Is the complement of Cantor set C still measure zero? Meanwhile, I know its accumulation point is C itself (right?). So its closure would be C, correct? Why??? Notice: I am asking for the complement of Cantor set C over closed interval [0,1].
Answer
Let C be the Cantor set and X=[0,1]∖C its complement in the unit interval. Then m(X)=m([0,1])−m(C)=1, so the complement has measure 1. Furthermore the Cantor set is nowhere dense, so its complement must be dense and thus ¯X=[0,1].
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