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.

Let $S\subset \mathbb{R}\setminus \mathbb{Q}$ be any non empty subset of the irrational . Choose any point $x\in S$, let $(a,b)$ be any open interval in $\mathbb{R}$ containing the point $x$. Then $(a,b)$ cannot be contained inside $S$ because of the fact that every interval contains atleast one rational number. Therefore $x$ cannot be an interior point of $S$, and hence $S$ is not open in $\mathbb{R}$.
