Does there exist a bijective, continuous map from the irrationals onto the reals? 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.
 A: Theorem:$[\textbf{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 a Baire space.
$X=\mathbb{R}$ : set of reals is a perfect Polish space.
Hence it is possible to find a continuous bijection from $\mathcal{P}$ onto $\mathbb{R}$
A: 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.
A: 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.
