alternative proof of sup (S + T) = sup (S) + sup (T) I am not sure why in my textbook there is a long proof for this because on the pages before this was prooven:
Let $S = \{ x \mid a \leq x \leq b \hbox{ and } x \in \mathbb{R} \}$. Then $\sup S = b$ must hold.
EDIT: also $\inf S=a$ follows. But the proof for $\sup S$ does not depend on there being a lower bound to $S$ this can be modified to be:
Let $S = \{ x \mid x \leq b \hbox{ and } x ∈ \mathbb{R} \}$. Then $\sup S = b$ must hold.
With this you can  instantly proove $\sup (S + T) = \sup (S) + \sup (T)$ 
since from $s \leq \sup  S$  and $t\leq  \sup  T$ we get $s+t \leq \sup S + \sup T$ and so we get $\sup (S+T)=\sup S + \sup T$.
I just don't see why the book goes the long way when we already just did the work. Or am I just getting some logic completely wrong here?
 A: If $S=\{x\in\mathbb{R}:a\le x\le b\}$, then it's of course true that $b=\sup S$: indeed, $b\le x$, for all $x\in S$ and $b\in S$, so $b$ is actually the maximum of $S$.
When you have two sets $S$ and $T$ (subsets of $\mathbb{R}$), then
$$
S+T=\{x+y:x\in S,y\in T\}
$$
is the set of all numbers that can be expressed as the sum of an element in $S$ with an element in $T$.
Assume both sets are bounded (and non empty). We wish to prove that
$$
\sup(S+T)=\sup S+\sup T
$$
If $x\in S$ and $y\in T$, then $x\le\sup S$ and $y\le\sup T$, so $x+y\le\sup S+\sup T$. Therefore $\sup S+\sup T$ is an upper bound for $S+T$ and, by definition of supremum, $\sup(S+T)\le\sup S+\sup T$.
Conversely, we want to see that, for every $\varepsilon>0$, $\sup S+\sup T-\varepsilon$ is not an upper bound for $S+T$, so it will follow that $\sup S+\sup T$ is the least upper bound of $S+T$.
Since $\sup S-\frac{\varepsilon}{2}$ is not an upper bound of $S$, there exists $x\in S$ such that $\sup S-\frac{\varepsilon}{2} < x$. Similarly, there exists $y\in T$ such that $\sup T-\frac{\varepsilon}{2}<y$.
It follows that
$$
\sup S+\sup T-\varepsilon<x+y
$$
Since $x+y\in S+T$, we have what we wanted.

Note that the set $(S+T)'=\{x\in\mathbb{R}:x\le\sup S+\sup T\}$ can be very different from $S+T$. Just take $S=\{1\}$ and $T=\{2\}$; then $S+T=\{3\}$ (one element set), while $(S+T)'$ contains infinitely many elements.
A: The problem is that $S+T$ is not defined as
$$S+T=\{x\in\Bbb R\mid\inf S + \inf T\le x\le\sup S+\sup T \}$$
It is defined as
$$S+T=\{s+t\mid s\in S,t\in T\}$$
So you can not use your previous result that if $X=\{x\in\Bbb R\mid a\le x \le b\}$ then $\sup X=b$. While it is true that $s+t\le \sup S+\sup T$, this is not how the set is defined.
