Proving that $\sup (A + B) = \sup (A) + \sup (B)$ through two contradictions

Question

There are certainly other posts on this website asking for help proving that $\sup (A + B) = \sup (A) + \sup (B)$. I would like to ask whether my proof, which I haven't seen elsewhere on this site after a brief search, is valid.

Exercise:

For nonempty sets $A$ and $B$ of $\mathbb R$, prove that $\sup (A + B) = \sup (A) + \sup (B)$.

My attempt:

Suppose that $\sup (A + B) > \sup (A) + \sup (B)$. Then there exist $a \in A, b \in B$ such that $a + b > \sup (A) + \sup (B)$. Since we know that we must have $a \leq \sup (A)$ and $b \leq \sup (B)$, we see that this is not possible.

Now suppose that $\sup (A + B) < \sup (A) + \sup (B)$. That means that, for all $a \in A, b \in B$, we must have $a + b \leq \sup (A + B) < \sup (A) + \sup (B)$. However, let $A = B = [0, 1]$ to see that this clearly cannot hold, so we conclude that $\sup (A + B) = \sup (A) + \sup (B)$ as desired.

Answer 1

Consider the statement

$P$ : for all $A,B\subset \Bbb{R}$ which are non-empty and bounded above, $\sup(A+B) = \sup(A)+ \sup(B)$.

Its negation is

$\neg P$ : There exist $A,B\subset \Bbb{R}$ which are non-empty and bounded above, such that $\sup(A+B) \neq \sup(A)+ \sup(B)$.

You're trying to prove that $P$ is true by method of contradiction. What this means is that you have to assume the negation $\neg P$ is true, and then show that you arrive at a contradiction (a false statement).

Ok, so let's suppose that $\neg P$ is true: so there exist $A,B\subset \Bbb{R}$ which are non-empty and bounded above, such that $\sup(A+B) \neq \sup(A)+ \sup(B)$. Now, choose such sets $A,B$ (and we have to keep them fixed throughout the remainder of the discussion). Now, there are two possibilities.

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Case $1$: $\sup(A+B) > \sup A + \sup B$.

You wrote

Suppose that $\sup (A + B) > \sup (A) + \sup (B)$. Then there exist $a \in A, b \in B$ such that $a + b > \sup (A) + \sup (B)$. Since we know that we must have $a \leq \sup (A)$ and $b \leq \sup (B)$, we see that this is not possible.

This is correct. You have shown that by assuming the false hypothesis of Case $1$, we arrive at a false statement.

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Case $2$: $\sup (A+B) < \sup A + \sup B$.

Now you say take $A=B=[0,1]$. This is logically meaningless, because the $A$ and $B$ have already been fixed in the beginning of this discussion (we fixed the sets $A,B$ above, see the bold statement). You do not know that the $A,B$ which come from your hypothesis of $\neg P$, "there exist $A,B$..." are equal to $[0,1]$.

You can't change things in the middle of a discussion. So, really all you have done in your final paragraph is say that

It is not true that $\sup([0,1]+ [0,1]) < \sup[0,1] + \sup [0,1]$.

which is not at all what was required to complete the proof by contradiction (just to reiterate, you have to show that for the "unknown" $A,B$ from the beginning, whose existence you already know of, that they cannot actually satisfy $\sup(A+B)< \sup(A)+\sup(B)$).