Set theory proof problem about bounds Suppose $A \ne \emptyset$ is bounded below. Let $-A$ denote the set of all $-x$ for $x$ in $A$. Prove that $-A \ne \emptyset$ that $-A$ is bounded above, and that $-\sup(-A)$ is the greatest lower bound of $A$.
I have absolute no clue, what I thought was:
Let $B = \inf A$
We assume $-A = \emptyset$ then we recognize that $A = x \in $R
If elements of $-A = -($elements of A) then we reach a contradiction because $-x$ exists as $A \ne \emptyset$. 
I really cant prove anything else. This problem has me stumped. Anything is appreciated. 
 A: Background ideas.
Notation. If $x \in \mathbb{R}$ and $A \subseteq \mathbb{R}$, write $x \leq A$ to mean that $x$ is a lower bound for $A$, meaning that $x \leq a$ for all $a \in A$. Similarly, write $A \leq x$ to mean that $x$ is an upper bound for $A$, meaning that $a \leq x$ for all $a \in A$.
Exercise. Show that if $f : \mathbb{R} \rightarrow \mathbb{R}$ is an order reversing function, then $x \leq A$ implies $f(A) \leq f(x)$.
Anyway, in the above notation, we have the following "fundamental theorem" (that no one ever tells you about) which provides the easiest way to prove anything about $\mathrm{inf}$. There is a version for $\mathrm{sup}$, too.
Fundamental Theorem Of Infima. Suppose $k$ is an element of $\mathbb{R}$ and that $A \subseteq \mathbb{R}$ is non-empty and bounded below. Then $\mathrm{inf}(A) = k$ iff for all $x \in \mathbb{R}$, we have $x \leq A$ iff $x \leq k$.
Corollary. $x \leq A$ iff $x \leq \mathrm{inf}(A)$.

The actual problem.
You want to prove that
$$\mathrm{inf}(A) = -\mathrm{sup}(-A)$$
Its no harder to prove a more general theorem. An involution on $\mathbb{R}$ is a function $f : \mathbb{R} \rightarrow \mathbb{R}$ such that $f(f(x))=x$ for all $x \in \mathbb{R}$. We may as well prove the more general case that for any order-reversing involution $f : \mathbb{R} \rightarrow \mathbb{R}$, we have
$$\mathrm{inf}(A) = f(\mathrm{sup}(f(A)))$$
By the fundamental theorem of infima, to the prove the statement of interest, it suffices to let $x$ denote an arbitrary element of $\mathbb{R}$, and to prove that $x \leq f(\mathrm{sup}(f(A)))$ iff $x \leq A$. (Note that we're using the full theorem, not just the corollary.) So observe that the following are equivelent.


*

*$x \leq f(\sup(f(A)))$

*$\sup(f(A)) \leq f(x)$

*$f(A) \leq f(x)$

*$x \leq A$


Note that to get from Line 2 to Line 3, we're using the "fundamental theorem of suprema." Or more precisely, a corollary thereof.
Fundamental Theorem Of Suprema. Suppose $k$ is an element of $\mathbb{R}$ and that $A \subseteq \mathbb{R}$ is non-empty and bounded above. Then $\mathrm{sup}(A) = k$ iff for all $x \in \mathbb{R}$, we have $A \leq x$ iff $k \leq x$.
Corollary. $A \leq x$ iff $\mathrm{sup}(A) \leq x$.
A: Hint: (1). $x \in A \Rightarrow -x \in -A \Rightarrow -A$ is non-empty. 
(2). First note that $-(-A) = A$. So for every element $y \in -A, -y \in A.$ Since $A$ is bounded below, $-A$ is bounded above. If $u$ is an upper bound of $-A$ then $-u$ is a lower bound of $A.$ So...?
