Real Analysis Chapter 1 Solutions

3. (a) Let M be an infinite σ-algebra of subsets of some set X. There exists a countably infinite subcollection C ⊆M, and we may choose C to be closed under taking complements (adding in missing complements if necessary). For each x ∈ X, define Dx := ∩{C ∈ C | x ∈ C}, so that Dx ∈ M. Let x, y ∈ X and suppose y ∈ Dx. Then y ∈ C for all C ∈ C with x ∈ C. Moreover, if y ∈ C for some C ∈ C, then y / ∈ Cc ∈ C, so x / ∈ Cc and hence x ∈ C. This implies that, if C ∈ C, then x ∈ C iff y ∈ C. In particular Dx = Dy. If x, y ∈ X and there exists z ∈ Dx ∩ Dy, then Dx = Dz = Dy, so the collection D := {Dx | x ∈ X} is pairwise disjoint. If C ∈ C, then C = ∪{Dx | x ∈ C}, since x ∈ Dx ⊆ C for all x ∈ C. Therefore, the image of the map ∪ : 2 → 2X contains C, so there exists a surjection from some subset of 2 onto C. If D were finite, then 2 would be as well, which is a contradiction because C is infinite. Therefore D is an infinite, pairwise disjoint subcollection of M.