2
$\begingroup$

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 $1-\alpha$?

$\endgroup$
  • 1
    $\begingroup$ 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
4
$\begingroup$

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 $1-\alpha$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.