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We want to show that $\sup(A)+\sup(B)$ is the least upper bound of the set $A + B$. First, we need to show that $\sup(A) + \sup(B)$ is an upper bound for the set $A + B$. Indeed, if $z\in A + B$, then there exists $a\in A$ and $b\in B$ such that $z = a + b$. But by definition of $\sup(A)$ and $\sup(B)$, $a \leq \sup(A)$ and $b \leq \sup(B)$, so $z = a + b \leq \sup(A) + \sup(B)$. So, $\sup(A) + \sup(B)$ is an upper bound for $A + B$.

We now wish to show that $\sup(A) + \sup(B)$ is the least upper bound for the set $A + B$. So, if $u$ is an upper bound for $A + B$, we need to show that $\sup(A)+\sup(B) \leq u$. We will use part (i): that is, we need to show that there exists $\varepsilon > 0$, $\sup(A) + \sup(B) < u + \varepsilon$.

To do this, note that since $\sup(A)$ is the least upper bound for $A$, $\sup(A) - \varepsilon/2$ is not an upper bound for A, so there exists an $a\in A$ so that $\sup(A) - \varepsilon/2 < a$. Similarly, there is a $b\in B$ so that $\sup(B) - \varepsilon/2 < b$.

Adding these two inequalities gives $$ \sup(A) + \sup(B) - \varepsilon < a + b; $$ in other words $$ \sup(A) + \sup(B) < a + b + \varepsilon. $$ But $u$ is an upper bound for the set $A + B$, so $a + b \leq u$, and hence we have $$ \sup(A) + \sup(B) < u + \varepsilon. $$ Thus, by part (i), $\sup(A) + \sup(B) \leq u$, so $\sup(A) + \sup(B)$ is the least upper bound for $A + B$, as required.

Now how do I show that $\sup(A+B)$ exists?

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Show $A+B$ is nonempty and bounded above.

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  • $\begingroup$ i)A + B is non-empty since a + b E A + B. ii) If A and B are bounded, then so is A + B because by assumption, there are positive numbers MsubA and MsubB with |a| <= MsubA and |b| <= MsubB for all aEA and bEB. Hence |a + b| <= |a| + |b| <= MsubA + MsubB for all a and b. Is that it? $\endgroup$ – user87274 Sep 15 '13 at 16:49
  • $\begingroup$ @mespebjidom Yes. $\endgroup$ – Pedro Tamaroff Sep 15 '13 at 16:56
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Assume $A,B$ are both nonempty. Otherwise you can run into some $\infty-\infty$ strangeness.

You have $a \le \sup A$ for all $a \in A$, and similarly $b \le \sup B$ for all $b \in B$. Hence we have $a+b \le \sup A + \sup B$ for all $a \in A, b \in B$. Hence we have $\sup (A+B) = \sup_{a \in A, b \in B} a+b \le \sup A + \sup B$.

Now let $\epsilon >0$, and choose $a \in A, b \in B$ such that $a > \sup A -\frac{\epsilon}{2}$ and $b > \sup B -\frac{\epsilon}{2}$. Then $\sup(A+B) \ge a+b > \sup A + \sup B - \epsilon$. Since $\epsilon>0$ was arbitrary, we are have the desired result.

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  • $\begingroup$ What's this for? I just wanted to know how I can show that sup(A + B) exists in simple terms. $\endgroup$ – user87274 Sep 15 '13 at 18:51
  • $\begingroup$ Your question reads 'Prove $\sup A + \sup B = \sup (A+B)$. The above shows, in simple terms, that $a+b \le \sup A + \sup B$ for all $a \in A, b \in B$, from which it follows that $\sup (A+B) \le \sup A + \sup B$. I'm not sure what you mean by 'exists', as the $\sup$ always exists (it may be $\pm \infty$, by convention $-\infty$ if the set is empty). $\endgroup$ – copper.hat Sep 15 '13 at 18:58
  • $\begingroup$ I already proved that sup(A+B) = sup(A) + sup(B). What I wanted to know was that if A is bounded and has an element a in A, and B is bounded and has an element b in B, then does this mean that sup(A+B) exits? As the above poster said, for sup(A+B) to exist we have to show that A + B is non-empty and bounded. That's all. $\endgroup$ – user87274 Sep 15 '13 at 19:30
  • $\begingroup$ I'm missing something here. The set needs to be bounded above, not necessarily bounded. If a non-empty set is bounded above, then it has a finite $\sup$. The set $A+B$ is clearly non-empty if both $A,B$ are non-empty. And you have shown that the sum of the $\sup$s is equal to the $\sup $ of the sum, which is presumed finite. $\endgroup$ – copper.hat Sep 15 '13 at 19:42
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    $\begingroup$ @Evinda: Actually, I wrote my answers :-). One definition of $\sup A$ is that it is an upper bound and for all $\epsilon>0$, there is some $a \in A$ such that $a > \sup A- \epsilon$. What definition are you using? $\endgroup$ – copper.hat Oct 5 '18 at 21:38

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