0
$\begingroup$

This question already has an answer here:

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.

$\endgroup$

marked as duplicate by Xander Henderson, Martin R, Theoretical Economist, Aweygan, Andrés E. Caicedo Sep 7 '18 at 18:19

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ 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. $\endgroup$ – user3496846 Sep 7 '18 at 18:21
  • 1
    $\begingroup$ 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 $s + t \le u$. $\endgroup$ – kobe Sep 7 '18 at 18:30
  • $\begingroup$ @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 $2b \le u - a + t$. $\endgroup$ – user3496846 Sep 7 '18 at 18:41
  • $\begingroup$ @kobe oh, wait, $t$ is the least upper bound by definition, I see now, thank you! $\endgroup$ – user3496846 Sep 7 '18 at 18:45
  • 1
    $\begingroup$ 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. $\endgroup$ – egreg Sep 7 '18 at 20:06

Browse other questions tagged or ask your own question.