Find the supremum and infimum of the set $E:=\left\{\frac{3}{n+2}+2\mid n \in \Bbb{N}\right\}$.
It's obvious that $$\frac{3}{n+2}+2 \leq 3, \forall n \in \Bbb{N}$$
So $3$ is an upper bound of $E$. Lets assume that $b$ is another upper bound for $E$, such that $b=3-\epsilon$, $\epsilon > 0$. If $b$ is an upper bound, then $b \geq x$ for all $x \in E$. However, $3 \in E$. Therefore $b$ is not an upper bound of $E$ and $\sup E=3$
As for the infimum. Since $$\frac{3}{n+2}+2 \geq 2, \forall n \in \Bbb{N}$$
So $2$ is a lower bound of $E$. Lets assume there is another lower bound $c=2+\epsilon$, where $\epsilon >0$. By the Archimedean property of real numbers, $$ \begin{split} \frac{3}{n+2} &< \epsilon\\ \frac{3}{n+2}+2 &< 2+\epsilon \end{split} $$
Thus $c$ is not a lower bound of $E$ and $\inf E=2$. Extending the idea of the Archimedean property to this expression seems intuitive to me ( we can always choose a natural number, such that the expression is smaller than a positive real number ) , but I would be hard pressed to find a precise explanation for its use here. What other flaws are in this proof?