The spaces $\mathbb{R}^n\setminus \mathbb{Q}^n$ and $(\mathbb{R}\setminus\mathbb{Q})^n$ with the Euclidean topology have the feeling of not being homeomorphic for $n>1$, because the "holes" in the former appear to be smaller in the former than in the latter, very informally speaking.


Is there any integer $n>1$ such that $\mathbb{R}^n\setminus \mathbb{Q}^n\cong(\mathbb{R}\setminus\mathbb{Q})^n$?



$\mathbb{R}^n\setminus \mathbb{Q}^n$ vs $(\mathbb{R}\setminus\mathbb{Q})^n$
For the usual Euclidean space $\mathbb{R}^n$, for the subset $X=\mathbb{R}^n\setminus \mathbb{Q}^n$ with the subspace topology, the space $X$ is path-connected. Indeed, for every $x=(x\_1,\dots,x\_n)\in X$, there exists some integer $1\leq i\leq n$ with $x\_i$ in $\mathbb{R}\setminus \mathbb{Q}$. Define the following function $$\alpha\_{i,x}:[0,1] \to \mathbb{R}, \ \ \alpha\_{i,x}(t) = (y\_1,\dots,y\_n),\ \ y\_j = \left\{ \begin{array}{cc} (1-t)x\_j + t\sqrt{2}, & j\neq i, \\ x\_i, & j=i \end{array}\right. $$ This is a continuous path whose image is contained in $X$, i.e., it is a continuous path in $X$. The final point is $\alpha\_{i,x}(1)=z = (z\_1,\dots,z\_n)$ where $z\_j$ equals $\sqrt{2}$ for $j\neq i$ and $z\_i$ equals $x\_i$. 


Now, for any $j$ with $1\leq j\leq n$ and $j\neq i$, the path $\alpha\_{j,z}$ is a continuous path in $X$ from $z$ to the point $w=(\sqrt{2},\dots,\sqrt{2})$. Therefore every point of $X$ is connected by a continuous path to $w$. Concatenating paths, every pair of points of $X$ is path-connected.


