Line 8: Line 8:
 
Intersection is distributive over union ⇔ A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)  <br/>
 
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</math> <br/>
 
<math>A\cap (B\cup C) = (A\cap B)\cup (A\cap C</math> <br/>
where <math>A</math>, <math>B</math> and <math>C</math> are events in a probability space.
+
where <math>A</math>, <math>B</math> and <math>C</math> are sets.
  
  
Line 22: Line 22:
  
 
Combining the two results we have that: <br/>
 
Combining the two results we have that: <br/>
A ∩ (B ∪ C) ⊂ (A ∩ B) ∪ (A ∩ C) and (A ∩ B) ∪ (A ∩ C) ⊂ A ∩ (B ∪ C).
+
A ∩ (B ∪ C) ⊂ (A ∩ B) ∪ (A ∩ C) and (A ∩ B) ∪ (A ∩ C) ⊂ A ∩ (B ∪ C)
 
⇔  A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)<br/>
 
⇔  A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)<br/>
 
<math>\blacksquare</math>
 
<math>\blacksquare</math>
  
 
----
 
  
 
----
 
----

Latest revision as of 11:13, 1 October 2013


Theorem

Intersection is distributive over union ⇔ A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
$ A\cap (B\cup C) = (A\cap B)\cup (A\cap C $
where $ A $, $ B $ and $ C $ are sets.



Proof

Let x ∈ A ∩ (B ∪ C). Then, x ∈ A and x ∈ (B ∪ C) ⇒ x ∈ A and at the same time, x ∈ B or x ∈ C, possibly both ⇒ either x ∈ A and x ∈ B or x ∈ A and x ∈ C (possibly both). Hence, x ∈ (A ∩ B) or x ∈ (A ∩ C), i.e. x ∈ (A ∩ B) ∪ (A ∩ C).
So we have that x ∈ A ∩ (B ∪ C) ⇒ x ∈ (A ∩ B) ∪ (A ∩ C), which is equivalent to saying that A ∩ (B ∪ C) ⊂ (A ∩ B) ∪ (A ∩ C).

Next we assume that x ∈ (A ∩ B) ∪ (A ∩ C). Then, x ∈ (A ∩ B) or x ∈ (A ∩ C)⇒ x ∈ A in addition to being in B, C or both ⇒ x ∈ A ∩ (B ∪ C).
This gives us that x ∈ (A ∩ B) ∪ (A ∩ C) ⇒ x ∈ (A ∩ B) ∪ (A ∩ C)A ∩ (B ∪ C), which is equivalent to saying that (A ∩ B) ∪ (A ∩ C) ⊂ A ∩ (B ∪ C).

Combining the two results we have that:
A ∩ (B ∪ C) ⊂ (A ∩ B) ∪ (A ∩ C) and (A ∩ B) ∪ (A ∩ C) ⊂ A ∩ (B ∪ C) ⇔ A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
$ \blacksquare $



References

  • W. Rudin, "Basic Topology" in "Principles of Mathematical Analysis", 3rd Edition, McGraw-Hill Inc. ch 2, pp 28.



Back to list of all proofs

Alumni Liaison

Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett