Complete the proof that  = fA + fB − 2fAfB [Theorem]. Suppose x  A and x  B. Then fA(x) =...

Complete the proof that  = fA + fB − 2fAfB [Theorem]. Suppose x  A and x  B. Then fA(x) = _____, fB(x) = _____, and fA(x)fB(x) = ______, so fA(x) + fB (x) − 2 fA(x) fB (x) = _____ Now suppose x  A and x  B. Then fA (x) = _____, fB (x) = _____, and fA (x) fB (x) = _____, so fA(x) + fB (x) − 2 fA (x) fB (x) = _____. The remaining case to check is x  . If x   then x  _____ and fA (x) + fB (x) − 2 fA (x) fB (x) = _____. Explain how these steps prove Theorem.

Theorem

Characteristic functions of subsets satisfy the following properties:

(a) fAB = fA fB; that is, fAB(x) = fA(x) fB(x) for all x.

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(b) fAB = fA+ fB− fA fB; that is, fAB(x) = fA(x)+ fB(x) − fA(x) fB(x) for all x.

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(c) = fA+ fB − 2 fA fB; that is,  (x) = fA(x) + fB(x) − 2 fA(x) fB(x) for all x.

Proof

(a) fA(x) fB(x) equals 1 if and only if both fA(x) and fB(x) are equal to 1, and this happens if and only if x is in A and x is in B, that is, x is in AB. Since fA fB is 1 on A  B and 0 otherwise, it must be f AB.

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(b) If x  A, then fA(x) = 1, so fA(x)+ fB(x) − fA(x) fB(x) = 1+ fB(x) − fB(x) = 1. Similarly, when x  B, fA(x) + fB(x) − fA(x) fB(x) = 1. If x is not in A or B, then fA(x) and fB(x) are 0, so fA(x) + fB(x) − fA(x) fB(x) = 0. Thus fA + fB − fA fB is 1 on A  B and 0 otherwise, so it must be f AB.

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(c) We leave the proof of (c) as an exercise.