I would like to verify my proof of the following:
Let $A=\{\frac{1}{n}-\frac{1}{m}:m, n \in \mathbb{N}\}$. I want to show that $-1$ and $1$ are the infimum and supremum respectively.
First I will show that $-1$ is the infimum. To show this I will first demonstrate that it is indeed a lower bound. Observe: $$\frac{1}{n}-\frac{1}{m} \geq \frac{1}{n}-1 \geq -1.$$
I now claim that $-1$ is the greatest lower bound. So, by the archimedean property, there exists an $x \in \mathbb{R}$, $x>0$ and $n' \in \mathbb{N}$ such that $1<n'x$. Thus, $\frac{1}{n'}<x$ and $$\frac{1}{n'}-1<x-1.$$ From the above it is clear that $\frac{1}{n'}-1 \in A$. Also, it is clear that $-1<-1+x$.
Let $y=\inf(A)$. So we let $$-1<y\leq-1+x.$$ By completeness there exists an $r\in\mathbb{Q}$ such that $$-1<r<y\leq-1+x.$$ Thus $y$ is not the infimum.
I was going to note that the supremum was 1 through the relationship between $$\inf(A)=-\sup(-A).$$