Inquiry about proofs involving families of sets

[Syrus]

Homework Statement

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).

Homework Equations

The Attempt at a Solution

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?

 

[lippyka]

The answer to your question is that it depends on the context. For example, in theorem 1, the hypothesis of F and G being nonempty is necessary because the union of empty sets is an empty set, so you would not get the desired result without at least one nonempty set. On the other hand, for theorem 2, the hypothesis of F and G being nonempty is not necessary because the intersection of empty sets is also an empty set, so no additional conditions on F and G are needed. Thus, it depends on the specific theorem you are trying to prove as to whether or not the "nonempty" condition is necessary.