Does there exist a bijective, continuous map from the irrationals onto the reals?

Question

Let $\mathbb{P}$ be the irrational numbers as a subspace of the real numbers. $\mathbb{P}$ is homeomorphic to $\mathbb{N}^\mathbb{N}$, which is also called the Baire space. It is well known, and fairly easy to see, that there is a continuous map from $\mathbb{P}$ onto the reals, or that there is a closed subset $A$ of $\mathbb{P}$ and a bijective, continuous map from $A$ onto the reals.

But does there exist a bijective, continuous map from $\mathbb{P}$ onto the reals?

I'm sure this question has been asked already and the answer is well-known. But I couldn't find a reference. Nor could I prove or disprove it.

Answer 1

Theorem: Classical Descriptive Set Theory,Alexander S. Kechris,p-$40$, Exercise $7.15$

$X$ be a non empty polish space. Then $X$ is perfect iff there is a continuous bijection $f:\mathcal{N}\to X$

$\mathcal{N}=\mathcal{P}$ : set of all irrationals is the Baire space $\omega^\omega$.

$X=\mathbb{R}$ : set of reals is a perfect Polish space.

Hence it is possible to find a continuous bijection from $\mathcal{P}$ onto $\mathbb{R}$

Answer 2

Yes, there is a continuous bijection from $\Bbb{N}^\Bbb{N}$ to $\Bbb R$. Quoting [1],

Waclaw Sierpinski proved in 1929 a remarkable theorem, which in its modern general form reads: if $X$ is any nonempty perfect Polish space, then there is a continuous bijection from $\Bbb{N}^\Bbb{N}$ to $X$. (See [2, pp. 40, 357]. A Polish space is a topological space which has a countable dense subset and which is complete with respect to a metric generating the topology. A space is perfect if it has no isolated points.) The spaces $\Bbb R^n$ and $I^n$ and the Cantor set are examples of perfect Polish spaces. Therefore $\Bbb{N}^\Bbb{N}$ can be mapped continuously and bijectively onto each of these spaces.

Thus the original reference is [3], although I was not able to access to this paper yet. Interestingly, there is no continuous bijection from $\Bbb R^2$ to $\Bbb R$, nor from $\Bbb R$ to $\Bbb R^2$, see for instance this answer.

[1] O. Deiser, A simple continuous bijection from natural sequences to dyadic sequences, Amer. Math. Monthly 116 (2009), no. 7, 643--646.

[2] A.S. Kechris, Classical descriptive set theory. Graduate Texts in Mathematics, 156. Springer-Verlag, New York, 1995. xviii + 402 pp.

[3] W. Sierpinski, Sur les images continues et biunivoques de l'ensemble de tous les nombres irrationnels, Mathematica 1 (1929) 18-21.

Answer 3

Here is a direct construction of a continuous bijection $f:\mathbb{N}^\mathbb{N}\to\mathbb{R}$. First, partition $\mathbb{R}$ into infinitely many left-closed right-open intervals $(I_n)_{n\in\mathbb{N}}$. Then partition each $I_n$ into infinitely many left-closed right-open intervals $(I_{nm})_{m\in\mathbb{N}}$, such that there is no rightmost $I_{nm}$ (i.e., there is no $I_{nm}$ whose right endpoint is the right endpoint of $I_n$). Continue this process recursively, defining for each finite sequence $s$ of natural numbers a left-closed right-open interval $I_s$ such that each $I_s$ is the disjoint union of all the intervals $I_{sn}$ and there is no rightmost $I_{sn}$. Also arrange that the length of each $I_s$ is at most $1/n$ where $n$ is the length of the sequence $s$.

Now for $\sigma\in\mathbb{N}^\mathbb{N}$, define $f(\sigma)$ to be the unique element of the intersection $\bigcap_s I_s$ where $s$ ranges over all initial segments of $\sigma$. Since the lengths of these $I_s$ go to $0$, it is clear there is at most one such point. Since the right endpoint of $I_{sn}$ is always strictly less than the right endpoint of $I_s$, $\bigcap_s I_s$ is actually equal to $\bigcap_s\overline{I_s}$ (where the closure just adds in the right endpoint) and so is nonempty by compactness. Thus $f$ is well-defined. It is easy to see that $f$ is continuous, and $f$ is bijective since each $I_s$ is partitioned by the sets $I_{sn}$ so each real number is in a unique $I_s$ for each length of $s$ which combine to form an infinite sequence.