My claim for this problem is that $Sup(S) = \frac{1}{2}$. To show this we must first show that $\frac{1}{2}$ is an upper bound for S. Proof: Since $\frac{1}{n+1} \leq \frac{1}{2}$, $\forall n\in \mathbb{N}$, $\frac{1}{2}$ is an upper bound for $S$. Do I need to expand this more? Or is this sufficient to show that $\frac{1}{2}$ is an upper bound? If not, I could rephrase this portion of the proof as Let $x\in S$ be given. Then $x \in \mathbb{N}$, so $x \geq 1$. Therefore, $x+1 \geq 2$ and thus $\frac{1}{x+1} < \frac{1}{2}$.
Furthermore, I would like to show that for any $x \in \mathbb{R}$, if $x$ is an upper bound of $S$, then $\frac{1}{2} \leq x$. Using contraposition of this statement: for any $x \in \mathbb{R}$, if $\frac{1}{2} > x$, then $x$ is not an upper bound of $S$. I am struggling to prove this portion of the proof. I have an idea of using the Archimedian Property, in which $\forall \varepsilon > 0$, $\exists n\in \mathbb{N}$ such that $\frac{1}{n} < \varepsilon$. But I have not been able to implement this into my proof. Any help and or hints would be much appreciated.