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Suppose sets $A$ and $B$ are not empty and are bounded from above. Prove that $\sup(A) + \sup(B) = \sup(A + B)$ without using $\epsilon$. This is exercise 1.3.6 from Understanding Analysis by Abbot, 2nd edition.

Here is my start:

Let $s = \sup(A)$, $t = \sup(B)$, then for any $a \in A, b \in B$ we have $a \le s$ and $b \le t$ and, consequently, $s + t \ge a + b$.

And here I get stuck:

Let $u$ be an arbitrary upper bound of $A + B \implies a + b \le u$. Show that $t \le {u - a}$.

$b \le {u - a}$ is promising, but I don't see how to proceed rigorously from here on.

EDIT

The other question has answers that prove the same thing (some of them w/out epsilon), but they don't do it the way the book proposes to. I am interested in the book's way as well.

user3496846
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  • The other question has answers that prove the same thing, but they don't do it the way the book proposes to. I am interested in the book's way as well. – user3496846 Sep 07 '18 at 18:21
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    For fixed $a$, $u - a$ is an upper bound for $B$, so $t \le u - a$. Since $a$ is arbitrary, $u - t$ is an upper bound for $A$. Thus $s \le u - t$, or $s + t \le u$. – kobe Sep 07 '18 at 18:30
  • @kobe in your reasoning, I don't see how $t \le u - a$ follows from $b \le u - a$. Having $b \le t$, the inequalities $b \le t$ and $b \le u - a$ can only yield $2b \le u - a + t$. – user3496846 Sep 07 '18 at 18:41
  • @kobe oh, wait, $t$ is the least upper bound by definition, I see now, thank you! – user3496846 Sep 07 '18 at 18:45
  • Regarding the edit: Your book’s only constraint seems to be not to use $\epsilon$, so I don’t understand why you think that the answers (which does not use $\epsilon$) “don't do it nearly the way the book proposes to”. – Jendrik Stelzner Sep 07 '18 at 18:59
  • @JendrikStelzner the book asks to show $t \le u - a$, the other answers only somewhat have a similar structure of the proof (as I see only now). The closest answer uses sup(A + B) instead of an arbitrary bound for A + B. Therefore, I said 'nearly', which probably is not a good term when discussing math, I will edit it out. Thanks. – user3496846 Sep 07 '18 at 19:09
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    You know that, for every $a\in A$ and $b\in B$, $a+b\le u$. You want to prove that, for every $a\in A$, $a+t\le u$. Otherwise $a+t>u$, that means $t>u-a$. As $t=\sup B$, there exists $b\in B$ with $u-a<b$, so $u<a+b$: contradiction. – egreg Sep 07 '18 at 20:06
  • @egreg makes sense, thanks for this demonstration! – user3496846 Sep 09 '18 at 11:09

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