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.

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



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