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.
Best Answer
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.