# Zero-dimensional separable metric spaces

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.

• How many elements does a zero-dimensional metric space have? Jan 18, 2016 at 13:43
• The Cantor set is zero-dimensional and has infinitely many elements.
– Wulf
Jan 18, 2016 at 13:51
• By the Baire space you mean $\mathbb N^\mathbb N,$ the product of countably many copies of the natural numbers, right?
– bof
Jan 18, 2016 at 13:55
• 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.) Jan 18, 2016 at 13:57
• Do not use the Baire category theorem. Jan 18, 2016 at 14:23

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.

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.

• 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. Jan 18, 2016 at 15:18
• 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... Mar 26, 2022 at 9:30
• @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! 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 $\{0,1\}^{\Bbb N}$.)