Erdős-Ko-Rado's theorem $:$
Let $\mathcal A$ be an intesecting family of $k$-subsets of $\{1,2, \cdots , n \}$ with $n \ge 2k$ and $|\mathcal A|=m$. Then $m \le \binom {n-1} {k-1}$.
A shorter proof of the above theorem was given by Katona by considering $n$-cycles. He first considered a collection $\mathcal F$ of $k$-intervals for the cycle $(1,2, \cdots , n)$ (say) of $\{1,2, \cdots ,n \}$ by taking the $k$-element subsets $F_i = \{i, i+1, \cdots , i+k-1 \}$ of $\{1,2, \cdots , n \}$ mod $n$ for $i=1,2, \cdots,n$. Then he claimed that for any intersecting family $\mathcal A$ of $k$-subsets of $\{1,2, \cdots , n \}$, $|\mathcal A \cap \mathcal F| \le k$.
Why is it so? Please help me in this regard.
Thank you very much.
I presume you want $k\le n/2$.
This argument comes from Bollobas's book.
Without loss of generality, assume that one of your sets in $\cal A\cap F$ is $\{1,2,\ldots,k\}$. Each other set in $\cal A\cap F$ is either $\{j-k+1,\ldots,j\}$ or $\{j+1,\ldots,j+k\}$ for $1\le j\le k-1$. But you cannot have both $\{j-k+1,\ldots,j\}$ and $\{j+1,\ldots,j+k\}$; they don't intersect. So you have at most one from each of these pairs of sets, so at most $k-1$ sets apart from the original, so at most $k$ sets in all.