We follow Lubell’s elegant proof via the LYM inequality. Consider all \(n!\) permutations \(\sigma \) of \(\{ 1,\dots ,n\} \), each generating a maximal chain \(\{ \sigma (1)\} \subset \{ \sigma (1),\sigma (2)\} \subset \cdots \subset \{ 1,\dots ,n\} \). A \(k\)-element set \(A\) appears in exactly \(k!(n-k)!\) of these chains. Since an antichain \(\mathcal{F}\) meets each chain in at most one set, summing over \(\mathcal{F}\) gives

i.e. \(\sum _{A \in \mathcal{F}} \frac{1}{\binom {n}{|A|}} \le 1\) (the LYM inequality). Since each summand is at least \(1/\binom {n}{\lfloor n/2\rfloor }\), we conclude \(|\mathcal{F}| \le \binom {n}{\lfloor n/2\rfloor }\).

Consider the \(n\) points arranged on a circle. Each arc of length \(k\) consists of \(k\) consecutive points. Two arcs of length \(k\) on a circle of \(n \ge 2k\) points are disjoint if and only if their starting points are at distance \(\ge k\). Since there are \(n\) possible starting points and each arc “blocks out” \(2k - 1\) starting points for intersecting arcs, the maximum number of pairwise intersecting arcs is at most \(k\). (This lemma is subsumed by the Kruskal–Katona machinery in Mathlib’s proof of EKR.)

We follow Katona’s cyclic permutation argument. Arrange \(\{ 1,\dots ,n\} \) on a circle. Among the \(n\) arcs of \(k\) consecutive elements, at most \(k\) can be pairwise intersecting (by the arc lemma). For each of the \((n-1)!\) circular permutations, an intersecting family \(\mathcal{F}\) contributes at most \(k\) arcs. Each \(k\)-set appears as an arc in exactly \(k!(n-k)!\) circular permutations. Double counting gives

so \(|\mathcal{F}| \le \binom {n-1}{k-1}\).

Necessity is clear: the \(m\) distinct representatives of any \(m\) sets must lie in their union.

For sufficiency, we use induction on \(n\). Case 1: If Hall’s condition holds strictly (\(|\bigcup _{i \in S} A_i| \ge |S| + 1\) for every proper nonempty \(S \subsetneq \{ 1,\dots ,n\} \)), pick any \(a_1 \in A_1\) as representative, remove \(a_1\) from all sets, and verify that Hall’s condition still holds for \(A_2 \setminus \{ a_1\} , \dots , A_n \setminus \{ a_1\} \).

Case 2: If some proper nonempty subset \(S\) is “tight” (\(|\bigcup _{i \in S} A_i| = |S|\)), first find an SDR for the subfamily \((A_i)_{i \in S}\) by induction. Then show that the remaining sets \((A_j)_{j \notin S}\), with the chosen representatives removed, still satisfy Hall’s condition (using tightness of \(S\)), and apply induction again.