I can't think of a counterexample to disprove this set theory theorem

[the baby boy]

I can't think of a counterexample to disprove this set theory "theorem"

Assume F and G are families of sets.

IF \cupF \bigcap \cupG = ∅ (disjoint), THEN F \bigcap G are disjoint as well.

 

[micromass]

Think of using the empty set efficiently.

 

[the baby boy]

Just so I understand, an empty set in a family of sets would be the following?:
Family of sets F = {{1,2,3},{4,5,6},{∅}}?

Suppose I included an empty set in both of the families, would the intersection still be a disjoint or would it be the set {∅}?

 

[mXSCNT]

It doesn't hold if F = G = {\emptyset}.

 

[the baby boy]

So empty sets are not counted in a union, but they are in an interception?

 

[mXSCNT]

So empty sets are not counted in a union, but they are in an interception?

No, that's not it.
\cup F \bigcap \cup G = \cup \{\emptyset\} \bigcap \cup \{\emptyset\}
= \emptyset \bigcap \emptyset
= \emptyset
However, F \cap G = \{\emptyset\}, which is a non-empty set, so F and G are not disjoint.
(Note it doesn't really make sense to say "F \cap G is disjoint" - it takes 2 sets to be disjoint, and F \cap G is only a single set. So I assume you meant "F and G are disjoint" and not "F \cap G is disjoint."

 

[the baby boy]

Could you further clarify how the union of a family set consisting of just {∅} becomes ∅?

From my previous example, what would ∪F be?
Family of sets F = {{1,2,3},{4,5,6},{∅}}
∪F = {1, 2, 3, 4, 5, 6} or {∅, 1, 2, 3, 4, 5, 6}?