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.
Best Answer
$A = \mathbb{R} \setminus \{0\}$ 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 $\epsilon$.