proving that sup A + sup B = sup C and inf A + inf B = inf C i have tried to answer this question on my own, but my solution isn't the same as others i've seen (they use the archimedian principle but i don't), so i want to see if my approach is right or not. i am more sure of my answer to part 1 than i am about part 2.
am i correct here? if not, where did i screw up? thanks in advance for any insights.
this is the question:
$A$ and $B$ are bounded and non-empty sets, and $C = [a + b: a \in A, b \in B]$. show that $C$ is a bounded set and that $\sup C = \sup A + \sup B$ and $\inf C = \inf A + \inf B$
this is my answer:
part 1: suprema

*

*let $\sup A = s$ and $\sup B = t$, so for every $x \in A, x \leq s$ and for every $y \in B, y \leq t$

*it follows that $x + y \leq s + t$, implying that $s + t$ is an upper bound for $C = A + B$

*now choose upper bound $u$ such that $u \geq s + t$ and fix $a \in A$, then $u \geq a + t \implies t \leq u - a$, and similar reasoning gives $s \leq u - b$

*adding the two results above gives $s + t \leq 2u - a - b \implies a + b \leq 2u - s - t$

*it follows that $u \leq 2u - s - t$, but recall that $u \geq s + t$, so:

$$a + b \leq s + t \leq u \leq 2u - s - t$$

*

*and our result of interest is $a + b \leq s + t$, which implies that $\sup C = \sup A + \sup B$
part 2: infima

*

*now let $\inf A = s$ and $\inf B = t$, so for every $x \in A, x \geq s$ and for every $y \in B, y \geq t$

*putting the information above together gives $x + y \geq s + t$, implying that $s + t$ is a lower bound for $C$

*now choose a lower bound $l$ such that $l \leq s + t$ and fix $a \in A$, which implies $t \geq l - a, s \geq l - b$

*adding the results above together gives us: $s + t \geq 2 \cdot l - a - b \implies a + b \geq 2 \cdot l - s - t$

*we also know that $l \geq 2 \cdot l - s - t$, which implies:

$$a + b \geq s + t \geq l \geq 2 \cdot l - s - t$$

*

*our result of interest is $a + b \geq s + t$, which tells us that $\sup C = \sup A + \sup B$ $\blacksquare$
 A: I will just comment on part $1$, I leave part $2$ to you. You are right when you write $x+y \leq s+t$ implies that $s+t$ is an upper bound of $C$. It then follows that $\sup C \leq \sup A + \sup B$. What you should do now is to prove that $\sup C \geq \sup A + \sup B$, but all you do is just showing that once you fix $a \in A$ and $b \in B$, then $a+b \leq s+t$, which you already know. So you are proving nothing...
A way out of this is the following. You have proved that $\sup C \leq \sup A + \sup B$. Suppose now $\sup C < \sup A + \sup B$. By definition of $\sup$, if you fix $\epsilon_1, \epsilon_2 > 0$ you will find $a \in A$ and $b \in B$ such that
\begin{align*}
\sup A - \epsilon_1 & < a < \sup A,\\
\sup B - \epsilon_2 & < b < \sup B.
\end{align*}
Now choose $\epsilon_1,\epsilon_2$ small enough such that $\sup C + \epsilon_1+\epsilon_2 < \sup A + \sup B$. Then
$$
\sup C < (\sup A - \epsilon_1) + (\sup B - \epsilon_2) < a+b,
$$
so you have found an element (i.e. $a+b$) which lies in $C$ but its greater than $\sup C$. This is a contradiction. So necessarily $\sup C = \sup A + \sup B$.
A: We have $C=\{a+b: a\in A\text{ and }b\in B\}$ and we are told $A$ and $B$ are bounded. Let $u_A, u_B$ be the least upper bounds of $A,B$, respectively.
Then for any $a,b$ we have $$a\le\sup{A}\implies a+b\le\sup{A}+b$$ and $$b\le\sup{B}\implies\sup{A}+b\le\sup{A}+\sup{A}$$
Therefore, $$a+b\le\sup{A}+\sup{B}$$ for all $a,b$ so $$\sup{A}+\sup{B}$$ is an upper bound. To show that it is the least upper bound we will show that any number less is not an upper bound for $C$.
Let $u=\sup{A}+\sup{B}-\epsilon$ for some positive $\epsilon$. Then $\sup{A}-\frac{\epsilon}{4}\in A$ and $\sup{B}-\frac{\epsilon}{4}\in B$ and $$\sup{A}-\frac{\epsilon}{4}+\sup{B}-\frac{\epsilon}{4}=\sup{A}+\sup{B}-\frac{\epsilon}{2}\gt\sup{A}+\sup{B}-\epsilon=u$$.
Therefore, $u$ is not an upper bound for $C$ and since all numbers less than $\sup{A}+\sup{B}$ can be expressed this way, $\sup{A}+\sup{B}=\sup{C}$.
