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Statement: Let $A \subset \mathbb{R}$ be bounded above. Prove that $\inf\{x\in\mathbb{R}: -x\in A\} = -\sup(A)$.

Proof: Since $A$ is (implicitly nonempty and) bounded above, $\sup(A)$ exists. Denote $S = \{x\in\mathbb{R}: -x\in A\}$. We have $\inf(S) \leq x$ for all $x\in S$, so $-\inf(S) \geq (-x)\in A$. Thus $-\inf(S)$ is an upper bound of $A$. If $l$ is any lower bound of $S$, then $l\leq x$, so $-l\geq (-x)\in A$. Therefore, $-l$ is an arbitrary upper bound of $A$. Since $\inf(S) \geq l$, we have $-\inf(S) \leq -l$, and so $-\inf(S) = \sup(A)$. Thus $\inf(S) = -\sup(A)$.

Question: Does the bold part follow from $l$ being an arbitrary lower bound? How do we know that $-l$ represents every upper bound of $A$? Is it sufficient to note that $-x$ is an arbitrary element of $A$, and $-l\geq (-x)\in A$ (the definition of an upper bound) is the only property of $l$? In other words, $l$ is the most general possible upper bound?

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