How I can find the supremum and infimum of $\{\frac{m}{2^n}: n,m\in\mathbb{N}\}$?

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.

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.
• For the upper bound , Can I use that $$0<\frac{m}{2^n}<\frac{m}{2}<m?$$ If this is correct then, with the same argument that the natural numbers are not bounded, I can conclude that there is no upper bound. Commented Nov 4, 2020 at 4:03
• I just realized $n > 0$ depending on your definition of natural numbers! But what you have there is good but might need to be a bit more detailed. I will edit my post. Commented Nov 4, 2020 at 4:03
• Yes, I do not consider $0$ as a natural number, but the test does not change much with this restriction. Commented Nov 4, 2020 at 4:14
• Yes, what you did is fine but it may be required that you now explicitly prove that $\mathbb {N}$ has no upper bound which you only need a part of my argument for. Commented Nov 4, 2020 at 4:16