I claim that supremum doesn't exist and infimum is $0$, but I want to prove this by definition. I can see that $$0<\frac{m}{2^n}<\frac{m}{2}$$ So, I claim that the supremum not exists and $0$ is a lower bound.
How to find the supremum and infimum of $ \{\frac{m}{2^n}: n,m\in\mathbb{N}\}$
real-analysissupremum-and-infimum
Best Answer
Yes, there does not exist an upper bound because otherwise you can take $n = 1$ and if we let $u$ be an upper bound for $m/2$ where $u$ is an integer then take $m = 2u+1$. If $u$ is a non-integer take $m = 2u_{1}+1$ where $u_{1} = \lceil u \rceil$ which denotes the smallest integer $u_{1}$ such that $u_{1} > u$.
Clearly $m/2^n > 0$ since $ m,2^n > 0$ which implies that $0$ is a lower bound. If there exists a lower bound $s > 0$ then take $m = 1$ and we must find an $n$ such that $0 < 1/2^n < s$ to arrive at a contradiction. One can choose $n$ such that $2^n > 1/s$ and more explicity, $n>log_2(1/s)$ and we are done.