Two questions:

(1.) Construct a subset of $[0,1]$ in the same manner as the Cantor set, except that at the $k$-th stage, each interval removed has length $\delta 3^{-k}$, $0<\delta <1$. Show that the resulting set is perfect, has measure $1-\delta$, and contains no intervals.

Showing that it's perfect is not difficult. The resulting set is an intersection of closed intervals, so is closed. Any point in the set is a limit point of endpoints of intervals (obviously I have more details in my written solution, but that's the idea.) But I don't see how the measure is $1-\delta$... At stage $k$, we remove $2^{k-1}$ intervals of length $\delta 3^{-k}$, so the measure of the resulting set is $$1-\sum\limits_{k=0}^{\infty}{2^{k}\cdot\dfrac{\delta}{3^{k+1}}} = 1-\dfrac{\delta}{3}\cdot\sum\limits_{k=0}^{\infty}{\dfrac{2^k}{3^k}} = 1-\dfrac{2\delta}{3} .$$


(2.) Construct a Cantor-type set subset of $[0,1]$ by removing from each interval remaining at the $k$-th stage a subinterval of relative length $\theta_k$, $0<\theta_k<1$. Show that the remainder has measure zero if and only if $\sum{\theta_k}=\infty$. (Use the fact that for $a_k>0$, $\prod\limits_{k=1}^{\infty}{a_k}$ converges, in the sense that $\lim\limits_{N\to\infty}{\prod\limits_{k=1}^N{a_k}}$ exists and is not zero, if only if $\sum\limits_{k=1}^{\infty}{\log{a_k}}$ converges. )

I don't really understand the construction. Suppose we have our $\theta_1$ and we remove an interval of that length. So then we choose $\theta_2$ so that $\theta_2 < \dfrac{1-\theta_1}{2}$, or something like that? Or... ? The phrasing is weird to me. Secondly, the hint does not help at all; it is complete opaque to me, so any insight you can offer would be great.


1 Answer 1

  1. Your last step in the computation is incorrect. The sum $\sum_{n=0}^{\infty}\frac{2^k}{3^k}$ is: $$\sum_{n=0}^{\infty}\frac{2^k}{3^k} = \sum_{n=0}^{\infty}\left(\frac{2}{3}\right)^k = \frac{1}{1-\frac{2}{3}} = 3$$ so when you multiply it by $\frac{\delta}{3}$ you get $\delta$, exactly what you expect it to be. I'm not sure why you thought you get $1- \frac{2\delta}{3}$. (Recall that if $|x|<1$, then $\sum_{n=0}^{\infty}x^n = \frac{1}{1-x}$).

  2. Notice that it says "relative length", not "length". So at each step, you are taking out an interval which is $\theta_k\cdot\ell$, where $\ell$ is the length of the interval you are looking at. In the Cantor set, you take $\theta_k=\frac{1}{3}$ for every $k$; in the example in (1), you are setting $\theta_k = \frac{\delta}{3}$ for every $k$. If you take out the central $\theta_k\ell$ portion of an interval of length $\ell$, then you are left with two intervals of length $\frac{\ell}{2}(1-\theta_k)$.

    Added: As per your question in the comments, in this question, all the $\theta_k$ are constant and equal to $\theta$, much like in the regular Cantor set all $\theta_k$ are equal to $\frac{1}{3}$. To see an example, suppose that you set $\theta_k = \frac{1}{2^{k+1}}$, $k=0,1,\ldots$. First you remove $\theta_0 = \frac{1}{2}$ of the interval, leaving you with $[0,\frac{1}{4}]\cup[\frac{3}{4},1]$. Then you remove $\theta_1=\frac{1}{4}$th of each remaining interval; each interval is of length $\frac{1}{4}$, so you are removing the central $\frac{1}{16}$th part of the interval; this leaves you with $[0,\frac{3}{32}]\cup[\frac{5}{32},\frac{1}{4}]\cup[\frac{3}{4},\frac{27}{32}]\cup[\frac{29}{32},1]$. At the next step, you are removing the central $\theta_2=\frac{1}{8}$th of each interval; the intervals are of length $\frac{3}{32}$, so you are removing the central $\frac{3}{256}$th portion. Etc.

  • $\begingroup$ Oh! Whoops. I somehow thought that the first term of the sum in (1) was 2/3... Simple mistake. Concerning (2): perhaps you can help me understand the difference between the construction posted here and the construction in a previous question -- math.stackexchange.com/questions/6643/cantor-like-construction $\endgroup$
    – Bey
    Oct 26, 2010 at 4:11
  • $\begingroup$ I've added to the answer to address this. $\endgroup$ Oct 26, 2010 at 4:37

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