Thursday 12 January 2017

real analysis - The complement of Cantor set over closed interval 0 to1. What is its measure and closure??



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 $X=[0,1] \setminus C$ 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 $\overline{X}=[0,1]$.


No comments:

Post a Comment

real analysis - How to find $lim_{hrightarrow 0}frac{sin(ha)}{h}$

How to find $\lim_{h\rightarrow 0}\frac{\sin(ha)}{h}$ without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...