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## Homework Statement

Let A ⊂ R be nonempty. Define −A = {−x / x ∈ A}. Show that sup(−A) = − inf A and inf(−A) = − sup A.

So I thought the definition of a, say, least upper bound, was that [tex]\alpha[/tex] is a least upper bound of

*E*if it is an upper bound of

*E*and if [tex]\gamma < \alpha[/tex] then [tex]\gamma[/tex] is not an upper bound of

*E*. But in the solution, they almost seem to be using a different definition of least upper bound / greatest lower bound.

**3. (part of the solution)**

This is just part of the solution. I'm not really confused about the actual answer to the question, I'm just confused about how they define the least upper bound.

In the solution, in a case where they assumed A had a lower bound:

Let us define r := inf A. Then r ∈ R. Since r is a lower bound for

A,we know that x ≥ r, ∀x ∈ A. It follows that: −x ≤ −r, ∀x ∈ A.

Hence, −r is an upper bound for A. We must show that it is the least

upper bound for A. To see this, let e > 0 be given. Since r = inf A,

it follows that ∀ e > 0, ∃a ∈ A such that a < r + e. So, ∀ e > 0, ∃a ∈ A

such that −a > −r −e . Hence, since such an element of −A can be

found ∀ e > 0, it follows that −r is the least upper bound.

...What they were doing with e seems almost more like something with a limit point...So I got confused...