User:Michellepoliseno/Sandbox

Math 70 assignment 1
1. Let A and B be bounded, non-empty subsets of \mathbb{R} \. Define A+B = \left\{{ a+b : a\in A \land b \in B }\right\} \.

Suppose x, y \ are upper bounds for A, B \ respectively. Then for all elements a\in A, b \in B \, we have a<x, b \delta \.

Let x\in A \ be such that \text{sup}(A)-x<\delta/4 \ ; we know such a number exists because the alternative is that \forall x \in A, \text{sup}(A)-x\geq \delta/4 >0 \, and so \text{sup}(A)-\delta/5 \ would be a lower bound for the set which is less than \text{sup}(A) \ , which contradicts the definition of supremum.

Similarly, we can find y\in B \ such that \text{sup}(B)-y<\delta/4 \.

Then we have \delta<\text{sup}(A)+\text{sup}(B)-\text{sup}(A+B)<x+y+\delta/2 -\text{sup}(A+B) \.

This leads to \text{sup}(A+B)+\delta/2 <x+y \. But x+y \in A+B \, and so we must also have \text{sup}(A+B) \geq x+y \. Since we have reached a contradiction, our assumption \text{sup}(A+B) < \text{sup}(A)+\text{sup}(B) \ must have been wrong. Since we have demonstrated that \text{sup}(A+B) \ is an upper bound for A+B \, we must conclude that \text{sup}(A+B)=\text{sup}(A)+\text{sup}(B) \.

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