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Homework Statement
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)##.
Homework Equations
Definition:
Suppose ##S## is an ordered set, ##E\subset S##, and ##E## is bounded above. Suppose there exists an ##\alpha \in S## with the following properties:
(i) ##\alpha## is an upper bound of ##E##.
(ii) If ##\gamma < \alpha## then ##\gamma## is not an upper bound of ##E##.
Then ##\alpha## is called the supremum of ##E## and we write ##\alpha = \sup E##.
(Equivalently for infimum)
The Attempt at a Solution
From the definition of supremum ##\exists \alpha > y, \forall y \in -A## or equivalently ##\exists \alpha > -x, \forall x \in A##.
Then ##-\alpha < x, \forall x \in A##, hence ##-\alpha = -\sup(-A)## is a lower bound of ##A##.
It's left to show that if ##\gamma > -\alpha## then ##\gamma## is not an lower bound of ##A##.
Suppose ##\gamma## is a lower bound of ##A## with ##\gamma > -\alpha##. Then ##\gamma < x \forall x\in A## or equivalently ##-\gamma > -x \forall x \in A##. But this means that ##\gamma## is an upper bound of ##-A## and since ##\sup(-A) = \alpha## we have that ##-\gamma \ge \alpha## or equivalently ##\gamma \le -\alpha## a contradiction! Hence ##\inf A = -\sup(-A)##.
Is the above correct? Anything I could do to improve it? I'm quite new to proofs so I'm not sure if I'm doing this right.