Greatest Lower Bound of A - Proving it with an Axiom

[steven187]

hello all

I know this might be a simple question to ask, but i want to find other ways of proving it anyway here we go
propve that if A is a subset of R and is non empty and bounded below, then it has a greatest lower bound.

This is how i did it:

let b be a lower bound of A. then for every a an element of A b<=a so -a<=-b. now we notice that -b is an upper bound for -A, where -A={-x,x an element ofA}. since -A is non empty and it is bounded above then by the least upper bound axiom -A has a least upper bound , let's call this least upper bound -L , since -L is the least upper bound of -A, we must have -L<=-b or b<=L. and so this is true for any b which is a lower bound of A and hence L must be the greatest lower bound of A

is there another method to do this problem if not is there anyway of simplyfying this proof?

thanxs

 

[rachmaninoff]

It's the shortest proof I can think of, if you're already given the greatest upper bound axiom.

 

[M98Ranger]

Hi there,

Thank you for sharing your proof with us. Your approach is a valid and logical way to prove the existence of a greatest lower bound for a non-empty and bounded below set A. However, there is another way to prove this using the completeness axiom for the real numbers.

The completeness axiom states that every non-empty subset of the real numbers that is bounded above has a least upper bound. We can use this axiom to prove the existence of a greatest lower bound for a non-empty and bounded below set A.

Let b be a lower bound of A. Then for every a ∈ A, b ≤ a. This means that b is an upper bound for the set B = {-x | x ∈ A}. Since B is non-empty and bounded above by b, by the completeness axiom, B has a least upper bound, let's call it L.

Now, we need to show that L is the greatest lower bound of A. This can be done by contradiction. Assume that there exists a lower bound b' of A such that b' > L. This means that -b' < -L and since -L is the least upper bound of B, we have -L ≤ -b'. This leads to b' ≤ L, which contradicts our initial assumption. Therefore, L must be the greatest lower bound of A.

In summary, we can prove the existence of a greatest lower bound of A by using the completeness axiom for the real numbers. This proof is similar to yours, but it might be viewed as simpler and more direct. I hope this helps!