The set of irrationals in the reals is open or closed? 
The set of irrationals in the reals is open or closed?

I only know that the set of irrational numbers neither closed nor open so this set is neither closed nor open. Am I right?
 A: Recall that the set of irrational numbers is dense in $\mathbb{R}$, meaning that between any two distinct real numbers there exists an irrational number (that is, every open interval contains an irrational number). The same is true for the rationals. 
Let $x\in \mathbb{R}\backslash \mathbb{Q}$. Let $B=(x-\varepsilon, x+\varepsilon)$ be an pen ball around $x$. Then there exists a rational number in $B$, since the rationals are dense in $\mathbb R$. Since every open ball centered at a point in $\mathbb{R}\backslash \mathbb{Q}$ contains a point not in $\mathbb{R}\backslash \mathbb{Q}$, it follows that $\mathbb{R}\backslash \mathbb{Q}$ is not open. The same reasoning shows that $\mathbb Q$ is not open, and so $\mathbb{R}\backslash \mathbb{Q}$ is not closed. 
A: You are Correct! 
Here's my reason: Pick $0 \in \mathbb{R} \setminus I$. For any $\epsilon>0$, $(0-\epsilon,0+\epsilon)$ contains irrational points such as $\frac{\sqrt{2}}{n}$ for large $n$. So $\mathbb{R} \setminus I$ is not open and hence $I$ is not closed.   
