I wanted to check my reasoning for the following question:
Determine minimum, maximum, supremum and infimum of the set: $$B=\left\{ -\frac{1}{n} \in \mathbb{Q}: n \in \mathbb{N}_+ \right\}$$
Whenever $n=1$ we have that $-\frac{1}{1}=-1$, this is the minimum element of the set, since for all $n$ we know that $n\geq 1$ and thus $\frac{1}{n} \leq 1$, this gives $- \frac{1}{n} \geq -1 $. As the value of $n$ gets bigger (we will later define limits), we notice that $-\frac{1}{n}$ will approach zero, but zero is not contained in the set. We realise that $\sup(B)=0$ and summarise: $$\min(B)=\inf(B)=-1$$ $$\sup(B)=0 $$ $\max(B) $ does not exist.
You are correct!
Indeed, if you feel uncomforatble to work with negative signs, just compute $\inf A$ and $\sup A$ where $A=\{1/n:n \in \Bbb N\}$ and then use $$\sup(-A)=-\inf A\\\inf(-A)=-\sup A$$