# Proving that $\sup(A+B)=\sup(A)+\sup(B)$, $A,B\subseteq\mathbb{R}$. [duplicate]

Possible Duplicate:
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!

## marked as duplicate by Jonas Meyer, Asaf Karagila♦, Arturo Magidin, Akhil MathewApr 5 '11 at 6:16

• 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 4 '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 4 '11 at 16:33
• @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 4 '11 at 16:37