There's something that I don't get in this exercise:
Let $A\subseteq \mathbb{R}$ be nonempty and bounded above. Prove that $\operatorname{inf}\lbrace x\in \mathbb{R}:-x\in A\rbrace = -\operatorname{sup}(A)$.
I can assume e.g. $A = \lbrace1,2,3 \rbrace$ in which case $\operatorname{sup}(A)=3$ but $\operatorname{inf}\lbrace x\in \mathbb{R}:-x\in A\rbrace = \operatorname{inf}\lbrace \rbrace$ is empty. Is this correct or am I doing something wrong?
If $A = \{1,2,3\}$, then
$$\lbrace x\in \mathbb{R}:-x\in A\rbrace = \{-1,-2-3\}$$
as $-1,-2,-3 \in \mathbb{R}$, and each of their negations are in $A$.