Proving Existence and Uniqueness of Cut C for A+C=B

[jawad1]

Homework Statement

Show that for any two Dedekind cuts A,B, there exists a unique cut C such that A+C=B

2. The attempt at a solution

In order to prove this, I need to prove the existence and uniqueness of such a cut.

For the existence, I started by considering a cut for which this works: C={b-a, b \in B, a \in \mathbb{Q}-A} but I am having trouble showing it is a cut.

For the uniqueness, I want to consider two cuts C and C' such that C≠C' and I wanted to show that C is not a subset of C' and C' is not a subset of C. However, could someone help me to prove this ?

Thank you in advance

 

[mighty2000]

.
Thank you for your post. I will now provide a proof for the existence and uniqueness of the cut C that satisfies the given condition.

Existence:
Let A and B be two Dedekind cuts. We will construct the cut C={b-a, b\in B, a\in \mathbb{Q}-A}. First, we will show that C is a cut.
1. Non-emptiness: Since A and B are cuts, there exist rational numbers a and b such that a∈A and b∈B. Then, b-a∈C, which proves that C is non-empty.
2. Closure under subtraction: Let c∈C and d∈\mathbb{Q} be such that d<c. Since c=b-a for some b∈B and a∈\mathbb{Q}-A, we have d<b and d>a. Since A is a cut, d∈A. Thus, d∈\mathbb{Q}-A and b-d∈C. Therefore, C is closed under subtraction.
3. No largest element: Suppose there exists a largest element, c∈C. Then, c=b-a for some b∈B and a∈\mathbb{Q}-A. But since A is a cut, there exists a rational number d such that a<d and d∈A. Then, d∈\mathbb{Q}-A and b-d∈C. But b-d<c, which contradicts the assumption that c is the largest element. Therefore, C has no largest element.
4. No smallest element: Similarly, we can show that C has no smallest element.
Therefore, C is a cut.

Now, we will show that A+C=B.
Let x∈B. Since B is a cut, there exists a rational number b such that b>x and b∈B. Then, b-x∈\mathbb{Q}-A and b-x∈C. Therefore, x+(b-x)=b∈A+C. This shows that B⊆A+C.

Now, let x∈A+C. Then, x=a+c for some a∈A and c∈C. Since C={b-a, b∈B, a∈\mathbb{Q}-A}, we have c=b-a for some b