Lemma 5. There is an ultrafilter U on D disjoint from K , containing every G_t , and satisfying X∈ U t< c:X⊆_ K G_t is finite.

Call Z⊆D forbidden if it is contained modulo K in infinitely many of the G_t . Require a filter to contain all complements of members of K , all G_t , and all complements of forbidden sets. We verify the finite-intersection property. Combine the finitely many K -small exclusions into one set N∈ K . Let F be the finite set of required positive G -indices, and let Z_1,…,Z_q be the forbidden sets to exclude. For each j , choose an index t_j such that Z_j G_ t_j ∈ K , with the t_j distinct and outside F . Each Z_j has infinitely many witnessing indices, so these choices are possible. The pattern P=⋂_ t∈F G_t∩⋂_ j=1 ^qG_ t_j ^c is K -positive. Moreover P∩Z_j⊆Z_j G_ t_j ∈ K . It follows that P(N∪⋃_ j=1 ^qZ_j) is still K -positive: otherwise P would be a finite union of members of K . This nonempty set lies in the required finite intersection. Extend the resulting proper filter to an ultrafilter by the ultrafilter lemma. It contains no member of K and no forbidden set, as required.