Theorem

A\(B ∩ C) = (A\B) ∪ (A\C)
where A, B and C are sets.



Proof

If x ∈ A\(B ∩ C), then x is in A, but x is not in (B ∩ C). Hence, x ∈ A and both x ∉ B and x ∉ C. So either x ∈ A and x ∉ B or x ∈ A and x ∉ C ⇒ x ∈ (A\B) or x ∈ (A\C) ⇒ x ∈ (A\B) ∪ (A\C).

If x ∈ (A\B) ∪ (A\C), then x ∈ (A\B) or x ∈ (A\C).
If x ∈ (A\B), then x ∈ A and x ∉ B. But (B ∩ C) ⊆ B (proof). Therefore, x ∉ (B ∩ C) ⇒ x ∈ A\(B ∩ C).
Likewise, if x ∈ (A\C), then x ∈ A and x ∉ C. But (B ∩ C) ⊆ C (proof). Therefore, x ∉ (B ∩ C) ⇒ x ∈ A\(B ∩ C).

Since the sets A\(B ∩ C) and (A\B) ∪ (A\C) contain the same elements, A\(B ∩ C) = (A\B) ∪ (A\C).
$ \blacksquare $


Note
Using the above result, we can prove that (A ∩ B)' = A' ∪ B' because:
(A ∩ B)' = S\(A ∩ B) = (S\A) ∪ (S\B) = A' ∪ B'.



References

  • R. G. Bartle, D. R. Sherbert, "Sets and Functions" in "Introduction to Real Analysis", 3rd Edition, John Wiley and Sons, Inc. 2000. ch 1, pp 3.



Back to list of all proofs

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang