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

A
∪
B = A + B
– 
A
∩
B
.
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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View Full DocumentMore precisely, we first claim that the following is a disjoint
union:
A
= (
A
–
B
)
∪
(
A
∩
B
)
(1)
Thus, by the definition of set equality, we want to prove that
A
⊆
(
A
–
B
)
∪
(
A
∩
B
)
(2)
(
A
–
B
)
∪
(
A
∩
B
)
⊆
A
(3)
and (
A
–
B
)
∩
(
A
∩
B
)
=
∅
(4)
To prove (2), let
x
∈
A
.
Since either
x
∈
B
or
x
∉
B
is true: in
the former case,
x
∈
A
∩
B
by definition, and in the latter case,
we have
x
∈
A
and
x
∉
B
, which means
x
∈
A
–
B
by
definition.
To prove (3), 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 (4), 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
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 Spring '09
 Set Theory, Empty set, Basic concepts in set theory, Disjoint union

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