In the dynamical system, there is one map called Doubling Map, defined as
\begin{align} f:[0,1) & \rightarrow [0,1) & \\ f(x)&= \ \begin{cases} 2x & x< \frac{1}{2} \\ 2x-1 & \frac{1}{2} \leq x<1 & \end{cases} \\ \end{align} Equivalently $f(x) = 2x$mod$1$. The iteration of this map for any $x \in [0,1)$ is given by $f^{n}(x) = 2^{n}x$mod$1$. I found that the sequence of iterates,$f^{n}(x)$, for $x \in \{ \frac{r}{2^m}: r,m \in \mathbb{Z}^+, 0<r<2^m\}$ converges to $0$.
I want to prove that these,$\{\frac{r}{2^m}: r,m \in \mathbb{Z}^+, 0<r<2^m\}$, are the only points for which the sequence of iterates converges.
And also I want to do characterization of converges/behaviour of this sequence in the more general setting i.e.
(1) What is the behaviour of the sequence $f^{n}(x)$ when $x$ is irrational nummber in $[0,1)$ ?
$(2)$ What is the behaviour of the sequence $f^{n}(x)$ when $x$ is rational nummber in $[0,1)$
The best way to think of this dynamical system is via binary expansion. If $x=0.a_1 a_2 \ldots$, where $a_i \in \{0,1\}$ is the binary expansion of $x \in [0,1)$, then the binary expansion of $f(x)$ is $0.a_2 a_3 \ldots$. In other words, your map $f$ is conjugated to the shift operator $\sigma (a_1, \ldots) = (a_2, \ldots)$ acting on $X=\{0,1\}^\mathbb{N}$.
Then for example the orbit of $x$ converges if and only if it's binary expansion is eventually either all zeroes or all ones. Now to address your questions.
is much to broad to hope for a simple answer. The assumption that $x$ is irrational means that its binary expansion is not eventually periodic; but other than that is can be pretty much anything. In particular, the sequence $\{f^n(x)\}$ may or may not be dense in $[0,1)$, depending on whether all finite binary words eventually appear or not in the binary expansion of $x$. This is often open for specific irrationals, such that for instance $\pi$ mod 1 = $\pi -3$.
is a lot simpler. If $x$ is rational, then its binary expansion is eventually periodic, and so the sequence $f^n(x)$ is also eventually periodic.