I know that the question has been already posted and I read the answers but, since I don't have experience in proofs, I post it again, hoping you can help me.
With some help of a friend, I before proved that $\sup(A+B) = \sup A + \sup B$. I know that the proof of this question is similar but, as I said, I really don't how to get in such proofs. That's my proof of $\sup(A+B) = \sup A+\sup B$
From the definition of least upper bound (supremum), we have:
$ \sup A \geq a$, for every $a \in A$ and $\sup B \geq b$, $b \in B$
Thus, $\sup A + \sup B \geq a + b$
We have now satisfied the first requirement for a supremum. i.e. being an upper bound. We want to satisfy the second requirement, i.e. the supremum is the least upper bound. Thus:
it doesn't exist $x < \sup A$ s.t. $x \geq a$, $a \in A$ and it doesn't exist $y < \sup B$ s.t. $y\geq b$, $b \in B$
Then, it doesn't exist $x,y$ s.t. $x+y \geq a + b$
Now, let $z = x+y$, then it doesn't exist $z < \sup A + \sup B$ s.t. $z \geq a + b$
Thus, $\sup A + \sup B = \sup (A+B)$
I have understood this procedure and, someone told me, it works the same for proving that $\sup (A−B) = \sup A − \inf B$.
However, I would like to try another thing for proving this. Since we know that a property of supremum / infinum consists on the fact that $\sup (−B) = − inf B$, plus the result of the previous proof, we may arrive at the conclusion that $$\sup (A−B) = \sup A − \inf B$$. Am I right?
However, I believe that in my exercise I should not take for granted the property that I have mentioned and I should prove it too, before.
Could you give me some help in this direction, please?
I would like to mantain the proof simple as it has been showed (if it is correct), without adding any delta and fractions, as in the classical examples I find on the web. The simpler, the better.