Proof for $\max(A+B) = \max(A) + \max(B)$ correct?

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Hi I'm trying to prove the following: Let $A, B \subset \mathbb {Q}$ and $A+B = \{a+b: a\in A, b \in B\}$.

Suppose $\max(A)$ and $\max(B)$ exist, show that $\max(A+B)$ also exists and that

$$\max(A+B)= \max(A) + \max(B)$$

I have the following proof I'm not sure if it's correct:

The real number $\max(A) + \max(B) \geq x$ for all $x \in A+B$

Since $\max(A) \in A$ and $\max(B) \in B$ the definition of $A+B$ that $\max(A+B) \in A+B$ holds.

Now let $x \in A+B$. $\Rightarrow$ there exists $a \in A$ and $b \in B$ such that $x = a+b$

According to the definition of $\max$ however, $\max(A) \geq a$ and $\max(B) \geq b$.

Therefore $\max(A) + \max(B) \geq a+b = x$

Is this proof correct?

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Your proof is correct, but a bit redundant. You showed that $\max(A)+\max(B) \ge x$ for all $x \in A+B$, and you explained how $\max(A)+\max(B) \in A+B$. This suffices to show what we need, as we've found an upper bound for $A+B$ that also belongs in $A+B$, hence it's the maximum.