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Supremums 'alpha' and 'beta' problem

by kmeado07
Tags: supremum
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kmeado07
#1
Dec27-08, 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...
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mutton
#2
Dec27-08, 06:51 AM
P: 179
To show that alpha+beta is the supremum of A+B, use the definition of supremum.
boaz
#3
Dec27-08, 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+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
#4
Dec27-08, 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...
boaz
#5
Dec27-08, 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+supB-epsilon. and that has brought me to the conclusion that sup(A+B)=supA+supB.
HallsofIvy
#6
Dec27-08, 07:44 AM
Math
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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 [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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