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# lec0919 - The Inclusion-Exclusion Principle Let A and B...

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The Inclusion-Exclusion 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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Here is a more rigorous proof of the inclusion-exclusion 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.
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