Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|cite|improve this question

2 Answers 2

up vote 4 down vote accepted

Define a mapping $f: \mathbb{N}^\mathbb{N} \to \{ 0,1 \}^\mathbb{N}$ by $$f : \langle x_n \rangle_{n \in \mathbb{N}} \mapsto 0^{x_0} 1 0^{x_1} 1 \cdots$$ where by $0^k$ we mean the length $k$ sequence consisting of only zeroes. It is fairly straightforward to show that this is a homeomorphic embedding of $\mathbb{N}^\mathbb{N}$ into $\{ 0,1 \}^\mathbb{N}$. It is also not too difficult to show that the range of $f$ is $$\{ \langle y_n \rangle_{n \in \mathbb{N}} \in \{ 0,1 \}^\mathbb{N} : y_n = 1\text{ for infinitely many } n \},$$ which is dense in $\{ 0,1 \}^\mathbb{N}$. (Use the fact that basic open sets in $\{ 0,1 \}^\mathbb{N}$ are determined by finite binary sequences, and then extend that finite binary sequence in any way to have infinitely many $1$s.)

share|cite|improve this answer

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 $\{0,1\}^{\Bbb N}$.
share|cite|improve this answer
I must admit it seemed a bit obfuscated to me until I realised that "separable + Čech-complete + metrizable" = Polish. – Arthur Fischer Dec 9 '12 at 15:22

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.