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

  • #1

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 [itex]\cup[/itex]F [itex]\bigcap[/itex] [itex]\cup[/itex]G = ∅ (disjoint), THEN F [itex]\bigcap[/itex] G are disjoint as well.
 

Answers and Replies

  • #2
22,097
3,281


Think of using the empty set efficiently.
 
  • #3


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:
  • #4
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It doesn't hold if F = G = {[itex]\emptyset[/itex]}.
 
  • #5


So empty sets are not counted in a union, but they are in an interception?
 
  • #6
315
1


So empty sets are not counted in a union, but they are in an interception?
No, that's not it.
[itex]\cup F \bigcap \cup G = \cup \{\emptyset\} \bigcap \cup \{\emptyset\}[/itex]
[itex]= \emptyset \bigcap \emptyset[/itex]
[itex]= \emptyset[/itex]
However, [itex]F \cap G = \{\emptyset\}[/itex], which is a non-empty set, so F and G are not disjoint.
(Note it doesn't really make sense to say "[itex]F \cap G[/itex] is disjoint" - it takes 2 sets to be disjoint, and [itex]F \cap G[/itex] is only a single set. So I assume you meant "F and G are disjoint" and not "[itex]F \cap G[/itex] is disjoint."
 
  • #7


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}?
 

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