lec0201 - 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 |. 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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More 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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lec0201 - The Inclusion-Exclusion Principle Let A and B...

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