# supremum

Tags: supremum
 P: 41 1. The problem statement, all variables and given/known data Let A, B,be two non empty sets of real numbers with supremums 'alpha' and 'beta' respectively, and let the sets A+B and AB be defined by: A+B={a+b / a belongs to A, b belongs to B}, AB= {ab / a belongs to A, b belongs to B}. Show that alpha+beta is a supremum of A+B. 2. Relevant equations 3. The attempt at a solution Im not really sure how to go about this question. Maybe use the completness axion, all real numbers satisfy it, which means that A+B must have a supremem...
 P: 179 To show that alpha+beta is the supremum of A+B, use the definition of supremum.
 P: 36 a = supA -> a>=x1 when x1 in A b = supB -> b>=x2 when x2 in B a+b>=x1+x2 -> x1+x2 in A+B -> a+b is an upper bound. thus sup(A+B) <= a+b. lets say a+b-sup(A+B)=epsilon which is , sup(A+B)=a+b-epsilon there is x in A, which is smaller than a, and x=a-epsilon/2 there is also y in B, which is smaller than a, and y=b-epsilon/2 there is x1 in A which is bigger than x (cuz x is smaller than a [which is supA]) and there is y1 that is bigger than y too. if that so, we can say : x1 > x -> x1>a-epsilon/2 y1>y -> y1>b-epsilon/2 x1+y1 is in A+B and is also x1+y1 > a+b-epsilon = sup(A+B) contrary to the definition of sup(A+B). in conclusion, sup(A+B)=a+b=supA+supB
P: 41

## supremum

Well the definition of supremum is that it's the least upper bound, it is greater than or equal to each element in the set.

I still don't know how so start off showing that alpha+beta is the sup of A+B...
 P: 36 i probably haven't made myself clear. first of all, i have shown that supA+supB is an upper bound of the set A+B : supA+supB >= sup(A+B) later, i've proofed by contradiction that supA+supB>sup(A+B) so i've assumed that there is a number, epsilon, which is bigger than zero, that : sup(A+B)=supA+supB-epsilon. and that has brought me to the conclusion that sup(A+B)=supA+supB.
 Math Emeritus Sci Advisor Thanks PF Gold P: 38,879 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?

 Related Discussions Calculus & Beyond Homework 4 Calculus 1 Calculus & Beyond Homework 10 Calculus 1 Calculus 4