Let's consider a map $\varphi: [0, 1] \rightarrow [0, 1]$ so that $x \mapsto \{2x \}$. I would like to find a point $x$ so that its trajectory is everywhere dence in $[0,1]$.

Firstly, the basic idea is to encode the reals from $[0,1$] with the infinite sequences of $0$ and $1$ (using recursive algorithm, which includes partioning on every step and deciding whether to put $0$ or $1$ according to the location of the point -- left or right to the center of partition). Then, the image of a sequence $l_{1}l_{2}l_{3} \ldots $ under the map $\varphi$ is $l_{2}l_{3}l_{4} \ldots $, where $l_{i} \in \mathbb{Z}_{2}$. It can be seen that the rational number has the expansion, consisting of a finite number of non-zero elements, so the rationals are not the seeking set.

How to find such a point? Any help would be much appreciated.


1 Answer 1


So, I see that you've noticed that $\varphi$ is essentially a left-shift. So go ahead and write out all finite sequences of 0's and 1's. These are the dyadic rationals so they are dense in [0, 1]. Now this is a countable set of elements since $2^{<\omega}$ is countable. I'll begin the list here


Choose your element $x=0100011011000001 \ldots$

This element will have a dense orbit since it will get arbitrarily close to every dyadic rational since all we're doing is a left-shift.


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