Principles of Mathematical Analysis, Rudin, 3th ed, Chapter 1, Problem 5

Problem

Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that

$$\inf A = -\sup(-A).$$

Answer

Let $\alpha = \inf A$. For all $x \in A$, $\alpha \geq x$  $\Longrightarrow$  $-\alpha \geq -x$. Therefore, $-\alpha$ is upper bound of $-A$.

To prove $-\alpha$ is the supremum of $-A$, suppose that there exists another upper bound $\beta$ of $-A$. It remains to show that $\beta \geq -\alpha$. For all $x \in A$, $-x \leq \beta$  $\Longrightarrow$ $x \geq -\beta$, which means $-\beta$ is lower bound of $A$. Since $\alpha=\inf A$, $\alpha \geq -\beta$ $\Longrightarrow$ $\beta \geq -\alpha$.

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