Fat Cantor Set with large complement???

by tyo   Last Updated October 10, 2019 05:20 AM

I'm encontering fat Cantor sets for the first time, and I found the formula for the length of the complement on Wikipedia (you know, like you do) as

$\mu(I \setminus C_\alpha) = \alpha \sum_{n=1}^\infty (2\alpha)^n = \frac{\alpha}{1-2\alpha}$.

This makes sense and is a pretty intuitive extension of the normal Cantor set formula, but also it seems like I can make it arbitrarily large for an $\alpha$ arbitrarily close to $\frac{1}{2}$, which doesn't make much sense for a subset of $[0,1]$. That said, given the construction of the set, I'm not sure why I'm not allowed to take intervals of length $3^{-n} < \alpha^n < 2^{-n}$ and still use this formula to compute the length of the complement.

I feel like I'm missing someting small and silly, so if anyone can set me right on this, I'd appreciate it.



Answers 2


The set $C_\alpha$ is defined by saying that at the $n$th step, you remove a middle interval of length $\alpha^n$ from each of the remaining intervals. This only makes sense if each of the remaining intervals actually has length at least $\alpha^n$. That will be false for sufficiently large $n$ for any $\alpha>1/3$, and so $C_\alpha$ is only defined for $\alpha\leq 1/3$.

For an explicit example, if $\alpha=0.4$, then we first remove an interval of length $0.4$ to leave $[0,0.3]$ and $[0.7,1]$. For subsequent steps, let's keep track of just the leftmost of our intervals. We next remove intervals of length $0.16$ to leave $[0,0.07]$ as our leftmost interval. Then we remove intervals of length $0.064$ to leave $[0,0.003]$ as our leftmost interval. Next we need to remove intervals of length $0.0256$, but we can't, because our remaining intervals have length $0.003$!

Note that in fact you can see that the intervals will always be long enough iff the sum $ \sum_{n=1}^\infty \alpha(2\alpha)^n$ is at most $1$, since each new term we add represents the total length of intervals we need to remove at the next step, and there is enough length left iff the resulting partial sum is still at most $1$. So that's where the bound $\alpha\leq 1/3$ comes from: when $\alpha=1/3$, the sum converges to $1$, and any larger $\alpha$ makes it greater than $1$ so the intervals eventually become too small.

Eric Wofsey
Eric Wofsey
October 10, 2019 04:53 AM

The formula doesn't hold for $a > 1/3$. If you try to remove more than $2^n (1/3)^{n+1}$ at each step $n$, then you will remove the entire interval $[0,1]$ after finitely many steps.

For example, consider $a = 2/5$.

At $n=0$ we remove $2^0 (2/5)^1 = 2/5$. The remaining length is $3/5 = 0.6$.

At $n=1$ we remove $2^1 (2/5)^2 = 0.32$. The remaining length is $0.6 - 0.32 = 0.28$.

At $n=2$ we remove $2^2 (2/5)^3 = 0.256$. The remaining length is $0.28 - 0.256 = 0.024$.

At $n=3$ we remove $2^3 (2/5)^4 = 0.2048$. Oops, no we don't, because that's more than the remaining length.

Bungo
Bungo
October 10, 2019 04:54 AM

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