Hausdorff dimension of a subset of Cantor set

What is the Hausdorff dimension of the subset $$F := \{ x = \sum^\infty_{n=1} \frac{2 x_n}{3^n} \in [0,1] : x_n \in \{ 0 , 1 \} , x_n = 1 \Rightarrow x_{n+1}=0 \}$$ of the Cantor set? Is it known already?

As far as I know, this set can be corresponded to a binary tree related to the Fibonacci sequence. (I don't know how to call that tree.)

• Presumably, the subscript $n$ on the last $x$ should be $n+1$. Then, at stage $n$ of the usual middle-thirds construction of the Cantor set, when you have $2^n$ intervals of length $1/3^n$ each, the number of those intervals that meet $F$ is the $n$-th Fibonacci number (if you index the Fibonacci numbers appropriately), which is asymptotically a constant time $\phi^n$, where $\phi$ is the golden ratio $(1+\sqrt 5)/2$. So I'd expect the Hausdorff dimension of $F$ to be $(\ln\phi)/(\ln 3)$. Commented Feb 15, 2012 at 23:31
• If so, then you can brake your set in two parts, those which start with $1,0$ and the rest. These parts are rescalings of the original set with coefficients $\tfrac19$ and $\tfrac13$ therefore dimension is the number $\alpha$ such that $$\tfrac 1{3^\alpha}+\tfrac1{9^\alpha}=1.$$ Commented Feb 15, 2012 at 23:35
• Edit: $x_n=1 \Rightarrow x_{n+1}=0$. Commented Feb 15, 2012 at 23:59
• See related math.SE question: math.stackexchange.com/questions/73547/… Commented Feb 16, 2012 at 1:43
• Thanks for your replies~ @Anton, I'm still uncertain about one point. At step n+1, the part starting with 1 doesn't seem to be exactly a self-similar part of the set at step n. And why is the ratio $\frac{1}{9}$ for this part? Commented Feb 16, 2012 at 4:19

The condition that $x_n = 1 \Rightarrow x_{n+1} = 0$ is a Markov condition: the value of $x_{n+1}$ is restricted by the value of $x_n$. In your case you are considering all sequences in $\{0,1\}^\mathbb{N}$ such that the symbol $1$ cannot follow itself; one could also consider more symbols and more complicated restrictions, such as "every occurrence of $2$ can only be followed by $0$ or $2$, but not $1$". See http://en.wikipedia.org/wiki/Subshift_of_finite_type for more details.
When you construct a subset of the Cantor set as in your question, the topological entropy turns out to be directly related to the Hausdorff dimension: namely Hausdorff dimension is topological entropy divided by $\log \lambda$, where $\lambda$ is the contraction ratio at each step of the construction of the Cantor set.
• Would you mind clarifying whether the set $F$ is self-similar or not? Commented Feb 25, 2012 at 4:53