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Originally posted by kel
The formal way to prove this is if you take the opposite of union (I forgot the formal name for this operation) of the two sets then it should the empty set. This means they are equivalent (I don't know what conformally equivalent is).
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at first i thought you meant intersection, but now i know you must mean the minus, ie: for two sets A and B, A\B={x in A: x not in B}. therefore, if the sets are equivalent, then A\B=null.
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Don't know if this proof holds water in the crazy set universe.
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well, at least for standard axioms (ZF?), it works.. to recap your steps:
if we name the two sets A and B respectively, then, if R is less than infinity, then we must have |A\B|>0 because some elements (an infinite number) will not be removed at the positive tail.
if r>0, then once again, |A\B|>0 since some elements (possibly infinite again if we are talking about say hte real numbers) will not be removed about the point 0 (zero).
therefore, A\B=null only if A=B. if A=B, then r=0 and R=infinity.
that's a clever way to prove seemingly obvious facts about sets, i'll use it some day, thx..
i wonder if the same would hold for complex numbers.. since there is no actual lessthan/greaterthan relation on complex numbers, only on their real moduli, could we still use this argument (putting aside "conformally" for a moment and just talking about actual set equivalence)?