# Proving $x+y =\sup\left\{ r+s: r,s \in \mathbb{Q}, r\leq x, s\leq y\right\}$ [duplicate]

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I am trying to show that the following holds: $$x+y =\sup\left\{ r+s: r,s \in \mathbb{Q}, r\leq x, s\leq y\right\}.$$

I have been able to show that

$$x=\sup\{q\in \mathbb{Q}: r\leq x\}$$ and similarly that $$y=\sup\{q\in \mathbb{Q}: r\leq y\}.$$

Thus, I am essentially trying to show that $$\sup\left\{ r+s: r,s \in \mathbb{Q}, r\leq x, s\leq y\right\}=\sup\{q\in \mathbb{Q}: r\leq x\}+\sup\{s\in \mathbb{Q}: s\leq y\}.$$

One direct is trivial, namely that $$\sup\left\{ r+s: r,s \in \mathbb{Q}, r\leq x, s\leq y\right\}\leq\sup\{q\in \mathbb{Q}: r\leq x\}+\sup\{s\in \mathbb{Q}: s\leq y\}=x+y.$$

The other direction is tricker, though. I was trying to approach this by contradiction: Let $A=\{ r+s: r,s \in \mathbb{Q}, r\leq x, s\leq y\}$ and suppose that $\sup A < x+y$. I was trying to use the fact that $\mathbb{Q}$ is dense in $\mathbb{R}$, but I'm still not seeing how to get the proof to work out. Suggestions?? Thank you!!

## marked as duplicate by Community♦Sep 17 '16 at 22:50

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• @iamvegan Not a duplicate. But a corollary. – 6005 Sep 17 '16 at 20:02