# Least Upper Bounds

• Oct 3rd 2010, 06:11 PM
Syia
Least Upper Bounds
Problem:

Suppose that A is not an empty set and is bounded below, let B be the set of all lower bounds of A.

1.) Show that B is not an empty set

By the definition of bounded below, there exists an element b in the set B where b <= a for every a in A. Thus, B is not an empty set.

2.) Show that B is bounded above

I'm a bit lost here; I believe that I have to prove that there exists some number n such that for every b in the set B, m has to be greater than or equal to b. But I'm not sure how to prove this.

3.) Show that that sup B is the greatest lower bound of A.

In essence, I'm trying to prove that sup B = inf A. Which intuitively I understand, but I'm not sure how to approach showing this.

I am really struggling with conceptually following proofs of least upper bounds. Any help would be greatly appreciated!

• Oct 3rd 2010, 06:59 PM
DontKnoMaff
A is yourset of upper bounds. Try from there.

For 3 i would try and prove that supB < infA and infA< supB both lead to contradictions
• Oct 3rd 2010, 07:15 PM
elim
1.) Show that B is not an empty set

By the definition of bounded below, A has a lower bound b and by the definition of B, b is in B and so B is not empty

2.) Show that B is bounded above

Since A is not empty, we can pick an a in A. For any b in B, b is a lower bound of A and so b ≤ a. Hence a is an upper bound of B. Hence B is bounded above.

3.) Show that that sup B is the greatest lower bound of A.

Let t = sup B, a ∈ A, by the proof of 2), a is an upper bound of B and since t is the least upper bound of B, we have t ≤ a ans so
(1) t is a lower bound of A.
If T > t = sup B, then T is not a element of B hence not a lower bound of A and so a < T for some a ∈ A and so T is not a lower bound of A. In short,
(2) T > t implies T is not a lower bound of A.

Combine (1) and (2) we see that t = sup B = inf A

Now we see that