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If I were to construct a Cantor Set by removing the middle $\alpha$ open interval, where $0 <\alpha < 1$, from each closed interval starting with $[0,1]$, how would I go about finding the total sum of the lengths of the deleted intervals?

My thought so far is that the length of the removed interval at the $n^{th}$ step is $\alpha(1-\alpha)^n$ and the sum is a geometric series resulting in $\frac{\alpha}{1-(1-\alpha)} = 1$. I would appreciate any help.

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  • $\begingroup$ Almost: the total length of the removed intervals at the $n$'th step is $\alpha (1-\alpha)^n$. Yes, the answer is $1$. $\endgroup$ – Robert Israel Mar 29 '18 at 7:38
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Step 1. Remove the middle interval of length $\alpha$, and you are left with two intervals $I_{1,1}$, $I_{1,2}$, each of length $(1-a)/2$.

Step 2. Remove the middle interval of length $\alpha\times \frac{1-a}{2}$, from each of the two intervals and you are left with four intervals, $I_{2,j}$, $j=1,2,3,4$, each of length $(1-a)^2/4$.

Step $n$. Remove the middle interval of length $\alpha\times \frac{(1-a)^{n-1}}{2^{n-1}}$, from each of the $2^{n-1}$ intervals and you are left with $2^n$ intervals, $I_{n,j}$, $j=1,\ldots,2^n$, each of length $(1-a)^n/2^n$.

Total length of removed intervals $$ \sum_{n=1}^\infty 2^{n-1}\frac{a(1-a)^{n-1}}{2^{n-1}}=1 $$

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