Dense subset of Cantor set homeomorphic to the Baire space

Question

Does anyone know a proof that the Cantor set, $\{0,1\}^{\mathbb{N}}$, has a dense subset homeomorphic to the Baire space, $\mathbb{N}^{\mathbb{N}}$? Thank you.

Answer 1

Hint: If $x = ( x_n )_{n \in \mathbb{N}}$ is an element of the Baire space $\mathcal{N} = \mathbb{N}^\mathbb{N}$, map it to $$0^{x_0} 1 0^{x_1} 1 \cdots$$ where by $0^k$ we mean the length $k$ sequence consisting of only zeroes. You then show that this is a homeomorphic embedding of $\mathcal{N}$ into $2^{\mathbb{N}}$. There is an nice description of the range of this function that makes its denseness in $2^{\mathbb{N}}$ quite easy to demonstrate.

Answer 2

If you happen to have it handy, there’s a pile-driver that takes care of the problem in short order: a characterization of the irrationals due originally to Aleksandrov, if I’m not mistaken. The space of irrationals is (up to homeomorphism) the unique zero-dimensional, separable, Čech-complete metrizable space that is nowhere locally compact. A Tikhonov space is Čech-complete iff it’s a $G_\delta$ in some (and in fact in any) compactification. Let $$X=\left\{x\in\{0,1\}^{\Bbb N}:x\text{ is not eventually constant}\right\}\;.$$ ($X$ corresponds to the points of the middle-thirds Cantor set that are not endpoints of removed intervals.)

• $X$ and $\{0,1\}^{\Bbb N}\setminus X$ are both dense in $\{0,1\}^{\Bbb N}$, which is therefore a compactification of $X$.
• $\{0,1\}^{\Bbb N}\setminus X$ is countable, so $X$ a $G_\delta$ in its compactification $\{0,1\}^{\Bbb N}$ and is therefore Čech-complete.
• $\{0,1\}^{\Bbb N}$ is a zero-dimensional, separable metrizable space, so $X$ is as well.
• That $X$ is nowhere locally compact follows easily from the fact that its complement is dense in $\{0,1\}^{\Bbb N}$.