(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Let A and B be two nonempty sets of numbers which are bounded above, and let A+B denote the set of all numbers x+y with x in A and y in B. Prove that sup(A+B) = sup(A) + sup(B).

Hint: The inequality sup(A+B) <= sup(A) + sup(B) is easy. To prove that sup(A) + sup(B)<= sup(A+B) it suffices to prove that sup(A) + sup(B) <= sup(A+B) + E for all E > 0; begin by choosing x in A and y in B with sup(A) - x < E/2 and sup(B) - y < E/2.

2. Relevant equations

N/A

3. The attempt at a solution

My attempt was not this complicated, so it's pretty surely wrong, yet I'm having trouble seeing where.

Since A is a nonempty set that is bounded above, then it must have a least upper bound. Call this least upper bound x_{0}. Then sup(A) = x_{0}, and x_{0}is the greatest possible value for x. Similarly, since B is a nonempty set that is bounded above, then it must have a least upper bound. Call this least upper bound y_{0}. Then sup(B) = y_{0}and y_{0}is the greatest possible value for y. Thus sup(A) + sup(B) = x_{0}+ y_{0}.

sup(A+B) is the least upper bound for A+B, and therefore occurs where A+B is its highest. Then the least upper bound for A+B is wherever x+y is highest. Since we already determined that x_{0}is the highest x and y_{0}is the highest y, then x_{0}+ y_{0}is the highest x+y and thus sup(A+B) = x_{0}+ y_{0}.

Therefore, sup(A+B) = x_{0}+ y_{0}= sup(A) + sup(B).

Also, I don't see how in Spivak's hint it's possible for him to simply pick an x in A such that sup(A) - x < E/2 if E>0. What if A is just the set {...-3, -2, -1, 0}? Then sup(A) - x >= 1 unless x is sup(A). And if E is, say, 1, then the inequality seems to fail. What's up, here?

Thank you.

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# Homework Help: Question About Spivak Proof Of Bounds

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