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The Erdős-Ko-Rado theorem asserts that, for $1\le r<n/2$, every intersecting family $\mathcal{A}\subseteq \left(\genfrac{}{}{0ex}{}{[n]}{r}\right)$ satisfiesThe bound is attained by the set family of all $r$-subsets containing one prescribed point. Here lexicographic order means that $X<Y$ when the smallest element of their symmetric difference belongs to $X$.

The first $M=\left(\genfrac{}{}{0ex}{}{n-1}{r-1}\right)$ members in lexicographic order are exactly those containing $1$. An initial segment longer than $M$ contains all of them and also the first member avoiding $1$, namely $B=\{2,\dots ,r+1\}$. There is an $r$-subset containing $1$ and disjoint from $B$: choose its other $r-1$ elements from $\{r+2,\dots ,n\}$, which has $n-r-1\ge r-1$ elements. Consequently any longer lexicographic initial segment fails to be an intersecting family. Thus preservation of the intersection property upon replacing a set family by the equally large lexicographic initial segment proves the displayed bound.

Here is a direct UV-compression proof of that preservation statement. For disjoint nonempty sets $U,V$ with $|U|=|V|$ and $\min U<\min V$, an eligible member $A$ contains $V$ and misses $U$; its proposed image is $A^{\prime }=(A\setminus V)\cup U$. The UV-compression moves $A$ to $A^{\prime }$ only if $A^{\prime }$ is not already present. Each eligible pair is either kept as a pair or moved from its later to its earlier member, so this operation preserves cardinality and uniformity and strictly decreases the sum of lexicographic order positions whenever it changes the set family.

The essential intersection-preserving lexicographic UV-compression lemma is that this operation preserves an intersecting family whenever all smaller lexicographically improving UV-compressions already leave that set family unchanged. Suppose instead that the compressed set family has disjoint members. At least one, say $A^{\prime }$, must be newly moved. Both newly moved members would contain $U$, so the other, $B$, is a retained old member. Write $A=(A^{\prime }\setminus U)\cup V$ for the old preimage. Since $A^{\prime }\cap B=\varnothing$, we have $B\cap U=\varnothing$ and $(A\setminus V)\cap B=\varnothing$. Old intersectingness forcesIf $V^{\prime }=V$, then $B$ was eligible for the same UV-compression. Since it was retained, $B^{\prime }=(B\setminus V)\cup U$ was already present. But $A\cap B^{\prime }=\varnothing$, contradicting old intersectingness.

Otherwise $0<|V^{\prime }|<|V|$. Choose $U^{\prime }\subset U$ with $|U^{\prime }|=|V^{\prime }|$ and $\min U\in U^{\prime }$. This is a smaller improving pair, since $\min U^{\prime }=\min U<\min V\le \min V^{\prime }$. The old member $A$ is eligible for its UV-compression. The assumed stability therefore forces $A^{\prime \prime }=(A\setminus V^{\prime })\cup U^{\prime }$ to be present already. Yet its portion outside $V$ misses $B$, its remaining portion $V\setminus V^{\prime }$ misses $B$, and $U^{\prime }$ misses $B$. Thus $A^{\prime \prime }\cap B=\varnothing$, the required contradiction. This proves the lemma, including the size-one case, where only the first alternative can arise.

Starting with $\mathcal{A}$, repeatedly choose a changing improving UV-compression with the smallest possible $|U|$. The lemma applies at each step because every smaller improving UV-compression is then unchanged. Thus every intermediate set family is intersecting. The strictly decreasing nonnegative integer sum of lexicographic order positions ensures termination. If the terminal set family were not an initial segment, there would be a missing $X$ earlier than a present $Y$. Taking $U=X\setminus Y$ and $V=Y\setminus X$ gives equally sized nonempty disjoint sets with $\min U<\min V$, and moves $Y$ to the missing $X$. That contradicts termination. The terminal set family is therefore exactly the equally large lexicographic initial segment, and it is intersecting. This proves the required lexicographic statement directly, and hence the Erdős-Ko-Rado theorem.