Non-trivial open dense subset of $\mathbb{R}$. I recently found the following exercise real analysis:

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

I think it is not too hard to prove. But do we have a non-trivial example of such a set? So my question is:

Can we find an example of a subset $A\subset\mathbb{R}$ that is open and dense, but $A\neq \mathbb{R}$?

I don't know a lot of examples of dense subset of $\mathbb{R}$. The rational and irrational numbers are dense, but clearly not open.
 A: $A = \mathbb{R} \setminus \{0\}$ works for this purpose, and isn't equal to $\mathbb{R}$. But that's still fairly trivial.
For a less trivial example, fix an enumeration $\{r_n\}_{n = 0}^{\infty}$ of rational numbers and a positive number $\epsilon$. Define open intervals
$$\mathcal{O}_n = \left(r_n - \frac{\epsilon}{2^{n + 2}}, r_n + \frac{\epsilon}{2^{n + 2}}\right)$$
and define $\mathcal{O} = \bigcup_n \mathcal{O}_n$. Then $\mathcal{O}$ is an open, dense subset of $\mathbb{R}$ with Lebesgue measure at most $\epsilon$.
In fact, we could (by dilating one of our intervals) make the measure of $\mathcal{O}$ equal to any given positive number $\epsilon$.
A: Take $A=\mathbb{R}-\{0\}$. This clearly open and dense.
A: An important non-trivial example is the middle-thirds Cantor set $C$: it’s a subset of $\Bbb R$ of cardinality $2^\omega=\mathfrak{c}=|\Bbb R|$, and since it is closed and nowhere dense in $\Bbb R$, its complement is open and dense. With a fairly small change in the construction one can construct a ‘fat’ Cantor set that has all of the properties of $C$ that I just mentioned and in addition has positive Lebesgue measure; its complement is again a dense open subset of $\Bbb R$.
A: If you remove any finite number of points from $\mathbb{R}$ then the remaining set will be open and dense in $\mathbb{R}$.
Think a open set $A \in \mathbb{R}$ which is closed under addition. Then what is $A$  in $\mathbb{R}$?
