Ytukyg

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.

Your set contains $$frac34,frac78,frac{15}{16},ldotsto 1$$so $1$ is a limit point.

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 $qin mathbb{N}setminus {0,1}$.

From
$$color{red}{0leq m}=left lfloor rcdot q^n right rfloor leq rcdot q^n < left lfloor rcdot q^n right rfloor +1=m+1 Rightarrow
frac{m}{q^n} leq r < frac{m}{q^n}+frac{1}{q^n}$$
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 $M_q=left{frac{m}{q^n} mid m,ninmathbb{N}right}$, $qin mathbb{N}setminus {0,1}$, is dense in $[0,infty)$ or any point in $[0,infty)$ is a limit point for $M_q$.

Proposition 2 $left{left{frac{m}{q^n}right} mid m,ninmathbb{N}right}$, for a fixed $qin mathbb{N}setminus {0,1}$, is dense in $[0,1]$.

(the second ${}$ is the fractional part).

For $forall r in [0,1)$ and $forall varepsilon >0$ s.t. $0leq r < r+ varepsilon < 1$, we always have (from proposition 1) $m,n inmathbb{N}$ 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 $m_k,n_k inmathbb{N}$ s.t.
$$r_k=1-frac{1}{k} leq frac{m_k}{q_k^n} < 1-frac{1}{k} + varepsilon < 1$$
or
$$left|1-frac{m_k}{q_k^n}right|leq left|1-r_kright|=frac{1}{k}$$
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.