Prove that sup(a, b)=sup[a, b]=b and inf(a, b)=inf[a, b]=a. Is completeness relevant?

Question

Denoting, as usual, by (a, b) an ”open interval”, {x : a<x< b} and by [a, b] the corresponding ”closed interval”, {x : a≤x≤b} of real numbers, prove that sup(a, b)=sup[a, b]=b and inf(a, b)=inf[a, b]=a.

Is completeness relevant?

In other words, does the proof work for any ordered field?

Answer 1

There are several ways to show that. The direct way is just to use definition of sup and inf.

1) We'd like to show that

sup(a, b)=b, (a<b).

If a is a minimal element from (a, b), then by the definition of inf this means that ∀x∈ (a, b)

 

i)x≥a

ii) ∀ε>0 ∃x' ∈ (a, b) such that

a+ε<x'.

Now, x was the minimal element but we found another one x'-ε, which is smaller then x. This proves that a inf(a,b). Similar one can show that a=inf[a,b].

As for the sup, you can use the fact sup{-x}=-inf{x}, where {-x} is the set elements, opposite to x sign.