(This question actually arose in some research on number theory.)

I once learned that any countable dense subspace of any Euclidean space $\mathbb R^n$ is homeomorphic to the rationals $\mathbb Q$.

Now I wonder if something similar is true for the irrationals $J \mathrel{:=} \mathbb R \setminus \mathbb Q$ (with the subspace topology from $\mathbb R$).

Let $\mathfrak c$ denote the cardinality of the continuum.

>**I**.    Is each cartesian power $J^n$ homeomorphic to $J$?

Also, how far can this be pushed?

>**II**.   Let $X$ be a dense totally disconnected subspace of $\mathfrak R$ such that every neighborhood of each point of $X$ contains $\mathfrak c$ points.  Is $X$ homeomorphic to $J$?

What about for such subspaces of fairly nice subspaces of $\mathbb R^n$?

>**IIa**.  Let $X$ be any subspace of $\mathbb R^n$ as described in **II**., and let $B$ denote any subspace of $\mathbb R^n$ homeomorphic to [the open unit ball in $\mathbb R^n$ $\cup$ any subset of its boundary].   Then is $X \cap B$ homeomorphic to $J$?

And what about greater generality?

>**III**.  Is there a simple set of conditions that describe exactly all spaces (or subspaces of $\mathbb R^n$) that are homeomorphic to $J$?  What about $J^n$? (Perhaps the word *homogeneous* or *metric* needs to be included.)

(I found nothing relevant via Google, in MathSciNet, or here on MathOverflow.)