Let \(A\) be a nonempty set of real numbers which is bounded below. Let \(-A\) be the set of all numbers \(-x\), where \(x \in A\). Prove that
\[
\inf A = \,-\sup(-A)
\]
To show this we first observe that if \( x < y\) then \( -y < -x \): [latex] \begin{align} x &< y \\ x-y &< y-y \\ x-y &< 0 \\ -x + ( x - y ) &< -x + 0 \\ (-x + x) -y &< -x \\ 0 - y &< -x \\ -y &< -x \end{align} [/latex] Since \(A\) is nonempty, we know that there exists \(x \in A\) and for any \(x \in A\) we have \(-x \in -A\). It should be clear that there are no other elements of \(-A\). And likewise for any element \(y \in -A\), \(-y \in A\) (and there are no other elements of \(A\).) Let \(x\) be any member of \(A\) then \(\inf A \le x \) and from our note above, \( -x \le -\inf A\). Therefore for all \(-x \in -A\), \(-x \le -\inf A\) thus \(-A\) is bounded above by \(-\inf A\). So \(-\inf A\) is an upper bound of \(-A\). We now need to show that it is the least upper bound, i.e. \(-\inf A = \sup(-A)\).
Suppose that this is not the case, that \( -\inf A \ne \sup(-A)\), then \(\sup (-A) < -\inf A\). Then there exists \(\beta < -\inf A \) such that for all \(y \in -A\), \(y \le \beta\). If \(y \in -A\) then \(-y \in A\) and \(-\beta \le -y\) thus \( -\beta\) would be a lower bound of \(A\). But because \(\beta < -\inf A \) then \( \inf A < -\beta\). In other words we have \(\beta\) such that for all \(-y \in A\), \(\inf A < \beta \le -y\). This is contrary to the assumption that \(\inf A\) is the infimum or greatest lower bound of \(A\). Thus \(\beta\) does not exist, so \(-\inf A = \sup(-A)\), i.e. \(\inf A =\,-\sup(-A)\).