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How can I prove sup(A+B)=supA+supB if A+B={a+b}

How would I go about proving that the supremum of A + B (where A and B are each subsets if $\mathbb{R}$) is equal to the supremum of A plus the supremum of B?

I was thinking of using variables to represent the sup of A and the sup of B (say a & b), showing that a + b is an upper bound of A + B, and finally showing that no matter how small some other variable, e, is, that a + b - e cannot represent an upper bound for A + B.

Any idea how I might go about formalizing the proof? Thanks!

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  • Sounds like you're right on track. To formalize the proof, just consider what would happen if $a + b - \varepsilon$ indeed was an upper bound for $A+B$. – JavaMan Apr 04 '11 at 16:31
  • By the way, I'm assuming $A+B$ is meant to be the Minkowski Sum: $$ A+B = {a + b : a \in A , b \in B}?$$ – JavaMan Apr 04 '11 at 16:33
  • @DJC: Why do a proof by contradiction, when a simple direct proof will do? – Arturo Magidin Apr 04 '11 at 16:35
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    @Jim T.: You are on exactly the right track. As for formalizing, you just need to show that every element of $A+B$ is less than or equal to $a+b$ (use the fact that $a=\sup(A)$ and $b=\sup(B)$). To show that $a+b-e\leq \sup(A+B)$, write $a+b-e$ as $(a-(e/2)) + (b-(e/2))$ and again use the properties of the supremum to find elements $x$ of $A$ and $y$ of $B$ such that $a+b-e \lt x+y$. Remember that being the supremum means satisfying two conditions; you will use one condition for the first inequality, and the other for the second indequality. – Arturo Magidin Apr 04 '11 at 16:37
  • @Arturo That's a fair point. I suppose it was just the first place my mind went when proving such an inequality. – JavaMan Apr 04 '11 at 16:41
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    @Jim T. Once you write down an argument, you can try posting it as an answer to your own question and people can comment and let you know it's correct, or point out mistakes. – Arturo Magidin Apr 04 '11 at 16:50
  • I'll try! Thanks for the input guys! Great resource we've got here. –  Apr 04 '11 at 17:07

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