Let $-S$ be the set $\left\{ -s:s \in S \right\}$ where $-S$ is the set that contains negatives of the members of $S$. We want to prove that $\inf(S) = -\sup(-S)$
Here is how I proved it
Let $s_0= \sup(-S)$. That is for all $-s_1\in -S$ then $-s_1 \leq s_0$. Multiplying both sides by $-1$ we get $-s_0 \leq s_1$ for all $s_1 \in S$. So $\inf(S)=-s_0=-\sup(-S)$ It looks short and sweet. Not sure if its right though.
Best Answer
It’s fine as far as it goes, but it’s incomplete. You’ve shown that $-\sup(-S)$ is a lower bound for $S$, but not that it’s the greatest lower bound. If you assume that it’s not the greatest lower bound, you can use the same basic idea to get a contradiction; I’ll leave it to you to try that on your own, but I’ll be happy to say more if necessary.