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Prove $\sup \left( {A + B} \right) = \sup A + \sup B$ using the definition given in this problem (link).
I was able to prove using lemma of the given definition.
My attempt is here,
let $s = \sup A$, and $t = \sup B$
Now, to show $s + t$ is least upper bound for $A + B$ I need to show that for all $\varepsilon > 0$ there is $c \in A + B$ such that $s + t \le c$.
Since,$s = \sup \left( A \right)$ , therefore, there is some $a \in A$ such that $s - {\varepsilon \over 2} \le a$ . Similarly, there is $b \in B$ such that$t - {\varepsilon \over 2} \le b$ . Hence, for any $c \in A + B$,
$s - {\varepsilon \over 2} + t - {\varepsilon \over 2} \le a + b = c$ which completes the proof.

My question is using the usual definition rather than lemma to prove that $\sup \left( {A + B} \right) = \sup A + \sup B$
I could show $s + t$ is upper bound for $A + B$. Then lets chose $x$ be an arbitrary upper bound for $A+B$ and temporary fix $a \in A$. How do I show this,
$t \le x- a$,
and then finally conclude that $\sup \left( {A + B} \right) = \sup A + \sup B$.
Please, if possible, explain each step, and why that step is taken and how the definition is used in steps.
EDIT:
My question is specific to using the definition. Here is what I have tried, since $x$ is upper bound of $A+B$, therefor, I can write, $a+b \le x$, for all $a \in A$ and $b \in B$. For fixed $a \in A$ I can proceed further and can write, $b \le x - a$, giving $x-a$ as an upper bound of $B$. Now I know that $t$ is least upper bound of $B$, so $t \le x-a$ which proves part of my problem. Going by similar argument it can be achieved that $s \le x-b$.
Adding both the inequality gives,
$s+t \le 2x-(a+b)$
Since, it is already known that $a+b \le x$, therefore, the maximum value of $a+b$ will be $x$. Which gives,
$s+t \le 2x-(x)$
$s+t \le x $
Now, since $t+s \le x$, therefore $t+s$ is least upper bound of $A+B$.
$\sup \left( {A + B} \right) = \sup A + \sup B$.
Proved
Please check and suggest if there is something wrong with the solution or its complete.

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  • $\begingroup$ I take it you are considering sets ($A$ and $B$) of positive reals? $\endgroup$ – Chris Jan 22 '17 at 18:42
  • $\begingroup$ Yes, indeed. Please explain what happens if its not positive reals. $\endgroup$ – user405401 Jan 22 '17 at 18:43
  • $\begingroup$ Actually I realized it (should be) irrelevant. I was thinking of multiplication there.... $\endgroup$ – Chris Jan 22 '17 at 18:45
  • $\begingroup$ You're gonna want $s + t < c + \epsilon$ for every $\epsilon > 0 $ (and some $c \in A + B$), so there's a typo there $\endgroup$ – Chris Jan 22 '17 at 18:46
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We want to show that for every $\epsilon > 0$, $\sup(A + B) > \sup(A) + \sup(B) - \epsilon$. But $(\sup(A) - \epsilon/2) < a$ for some $a \in A$ by the definition of $\sup(A)$, and similarly $(\sup(B) - \epsilon/2) < b$ for some $b \in B$. Thus $\sup(A) + \sup(B) - \epsilon < a + b \le \sup(A+B)$, by the definition of $\sup(A + B)$. Because $\sup(A + B) > \sup(A) + \sup(B) - \epsilon$ for every $\epsilon > 0$, $\sup(A+B) \ge \sup(A) + \sup(B).$

The other direction, $\sup(A + B) \le \sup(A) + \sup(B)$, follows immediately from the fact that $\sup(A) + \sup(B) \ge a + b$ for any $a \in A, b \in B$, because $\sup(A) + \sup(B)$ is therefore an upper bound on $A + B$, and so is greater than or equal to the least upper bound, $\sup(A+B)$.

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  • $\begingroup$ My question is specific to using the definition. Here is what I have tried, since $x$ is upper bound of $A+B$, therefor, I can write, $a+b \le x$, for all $a \in A$ and $b \in B$. For fixed $a \in A$ I can proceed further and can write, $b \le x - a$, giving $x-a$ as an upper bound of $B$. Now I know that $t$ is least upper bound of $B$, so $t \le x-a$ which proves part of my problem. How do I prove, going by the line I am following, that $\sup(A+B)$=$s+t$. If I could show somehow that $t+s \le x$, then my proof will be complete. But I do not know how to proceed further. $\endgroup$ – user405401 Jan 22 '17 at 20:06
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    $\begingroup$ Well, since you have $t \le x - a$ for all $a \in A$ (since the fixed $a$ you picked was arbitrary), actually you have $t \le x - s$, where $s = \sup(A)$, and this solves your problem. The reason is that if we were to have $t > x - s$, then $t = (x - s) + \epsilon$; then we can pick a particular $a \in A$ such that $s - a < \epsilon/2$, and $x - a$ for this $a$ equals $(x - s) + (s - a) = [t - \epsilon] + (s - a) < t - \epsilon/2$ - so we would have $t > x - a + \epsilon/2$, and thus $t > x -a$ for that $a$, which you already proved is impossible. $\endgroup$ – Chris Jan 22 '17 at 20:49

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