I have to prove that every separable metric space which is zero-dimensional is isomorphic to a closed subset of the Baire space. Maybe I can use the Baire category theorem, but I don't know how.

  • $\begingroup$ How many elements does a zero-dimensional metric space have? $\endgroup$
    – Roland
    Jan 18, 2016 at 13:43
  • $\begingroup$ The Cantor set is zero-dimensional and has infinitely many elements. $\endgroup$
    – Wulf
    Jan 18, 2016 at 13:51
  • $\begingroup$ By the Baire space you mean $\mathbb N^\mathbb N,$ the product of countably many copies of the natural numbers, right? $\endgroup$
    – bof
    Jan 18, 2016 at 13:55
  • $\begingroup$ You were talking about dimensions, I was assuming that you'd be considering vector spaces. Which notion of dimension are you referring to? (Obviously not the maximal number of linear independent elements in the space.) $\endgroup$
    – Roland
    Jan 18, 2016 at 13:57
  • 2
    $\begingroup$ Do not use the Baire category theorem. $\endgroup$
    – GEdgar
    Jan 18, 2016 at 14:23

2 Answers 2


HINT: Suppose that $X$ is a separable, zero-dimensional metric space. Then $X$ has a countable base. It also has a base of clopen sets.

  • Prove that $X$ has a countable base $\mathscr{B}$ of clopen sets. This is a special case of the result proved in this answer, if you get stuck.

Let $\mathscr{B}_0=\{B\in\mathscr{B}:\operatorname{diam}(B)<1\}=\{B(0,k):k\in\Bbb N\}$.

  • Recursively construct a pairwise disjoint countable clopen refinement $\mathscr{R}$ of $\mathscr{B}_0$ that covers $X$. If $\mathscr{R}$ is infinite, let $\mathscr{R}=\{R(k):k\in\Bbb N\}$; if $|\mathscr{R}|=m\in\Bbb N$, let $\mathscr{R}=\{R(k):0\le k<m\}$.

The idea is that each set $R(k)$ will map to

$$\left\{\langle n_i:i\in\Bbb N\rangle\in\Bbb N^{\Bbb N}:n_0=k\right\}\;.$$

Now for each $R(k)\in\mathscr{R}$ repeat the process. Start with $$\{B\in\mathscr{B}:B\subseteq R(k)\text{ and }\operatorname{diam}(B)<2^{-1}\}$$ and recursively construct a pairwise disjoint countable clopen refinement $\mathscr{R}(k)$ covering $R(k)$. Index the members of $\mathscr{R}(k)$ as $R(k,\ell)$, where $\ell$ ranges over $\Bbb N$ if $\mathscr{R}(k)$ is infinite, and over some initial segment of $\Bbb N$ otherwise. The idea is that $R(k,\ell)$ will map to

$$\left\{\langle n_i:i\in\Bbb N\rangle\in\Bbb N^{\Bbb N}:n_0=k\text{ and }n_1=\ell\right\}\;.$$

Keep going, cutting the bound on the diameter in half at each level of the construction. In the end you have clopen sets $R(k_0,\ldots,k_m)$ for certain finite sequences $\langle k_0,\ldots,k_m\rangle$ of natural numbers, and the idea is that $R(k_0,\ldots,k_m\rangle$ maps to

$$\left\{\langle n_i:i\in\Bbb N\rangle\in\Bbb N^{\Bbb N}:n_i=k_i\text{ for }i=0,\ldots,m\right\}\;.$$

To construct the homeomorphism, use the fact that for each $x\in X$ there is a unique $\langle k_i:i\in\Bbb N\rangle\in\Bbb N^{\Bbb N}$ such that $$\bigcap_{m\in\Bbb N}R(k_0,\ldots,k_m)=\{x\}\;.$$

Showing that the image of $X$ is closed in $\Bbb N^{\Bbb N}$ is very easy if you've done everything right.

Added 26 March 2022: In fact the desired result is false as stated: $X$ is always homeomorphic to some subset of the Baire space, but that subset need not be closed. $\Bbb Q$ is a counterexample: it is certainly separable and zero-dimensional, but it is not completely metrizable, so it cannot be homeomorphic to a closed subset of the completely metrizable Baire space.

  • $\begingroup$ This was written on a Kindle, so it's a bit less polished and detailed than I'd prefer; I may well revise it a bit when I'm on a real computer again, but all of the essential ideas are here. $\endgroup$ Jan 18, 2016 at 15:18
  • $\begingroup$ Take $\mathbb{Q}$, this is separable, metrizable and 0-dimensional, but it's not completely metrizable, hence it cannot be homeomorphic to a closed subsets of the Baire space... $\endgroup$
    – Lorenzo
    Mar 26, 2022 at 9:30
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    $\begingroup$ @Lorenzo: You are of course correct. I was in bed at the time (hence working on a Kindle) and must have been sleepier than I thought, but I’m surprised that it’s taken this long for anyone to notice. I’ll add a note explaining the true situation. Thanks! $\endgroup$ Mar 26, 2022 at 19:28

Hint: Note that this is far from being a solution; it's just an indication of a natural way to define a mapping from your space $X$ into $\Bbb N^{\Bbb N}$. You need to find extra conditions on the construction below to make that mapping a homeomorphism onto a close set.

Partition $X$ into (finitely many? countably many?) sets $X_{j_1}$, $j_1=1,2\dots$ such that, well you have to figure out what the "such that" should be to make this work. For each $j_1$, partition $X_{j_1}$ into sets $X_{j_1,j_2}$. Etc.

Now if $x\in X$ there exists a unique $j_1$ with $x\in X_{j_1}$. And then there exists a unique $j_2$ with $x\in X_{j_1,j_2}$. Etc. Map $x$ to the sequence $j_1,j_2,\dots$.

Probably at least you want all these sets to be clopen, and to make sure the diameter tends to zero as you proceed down the tree...

(If $X$ is the Cantor set and you do this in the obvious way, so each partition is a partition into two subsets, you do get a homeomorphism onto $\{0,1\}^{\Bbb N}$.)


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