Open dense subset of $\mathbb{R}$.

Question

I found this question in one of the posts

Let $A\subseteq\mathbb{R}$ be open and dense. Show that $$\mathbb{R}=\{x+y:x,y\in A\}$$

The poster stated that it is not too hard to prove but I can't seem to figure out the proof. Can someone help me out.

Answer 1

Since $A$ is open and nonempty, it contains some open interval $J = (a,b)$ with $a < b$. Consider any $z \in \mathbb R$. Since $A$ is dense, $z - J = (z-b, z-a)$ contains some member $x$ of $A$. Thus $x = z - y$ where $a < y < b$, so $z = x+y$ where $x \in A$ and $y \in A$.