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**novice** Let $\displaystyle S$ be a nonempty subset of $\displaystyle \mathbb{R}, $ and let $\displaystyle -S = \{-s: s \in S\}.

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Prove that $\displaystyle inf S = -sup (-S). $ [Hint: $\displaystyle -\infty < inf S]$

I wrote the proof as follows:

$\displaystyle s \in S \Longrightarrow -s\in -S, \forall -s \in -S $

$\displaystyle \Longrightarrow -s \leq sup(-S) $

$\displaystyle \Longrightarrow -sup(-S) \leq s, \forall s \in S.$

$\displaystyle \Longrightarrow -sup(-S)$ is a lower bound of $\displaystyle S.$

$\displaystyle s \in S \Longrightarrow \exists m \in \mathbb{R}$ such that $\displaystyle m \leq s.$

$\displaystyle \Longrightarrow -s \leq -m $, so $\displaystyle -m$ is an upper bound, and $\displaystyle sup(-S) \leq -m$

$\displaystyle \Longrightarrow m \leq -sup(-S)$

Since $\displaystyle m$ is a lower bound of $\displaystyle S$, $\displaystyle -sup(-S)$ is the greates lower bound of $\displaystyle S$.

Therefore, $\displaystyle inf(S)=-sup(-S).$

Question: Could someone please show me how to use the hint?