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prove that [if (A)=n then the power set of A=2 tothe power n]. i've been doing this since yesterday
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try by induction
if (A) = 1 => (P(A)) = 1; Let's (A) = k and (P(A))=2^k. Add one element x to A, P(A and x) = P(A) + {B and x}, where B in P(A), so ({B and x}) = 2^k And (P(A and x)) = 2^k + 2^k= 2^(k+1)
Imagine all n elements are sitting in a pile next to the set. Pick up the first element, you can put it in the set or not, 2 choices. Ditto all the other elements. So 2*2*2*......=2^n
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