How to show $-\sup(-A)=\inf(A)$?

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.

Solution 1:

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 $\inf(A)$.

Solution 2:

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.

Solution 3:

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 $-\inf(-B)=\sup(B)$.