A property of Sturmian sequence I need the following simple property of  Sturmian sequence: 
Let $\omega\in \{0,1\}^{\mathbb{N}}$ be a  Sturmian sequence, we define the orbits space $\overline{O}=\overline{\{\sigma^{n}{(\omega)}:n\in\mathbb{N}\}}$, where $\sigma$ denotes the left shift, I want to prove that $\sigma:\overline{O}\to \overline{O}$ is a bijection. 
 A: I'll use the definition of $\overline{O}$ which I find most intuitive and which is convenient for proving this statement is false. If you need me to explain why this definition of $\overline{O}$ is equivalent to the orbit closure of the left shift of a particular sequence feel free to ask in the comments, though it's a fairly common definition.
Let $\omega$ be a right-infinite Sturmian sequence. So $\omega$ is an element of $\{0,1\}^{\mathbb{N}}$ and not a bi-infinite sequence which would be an element of $\{0,1\}^{\mathbb{Z}}$ - this is an important distinction! To $\omega$ we associate the real irrational number $\alpha$ which is found using the usual continued fraction argument (See Series 1985 - The Geometry of Markoff Numbers).
Now, define the space $O_\alpha$ to be the set of all 'cutting sequences' of slope alpha. We generate a cutting sequence in the following way.
Start with a point $x\in\mathbb{R}^2$ and fire off a projectile from $x$ at slope $\alpha$ in the plane. Whenever the projectile meets a horizontal line integral segment (so its $y$-coordinate is an integer) write down a $0$. When the projectile meets a vertical integral line segment (so the $x$-coordinate is an integer) write down a $1$. If the projectile meets a lattice point (point in $\mathbb{Z}^2$) we generate two cutting sequence - one where we write down $01$ and one where we write down $10$. We call such a pair of sequence a singular pair.
If $x$ happens to have an integer coordinate then we take the convention that we include the relevant symbols in the sequence (equivalent to moving $x$ backwards slightly along the slope $-\alpha$).
Now, I claim without proof that $\overline{O}$ is equal to $O_\alpha$ and so the left shift on $\overline{O}$ is a bijection if and only if the left shift on $O_\alpha$ is a bijection.

Proposition. The left shift $\sigma\colon O_\alpha\to O_\alpha$ is not injective.

Proof. Let $\omega_1$ and $\omega_2$ be the singular pair of cutting sequences associated to the starting point $x=(0,0)$ with $01$ being the choice of the first two symbols for $\omega_1$ and $10$ for $\omega_2$. It is clearly the case that $\omega_1\neq\omega_2$, but $\sigma^2(\omega_1)=\sigma^2(\omega_2)$ because the line of slope $\alpha\in\mathbb{R}\setminus\mathbb{Q}$ does not intersect any lattice points with positive coordinates. This can not be the case if $\sigma$ is injective.
