Zero-dimensional separable metric spaces

Question

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.

Answer 1

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.

Answer 2

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 $\{0,1\}^{\Bbb N}$.)