Inquiry about proofs involving families of sets
This post does not concern a particular problem or exercise, but instead a peculiarity (for me) in one genre: proofs involving families of sets (that is, sets containing sets as elements). I have noticed that in some statements of theorems which involve families of sets, the hypothesis includes " let F and G be families of sets," whereas in others, the hypothesis is slightly altered to: "let F and G be nonempty families of sets." I have included two theorems (which I have already proven) as examples of this:
1. Suppose F and G are nonempty families of sets. Prove that U(F U G) = (UF) U (UG).
2. Suppose F and G are families of sets. Prove that U(F ∩ G) ⊆ (UF) ∩ (UG).
The difference obviously regards some property about the empty set. My original thought was that being nonempty allowed for the assertion of F or G containing some set- which arises during the course of the proof of 1. That is, x ∈ U(F U G) means there is some A ∈ F U G for which x ∈ A. But then, however, i noticed that the proof of theorem 2. also asserts the existence of some set which is an element of F (and G), without the "nonempty" portion of the hypothesis being present. Could this simply be a mistake on the part of the author, or am i missing a notion here?
Re: Inquiry about proofs involving families of sets
By definition, x ∈ ⋃F iff there exists an A ∈ F such that x ∈ A. Therefore, ⋃∅ = ∅. So, both claims above are true regardless whether F and G are empty. On the other hand, x ∈ ⋂F iff for all A ∈ F it is the case that x ∈ A. If there is an assumed universal set U, then ⋂∅ = U. However, in general ⋂∅ is not defined because there is no set of all sets. Thus, the claim ⋂(F ∪ G) ⊆ ⋂F ∪ ⋂G makes sense only for nonempty F and G.
One guess why the first statement in the OP requires nonempty sets is that the author knew that either general intersections or general unions are not defined for empty families but did not want to remember which one is not defined.