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.

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    $\begingroup$ It's not particularly natural, but I'm thinking the set of irrational numbers whose continued fraction expansion $[a_0; a_1, a_2, \ldots]$ is such that $[a_0; a_2, a_4, \ldots]$ is quadratic irrational should be an example. $\endgroup$
    – Todd Trimble
    Jun 15 '20 at 2:41

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).


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.

  • $\begingroup$ why is e.g. $T_e$ continuous? For example $a_n=1-2^n=0,1...1$ converges to one, but $T_e(a_n)$ converges to $0.0101010101...\neq T_e(1)=1$? $\endgroup$ Jan 18 at 13:14
  • $\begingroup$ @HenrikRüping We require that $M(p,q)=T_e(q)+T_o(p)$ is continuous not $T_e$ or $T_o$ individually. $\endgroup$
    – Ivan Meir
    Jan 19 at 0:32
  • $\begingroup$ I still don't get it. If $M(p,q)$ is continuous, then $M(0,q)=T_e(q)$ would also be continuous. $\endgroup$ Jan 21 at 9:35
  • $\begingroup$ @HenrikRüping (I think you mean $M(q,0)=T_e$) The problem with setting $p=0$ is that $p\notin \mathbb{P}$, the set of irrational numbers. However I certainly do appreciate your point which is relevant for fractions which have a finite binary representation and when you use this finite form in the construction. I believe you simply need to ensure you always use the infinite binary representation when you construct the mapping and the dense subset. I have updated my answer with this clarification. Thank you for your interesting observation. $\endgroup$
    – Ivan Meir
    Jan 21 at 14:37
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    $\begingroup$ Thank you Henrik you are definitely correct but I believe we can just replace $\mathbb{Q}$ by say the quadratic irrationals or even simpler $\sqrt{2} \mathbb{Q}$ and this should work. Let me know what you think - I have updated my answer to reflect this. $\endgroup$
    – Ivan Meir
    Jan 22 at 11:41

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