Let $a,b\in\mathbb{R}$. Is $(a,b)\setminus \mathbb{Q}$ homeomorphic to $\mathbb{R}\setminus \mathbb{Q}$?
It seems true to me from this fact: $(a,b)$ is homeomorphic to $\mathbb{R}$. Let $f$ be this homeomorpshim then $f|_{(a,b)\setminus\mathbb{Q}}$ is a homeomorphsim from $(a,b)\setminus\mathbb{Q}$ to it's image under $f$. Of course, the image does have to be made up of just irrational, but is that possible?
The set $(a,b)\cap \Bbb Q$ is order-isomorphic to $\Bbb Q$, since both are countable, densely ordered sets with no endpoints. Pick an order isomorphism $F:(a,b)\cap \Bbb Q\to\Bbb Q$. This extends uniquely to an order isomorphism $G:(a,b)\to\Bbb R$: for $x\in(a,b)$ define $G(x)=\sup\{F(t):t\in(a,b)\cap\Bbb Q,t\le x\}$. As $(a,b)$ and $\Bbb R$ have the order topology, then $G$ is a homeomorphism. Its restriction to $(a,b)\setminus\Bbb Q$ is a homeomorphism to $\Bbb R\setminus\Bbb Q$.