Polish Spaces

The proofs in the introduction section are quite general, that is, they make little use of specific properties of $\Real$. If we scan the arguments carefully, we see that we can replace $\Real$ by any metric space that is complete and contains a countable basis of the topology.

Review of some concepts from topology¶

Basis¶

Let $(X, \mathcal{O})$ be a topological space. A family $\mathcal{B} \subseteq \mathcal{O}$ of subsets if $X$ is a basis for the topology if every open set from $\mathcal{O}$ is the union of elements of $\mathcal{B}$. For example, the open intervals with rational endpoints form a basis of the standard topology of $\Real$. A family $\mathcal{S} \subseteq \mathcal{O}$ is a subbasis if the set of finite intersections of sets in $\mathcal{S}$ is a basis for the topology.

Finally, if $\mathcal{S}$ is any family of subsets of $X$, the topology generated by $\mathcal{S}$ is the smallest topology on $X$ containing $\mathcal{S}$. It consists of all unions of finite intersections of sets in $\mathcal{S} \cup \{X,\emptyset\}$.

Density¶

A set $D \subset X$ is dense if for any open $U \neq \emptyset$ there exists $z \in D \cap U$. If a topological space $(X, \mathcal{O})$ has a countable dense subset, the space is called separable.

Products¶

If $(X_i)_{i \in I}$ is a family of topological spaces, one defines the product topology on $\Pi_{i \in I} X_i$ to be the topology generated by the sets $\pi_j^{-1}(U)$, where $j \in I$, $U \subseteq X_j$ is open, and $\pi_j: \Pi_{i \in I} X_i \to X_j$ is the $j$th projection.

Subspaces¶

If $(X,\mathcal{O})$ is a topological space and $Y \subseteq X$, we can put a topology on $Y$ by intersecting it with all open sets in $\mathcal{O}$, that is, the open sets are

This is called the subspace topology on $Y$.

Now suppose $(X,d)$ is a metric space. With each point $x \in X$ and every $\eps > 0$ we associate an $\eps$-neighborhood or $\eps$-ball

The topology generated by the $\eps$-neighborhoods is called the topology of the metric space $(X,d)$. If this topology agrees with a given topology $\mathcal{O}$ on $X$, we say the metric $d$ is compatible with the topology $\mathcal{O}$. If for a topological space $(X, \mathcal{O})$ there exists a compatible metric, $(X, \mathcal{O})$ is called metrizable.

If a topological space $(X,\mathcal{O})$ is separable and metrizable, then the balls with center in a countable dense subset $D$ and rational radius form a countable basis of the topology.

Polish spaces – the basics¶

There may be many different compatible metrics that make $X$ complete. If $X$ is already given as a complete metric space with countable dense subset, then we call $X$ a Polish metric space.

The standard example is, of course, $\Real$, the set of real numbers. One can obtain other Polish spaces using the following basic observations. (We leave the proof as an exercise.)

We conclude that $\Real^n$, $\C$, $\C^n$, the unit interval $[0,1]$, the unit circle $\Ci = \{z \in \C \colon |z| = 1\}$, and the infinite dimensional spaces $\Real^\Nat$ and $[0,1]^\Nat$ (the Hilbert cube) are Polish spaces.

Any countable set with the discrete topology is Polish, by means of the discrete metric $d(x,y) = 1 \: \Leftrightarrow \: x \neq y$.

Some subsets of Polish spaces are Polish but not closed.

We will later characterize all subsets of Polish spaces that are Polish themselves.

Product spaces¶

In a certain sense, the most important Polish spaces are of the form $A^\Nat$, where $A$ is a countable set carrying the discrete topology. The standard cases are

$\Cant$, the Cantor space, $\qquad$ and $\qquad$ $\Baire$, the Baire space.

We will, for now, denote elements from $A^\Nat$ by lowercase greek letters from the beginning of the alphabet. The $n$-th term of $\alpha$ we denote by either $\alpha(n)$ or $\alpha_n$, whichever is more convenient.

We endow $A$ with the discrete topology. The product topology on these spaces has a convenient characterization. Given a set $A$, let $\Astr{A}$ be the sets of all finite sequences over $A$. Given $\sigma, \tau \in A^{<\Nat}$, we write $\sigma \Sleq \tau$ to indicate that $\sigma$ is an initial segment of $\tau$. $\Sle$ means the initial segment is proper. This notation extends naturally to hold between elements of $\Astr{A}$ and $A^\Nat$, $\sigma \Sle \alpha$ meaning that $\sigma$ is a finite initial segment of $\alpha$.

A basis for the product topology on $A^\Nat$ is given by the cylinder sets

that is, the set of all infinite sequences extending $\sigma$. The complement of a cylinder is a union of cylinders and hence open. Therefore, each set $\Cyl{\sigma}$ is clopen.

A compatible metric is given by

The representation of the topology via cylinders (which are characterized by finite objects) allows for a combinatorial treatment of many questions and will be essential later on.

Via the mapping

$\Cant$ is homeomorphic to the middle-third Cantor set in $\Real$, whereas the continued fraction mapping

provides a homeomorphism between $\Z \times (\N\setminus\{0\})^\N$ and the irrational real numbers.

The universal role played by the discrete product spaces is manifested in the following results. A Polish (or, more generally, topological) space is perfect if it does not have any isolated points (no singleton set $\{x\}$ is open).

The proof is very similar to the proof of Theorem (Cantor, 1884). Note that this proof actually constructs an embedding of $\Cant$. The continuity of the mapping is straightforward.

In a similar way we can adapt the proof of Cantor-Bendixson Theorem to show that the perfect subset property holds for closed subsets of Polish spaces.

The special role of Baire space $\Baire$ is underlined by the following fact.

Continuous mappings on product spaces¶

Let $f: A^\Nat \to A^\Nat$ be continuous. We define a mapping $\phi: \Astr{A} \to \Astr{A}$ by setting

This mapping has the following properties:

1. It is monotone, i.e. $\sigma \Sleq \tau$ implies $\phi(\sigma) \Sleq \phi(\tau)$.
2. For any $\alpha \in A^\Nat$ we have $\lim_n |\phi(\alpha\Rest{n})| = \infty$. This follows directly from the continuity of $f$: For any neighborhood $\Cyl{\tau}$ of $f(\alpha)$ there exists a neighborhood $\Cyl{\sigma}$ of ${}\alpha$ such that $f(\Cyl{\sigma}) \subseteq \Cyl{\tau}$. But ${}\tau$ has to be of the form $\tau = f(\alpha)\Rest{m}$, and ${}\sigma$ of the form $\alpha\Rest{n}$. Hence for any $m$ there must exist an $n$ such that $\phi(\alpha\Rest{n}) \Sgeq f(\alpha)\Rest{m}$.

On the other hand, if a function $\phi: \Astr{A} \to \Astr{A}$ satisfies (1.) and (2.), it induces a function $\phi^*: A^\Nat \to A^\Nat$ by letting

This $\phi^*$ is indeed continuous: The preimage of $\Cyl{\tau}$ under $\phi^*$ is given by

which is an open set.

We have shown the following.

Note that we can completely describe a topological concept, continuity, through a relation between finite strings.