A natural $\mathbb Q\times \mathbb P$ subset of $\mathbb R$?

Question

I would like a simple description of a dense subset of $\mathbb R$ which is homeomorphic to $\mathbb Q\times \mathbb P$. Preferably the description will be of an algebraic nature, and perhaps the set will even be a subgroup of $\mathbb R$. Here $\mathbb R$ is all real numbers, $\mathbb Q$ is all rational numbers, and $\mathbb P$ is all irrational numbers.

The following characterization by Jan van Mill may be useful: $\mathbb Q\times \mathbb P$ is the unique zero-dimensional separable metrizable space which is strongly $\sigma$-complete, nowhere complete, and nowhere $\sigma$-compact.

Answer 1

I made a comment earlier, but let me try converting it to an answer. It's similar in flavor to Ivan's.

By a back and forth argument, all countable dense subsets of $\mathbb{R}$ are homeomorphic to $\mathbb{Q}$, which allows one to replace $\mathbb{Q}$ with the space of quadratic irrationals, which I'll denote by $Q'$. The (regular) continued fraction expansions of elements of $Q'$ are precisely infinite continued fractions that are eventually periodic. By taking continued fractions, we have a homeomorphism $\mathbb{Z} \times \mathbb{N}^\mathbb{N} \cong \mathbb{P}$ defined by

$$(a_0, a_1, a_2, \ldots) \mapsto a_0 + \frac1{a_1 + \frac1{a_2 + \ldots}}$$

and from there we easily get a homeomorphism $\mathbb{N}^\mathbb{N} \cong \mathbb{P}$. Since $\mathbb{N}^\mathbb{N}$ is homeomorphic to its square via the interleaving map

$$((a_0, a_2, \ldots), (a_1, a_3, \ldots)) \mapsto (a_0, a_1, a_2, a_3, \ldots)$$

we get a homeomorphism $\mathbb{P} \times \mathbb{P} \to \mathbb{P}$ by interleaving continued fractions. The subset $Q' \times \mathbb{P}$ maps homeomorphically onto its image under this map, and this image is of course dense (it contains for example the dense set $Q'$ of numbers with eventually periodic cf's).

Answer 2

Take the dense subset to be those irrationals whose binary expansion $a_m 2^m+\cdots+a_n2^n$ is an irrational quadratic for the even powers (for alternatives see below) the binary number formed by the coefficients of the even places is irrational quadratic and the odd places irrational.

As observed by Henrik in the comments there are continuity issues mapping directly from $\mathbb{Q}$ so we replace it by any irrational homeomorphic set $\mathbb{Q}'$ which could be the quadratic irrationals as in Todd's answer or simply $\lambda \mathbb{Q}$ for some $\lambda \notin \mathbb{Q}$.

Then the map M from $\mathbb Q'\times \mathbb P$ to a dense subset of $\mathbb R$ given by interleaving the binary expansions should work. More formally $M(q,p)=T_e(q)+T_o(p)$ where $T_e(r)$ and $T_o(r)$ are defined by taking the binary expansion of $r$. Then in the binary form you map $2^n\rightarrow 2^{2n}$ and $2^n\rightarrow 2^{2n+1}$ respectively. i.e.

$T_e(11)=T_e(2^0+2^1+2^3)=2^0+2^2+2^6=69$, $T_o(11)=T_e(2^0+2^1+2^3)=2^1+2^3+2^7=138$.

This map is clearly continuous both ways, 1-1 and maps to a dense subset since any real number has an arbitrarily close rational approximation.

There are of course many possible such examples as for any infinite fixed subset S of the integers we can take the dense subset to be those irrational numbers whose binary expansion over $S$ or $\mathbb Z\setminus S$ is infinite and represents a value in $\mathbb{Q}'$ or is irrational respectively.