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.
 A: Let $\Bbb P=\Bbb R\setminus\Bbb Q$ be the set of irrationals. Let $U$ be a non-empty open set in $\Bbb R$; then there are $a,b\in\Bbb R$ such that $a<b$ and $(a,b)\subseteq U$. As you say, the rationals are dense in $\Bbb R$, so there is a rational $q\in(a,b)$, and it follows that $$q\in(a,b)\setminus\Bbb P\subseteq U\setminus\Bbb P$$ and hence that $U\nsubseteq\Bbb P$. Thus, as you suspected, the irrationals do not contain any non-empty open subset of $\Bbb R$.
The fact that $\Bbb Q$ is countable and therefore can be covered by an open set of small radius is simply irrelevant: there is no reason to think that in general an open set covering a dense set should cover the whole space. A dense set can even be open. For a very simple example, consider the subspace
$$X=\{0\}\cup\left\{\frac1n:n\in\Bbb Z^+\right\}$$
of $\Bbb R$: the set $X\setminus\{0\}$ is a dense, open subset of $X$, so it is an open cover of itself that covers only itself, not all of $X$.
A: 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.
A: 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 $\mathbb{R}$.
