Let $\alpha$ be a fixed real number s.t. $0<\alpha<1$.
In stage one of the construction, remove the centrally situated open interval in $[0,1]$ of length $\alpha$. In stage 2, remove two central intervals each of relative length $\alpha$, one in each of the remaining intervals after stage 1, and so on.
Let $C_\alpha$ denote the set which remains after applying the above procedure indefinitely.
So that's the set up. I am now trying to prove that the complement of $C_\alpha$ in $[0,1]$ is the union of open intervals of total length equal to 1. I'm having trouble understanding how there can be an open interval at all in the complement, because to me it seems like any open interval would have to contain a element of the cantor set. But then it's common knowledge that cantor sets are nowhere dense, and therefore every open interval on $[0,1]$ must be contained in it's complement?
If anyone could shed some light on this for me I'd really appreciate it!
I believe I have proven that $m_*(C_\alpha)$=0.
To see this, I will prove that after the k'th stage, the remaining set has total length $(1-\alpha)^k$.
The base case is trvial, assume by induction that this holds for the case $n=k$.
To get to the kth step, we removed $2^{k-1}$ intervals, and so at this step we have $2^k$ intervals total. I want to calculate $\alpha_k$.
$((1-\alpha)^k/2^k)$/$(\alpha_k)$ = $1/\alpha$
Because on step k the total length is $(1-\alpha)^k$, and this total length is divided amongst $2^k$ intervals, and so this equation makes sense because $\alpha_k$ is defind as being relative to $\alpha$.
So $\alpha_k=\alpha(1-\alpha)^k/(2^k)$
The length at the k+1 stage will be the length of the kth stage minus the total length of the $2^k$ segments we remove, i.e.
$(1-\alpha)^k-2^k\alpha_k=(1-\alpha)^k-2^k\alpha(1-\alpha)^k/(2^k)=(1-\alpha)^k-\alpha(1-\alpha)^k=(1-\alpha)^k(1-\alpha)=(1-\alpha)^{k+1}$
The length of this interval will go to zero as $k \to \infty$
Best Answer
Let $C_0=[0,1.$ For $n\geq 0$ we have $C_n=\cup F_n$ where $F_n$ is a finite set of pairwise-disjoint closed intervals, each of positive length. And the measure $m(C_n)$ is $\sum_{f\in F_n}m(f).$
For $n\geq 1,$ at stage $n$ we remove an open interval $I_f$ from each $f\in F_{n-1},$ with $m(I_f)=\alpha \cdot m(f).$ So the measure $$m( C_n)=\sum_{f\in F_{n-1}}(1-\alpha)m(f)=(1-\alpha)m(C_{n-1}).$$ So by induction on $n$ we have $m(C_n)=(1-\alpha)^n\cdot m(C_0)=(1-\alpha)^n. $
Therefore $m(\cap_{n\geq 0}C_n)\leq \inf_{n\geq 0}m(C_n)=0.$
For $n\geq 0$ let $L(n)$ be the length of the longest member of $F_n.$ Each $f\in F_n$ has a central open piece $I_f$ of length $\alpha \cdot m(f)$ removed at stage $n+1,$ resulting in $2$ members $f',f''$ of $F_{n+1},$ each of length $\frac {1-\alpha}{2}m(f).$ Therefore by induction on $n$ we have $$L(n)= ((1-\alpha)/2)^{-n}L(0)<2^{-n}L(0)=2^{-n}.$$ So if $K$ is an interval in $[0,1]$ of length $m(K)>0,$ then for any $n$ large enough that $2^{-n}<K$ we have $K\not \subset C_n,$ and a fortiori we have $K \not \subset \cap_{n\geq 0}C_n.$
The open intervals removed at stage $n\geq 1$ are disjoint from each other and are disjoint from the open intervals removed at any previous stage. So the family $I$ of all open intervals that are removed is a pair-wise disjoint family, so $$\sum_{i\in I}m(i)=m(\cup_{i\in I}\;i)=m(\cup I)=m([0,1]\backslash \cap_{n\geq 0}C_n)=1.$$
BTW some students erroneously suppose that $C=\cap_{n\geq 0}C_n$ consists entirely of the end-points of the members of $I$. But all limit-points of all convergent sequences of those end-points also belong to $C$.