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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.

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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.

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  • $\begingroup$ 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. $\endgroup$
    – randal
    Commented Nov 4, 2020 at 4:03
  • $\begingroup$ 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. $\endgroup$
    – Derek Luna
    Commented Nov 4, 2020 at 4:03
  • $\begingroup$ Yes, I do not consider $0$ as a natural number, but the test does not change much with this restriction. $\endgroup$
    – randal
    Commented Nov 4, 2020 at 4:14
  • $\begingroup$ 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. $\endgroup$
    – Derek Luna
    Commented Nov 4, 2020 at 4:16

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