# Supremums 'alpha' and 'beta' problem

 Math Emeritus Sci Advisor Thanks PF Gold P: 39,552 If "x" is any member of A+ B, then a= a+ b for some a in A, b in B. Since $\alpha$ is an upper bound on A, $x\le \alpha$. Since $\beta$ is an upper bound on B, $b\le\beta$. Therefore, $a+ b\le$ ? That shows that $\alpha+ \beta$ is an upperbound on A+ B. Now you need to show it is the smallest upper bound and I recommend you use an "indirect proof" or "proof by contradiction" to do that: suppose $\alpha+ \beta$ is NOT the least upper bound of "A+ B". Suppose there exist some lower bound, $\gamma$ smaller than $\alpha+ \beta$. Let $\epsilon= (\alpha+ \beta)- \gamma$. Can you find some "a" in A so that $x> \alpha+ \epsilon/2$? Can you find some "b" in B so that $y> \beta+ \epsilon/2$? If so, what can you say about a+ b?