Theorem
Let $ A $ and $ B<math> be sets. Then <br/> '''(a)''' (A ∩ B) ⊂ A '''(b)''' A ⊂ (A ∪ B) and Intersection is distributive over union ⇔ A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) <br/> <math>A\cap (B\cup C) = (A\cap B)\cup (A\cap C $
where $ A $, $ B $ and $ C $ are events in a probability space.
Proof
(a) let x ∈ (A ∩ B) ⇔ x ∈ A and x ∈ B ⇒ x ∈ A ⇒ (A ∩ B) ⊂ A.
(b) let x ∈ A. Then it is true that x is either in A or in B ⇔ x ∈ (A ∪ B) ⇒ A ⊂ (A ∪ B).
$ \blacksquare $
References
- B. Ikenaga, "Set Algebra and Proofs Involving Sets" March 1st, 2008, [October 1st, 2013]
