Consider $C$ the classic Cantor ternary set in $[0,1]$. I am interested in the following problem:

Find the largest constant $0<k<1$ such that it is true that any interval $[a,b] \subseteq [0,1]$ contains an open interval $(\alpha,\beta)$ with $k(b-a)\leq (\beta - \alpha)$ and disjoint from $C$.

My thoughts:

Note the inequality condition is just saying in terms of measure/length that $k \cdot \lambda([a,b]) \leq \lambda((\alpha,\beta))$.

I believe much information can be obtained by looking at big intervals such as $[a,b]=[0,1]$ itself. This should be true from the self-similarity of Cantor, meaning it is the same as considering any interval $[a,b]$ where $a$ and $b$ are endpoints from intervals that were removed in the construction of $C$. From this case we can readily see that $k \leq 1/3$ since certainly we will not find any bigger interval disjoint from $C$ than the removed middle-third.

Actually, trying different examples I had almost convinced myself that $k=1/3$ worked for any interval (in some cases the inequality being true by a wide margin), but after more work finally found counterexamples. The following gives one of the better bounds I have found (which I think can still be improved):

Let $[a,b]=[2/9,3/8]$, then I claim the largest disjoint interval is of length $1/24$, corresponding to $(\alpha,\beta)=(1/3,3/8)$, since the disjoint interval $(7/27,8/27)$ has length $1/27<1/24$ and on the other hand since $1/4 \in C$ (see for example 1/4 is in the Cantor set?), it restricts the possibility of the left interval $(2/9,1/4)$ which has length $1/36<1/24$. All in all, we get that $k \cdot 11/72 \leq 1/24$, which gives $k\leq 3/11$.

I also think a lower bound of for example $k \geq 1/9$ should be easy to see, since by going one step forward into the removal of intervals after $\frac{1}{3^{m}}\leq b-a$ happens for the first time, we should get one from size $\frac{1}{3^{m+1}}$ inside $(a,b)$.

What is the actual value of $k$? (consider it to be a supremum just in case it is not achieved)

  • 1
    $\begingroup$ $k\leqslant1/5$, see $(a,b)=(5/27,10/27)$. $\endgroup$ – Did Oct 13 '14 at 23:41
  • $\begingroup$ @Did thank you! Nice example...with that I actually conjecture $k=1/5$ now. $\endgroup$ – Luke Skywalker Oct 14 '14 at 0:23

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