Proving $\sup \left( {A + B} \right) = \sup A + \sup B$ using the usual definition.

Question

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.

Answer 1

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, $\sup(A+B)$.

Answer 2

In order to show $\sup \left( {A + B} \right) = \sup A + \sup B$ . We need to verify the following:

1. $(s+t)$ is an upper bound for the set $A+B$ , where s = $ \sup A$ and t = $ \sup B$
2. if $b$ is any upper bound for set $A+B$, then $(s+t)≤b $

Proof: (Direct)

Lets try to verify (1) first:

Since $s$ is the $\sup A$

So, $a ≤ s$, for all $a \in A$ . We can write:

$a+b≤(s+b)$ for all $ a \in A$ and $b \in B$

Similarly, since $t$ is the $\sup B$ .

So , $b ≤ t$, for all $b \in B$ . Hence

$(s+b)≤(s+t) $ for all $b \in B$ . So,

$(a+b)≤ (s+b)≤(s+t) $

$(a+b)≤ (s+t) $ for all $ a \in A$ and $b \in B$

Thus, $(s+t)$ is the upper-bound of set $A+B$

So (1) is verified.

To verify (2):

Assume $u$ be an arbitrary upper-bound for set $A+B$. Let's temporarily fix $a \in A$.

So, $(a+b)≤ u$ , for all $b \in B$ .

$b≤ (u-a)$ , for all $b \in B$ .

This means, $(u-a)$ is an upper-bound for set $B$.

Since t = $ \sup B$. So, $t≤ (u-a)$.

Rearranging, $a≤ (u-t)$ for all $a \in A$ .

This means, $(u-t)$ is an upper-bound for set $A$.

Since s = $ \sup A$. So, $s≤ (u-t)$.

Therefore, $ (s+t) ≤ u$

This means $(s+t) $ is smaller than an arbitrary upper-bound of set $A+B$

Hence (2) is verified.

In conclusion, we have shown :

1. $(s+t) $ is an upper-bound for set $A+B$
2. $(s+t)$ is smaller than (or equal to) to any arbitrary upper-bound for set $A+B$

So, $s+t= \sup(A+B)$

$\sup(A) + \sup(B)= \sup(A+B) $