Question

Let E⊆R be nonempty. Prove that:

(i)E has an infimum if and only if –E has a supremum, in which case

sup⁡(-E)=-inf⁡E.

Answer 1

Question 1. Let $E \subseteq \mathbb{R}$ be nonempty. Prove that $E$ has an infimum if and only if $E$ has a supremum, in which case $\sup(-E) = -\inf E$.

Solution. Recall that $-E = \{-x \mid x \in E\}$. The set $E$ has an infimum if and only if there is $a \in R$ such that $x \geq a$ for all $a \in E$, or, equivalently, $-x \leq -a$ for all $a \in E$. The latter means that $-E$ is bounded from above by $-a$, hence $-E$ has a supremum. We prove that $\sup(-E) = -a$. By definition of an infimum, for any $\varepsilon > 0$ there is $x_{\varepsilon} \in E$ such that $a \leq x_{\varepsilon} < a + \varepsilon$. Multiplying by $-1$, we get the inequality $-a \geq -x_{\varepsilon} > -a - \varepsilon$. Thus, for any $\varepsilon > 0$ there is $y_{\varepsilon} = -x_{\varepsilon} \in -E$ such that $-a - \varepsilon < y_{\varepsilon} \leq -a$. This is exactly the assertion that $-a = \sup(-E)$.

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