Consider the expanding map $E_3\colon \mathbb{S}^1 \to \mathbb{S}^1, x\mapsto 3x \mod 1$. I want to find a $x$ on the circle such that the forward orbit of $x$ never ends up in the standard middle third Cantor set and the closure of the orbit of $x$ is equal to the middle third Cantor set (and the orbit itself).
I'm really stuck on this. To ensure that $E_3^t(x)$ is never in the Cantor set, for all $t\geq 0$, $E_3^t(x)$ must always have a digit 1 when represented in base $3$. So I thought about using $x$ such that when converted to base $3$ it is equal to the concatenation of all finite binary sequences in lexicographic order, i.e. $(x)_3 = 0.01 00 01 10 11\ldots$ since $E_3$ acts on $x$ by left shift I think this would be ok.
Assuming I'm on the right track how I would I then go about showing that $\overline{E_3^t(x)} = C\cup E_3^t(x)$, where $C$ is the above mentioned Cantor set. If I understand things correctly this amounts to showing that, since we are in a metric space, that all convergent sequences in the orbit of $x$ converges to an element in $C$. So if we consider a sequence $(x_i)_{i=0}^{\infty},x_i \in E_3^t(x)$, then $x_i$ gets closer and closer to a number which when represented in base $3$ has no $1$:s in its digits. But how can I be sure that the sequence converges to such a number?
I have pretty much no idea what I'm doing so any solution, pointers, hints or suggestions would be highly appreciated.