Limit points of $\left\{ \frac{m}{2^{n}} \right\}$ where $m,n$ are natural numbers. Let $E = \left\{\frac{m}{2^{n}}\right\}$ where $m,n$ are integers, then $0$ is the only limit point of this set.
$($true$/$false$) ?$
I think the $0$ is the only limit point of this set. Since, as $n \to \infty$ terms of this sequence come nearer to $0$. 
But the statement is false according to answer key.
 A: Your set contains $$\frac34,\frac78,\frac{15}{16},\ldots\to 1$$so $1$ is a limit point.
A: It is not too difficult to show that the set is dense, similar to this problem.
We can generalise the problem ...


Propositions 1. $\forall r \in [0,\infty)$ and $\forall \varepsilon>0, \exists m,n \in \mathbb{N}: \left|r-\frac{m}{q^n}\right|<\varepsilon$, for a fixed $q\in \mathbb{N}\setminus \{0,1\}$.

From
$$\color{red}{0\leq m}=\left \lfloor r\cdot q^n \right \rfloor \leq r\cdot q^n < \left \lfloor r\cdot q^n \right \rfloor +1=m+1 \Rightarrow 
\frac{m}{q^n} \leq r < \frac{m}{q^n}+\frac{1}{q^n}$$
we have
$$\left|r-\frac{m}{q^n}\right|<\frac{1}{q^n}$$
By "adjusting" $n>0$ we can find one such that $\frac{1}{q^n} < \varepsilon$. Also, $m,n \in \mathbb{N}$.
NOTE: This means that $M_q=\left\{\frac{m}{q^n} \mid m,n\in\mathbb{N}\right\}$, $q\in \mathbb{N}\setminus \{0,1\}$, is dense in $[0,\infty)$ or any point in $[0,\infty)$ is a limit point for $M_q$.


Proposition 2 $\left\{\left\{\frac{m}{q^n}\right\} \mid m,n\in\mathbb{N}\right\}$, for a fixed $q\in \mathbb{N}\setminus \{0,1\}$, is dense in $[0,1]$.
(the second $\{\}$ is the fractional part).

For $\forall r \in [0,1)$ and $\forall \varepsilon >0$ s.t. $0\leq r < r+ \varepsilon < 1$, we always have (from proposition 1) $m,n \in\mathbb{N}$ s.t. 
$$\color{red}{0}\leq r \leq \color{red}{\frac{m}{q^n}} < r+ \varepsilon < \color{red}{1}$$
(because if $M_q$ is dense in $[0,\infty)$, then it is also dense in $[0,1) \subset [0,\infty)$. But this means (see the parts highlighted in red) that $\frac{m}{q^n} = \left\{\frac{m}{q^n}\right\}$ ($\{\}$ is the fractional part). The only point left to check is $\{1\}$. For this purpose, we will be looking at 
$$r_k=1-\frac{1}{k} < 1-\frac{1}{k} + \varepsilon < 1$$
And again, (from proposition 1) we have $m_k,n_k \in\mathbb{N}$ s.t.
$$r_k=1-\frac{1}{k} \leq \frac{m_k}{q_k^n} < 1-\frac{1}{k} + \varepsilon < 1$$
or
$$\left|1-\frac{m_k}{q_k^n}\right|\leq \left|1-r_k\right|=\frac{1}{k}$$
which, again, can be made as small as we want and we covered $\{1\}$ too and thus the entire $[0,1] = [0,1) \cup \{1\}$

Summarising all these and making $q=2$ the answer is obviously false.
