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.

  • $\begingroup$ You are correct: these two exercises are equivalent since $\sup (-B) = - \inf B$ $\endgroup$
    – Crostul
    Oct 21 '15 at 18:34
  • $\begingroup$ yes, but how to complete the proof... ? $\endgroup$ Oct 21 '15 at 18:36
  • $\begingroup$ other answers, please? $\endgroup$ Oct 21 '15 at 18:59
  • $\begingroup$ I am relly dumb. I need simple explanations. Please, don't add other variables, letters and symbols. Try to write an answer which complete the proof I wrote before... $\endgroup$ Oct 21 '15 at 19:12

You have the right idea. Explicitly:

  • establish that $A-B=A+(-B)$. (this is immediate from the definition)
  • establish that $-\inf B=\sup(-B)$. [see below]
  • use your earlier result to conclude: $$\sup(A-B)=\sup(A+(-B))=\sup A + \sup (-B)= \sup A - \inf B$$

To establish the second bullet, it suffices to show that $- \sup(-B)$ is the greatest lower bound for $B$.

It is a lower bound for $B$ since for any $b \in B$ we have $-b \le \sup (-B)$ and thus $b \ge -\sup(-B)$.

If there were a greater lower bound $L>-\sup(-B)$ for $B$, then $-L < \sup(-B)$ so there exists some $b \in B$ such that $-L < -b \le \sup(-B)$ by the definition of supremum. But then $b>L$, contradicting the fact that $L$ is a lower bound for $B$

  • $\begingroup$ I am not sure that I get your proof of the second bullet. Maybe I need an explanation by words, also. $\endgroup$ Oct 21 '15 at 18:54
  • $\begingroup$ why to use contradiction in the last part? Can't we continue to prove everything in direct way? $\endgroup$ Oct 21 '15 at 19:17
  • $\begingroup$ @MarioVald In the proof of the second bullet, I show that $- \sup(-B)$ is the infimum of the set $B$. To do this, I need to show that a) it is a lower bound for $B$, and b) it is greater than any other lower bound, which is what I have done. $\endgroup$
    – angryavian
    Oct 23 '15 at 5:38

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