= Solution
\b[A repeated trace produces the common set.] If $B_j=B_\ell$ for two distinct remaining indices, their common value $S$ has size $\lambda$. For any other remaining index $h$, the identity $|B_h\cap B_j|=\lambda=|S|$ forces $S\subseteq B_h$. The inclusion also holds for $j,\ell$ themselves. Since $S\subseteq K$, it lies in every $A_i$ indexed by $I$ as well. Consequently
$$
\boxed{|S|=\lambda,\qquad S\subseteq A_i\text{ for every }i.}
$$
This proves the common-set alternative of the <constant t-wise intersection dichotomy>. The conclusion only requires a common subset of size $\lambda$; for $m\geq t$, the full common intersection in fact also has size $\lambda$, because it is contained in a $t$-fold intersection of that size.
Back to article page
