This completes the proof in the case .

Inductive Step: Suppose and that the statement is true for . The argument above shows that any subcollection of sets will have nonempty intersection. We may then consider the collection where we replace the two sets and with the single set . In this new collection, every subcollection of sets will have nonempty intersection. The inductive hypothesis therefore applies, and shows that this new collection has nonempty intersection. This implies the same for the original collection, and completes