Cantor-like functions for $\xi\neq \frac{1}{3}$ Let $\xi \in (0,\frac{1}{2})$. Let $C_\xi$ be the perfect symmetric set built by iterating the transformation
$$[0,1] \to [0,\xi]\cup [1-\xi, 1].$$
The set $C_\frac{1}{3}$ would then correspond to the Cantor set. Define the $\xi$-Cantor-like function as the limit of the piecewise linear functions $f_i$ which are constant on the intervals removed in the $i$-th iteration for constructing $C_\xi$, see Wikipedia [1] for a pictorial analysis.
Can we show that the procedure for constructing the $\xi$-Cantor-like function will still yield a well-defined limit function, the $\xi$-Cantor-like function, which will be surjective and (since its monotone) continuous?
[1] https://en.m.wikipedia.org/wiki/Cantor_function, last accessed 28/02/2022 at 11:35 PM GMT.
 A: HINT: For the original Cantor function with $\xi=1/3$, surjectivity can be proven by relying on the uniqueness of the base-$3$ representation of real numbers in $[0,1]$, and the fact that the Cantor function can be described as a transformation from base-$3$ digit sequences to base-$2$ digit sequences. This type of argument can actually be generalized to arbitrary bases $\xi\in (0,1/2)$, although it's not immediately clear how.
Define $3$ functions $f_1,f_2,f_3$ as follows:
$$f_1(x) = \xi x \\ f_2(x) = \xi + (1-2\xi)x \\ f_3(x)=(1-\xi) +\xi x$$
It happens that every real number $x\in[0,1]$ can be expressed as the limit of a sequence $(d_n)$ taking the form
$$d_1 = f_{n_1}(0) \\ d_2 = f_{n_1}\circ f_{n_2}(0) \\ d_3 = f_{n_1}\circ f_{n_2}\circ f_{n_3}(0) \\ \cdots$$
where each $n_i$ equals $1,2,$ or $3$. When $\xi=1/3$, the functions $f_i$ are given by
$$f_1(x) = \frac{1}{3} x \\ f_2(x) = \frac{1}{3} + \frac{1}{3}x \\ f_3(x)=\frac{2}{3} +\frac{1}{3} x$$
so that $f_1,f_2,f_3$ can be seen as shifting the base-$3$ representation of $x$ one digit to the right and appending a $0,1,$ or $2$ respectively, and the approximations $d_1,d_2,d_3,\cdots$ are $n$-digit base-$3$ numbers whose $n$ digits agree with the most significant $n$ digits of $x$. Hence the analogy with the $\xi=1/3$ case - we are generalizing the base-$3$ representation to values $\xi\ne 1/3$. The values of $n_1,n_2,\cdots$ are analogous to the "digits" of $x$ under this alternative base representation system.
My suggestion is the following: show that any $x\in [0,1]$ has a unique representation in the described form for any $\xi\in (0,1/2)$, determine how the Cantor function transforms these sequences into base-$2$ sequences, and deduce that the Cantor function is surjective from the fact that every base-$2$ sequence has a preimage.
