Let $F$ be the set of of numbers $x\in [0,1]$ with base 3 expansions $0.a_1a_2...$ for which there exists an integer k such that $a_i\neq 1$ for all $i\geq k$. Find the Hausdorff dimension of $F$.

Note that $F$ is the countable union of $F_k$ where $F_k$ is the points $0.a_1a_2...a_k..$ such that for all $i\geq k$ we have $a_i\neq 1$.

So it suffices to find that $supdim_HF_k$. Note that $F_1$ is the cantor set, with Hausdorff dimension $log2/log3$.

I'm not sure how to approach this. Im supposed to use self similarity, but I'm having a hard time understanding which sets are self similar to others in this case, $I$ suspect each of the $F_ks$ are self similar to $F$ , but im not sure how or why.

May someone elaborate?


Look at $F_2$ this way: divide the unit interval in three intervals of length $1/3$ according to the first digit of the base 3 expanaion. In each of them, you look at the points without a 1 in the rest of their expansion. This translates to seeing a cantor set in each of the three intervals. So $F_2$ is essentially three scaled down copies of $F_1$. Can you finish it from that?

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  • $\begingroup$ Perez, what do you mean by "according to the first digit of the base 3 expansion"? $\endgroup$ – orientablesurface Mar 9 at 21:56
  • $\begingroup$ Just meant to divide $[0,1]$ in the intervals $[0,1/3]$, $[1/3,2/3]$ and $[2/3,1]$. Depending on which interval a point is, you have different first digit in base 3 (unless you fall in the boundaries, but you can forget about those because they're countably many points). $\endgroup$ – Felipe Pérez Mar 9 at 21:59
  • $\begingroup$ Perez, okay, so in $[0,\frac{1}{3}]$, I have 0.10000.. and 0.1222222... and points like that. I'm not sure what next. $\endgroup$ – orientablesurface Mar 9 at 22:07
  • $\begingroup$ In that interval the first digit is actually $0$. If you forget about the first digit for the numbers in that interval and think of the rest of the expansion, what do you have? Essentially the Cantor set. Can you compute the Hausdorff dimension of $F_2$ using that? $\endgroup$ – Felipe Pérez Mar 9 at 22:11
  • $\begingroup$ Essentially I have a shifted cantor set, right? I feel that should be the answer but I'm not sure why. If I forget about the first digit then the hasudroff dimension is that of the cantor sey $\endgroup$ – orientablesurface Mar 9 at 22:12

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