Quantum Gravity 343.8: C Such That P(CB ' ) = P(A) - P(B) is the "Real Conjugate" of the Symmetric Difference
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From: OsherD on 15 Nov 2009 16:54 From Osher Doctorow Edwin Hewitt and Karl Stromberg, "Real and Abstract Analysis," Springer-Verlag: N.Y. 1965 (pages 4 and 6), and Wikipedia's "Symmetric Difference" define the "Symmetric Difference" (Operation) DELTA by: 1) A DELTA B = (AB ' ) U (A ' B) for any sets A, B. Now recall from the last post: 2) P(CB ' ) = P(A) - P(B) But if A, B are bounded and intersect in a set of positive probability, then: 3) P(A) - P(B) = P(AB ' ) + P(AB) - [P(BA ' ] - P(BA) = P(AB ' ) - P (BA ' ) = P(AB ' ) - P(A ' B) On the other hand, taking probabilities on both sides of (1) (if defined) yields: 4) P(A DELTA B) = P(AB ' ) + P(A ' B) So the set CB ' , or more precisely the probability P(CB ' ), can be regarded as the "Real Conjugate" of P(A DELTA B) where A DELTA B is the symmetric difference of A and B. This isn't quite what the title of this post says, but the title was a "rough approximation". While this does not completely specify the set C, it does suggest that P(CB ' ) and CB ' is of more than usual interest, given some of the non-trivial properties of A DELTA B such as: 5) A(B DELTA C) = (AB) DELTA (AC) 6) A DELTA(B DELTA C) = (A DELTA B) DELTA C Osher Doctorow |