Theorem
Let $ A $ be a set in S. Then
A ∩ Ø = Ø
Proof
Let x ∈ S, where S is the universal set.
First we show that A ∩ Ø ⊂ Ø.
We know this is true because the set resulting from the union of two sets is a subset of both of the sets (proof).
Next, we want to show that Ø ⊂ A ∩ Ø.
Let x ∈ Ø. The antecedent (i.e. the "if") part is false by definition of the empty set. Then x ∈ Ø ⇒ x ∈ (A ∩ Ø) is true and we have that Ø ⊂ A ∩ Ø.
Since A ∩ Ø ⊂ Ø and Ø ⊂ A ∩ Ø, we have that A ∩ Ø = Ø.
$ \blacksquare $
References
- B. Ikenaga, "Set Algebra and Proofs Involving Sets" March 1st, 2008, [October 1st, 2013]
