(New page: Category:Set Theory Category:Math == Theorem == Let <math>A</math> and <math>B<math> be sets. Then <br/> '''(a)''' (A ∩ B) ⊂ A '''(b)''' A ⊂ (A ∪ B) and Intersection i...) |
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'''(a)''' let x ∈ (A ∩ B) ⇔ x ∈ A and x ∈ B ⇒ x ∈ A ⇒ (A ∩ B) ⊂ A. <br/> | '''(a)''' let x ∈ (A ∩ B) ⇔ x ∈ A and x ∈ B ⇒ x ∈ A ⇒ (A ∩ B) ⊂ A. <br/> | ||
| − | '''(b)''' let x ∈ A. Then it is true that x is either in A or in B ⇔ x ∈ (A ∪ B) ⇒ A ⊂ (A ∪ B). | + | '''(b)''' let x ∈ A. Then it is true that x is either in A or in B ⇔ x ∈ (A ∪ B) ⇒ A ⊂ (A ∪ B). <br/> |
<math>\blacksquare</math> | <math>\blacksquare</math> | ||
Revision as of 10:19, 1 October 2013
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]
