(New page: Category:ECE600 Category:Set Theory Category:Math == Theorem == A\(B ∪ C) = (A\B) ∩ (A\C) ---- ==Proof== First we show that every element in A\(B ∪ C) is contained...) |
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== Theorem == | == Theorem == | ||
| − | A\(B ∪ C) = (A\B) ∩ (A\C) | + | A\(B ∪ C) = (A\B) ∩ (A\C)<br/> |
| + | where A, B and C are sets. | ||
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Since the sets A\(B ∪ C) and (A\B) ∩ (A\C) contain the same elements, A\(B ∪ C) = (A\B) ∩ (A\C).<br/> | Since the sets A\(B ∪ C) and (A\B) ∩ (A\C) contain the same elements, A\(B ∪ C) = (A\B) ∩ (A\C).<br/> | ||
<math>\blacksquare</math> | <math>\blacksquare</math> | ||
| + | |||
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| + | ---- | ||
| + | |||
| + | '''Note'''<br/> | ||
| + | Using the above result, we can prove that (A ∪ B)' = A' ∩ B' because:<br/> | ||
| + | (A ∪ B)' = ''S''\(A ∪ B) = (''S''\A) ∩ (''S''\B) = A' ∩ B'. | ||
Latest revision as of 05:26, 6 October 2013
Theorem
A\(B ∪ C) = (A\B) ∩ (A\C)
where A, B and C are sets.
Proof
First we show that every element in A\(B ∪ C) is contained in both (A\B) and (A\C).
If x ∈ A\(B ∪ C), then x is in A, but x is not in (B ∪ C). Hence, x is in A and neither in B nor in C. So x is in A and not in B and x is in A but not in C. Therefore, x ∈ A\B and x ∈ A\C ⇒ x ∈ (A\B) ∩ (A\C). So we have that A\(B ∪ C) ⊂ (A\B) ∩ (A\C).
Next we show that if x is in (A\B) ∩ (A\C), then x is in A\(B ∪ C).
If x ∈ (A\B) ∩ (A\C), then x ∈ (A\B) or x ∈ (A\C). Hence x ∈ A and both x ∉ B and x ∉ C. So x ∈ A and x ∉ (B ∪ C) ⇒ x ∈ A\(B ∪ C). Therefore, (A\B) ∩ (A\C) ⊂ 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.
