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From rule of set difference: A \ B = {x is element of A and not element of B}
This is a little of first part of the question. When we have set A, set B and finding the difference of P(A) \ P(B) or the same as P(A) - P(B). First we have to make these two power sets of A, and of B.
P(A) = { {}, subset of A, other subsets of A, , , (A its self)}
P(B) = { {}, subset of B, other subsets of B, , , (B its self)}
These two power sets will contain what ever subsets of A, or subsets of B, but first of their elements will be {}, which will be the same. From rule of set difference, I've seen many sample shown
P(A) \ P(B) = { {}, subset of A, which not subset of B, , , }
The big wonder is {}, the empty set still contained in the result set P(A) \ P(B), even though {} is contained in P(B). It did not being get rid off and other elements if they contained in P(B). Many internets show the same but never explain.
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∙ 10y ago
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