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We know this is true because the set resulting from the union of two sets is a subset of both of the sets ([[Union_and_intersection_subsets_mh|proof]]). | We know this is true because the set resulting from the union of two sets is a subset of both of the sets ([[Union_and_intersection_subsets_mh|proof]]). | ||
| − | Next, we want to show that | + | Next, we want to show that Ø ⊂ A ∩ Ø.<br/> |
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. | 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. | ||
Revision as of 05:00, 6 October 2013
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]
