Let $\emptyset\neq A\subseteq\mathbb{R}$ a bounded set. Consider $-A=\{-a:a\in A\}$. I want to prove that $-\sup(-A)=\inf(A)$.
It is easy to see that $-\inf(A)$ is an upper bound of $-A$, so $\sup(-A)\le -\inf(A)$, then $-\sup(-A)\ge \inf(A)$.
How can we prove that $-\sup(-A)\le \inf(A)$?
Thanks.
On
The map $x \mapsto -x$ is an order-inverting bijection $\mathbb R \to \mathbb R$.
It sends lower bounds for $A$ to upper bounds for $-A$, and vice-versa, hence the result.
On
My approach starts out similar to @CliveNewstead's but then I let $-\gamma = \sup(-A)$ and used the fact that the infimum is unique.
Since $A$ is nonempty and bounded below, $A = \{a:a\in A\}$ and $\inf(A) = \alpha$. Now, $-A = \{-a:a\in A\}$ is also nonempty. Since $\alpha$ is the infimum of $A$, $\alpha\leq a$ for all $a\in A$. By multiplying by $-1$, we get the following inequality $$ -\alpha\geq -a. $$ That is, $-\alpha$ is an upper bound of $-A$. Suppose $-\gamma = \sup(-A)$ and $\varepsilon > 0$. Then $-\gamma + \varepsilon\not\in -A$ $$ -\alpha\geq -\gamma + \varepsilon\geq -\gamma\geq -a $$ Again, by multiplying by negative one, we have $$ \alpha\leq \gamma - \varepsilon\leq\gamma\leq a $$ but $\gamma - \varepsilon\notin A$ so $\gamma$ is a lower bound of $A$ which would contradict the fact that $\alpha$ is the greatest lower bound of $A$. In order for $\gamma$ to be the lower bound, $\gamma = \alpha$ since the infimum is unique. So $-\alpha$ is the supremum of $-A$. Therefore, $\alpha = \inf(A) = -\sup(-A) = -(-\alpha) = \alpha$.
On
You can round out the argument by using a nice symmetry. In particular, you already have that: $$-\sup(-A)\geq \inf(A)$$ and easily, by the same reasoning, that $$-\inf(-B)\leq \sup(B)$$ Now, if we set $A=-B$, then we get, from the first inequality, that: $$-\sup(B)\geq \inf(-B)$$ then negating both sides: $$\sup(B)\leq -\inf(-B)$$ but we can simply tack on the second inequality $$\sup(B)\leq -\inf(-B)\leq \sup(B)$$ which implies $-\inf(-B)=\sup(B)$.
On
Basiclly by using the definition of supremum and infimum we can easily see that: $$ \sup(A) \Longleftrightarrow \exists c_0 = \min\{c|\forall a \in A: a \le c\} $$ $$ \sup(-A) \Longleftrightarrow \exists c_1 = \min\{c|\forall \bar a \in -A: \bar a \le c\} = \min\{c|\forall a \in A: -a \le c\} $$ $$ -\sup(-A) \Longleftrightarrow \exists c_2 = \max\{c|\forall a \in A: a \ge c\} = \inf(A) $$
You should work directly from the definitions of $\sup$ and $\inf$. That is, prove that if $x = \inf(A)$ then $-x \ge b$ for all $b \in {-A}$ (i.e. $-x$ is an upper bound for $-A$) and that if $y \ge b$ for all $b \in {-A}$ then $y \ge {-x}$ (that is, $-x$ is a least upper bound). This verifies that $-x$ is the sup of $-A$, and its proof uses the (similar) definition of $\inf(A)$.