If $A$ is a subset of $\mathbb{R}$,denote by $−A={x|−x∈A}$. Assume that $A$ is bounded from below. Show that $−A$ is bounded from above and $−\inf(A) = \sup(−A)$.
In the solution, the first part states "To see the first part note that if $r$ is lower bound of $A$, then $−r$ is an upper bound of $−A$." and the proof follows from this. However, I don't understand where that statement comes from. Why is $-r$ an upper bound of $-A$?
On
For all $a$ in $A$ there is a corrsponding $-a$ in $-A$
$\forall a \in A, a\ge \inf A \implies -a \le \sup -A$
As $\inf A$ is the greatest lower bound, either $\inf A \in A$ or there is a sequence $\{a_n\}$ that converges to $\inf A$
If there is a sequence $\{a_n\}$ that converges to $\inf A$ the corresponding sequence $\{-a_n\}$ converges to $\sup -A$
If $x\geq r$ for all $x\in A$, then $-x\leq -r$ for all $x\in A$. But every element of $-A$ is of the form $-x$ for some $x\in A$ whence every element of $-A$ is at most $-r$.