Proving Sup(A+B) = Sup(A)+Sup(B), I got a question on something $A,B $ : Nonemtpty bounded sets on R
I'm proving on $\sup(A+B) \geq \sup (A)+\sup(B)$
prove by contradicting assumption: $\sup(A+B) < \sup(A) + \sup(B)$
Then $\sup(A+B) -\sup (B)< \sup (A)$
Then there exists $a \in  A$ such that  $\sup(A+B)-\sup(B)<a$
$\sup(A+B) -a <\sup(B)$...
(next steps are omitted)
Now, I have a question on this prove on existence of $a$
For exmaple, Let $\sup(A+B)=X, \sup(B)=Y$. Then $X-Y<\sup(A)$
For sure, $a\le\sup(A)$. Then, how can we know $X-Y<a?$
Or am I wrong on proving this?
 A: Yeah this is a good question to ask. The fact is that there is such an $a$ because the supremums $X$ and $Y$ are fixed, whereas $\sup(A)$ is a limit point (i.e. elements of $A$ get arbitrarily close to $\sup(A)$).
Let $\sup(A+B)=X, \sup(B)=Y$. Then $X-Y <\sup(A)$ is your assumption that you want to contradict.
Note that $\sup(A+B)$ and $\sup(B)$ are fixed. That is, $X-Y \in \mathbb{R}$ is fixed.  If $X-Y < \sup(A)$, (strict inequality), then there is a gap $r$ between them:
$$\sup(A) - (X-Y) = r > 0.$$
Recall that by definition, the supremum is a limit point of $A$: there exists a sequence $\{a_i\}_{i=1}^\infty \in A$ such that $a_i \rightarrow \sup (A)$. So there is an element of this sequence that is arbitrarily close to $\sup(A)$, i.e. some $a_k$ such that
$$\sup(A) - a_k < r.$$
Combine these two facts to see that $$ X-Y <a_k.$$
A: This is very simple using the definition of supremum. The actual definition, which says nothing about subsequences and limit points.
By definition $\sup(S)$ is the least upper bound of $S$. Which means two things: (i) $\sup(S)$ is an upper bound for $S$ and (ii) if $\alpha<\sup(S)$ then $\alpha$ is not an upper bound for $S$.
What you want ask about is this:


If $\alpha<\sup(S)$ there exists $a\in S$ with $\alpha<a$.


Proof: Since $\alpha<\sup(S)$, $\alpha$ is not an upper bound for $S$, which in turn means precisely that there exists $a\in S$ with $a>\alpha$.
