Supremums 'alpha' and 'beta' problem

[kmeado07]

Homework Statement

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.

Homework Equations

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...

 

[mutton]

To show that alpha+beta is the supremum of A+B, use the definition of supremum.

 

[boaz]

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

 

[kmeado07]

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...

 

[boaz]

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.

 

[HallsofIvy]

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?