I do not even have an idea how to prove the following. Could you please help me?
We say that a set of real numbers $A$ is bounded below if there is a number $d$ such that $d \leq x$ for all $x \in A$; then $d$ is called lower bound of $A$. Every nonempty bounded below set of real numbers has greatest lower bound which is denoted by $\inf A$. Prove that:
$\inf A = −\sup(−A)$, where $−A = \{−x : x \in A\}$.
Thanks in advance
Amadeus
Best Answer
Call $i=\text{inf}A$. Then, by virtue of being an infimum, $i$ is also a lower bound of $A$, thus $i\leq x$ for all $x \in A$. If $a$ is an element of $-A$, then by definition $-a$ is an element of $A$. This also means that $i\leq -a$ and thus $-i\geq a$. In other words, $-i$ is an upper bound of $-A$.
Now we have to show that $-i$ is the least upper bound. This is true if for every positive $\epsilon$ there exists an $a \in -A$ such that $-i-\epsilon<a$. Or equivalently, $i+\epsilon>-a$. Now, by virtue of $i$ being an infimum, there is for this positive $\epsilon$ a $x\in A$ such that $i+\epsilon>x$. But since $-x \in -A$, there thus does exist an $a$ satisfying our requirements, namely $-x$.