# Spivak Exercice, Least Upper Bounds Chapter, Proof

## Main Question or Discussion Point

Hi there!

I always think whether I am posting this correctly, or this belongs to the homework section. If so, my apologies.

I am trying to understand the solutions for a problem in Spicak Calculus, 3$^{rd}$ edition.

#8-13

The Problem:

"Let A and B bt two nonempty sets of numbers which are bounded above, and let A+B denote de set of all numbers x+y, with x in A and y in B. Prove that sup(A+B) = supA+supB. Hint: The inequality sup(A+B) ≤ sup A + sup B is easy. Why? To prove that supA + supB ≤ sup(A+B) it suffices to prove that supA + supB ≤ sup(A+B) + ε for all ε >0; begin by choosing x in A, and y in B with supA - x < $\frac{ε}{2}$ and supB -y <$\frac{ε}{2}$."

Well, proving sup(A+B) ≤ supA + supB seems to be straightforward, but it is proving that supA + supB ≤ sup(A+B) that bothers me.

The explanation in the "Answers" chapter of the book, doesn't explain very clearly anything in this exercice. So, I have tried to prove it on my own:

1 - First of all, why is it sufficient to prove that supA + supB ≤ sup(A+B) + ε for all ε >0?

supA + supB - sup(A+B) ≤ ε, for all ε > 0.
Lets denote X = { ε : ε > 0}.
Then X is bounded below, and it is nonempty. The greatest lower bound is 0. And, since supA + supB - sup(A+B) is also a lower bound for X, we should have
supA + supB - sup(A+B) ≤ 0 ⇔ supA + supB ≤ sup(A+B).

2 - How do you prove that supA + supB ≤ sup(A+B) + ε for all ε >0?

If supA and supB are the greatest upper bounds for the A and B sets, respectly, then, for every ε > 0, there is an x in A , and an y in B such that:

→supA - x < $\frac{ε}{2}$;
→supB - y < $\frac{ε}{2}$;

→supA < $\frac{ε}{2}$ + x;
→supB < $\frac{ε}{2}$ + y;

Which means : supA + supB < ε + (x+y) ⇔ (x+y) > supA + supB - ε.. All this means, that for every ε > 0 , there is an x in A and y in B such that (x+y) > supA + supB - ε.

And finally, since sup(A+B)≥ x+y, for every x in A and y in B, we have for every ε > 0

sup(A+B) ≥ supA + supB -ε. Which, according to the first point is suffient.

Well, I feel pretty confident about this. However I would really appreciate if you could check it. I have spent way to much time around this one, and wanted to move on. ;D

Regards,

c.teixeira

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Bacle2
I'm not sure I understand your first question, but if you show, in general, that:

x≤ y+1/n , for all n (subbing-in 1/n for ε ),

then it follows that x≤y

A way of showing this is that if x-y≤ 1/n for all n , then x≤y , by the

Archimedean Property. In a metric space,two elements cannot be

indefinitely-close ( using x-y as d(x,y) in ℝ ) , so we can conclude

x-y≤ 0 , so x≤y .