boolean lattice. For $k>2$, strict inequality between the complementary fibre weights forces this extremal family to be unique. For $k=2$, it is a largest family but need not be the only one; for example, all supports containing one fixed coordinate also attain the same size. For $k=1$, no word has a nonempty support, so the maximum is zero, again agreeing with the formula.