Let $F$ be the subset of $[0,1]$ constructed in the same manner as the Cantor set except that each of the intervals removed at $n$th iteration has a length $\frac{\alpha}{3^{n}}$ with $0<\alpha<1$ rather than $\frac{1}{3^{n}}$.

I've been told that on each iteration, each of the intervals removed has length $\frac{\alpha}{3^{n}}$, but that the total length of $F$ is $1-\alpha$.

I'm trying to show this by taking the limit as $n \to \infty$ of the length of the intervals left behind at each iteration. For example, on the second iteration, the way I see it, two intervals of length $\frac{\alpha}{3^{2}}$ should be removed, so the length left behind would be $1-\frac{\alpha}{3} - \frac{2\alpha}{9} = 1-\frac{5\alpha}{3^{2}} = 1-\left(\alpha-\frac{2\alpha}{3}\right)$.

So, my question is, a I correct in assuming that after the $n$th iteration, what is left behind will be $1-\left(\alpha - \left(\frac{2}{3}\right)^{n}\right)$, which, as $n \to \infty$, converges to $1-\alpha$?

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    $\begingroup$ Note that $\frac{5\alpha}{3^2} \neq \alpha - \frac{2\alpha}{3}$. Your guess at what's left behind is incorrect. Think about what happens when $n = 3, 4$. You should be able to see a different pattern emerging. $\endgroup$ – Michael Albanese Oct 21 '15 at 20:28

At the first step, as you say, we are removing an interval of length $\frac{\alpha}{3}$. At the second step we are going to remove two intervals of length $\frac{\alpha}{3^2}$, for a total length of $\frac{2\alpha}{3^2}$. In the same way, at the $n$-th step we will remove $2^{n-1}$ intervals of length $\frac{\alpha}{3^n}$, for a total length of $\frac{2^{n-1}\alpha}{3^n}$.

Now, to compute the total length of what's left behind, we can just compute the amount of length we removed and then subtract it from $1$. To do so, we need to sum all the lengths of those intervals we have removed:

$$ \sum_{n=1}^{\infty} \frac{2^{n-1}\alpha}{3^n}=\frac{\alpha}{3}\sum_{n=1}^{\infty}\left(\frac23\right)^{n-1}=\frac{\alpha}{3}\sum_{n=0}^{\infty}\left(\frac23\right)^n=\frac{\alpha}{3}\frac{1}{1-\frac23}=\frac{\alpha}{3}3=\alpha. $$

Then, the total length "left behind", or more precisely, the length of your generalized Cantor set is $1-\alpha$.


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