The InclusionExclusion Principle
Let
A
and
B
denote two finite sets.
Then, we have:

A
∪
B = A + B
– 
A
∩
B
.
This can easily be seen by a Venn Diagram:
Logically, we can argue that since each element of
A
∪
B
belongs to either
A
or
B
, the sum
A + B
includes a count for
each of the elements of
A
∪
B
, but those elements of
A
∩
B
are
counted twice.
Thus,
A + B
– 
A
∩
B
 counts each element of
A
∪
B
exactly
once, that is, it is equal to 
A
∪
B
.
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principle:
We first claim that the following is a disjoint union, meaning
that the two sets on the right of the equal sign have no elements
in common.
A = (A – B)
∪
(A
∩
B)
Thus, by the definition of set equality, we want to prove that
1. A
⊂
(A – B)
∪
(A
∩
B)
2. (A – B)
∪
(A
∩
B)
⊂
A
3. (A – B)
∩
(A
∩
B) =
∅
To prove 1, let x
∈
A.
Either x
∈
B or x
∉
B. In the first case,
x
∈
A
∩
B by definition, and in the second case, we have x
∈
A
and x
∉
B, which means x
∈
A – B by definition.
To prove 2, note that A – B
⊂
A because each x
∈
A – B must
also have x
∈
A by the definition of set difference.
Also, A
∩
B
⊂
A because each x
∈
A
∩
B must also have x
∈
A by the
definition of intersection. Thus, (A – B)
∪
(A
∩
B)
⊂
A by the
definition of set union and the subset relationship.
To prove 3, note that each x
∈
A – B must satisfy x
∉
B by the
definition of set difference.
Also, each x
∈
A
∩
B must satisfy
x
∈
B, by the definition of set intersection.
Thus, it is
impossible to have x
∈
(A – B)
∩
(A
∩
B), meaning that the set
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 Fall '09

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