Theorem
Let $ A $ be a set in S. Then
A ∩ Ø = Ø
Proof
First we show that A ∩ Ø ⊂ Ø.
We know this is true because the set resulting from the intersection of two sets is a subset of both of the sets forming the intersection (proof).
Next, we want to show that Ø ⊂ A ∩ Ø.
We know this is true because the empty set is a subset of all sets (proof).
Since A ∩ Ø ⊂ Ø and Ø ⊂ A ∩ Ø, we have that A ∩ Ø = Ø.
$ \blacksquare $
