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I have the set $$S = \left\{\frac{x}{2^y} | x, y \in \mathbb{N}\right\}$$ I want to find it's infimum. But there's a point in which I get stuck.

As $x,y \in \mathbb{N}$ we know that $x>0$ and $y>0$. Hence we have $\frac{x}{2^y} >0$ , $\forall x,y \in \mathbb{N}$. Hence $0$ is a lower boundfor $S$. Now take another lower bound $w$. I want to show that $w \leq 0$. I will construct a proof by contradiction. Suppose $ w> 0$. Then define $\epsilon : = w-0 = w >0$.

Now notice that the subset $S_1 = \left\{\frac{1}{2^y} | y \in \mathbb{N}\right\}\subset S$. Then by the Archimedean Property we have that $\exists n \in \mathbb{N}$ such that $\frac{1}{2^y} < \epsilon$

Now notice that $\frac{1}{2^y} \in S$ and so this is a contradiction as $w$ should be a lower bound. Hence $w \leq 0$ which means that $0 = \inf(S)$

The crucial point is the highlighted one. Is it correct? Particularly the AP part, because normally we say $\exists n\in\mathbb{N} : \frac{1}{n} < \epsilon$ But here we don't have just $\frac{1}{n}$ but rather $\frac{1}{f(n)} = \frac{1}{2^n}$.

Is there a better and more rigorous proof using AP? Is this correct?

Also, should I state which $n$ works? Cause I think I can take logarithm and find $\log_2(2^{-y}) < \log_2(w)$ and then $-y < \log(w)$ so I have $y > -log_2(w)$ ?

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    Your proof is correct. As $\frac1{2^n} < \frac1{n}$, we also have $\frac1{2^n} < \epsilon$. – 2017-01-22
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    @OpenBall is there a way I can show that with the tools of a basic Analysis course? – 2017-01-22
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    Induction ${}{}$ – 2017-01-22

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