Perfect Polish Spaces

Let \(\omega = \{0,1,\ldots\}\). The main business of descriptive set theory is the study of \(\omega,\R\) and their subsets, with particular emphasis on the definable sets of integers and reals. A fair name for it is definability theory for the continuum.

Perfect Polish Spaces#

The space of real numbers \(\R\) is a perfect Polish space. We get rid of the peculiarities of the reals and concentrate on the “essence”.

Definition. A perfect Polish space is a complete, separable metric space (Polish) with no isolated points (perfect).

The Baire space#

This is the set \(\mathcal{N}\) of all infinite sequence of natural numbers \(\mathcal{N} = {}^\omega\omega\) with the natural product topology taking \(\omega\) discrete. The topology of \(\mathcal{N}\) is generated by the complete metric

\[d(\alpha,\beta) = \begin{cases}0 & \alpha=\beta, \\ \frac{1}{\text{least }n[\alpha(n) \neq \beta(n)]+1} & \alpha\neq \beta.\end{cases}\]

and the set of ultimately constant sequences is countable and dense in \(\mathcal{N}\), making \(\mathcal{N}\) a perfect Polish space. In fact \(\mathcal{N}\) is homeomorphic to the set of irrational numbers topologized as a subspace of \(\R\). Hence we may call the members of \(\mathcal{N}\) irrationals.

This is a totally disconnected space. The neighborhood base consisting of basic neighborhoods of the form \(N(k_0,\ldots,k_n) = \{\alpha\in\mathcal{N}: \alpha(0) = k_0,\ldots,\alpha(n)=k_n\}\) consists of clopen sets.

Moreover, \(\mathcal{N}\) is terminal, in that

Theorem. For every Polish space \(P\), there is a continuous surjection \(\pi:\mathcal{N}\twoheadrightarrow P\).

Proof idea. Basically we want to approximate a point \(x \in P\) from a countable dense subset \(D= \{r_0, r_1, r_2,\ldots \}\subset P\) by referencing an \(\alpha\in\mathcal{N}\). \(D\) is dense in \(P\), hence every \(x\in P\) is the limit of a sequence \(\{r_{\alpha(i)}\}_i\), where \(\alpha(i)\) is the \(i\)-th component of \(\alpha\). Then by setting \(\pi(\alpha) = \lim_n r_{\alpha(n)} = x\) the function is given. The point is to specify how to choose this \(\alpha\). Given \(x\in P\), we may set \(\alpha(n)\) to be the least \(k\) s.t. \(d(x,r_k) < 2^{-n-1}\), then obviously \(\pi(\alpha) = \lim_n r_{\alpha(n)}= x\) and the map is surjective. The continuity can be easily checked by noticing that \(x_{n+1} = r_{\alpha(n+1)} = \begin{cases}r_{\alpha(n+1)} & d(x_n,r_{\alpha(n+1)})< 2^{-n},\\ x_n & d(x_n,r_{\alpha(n+1)})\geq 2^{-n}.\end{cases}\)

The Cantor Space#

This is the set of all infinite binary sequences \(\mathbb{C} = {}^\omega 2\) with the product topology. It is a compact subspace of \(\mathcal{N}\) naturally represented by the complete binary tree. It is homeomorphic to the classical Cantor set, and we abuse terminology by calling \(\mathbb{C}\) the Cantor set.

The Cantor set \(\mathbb{C}\) is in some sense initial, in that

Theorem. For every perfect Polish space \(P\) there is a continuous injection \(\pi:\mathbb{C}\hookrightarrow P\).

To prove the theorem we introduce some basic techniques and constructions. Each perfect Polish space \(P\) can be associated a fixed enumeration \(\{N(P,i)\}_{i\in\omega}\) of a countable set of open neighborhoods that generates the topology. The members of the enumeration can be denoted with \(N_0,N_1,\ldots\). With each \(N_i\), we assume that there is a center \(x_i\) and a radius \(p_i\), s.t.

\(x\in N_i\) provided that \(N_i\neq \emptyset\).
If \(x\in N_i\), then \(d(x,x_i) < p_i\).
If \(x \in P\), then for each \(n\) there is some \(N_i\) s.t. \(x\in N_i\) and \(p_i < 2^{-n}\).

Now let \(\overline{U}\) be the closure of \(U\). We may associate to each finite binary sequence an open neighborhood that “localizes the space” so that we get somthing that is similar to a filter:

Theorem. Let \(P\) be a perfect Polish space. For each finite binary sequence \(u=(t_0,\ldots, t_{n-1})\) (\(t_i= 0,1\)) an open neighborhood \(N_{\sigma(u)}\neq \emptyset\) in \(P\), s.t.
1. If \(u\) is a proper initial segment of \(v\), then \(\overline{N}_{\sigma(v)}\subseteq N_{\sigma(u)}\).
2. If \(u\) and \(v\) are incompatible (for some \(k\) that is smaller than the length of \(u\) and \(v\), \(u_k\neq v_k\)), then \(\overline{N}_{\sigma(u)}\cap \overline{N}_{\sigma(v)} = \emptyset\).
3. If \(u=(t_0,\ldots,t_{n-1})\), i.e. has length \(n\), then \(p_{\sigma(u)} \leq 2^{-n}\) (recall that \(p_{\sigma(u)}\) is the radius of \(N_{\sigma(u)}\)).

Proof idea. Simply construct by finding open balls that satisfy the conditions, starting from an \(N_{\sigma(\emptyset)}\) of radius \(\leq 1\).

Now we can prove the theorem.

Corollary. For every perfect Polish space \(P\) there is a continuous injection \(\pi:\mathbb{C}\hookrightarrow P\).

Proof idea. Given an infinite binary sequence \(\alpha\in\mathbb{C}\), put \(x^\alpha_n = \text{the center of }N(P,\sigma(\alpha(0),\ldots,\alpha(n-1)))\) and let \(\pi(\alpha) = \lim_{n} x^\alpha_n\). Verify that this is an injection and continuous.