Are these proofs regarding the infimum and supremum correct?

Question

I would like to know if I solved correctly the following (well known/easy) problems about infimums and supremums. I couldn't find these precise proofs anywhere and I wish to know if they are all right.
Problem 1. Let $A\subset R$ be a bounded non-empty set. Show that $\sup(-A)=-\inf(A)$, where $-A=\{-a | a\in A\}$.
Solution. For any $a\in A$ we have that $a\ge \inf A \iff -a\le -\inf A$ and from here it follows that $x\le -\inf A, \forall x\in -A$, so $-\inf A$ is an upper bound for $-A$.
We have a sequence $(x_n)_{n\ge 1}\subset A$ such that $x_n \to \inf A$. This means that $-x_n\to -\inf A$ and $(-x_n)_{n\ge 1}\subset -A$ so now we may say that $\sup(-A)=-\inf A$.
Problem 2. Find $\inf B$ and $\sup B$ if $B=\{\frac{m}{n} | m,n \in \mathbb{N}, m<2n\}$.
Solution. Obviously, $0<x<2$ for any $x$ in $B$, so $0$ is a lower bound and $2$ is an upper bound.
We have $(x_n)_{n\ge 1} \subset B$, $x_n=\frac{1}{n}\to 0$, so we may say that $\inf B=0$, and we also have that $(y_n)_{n\ge 1}\subset B$, $y_n=\frac{2n-1}{n}\to 2$, so we may say that $\sup B=2$.
I think that these proofs are all right, but most of the infimum supremum proofs I have seen used epislons. I just want to make sure that my solutions are correct too.

Answer 1

The proofs are correct. However, the first proof has a useless part. That occurs after you prove that, if $a\in A$, then $-a\leqslant-\inf A$. For some reason, you decided to deduce from this that, if $x\in A$, then $-x\leqslant-\inf A$. Why? It's the same assertion as the one that you had just proved. You simply replaced $a$ with $x$, but that changes nothing.