show that $\sup_n a_n = -\inf_n( -a_n)$ $s=\sup_n a_n \implies a_n \leq s \;\; \forall n$
$-i=\inf_n( -a_n) \implies -a_n \geq i \;\; \forall n$
it's clear that $i$ is an upper bound of $a_n$ and $-s$ is a lower bound of $-a_n$
so from that we can say that 
$s \leq i \; \text{&} -i \geq -s \Leftrightarrow  s \geq i $  
therefore $i=s \iff \sup_n a_n = -\inf_n( -a_n)$
I believe my reasoning is correct however I'd like to see a more formal proof involving epsilons
if anyone could provide that it'd be great.
thanks !
 A: Let $i=\inf_n(-a_n)$. This means that $i$ is a lower bound of the set $\{-a_n\,|\,n\in\mathbb{N}\}$ and, more precisely, that it is the greatest lower bound. So, if $n\in\mathbb N$, $i\leqslant-a_n$, which is equivalent to $a_n\leqslant -i$. Therfore, $-i$ is an upper bound of the set $\{a_n\,|\,n\in\mathbb{N}\}$. Is there a smaller lower bound $s'$? No, because$$(\forall n\in\mathbb{N}):a_n\leqslant s'\iff(\forall n\in\mathbb{N}):-s'\leqslant a_n$$and, since $i$ is the greatest lower bound of $\{-a_n\,|\,n\in\mathbb{N}\}$, $-s'\leqslant i$, which means that $s'\geqslant-i$.
Therefore, $-i$ is the least upper bound of the set $\{a_n\,|\,n\in\mathbb{N}\}$, which means that$$-i=\sup\{a_n\,|\,n\in\mathbb{N}\}.$$
A: Just use the definition $s=\sup_n a_n \Leftrightarrow (a_n \leq s \;\; \forall n$ and if $s' \ne s$ has the same property, then $s' \gt s)$, no need for epsilons
A: $S:=\sup_n a_n$, I.e $a_n \le S.$
$S$ is the least upper bound. Any $R$ with $R < S$ is not an upper bound.
Consider : $-a_n \ge -S$
(Multiply inequality by -1).
Means: $-S$ is a lower bound for $-a_n .$
Assume there is a greater lower bound $-L$ for $-a_n:$
Then $-a_n \ge -L \gt -S$, for all $n:$
This implies $a_n \le L \lt S$, for all $n.$
Contradiction, since $S$ is the least upper bound for $a_n.$
Hence: $-S = \inf (-a_n).$
