Let $A$ be a non-empty subset of the real numbers that is bounded below. Let $-A = \{ -x \mid x \in A\}$ Prove that $\text{inf}(A) = -\text{sup}(-A)$ Let $A$ be a non-empty subset of the real numbers that is bounded below. Let $$-A = \{ -x \mid x \in A\}$$ Prove that $$\inf(A) = -\sup(-A)$$
My attempt at a PROOF: 
Given that $$A \subset \mathbb{R},  A \ne \emptyset $$ and $A$ is bounded below, we have that there exists $\beta \in \mathbb{R} $ such that $\forall x \in A$ we have that $ x > \beta$. Also, $$ \forall x \in A, -x \in -A$$ This says that given $$ x > \beta$$ we have $$ -x < -\beta$$ Now, $-A \subset \mathbb{R}$ and $-A  \ne \emptyset$ and we have shown that $-A $ is bounded above. Therefore by the least-upper bound property on the real numbers there exists a least upper bound for $-A$, in particular we can say that $$\sup(-A) = - \beta$$ Now we look at the fact that given any $$ \alpha \in -A$$ if we have that $$ \alpha < -\beta$$ then $\alpha$ is not an upper bound for $-A$. So we have that $$\forall x \in A$$ $$-x \in -A$$ and so $$ -x \le \alpha < -\beta$$ Which gives that $$\beta < \alpha \le x$$ $\forall x, \alpha \in A$ hence we can see that $$\inf(A) = \beta$$ and this is exactly $$ -\sup(-A) = - (-\beta)= \beta$$
 A: First a proof of the result
Since $A$ is bounded below, $\inf A$ exists and we may suppose $a=\inf A$. 
For any $x\in -A$ we have $-x\in A$ and hence $a\le -x$ and so $x\le-a$. That establishes that $-a$ is an upper bound for $-A$. Now suppose that $b$ is any upper bound for $-A$.
For any $x\in A$ we have $-x\in -A$ and hence $-x\le b$ and so $x\ge-b$. Hence $-b$ is a lower bound for $A$ and so $-b\le a$. Hence $-a\le b$. So $-a$ is the least upper bound for $-A$.
So we have established that $-(\inf A)=\sup(-A)$ and hence $\inf A=-\sup(-A)$.
Turning to your proof
You start by saying (correctly) that $A$ has a lower bound $\beta$. But you then say that $-\beta=\sup(-A)$, which is not  necessarily true. However, suppose $-\beta$ is $\sup(-A)$.
You then say that if $\alpha\in(-A)$ satisfies $\alpha<-\beta$, then it is not an upper bound for for $-A$ (which is true). But you then say that for all $-x\in-A$ we have $-x\le\alpha$ (which is not true, as you just stated). You then conclude for reasons I don't follow that $\beta=\inf A$.
