SOLUTION: An indicator function IA satisfies the condition IA=0(X does not belong to A) IA=1(X belongs to A) How to prove the formula of an indicator function： I（A Union B）=Max(IA

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Question 1163527: An indicator function IA satisfies the condition
IA=0(X does not belong to A)
IA=1(X belongs to A)
How to prove the formula of an indicator function：
I（A Union B）=Max(IA,IB)=IA+IB-I（A intersection B）
Answer by solver91311(24713)   (Show Source): You can put this solution on YOUR website!

Case I:  I(A U B) = 0
If an element is not in the union of two sets, then it is not in either set, hence:

IA = 0, IB = 0, I(A ∩ B) = 0, IA + IB - I(A ∩ B) = 0, and Max(IA,IB) = 0,

therefore I(A U B) = Max(IA,IB) = IA + IB - I(A ∩ B)

Case II: I(A U B) = 1
If an element is in the union of two sets, then it is in one or the other or both sets

Case IIa: IA = 1, IB = 0 => I(A ∩ B) = 0, so Max(IA,IB) = IA = 1, IA + IB - I(A ∩ B) = 1 + 0 - 0 = 1

Case IIb: IA = 0, IB = 1 => I(A ∩ B) = 0, so Max(IA,IB) = IB = 1, IA + IB - I(A ∩ B) = 0 + 1 - 0 = 1

Case IIc: IA = 1, IB = 1 => I(A ∩ B) = 1, so Max(IA,IB) = IA or IB = 1, IA + IB - I(A ∩ B) = 1 + 1 - 1 = 1

Q.E.D by exhaustion.



John

My calculator said it, I believe it, that settles it
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