Open set in irrationals $A$ is the set of all irrationals in $\mathbb R$. Is there a non empty open set subset of $A$?
My hunch is that the irrationals can not contain an open set because rationals are dense in $\mathbb R$? But I am not able to reconcile to the fact that the rationals are countable and hence are contained in an open set of small radius. Please get me out of the muddled thinking.

Suppose a subset $B$ of $A$ is open. Then all interior points are contained in $B$. So for any point in $B$ , there is a neighborhood of that point which is contained in $B$.
But neighborhood is an open interval and by denseness property it must contain rational numbers which is not possible since $B$ contains only irrational numbers. So $B$ cannot be open.
