Theorem
Let $ A $ be a set in S. Then
A ∪ Ø = A
Proof
Let x ∈ S, where S is the universal set.
First we show that if A ∪ Ø ⊂ A.
Let x ∈ A ∪ Ø. Then x ∈ A or x ∈ Ø. by definition of the empty set, x cannot be an element in Ø. So by assumption that x ∈ A ∪ Ø, x must be in A. So A ∪ Ø ⊂ A
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 A ⊂ A ∪ Ø.
We know this is true because the set resulting from the union of two sets contains both of the sets forming the union (proof).
Since A ∪ Ø ⊂ A and A ⊂ A ∪ Ø, we have that A ∪ Ø = A.
$ \blacksquare $
References
- B. Ikenaga, "Set Algebra and Proofs Involving Sets" March 1st, 2008, [October 1st, 2013]
