Theorem

$\displaystyle E \subseteq \mathbb{R}, E \not = \varnothing$

sup $\displaystyle E$ exists $\displaystyle \Leftrightarrow$ inf $\displaystyle (-E)$ exists

Proof:

$\displaystyle "\Rightarrow"$

$\displaystyle supE$ exists

Now we show that $\displaystyle -supE=inf (-E).$

Show that

1. $\displaystyle -sup E$ is a lower bound of $\displaystyle -E$.

2. if $\displaystyle s$ is a lower bound of $\displaystyle -E \Rightarrow s \leq$ -sup $\displaystyle E.$

1.

$\displaystyle \because$ $\displaystyle sup E$ is an upper bound of $\displaystyle E$

$\displaystyle \therefore x \leq$ $\displaystyle sup E, \forall \in E \Rightarrow -x \geq$ $\displaystyle -sup E, \forall x \in E$

$\displaystyle \therefore -sup E$ is a lower bound of $\displaystyle -E$ (line added for correction 6:30 a.m. Sep 12)

2. Suppose that $\displaystyle s$ is a lower bound of $\displaystyle -E$

suppose not $\displaystyle \Rightarrow s >$ $\displaystyle -sup E \Rightarrow -s <$ $\displaystyle sup E$

on the other hand

$\displaystyle -x \geq s, \forall x \in E$ <------(Could someone tell me how this come about.)

$\displaystyle \therefore x \leq -s$

Hence, $\displaystyle -s$ is an upper bound of $\displaystyle E \rightarrow\leftarrow$

By 1. & 2, $\displaystyle inf (-E) \exists$ & $\displaystyle inf (-E)$ = $\displaystyle sup E$