User:J D Bowen/Math710 HW1

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

Suppose $$q, r \ $$ are upper bounds for $$A, B \ $$ respectively. Then for all elements $$a\in A, b \in B \ $$, we have $$a<q, b<r \ $$. We can add these two inequalities to get $$a+b <q+r \ $$, and so $$q+r \ $$ is an upper bound for $$A+B \ $$. Since $$\text{sup}(A)+\text{sup}(B) \ $$ is a sum of an upper bound for $$A \ $$ and an upper bound for $$B \ $$, it follows that it is an upper bound for $$A+B \ $$.

All that remains to be shown is that is it the least upper bound. Suppose to the contrary that $$\text{sup}(A+B) < \text{sup}(A)+\text{sup}(B) \ $$. By definition, this means we can find a positive number $$\delta \ $$ such that $$\text{sup}(A)+\text{sup}(B)-\text{sup}(A+B) > \delta \ $$.

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 $$\text{sup}(A+B)=\text{sup}(A)+\text{sup}(B) \ $$.

2. (a) (b) (c)