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I did the first one by contradiction. I noted that the first time I looked through it I forgot to verify their existance via the lub property of the reals, and I still wonder whether the proof needs cases by whether the supremum is infinite or finite.
Given $ A, B \neq \varnothing \subseteq \mathbf{R}, A+B = \{ a+b \vert a \in A, b \ in B \} $, and I'll let $ \alpha = \sup A, \beta = \sup B $
WTS: $ \sup(A+B) = \alpha + \beta $
First of all, by the lub property of $ \mathbf{R} $ we know that these exist.
Now we show that $ \alpha + \beta $ is an upper bound of $ A+B $ by contradiction. Thus $ \exist a,b \in A,B \ni a+b > \alpha + \beta $
$ (\alpha - a) + (\beta - b) < 0 $
WLOG assume $ (\alpha - a) < 0 $, then $ \exists a \in A \ni a > \alpha, but \alpha = \sup A $, contradiction.
$ \therefore \alpha + \beta $ is an upper bound of $ A+B $
Now we show that it is the least upper bound. Let $ \delta $ be an upper bound of $ A+B $, we need to show that $ \delta \geq \alpha + \beta $.
Assume not, then $ \delta < \alpha + \beta $, let $ c = \alpha + \beta - \delta > 0 $ and note $ \delta = \alpha + (\beta - c) $. But since $ \delta $ is an upper bound of $ A+B $ then $ \forall a,b \in A,B, a+b \leq \alpha + b \leq \delta = \alpha + \beta - c $. But then $ \forall b \in B, b \leq \beta - c $ so we have an upper bound for $ B $ that's smaller than the lub of $ B $, contradiction.
Therefore, $ \alpha + \beta $ is the lub of $ A+B $
Bobby's criticism: The sets A and B are not bounded above so you cannot apply the LUB property! If they are, your proof looks good. If not, your proof may have problems manipulating $ \infty=\sup A $. Break it into cases.
