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

## Main Question or Discussion Point

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

Assume F and G are families of sets.

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

Related Set Theory, Logic, Probability, Statistics News on Phys.org

Think of using the empty set efficiently.

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 {∅}?

Last edited:

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

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

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."

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}?