Is subtracting inequalities allowed? $\sup(A+B)= \sup(A)+\sup(B)$ - proof verification Given sets $A$ and $B$ define $A+B =\{ a+b:a \in A \  and\  b \in B \}$. If these sets are nonempty and bounded above show $ \sup(A+B)= \sup(A)+\sup(B)$.
My Attempt:
Let $\sup A= \alpha$ and $\sup B= \beta$. This means that for all $a \in A$, $a \le \alpha$ and for all $b \in B$, $b \le \beta$. Thus adding the inequalities it follows $a+b \le \alpha + \beta$ and hence $\alpha + \beta$ is an upperbound for $A+B$.
Let $\gamma $ be an upper bound for $A+B$. Thus $a+b= \gamma$. It follows that $a + \beta \le \gamma$. Thus (1) $\beta \le \gamma -a.$ Similarly $\alpha + b \le \gamma .$ Thus (2) $\alpha \le \gamma -b$.Adding (1) and (2) we get  $$\alpha + \beta\le 2\gamma -a -b$$ 
$\Rightarrow$ $$\alpha + \beta + a +b \le 2\gamma$$
Is the following valid: I took $a+b \le \gamma $ since $\gamma$ is an upper bound for $A+B$ and subtracted this from the inequlaity above to get $\alpha + \beta\le \gamma$. And thus proving the theorem. Only problem is that if take $-\gamma \le -a-b$ (equivalent to $a+b \le \gamma $) and add it to the inequality above we get $$\alpha + \beta + a +b -\gamma \le 2\gamma -a -b$$ which I was told by my tutor is wrong. If this wrong why so?? If so, why isn't subtracting inequalities allowed?
 A: It is wrong. It is not true that an upper bound $\gamma$ of $A+B$ can be written as $a+b$ with $a\in A$ and $b\in B$. Suppose, for instance, that $A=B=\{0\}$ and that $\gamma=10$.
However, you proved correctly that $\sup A+\sup B$ is an upper bound of $A+B$. Therefore, $\sup(A+B)\leqslant\sup A+\sup B$. What would happen if we had $\sup(A+B)<\sup A+\sup B$? In that case, take $\varepsilon=\sup A+\sup B-\sup(A+B)$, take $a\in A$ such that $a>\sup(A)-\frac\varepsilon2$ and take $b\in B$ such that $b>\sup(B)-\frac\varepsilon2$. What does this tell you about $a+b$?
A: We have e.g. $3<4$ and $1<2$, but subtracting would give $3-1 < 4-2 \implies 2<2$, which is wrong. In general, note that if $p<q$, we can write this as $p = q - \delta$ for some positive $\delta$, and similarly $r<s \implies r = s - \epsilon$ for some positive $\epsilon$. Subtracting would give $p - r < q - s \iff q - \delta - s + \epsilon < q - s \iff \epsilon < \delta$, which may or may not be true. In contrast, adding inequalities is allowed because $p + r < q + s \iff q - \delta + s - \epsilon < q + s \iff \delta + \epsilon > 0$, which is true as both $\delta$ and $\epsilon$ are positive.
In your last paragraph, your inequality is correct but totally useless for the proof, so it's "wrong" in that sense. Similarly if we take an example like $3<5$ and $1<2$, where we want to prove $3 - 1 < 5 - 2$, i.e. $2 < 3$, we can add $3<5$ and $-2<-1$ to get $1<4$, but this is a weaker inequality than the one we want to prove.
Considering your whole proof attempt, your first step of showing that $\sup(A) + \sup(B)$ (i.e. $\alpha + \beta$ in your notation) is an upper bound for $A + B$ is useful. Here's a hint for what to do next: this upper bound condition can be rewritten as $A + B \leq \sup(A) + \sup(B)$, so whatever $\sup(A+B)$ is, it must be less than or equal to this upper bound on $A + B$, i.e. we have $\sup(A + B) \leq \sup(A) + \sup(B)$. Now if we can also show that $\sup(A+B) \geq \sup(A) + \sup(B)$, the combination of the two inequalities will give $\sup(A+B) = \sup(A) + \sup(B)$, as required. In order to show $\sup(A+B) \geq \sup(A) + \sup(B)$, you should first prove that $\sup(A+B)$ is an upper bound for $A + \sup(B)$, so $\sup(A + \sup(B)) \leq \sup(A+B)$, or writing it the other way round, $\sup(A+B) \geq \sup(A + \sup(B))$. Finally, prove that $\sup(A + \sup(B)) \equiv \sup(A+B)$ in order to reach the desired result.
A: The example by NickD illustrates the problem with the present proof.
A quick fix: suppose that $\gamma\geq a+b$ for all $a\in A$ and $b\in B$. We want to show that $\gamma\geq\alpha+\beta$.
So suppose otherwise that $\gamma=\alpha+\beta-e$ for some $e>0$. Now $\alpha-e/2$ is strictly smaller than $\alpha$ so there is some $a^*\in A$ such that $a^*>\alpha-e/2$. Similarly, there is some $b^*\in B$ such that $b^*>\beta-e/2$. But then:
$$
a^*+b^*\in A+B\quad\text{and}\quad a^*+b^*>(\alpha-e/2)+(\beta-e/2)=\gamma.
$$
This is your desired contradiction.
A: Subtracting inequalities is wrong, for example  if you subtract $-5<2$ from $4<10$  you get $9<8$
For your proof of     $$ \sup(A+B)= \sup(A)+\sup(B)$$ where you have already shown that $$ \sup(A+B)\le \sup(A)+\sup(B)$$
you need to show that for any number $\gamma $ less than $ \sup(A)+\sup(B)$ you can find an element in A+B which is 
greater than $\gamma$
Note that $$\gamma= \sup(A)+\sup(B)-\epsilon = (\sup(A) - \epsilon /2) + (\sup(B) - \epsilon /2)$$
Now you can easily find   elements  $a\in A$ and $b\in  B$ such that $a+b >\gamma$
