I was recently asked the following question: Does the map $$\mu : [0, 1) \rightarrow [0, 1)\\ \hspace{105px}x \hspace{5px}\rightarrow \hspace{5px}2x \mod 1$$ have some point, say $\alpha$, whose orbit is dense in $[0, 1)$?
Once you consider representing $x$ as a binary string the solution comes fairly quickly, as $\mu$ can be seen to be just a shift to the left by one (dumping the digit past the decimal point.) I.e. $\mu(0.1010) = 0.010$
It follows that the number defined by concatenating all finite binary strings, $\alpha = 0.01000110110000…$ has a dense orbit in [0, 1). Having gotten to this point, I thought this screamed for a polynomial inspired generalization that I couldn’t solve (nor could the problem giver).
By Stone-Weierstrass polynomials are dense in $C[0, 1]$, but per Muntz-Szasz one does not need the entire family of monomials $\{1, x, x^2, … ,x^n, …\}$ to achieve density, and it suffices instead to just use any subset $\{1, x^{a_1}, x^{a_2}, … , x^{a_n}, …\}$ such that $\sum_{i=1}^\infty \frac{1}{a_i} = \infty$.
Likewise my solution to the problem clearly has a whole lot of redundancy: for any fixed integer $N$, I could for example concatenate just the strings of length greater than $N$, or more strongly just the strings whose length is a multiple of $N$. I was not however able to go any further.
The question if unfortunately a bit vague, but I’m looking for a more restrictive construction, perhaps in the spirit of Muntz-Szasz, of some point with a dense orbit under this map.