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Supremums 'alpha' and 'beta' problemby kmeado07
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#1
Dec2708, 06:01 AM

P: 40

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... 


#2
Dec2708, 06:51 AM

P: 179

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



#3
Dec2708, 07:01 AM

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+bsup(A+B)=epsilon which is , sup(A+B)=a+bepsilon there is x in A, which is smaller than a, and x=aepsilon/2 there is also y in B, which is smaller than a, and y=bepsilon/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>aepsilon/2 y1>y > y1>bepsilon/2 x1+y1 is in A+B and is also x1+y1 > a+bepsilon = sup(A+B) contrary to the definition of sup(A+B). in conclusion, sup(A+B)=a+b=supA+supB 


#4
Dec2708, 07:03 AM

P: 40

Supremums 'alpha' and 'beta' problem
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... 


#5
Dec2708, 07:10 AM

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+supBepsilon. and that has brought me to the conclusion that sup(A+B)=supA+supB. 


#6
Dec2708, 07:44 AM

Math
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P: 39,552

If "x" is any member of A+ B, then a= a+ b for some a in A, b in B. Since [itex]\alpha[/itex] is an upper bound on A, [itex]x\le \alpha[/itex]. Since [itex]\beta[/itex] is an upper bound on B, [itex]b\le\beta[/itex]. Therefore, [itex]a+ b\le [/itex] ? That shows that [itex]\alpha+ \beta[/itex] 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 [itex]\alpha+ \beta[/itex] is NOT the least upper bound of "A+ B". Suppose there exist some lower bound, [itex]\gamma[/itex] smaller than [itex]\alpha+ \beta[/itex]. Let [itex]\epsilon= (\alpha+ \beta) \gamma[/itex]. Can you find some "a" in A so that [itex]x> \alpha+ \epsilon/2[/itex]? Can you find some "b" in B so that [itex]y> \beta+ \epsilon/2[/itex]? If so, what can you say about a+ b? 


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