Bounded intervals in R and bisection method proof

[prettymidget]

Homework Statement

Let property 1 be : If [ai,bi] is a sequence of intervals that are closed such that for each i the interval [a(i+1), b(i+1)] is either the left half of [ai,bi] or the right half, then there exists precisely 1 number in all intervals sequence.


Show if a field f satisfies this, then it satisfies the least upper bound property.

Homework Equations

The Attempt at a Solution

Let F be an ordered field that satisfies property 1. Let [ai,bi] be a sequence of intervals in F such that for each i the interval [a(i+1), b(i+1)] is either the left half of [ai,bi] or the right half. Then we have that there exists precisely 1 number in all intervals in this sequence.

The least upper bound property as my professor stated was that for any bounded above subset T of R, sup(T) exists.

I get intuitively why this has to follow from the bisection property thing, but I'm baffled on how to start the proof. My professor gave a hint in form of "How does the archimedean property follow from the bisection principle" but I'm not seeing it.

 

[mighty2000]

First, we need to show that F has the Archimedean property. This property states that for any x,y∈F with x>0, there exists a positive integer n such that nx>y.

To prove this, let x,y∈F with x>0. We can construct a sequence of intervals [ai,bi] such that ai=0 and bi=nx for some positive integer n. By property 1, there exists a unique number c that is contained in all intervals [ai,bi]. Since c is contained in the intervals, it must be greater than or equal to 0 and less than or equal to nx. Therefore, we have nx>c.

Now, we need to show that F satisfies the least upper bound property. Let T be a bounded above subset of F. We can construct a sequence of intervals [ai,bi] such that ai=sup(T)-1 and bi=sup(T)+1. By property 1, there exists a unique number c that is contained in all intervals [ai,bi]. Since c is contained in the intervals, it must be greater than or equal to sup(T)-1 and less than or equal to sup(T)+1. Thus, c is an upper bound for T and sup(T)≤c.

Now, let d be any upper bound for T. Since sup(T) is the least upper bound, we have sup(T)≤d. Therefore, sup(T) exists and F satisfies the least upper bound property.