Continuity of $\mu(t)=\inf\{x \in \mathcal C : \kappa(x)=t\}$. Let $\Delta = \{ 0, 1\}^{\mathbb N}$ be a Cantor set. Define $\theta : \Delta \to [0,1]$ by the formula $$\theta(x_1,x_2,\dots) = \sum_{n=1}^\infty \frac{2x_n}{3^n}.$$ Denote $\mathcal C = \theta(\Delta)$. 
Define $\kappa : \mathcal C \to [0,1]$ by $$\kappa\left(\sum_{n=1}^\infty \frac{2x_n}{3^n}\right) = \sum_{n=1}^\infty \frac{x_n}{2^n}$$ and $\mu : [0,1] \to \mathcal C$ by $\mu(t)=\inf\{x \in \mathcal C : \kappa(x)=t \}$. Let $\displaystyle \mathcal D = \left\{ \frac{k}{2^n} : n \in \mathbb N, k \in \{0,1,...,2^n\}\right\}$
How can I show that $\mu$ is continuous on $[0,1] \backslash \mathcal D$?
I have shown that $\kappa$ is well defined and that it is a non-decreasing surjection, however I have trouble with dealing with $\mu$. Any help would be appreciated.
 A: If the binary representation of $x$ is unique, the fiber of $x$ under $\kappa$ is a singleton that you can explicit. Then you have to use the fact that $\mu$ is also non-decreasing.
If $a_n \xrightarrow[n\to\infty]{} x^+$, for $a_n$ near enough to $x$, as the binary representation of $x$ (written $(x_1,x_2,...)$) contains infinitely many 0, there is $\alpha(n)\in \mathbb{N}$ such as
$x_{\alpha(n)} = 0$, 
so that $x + \frac1{2^{\alpha(n)}} \in [0,1] \setminus \mathcal{D}$, 
$x \leq a_n \leq x + \frac1{2^{\alpha(n)}}$.
and $\alpha(n) \xrightarrow[n\to\infty]{} \infty$ 
Thus $\mu(x) \leq \mu(a_n) \leq \mu(x + \frac1{2^{\alpha(n)}}) = \mu(x) + \frac2{3^{\alpha(n)}}$
This double inegality allows to establish half of the result. The other half is pretty similar (use the infinite number of 1 in the binary representation of $x$).
A: Maybe you can start by showing that $\forall x \in [0,1] \setminus \mathcal{D}$, the binary representation of $x$ is unique and contains infinitely many $0$ and infinitely many $1$.
