Hausdorff dimension of $\lim_{n\to\infty}\sin(2^nx)$

Question

Calculate the Hausdorff dimension,$\dim_H$ of $$S=\{x\in(0,1):\lim_{n\to \infty}\sin2^nx=0\}$$

By definition We need to find the minimal $\alpha$ s.t $\sum_{i\in I}|U_i|^\alpha$ is minimal where $U_i$ is an open cover for the set.

On the other hand, I'm not sure it exists since the function diverges when $n\to\infty$. How can I find the dimension or prove that it doesn't exist?

Answer 1

Follow the hint by Daniel Fischer: $$ \lim_{n\to\infty} \sin 2^n x = 0 \iff \exists n\ \sin 2^n x=0 \tag{1} $$ The set of $x$ that satisfy the condition on the right is countable.

To prove (1), observe that if $|\sin y|\le 0.1$ then $|\cos y|\ge0.9$, so $|\sin 2y|\ge 1.8|\sin y|$. Thus, if for some $n$ the value of $|\sin 2^n x|$ is small but not zero, it will grow for a while and will not be that small.