Suppose $A \neq Ø$ is bounded below. Let $-A$ denote the set of all $-x$ for $x\in A$. Prove that $-A\neq Ø$, that $-A$ is bounded above, and that $-\sup(-A) =\inf(A)$.
Ok here is what I've done:
I) $-A\neq \emptyset$
We got that $-A=\left\{-x\,:\,x \in A\right\}$ Then as we have $x\in A \Longrightarrow -x\in-A$
II) $-A$ is bounded above
Here is where I'm having trouble, I have that $A$ is bounded below, so, let's say $y$ is a lower bound of $A$, then that then $-$y should be an upper bound of $-A$ but I don't know how to do this
III) $-\sup(-A) =\inf(A)$
This part is too long but I already did it.
How can I prove part II?
Suppose that $y$ is a lower bound of $A$. Take an element $b$ of $-A$. Then $b=-a$, for some $a\in A$, and $y\leqslant a\implies-y\geqslant-a=b$ .