nLab Polish space

Contents

Contents

Introduction

A Polish space is a topological space that’s homeomorphic to a separable complete metric space. Every second countable locally compact Hausdorff space is a Polish space, among others.

Polish spaces, or the measurable spaces that they define (standard Borel spaces) provide a useful framework for doing measure theory. As with any topological space, we can take a Polish space and regard it as a measurable space with its sigma-algebra of Borel sets. Then, there is a very nice classification of Polish spaces up to measurable bijection: there is one for each countable cardinality, one whose cardinality is that of the continuum, and no others.

Why are Polish spaces ‘not very big’? In other words, why are there none with cardinality exceeding the continuum? As with any separable metric space, it’s because any Polish space has a countable dense subset and you can write any point as a limit of a sequence of points in this subset. So, you only need a sequence of integers to specify any point in a Polish space. More sharply, see Lemma below.

Example

The primordial example (and in practice, one of the most convenient) is Baire space BB, viz. the space of irrational real numbers between 00 and 11, with the subspace topology inherited from the real line. This is obviously not complete with respect to the metric induced from the real line, but it is homeomorphic to the product space \mathbb{N}^\mathbb{N} via regular continued fraction expansions, and the latter is metrizable by a complete metric where the distance between two sequences a=(a 1,a 2,)a = (a_1, a_2, \ldots) and b=(b 1,b 2,)b = (b_1, b_2, \ldots) of positive integers is given by the formula d(a,b)=1/2 nd(a, b) = 1/2^n where nn is the least integer such that a nb na_n \neq b_n. A countable dense subset is given by continued fractions that eventually repeat (quadratic surds).

The convenience of Baire space is attested to by the fact that in descriptive set theory, a “real number” is often taken to mean just a point of Baire space. See also Theorem below.

Properties

Much of the material here is adapted from Marker, who in turn cites the text by Kechris as a main source.

  • A countable product X= nX nX = \prod_n X_n of Polish spaces is Polish. Indeed, each X nX_n is metrizable by a complete metric d nd_n that takes values in [0,1][0, 1], and then we may define another such complete metric on XX by
    d(f,g)= n12 n+1d n(f(n),g(n)).d(f, g) = \sum_n \frac1{2^{n+1}} d_n(f(n), g(n)).

    The metric topology on XX coincides with the product topology.

In particular, the Hilbert cube [0,1] [0, 1]^\mathbb{N} is Polish.

“Universal” Polish spaces

Proposition

Any Polish space is homeomorphic to a subspace of the Hilbert cube.

In fact, this is true of any separable metrizable space. Note that mere separability does not suffice, as there are separable spaces that are not first-countable, such as the Stone-Čech compactification of N\N, and hence cannot be subspaces of any metrizable space. What distinguishes Polish spaces is that they are, up to homeomorphism, precisely the G δG_\delta subsets of the Hilbert cube.

Lemma

Every inhabited Polish space XX admits a continuous surjection from Baire space.

Proof (sketch)

Construct by induction a collection of closed sets (balls) C sC_s indexed over finite sequences ss of positive integers, with the following properties:

  • For the empty sequence ee, C e=XC_e = X;

  • For nonempty sequences ss, diam(C s)1/ndiam(C_s) \leq 1/n where n=length(s)n = length(s);

  • Letting (s,k)(s, k) denote the extension of ss obtained by appending to ss the final element kk,

    C s= k=0 C (s,k);C_s = \bigcup_{k=0}^\infty C_{(s, k)};
  • If tt extends ss, then center(C t)C scenter(C_t) \in C_s.

Then define f:BXf: B \to X where for each infinite sequence a a \in \mathbb{N}^\mathbb{N},

f(a)={C s:aextendss}.f(a) = \bigcap \{C_s: a\; extends\; s\}.

One may check that ff is continuous and surjective.

Cantor-Bendixson rank

Let CXC \subseteq X be a closed subset of a Polish space XX. The following operation traces back to Cantor’s work in Fourier analysis, which in turn led to his study of countable ordinals and ordinal analysis.

For a subset AXA \subseteq X, recall that xAx \in A is a limit point if xCl(A{x})x \in Cl(A \setminus \{x\}). A point xAx \in A that is not a limit point of AA is called an isolated point of AA. Clearly each isolated point is open relative to AA, as is therefore the set of isolated points.

Definition

The Cantor-Bendixson derivative of CC is the set CCC' \subseteq C of limit points relative to CC. For each ordinal α\alpha the iterated derivative C αC^\alpha is defined by recursion: C 0=CC^0 = C, C α+1=(C α)C^{\alpha + 1} = (C^\alpha)', and C α= β<αC βC^\alpha = \bigcap_{\beta \lt \alpha} C^\beta if α\alpha is a limit ordinal.

Since β<α\beta \lt \alpha implies C βC αC^\beta \supseteq C^\alpha, it is clear that there is a least ordinal α\alpha for which C α=C α+1C^\alpha = C^{\alpha + 1}. This ordinal is called the Cantor-Bendixson rank of CC. A perfect set is a closed set CC such that C=CC = C'. (Some people insist that a perfect set also be nonempty; we do not.)

Proposition

For each nonempty perfect set PP in a Polish space XX, there is a continuous injection i:2 Pi: \mathbf{2}^\mathbb{N} \to P from Cantor space. In particular, the cardinality of PP is the continuum cc.

Proof

Let TT be the complete infinite binary tree, whose infinite paths from the root correspond to points in Cantor space 2 \mathbf{2}^\mathbb{N}. To each node ss in TT (a finite sequence of 00‘s and 11’s) we construct by induction an open set U sU_s with the following properties:

  • U e=XU_e = X for the empty sequence ee,

  • U t¯U s\widebar{U_t} \subseteq U_s if ss is an initial segment of tt,

  • For the two children (s,0)(s, 0) and (s,1)(s, 1) of ss, the sets U (s,0)U_{(s, 0)} and U (s,1)U_{(s, 1)} are disjoint,

  • diam(U s)1/ndiam(U_s) \leq 1/n where nn is the length of (nonempty) ss,

  • U sPU_s \cap P \neq \emptyset for each ss.

Indeed, if U sU_s has been constructed, and given xU sPx \in U_s \cap P, there are (at least!) two points x 0,x 1U sPx_0, x_1 \in U_s \cap P since PP is perfect. We can easily find disjoint neighborhoods U (s,0),U (s,1)U_{(s, 0)}, U_{(s, 1)} of these points respectively with the required properties.

Then for each path p=(a 0,a 1,)2 p = (a_0, a_1, \ldots) \in \mathbf{2}^\mathbb{N}, define

i(p)= spU si(p) = \bigcap_{s \preceq p} U_s

where sps \preceq p means ss is an initial segment of pp. The intersection consists of a single point because it equals the intersection of a decreasing chain of closed sets with shrinking diameter (thus closing in on a limit of a Cauchy sequence). The map ii is clearly injective by the disjointness of open sets of children, and it is easy to see ii is continuous.

Theorem

For CC a closed subset of a Polish space XX, the Cantor-Bendixson rank is a countable ordinal α\alpha. The complement CC αC \setminus C^\alpha is at most countable.

Proof

Let U iU_i be a countable basis of XX. For each β<α\beta \lt \alpha, each point xC βC β+1x \in C^\beta \setminus C^{\beta + 1} is an isolated point, so we can find an U i(x)U_{i(x)} in the basis such that U i(x)(C βC β+1)={x}U_{i(x)} \cap (C^\beta \setminus C^{\beta + 1}) = \{x\}. It is then clear that xi(x)x \mapsto i(x) is injective, so each C βC β+1C^\beta \setminus C^{\beta + 1} is countable. Similarly, whenever C βC β+1C^\beta \setminus C^{\beta + 1} is nonempty, we can find a basis element U j(β)U_{j(\beta)} that isolates one of its points (say xx), and this same U jU_j cannot isolate any point of an earlier C γC γ+1C^\gamma \setminus C^{\gamma + 1} since xx is a limit point of C γC^\gamma. It follows that βj(β)\beta \mapsto j(\beta) is an injective map, so that α\alpha must be a countable ordinal, and the collection FCC α= β<αC βC β+1F \coloneqq C \setminus C^\alpha = \bigcup_{\beta \lt \alpha} C^\beta \setminus C^{\beta + 1} is (at most) countable.

Corollary

A closed set in a Polish space is the disjoint union PFP \cup F of a perfect set PP and a finite or countable set FF. Hence an infinite closed subset in a Polish space has cardinality either 0\aleph_0 or the continuum c=2 0c = 2^{\aleph_0} (continuum hypothesis for closed sets).

Borel isomorphism between Polish spaces

Recall that a function f:XYf: X \to Y between topological spaces is Borel if f 1(V)f^{-1}(V) is a Borel set in XX for every open VV in YY. Topological spaces and Borel functions form a category.

Lemma

If A,BA, B are countably infinite T 1T_1-spaces, then any bijection f:ABf: A \to B is a Borel isomorphism.

For, the inverse image f 1(U)f^{-1}(U) of an open set, being countable, is an F σF_\sigma set. In particular, any two denumerable Polish spaces are Borel isomorphic.

Lemma

The unit interval [0,1][0, 1] and Cantor space 2 2^\mathbb{N} are Borel isomorphic.

Proof

Let E2 E \subseteq \mathbf{2}^\mathbb{N} be the set of (0,1)(0, 1)-sequences that are eventually constant. Then

f(a 1,a 2,)= na n2 nf(a_1, a_2, \ldots) = \sum_n \frac{a_n}{2^n}

maps 2 E2^\mathbb{N} \setminus E homeomorphically onto the space of non-(dyadic rational) numbers in [0,1][0, 1]. Pick any bijection g:E{dyadicrationals}g: E \to \{dyadic\; rationals\}. Then the union of ff and gg defines a Borel isomorphism h:2 [0,1]h: \mathbf{2}^\mathbb{N} \to [0, 1].

Proposition

Any Polish space XX is Borel isomorphic to a Borel subset of Cantor space.

Proof

There is a sequence of maps

X[0,1] h (2 ) j2 X \hookrightarrow [0, 1]^\mathbb{N} \stackrel{h^\mathbb{N}}{\to} (\mathbf{2}^{\mathbb{N}})^\mathbb{N} \stackrel{j}{\to} \mathbf{2}^\mathbb{N}

where the first map is an inclusion of a G δG_\delta set, the second is induced from a Borel isomorphism hh, and the third is a homeomorphism.

Theorem

Any two Polish spaces of the same cardinality are Borel isomorphic.

Proof

It suffices to prove this for uncountable Polish spaces (which have continuum cardinality, as we saw in Corollary ). We show that any such XX is Borel isomorphic to Cantor space. By Proposition , we have an inclusion i:X2 i: X \to \mathbf{2}^{\mathbb{N}} that maps XX Borel isomorphically onto its image, and by Proposition , we have an inclusion j:2 Xj: \mathbf{2}^{\mathbb{N}} \to X that maps Cantor space Borel isomorphically (even homeomorphically) onto its image. The rest is just a matter of checking that at least one of the proofs of the Cantor-Schroeder-Bernstein theorem applies to this Borel context.

Indeed, as explained here, we may consider

S= n0(¬ j¬ i) n(X)S = \bigcap_{n \geq 0} (\neg \exists_j \neg \exists_i)^n(X)

Each of the direct image maps i\exists_i and j\exists_j takes Borel sets to Borel sets, since their inverses i 1,j 1i^{-1}, j^{-1} are Borel functions on their domains. So each of the iterates (¬ j¬ i) n(X)(\neg \exists_j \neg \exists_i)^n(X) is a Borel set and so is their countable intersection SS. The set SS is a fixed point of ¬ j¬ i\neg \exists_j \neg \exists_i and so we construct a Borel isomorphism h:X2 h: X \to \mathbf{2}^{\mathbb{N}} as

x i(x) ifxS j 1(x) ifx¬S\array{ x & \mapsto & i(x) & if\; x \in S \\ & \mapsto & j^{-1}(x) & if\; x \in \neg S }

following the proof of the Cantor-Schroeder-Bernstein theorem.

For another proof, see theorem 3.1.1 of Berberian.

Further examples

  • The classical L pL^p-spaces for p<p \lt \infty are Polish spaces.

  • For a metric space XX, let K(X)K(X) be the set of nonempty compact subsets of XX, equipped with the Hausdorff metric. If XX is a separable complete metric space, then so is K(X)K(X).

  • A locally compact Hausdorff space is Polish iff it is second-countable.

  • If XX is Polish, then the space P(X)P(X) of Borel probability measures μ\mu equipped with the Prokhorov metric is also Polish. Definitions: for xXx \in X and AXA \subset X nonempty, put d(x,A)=inf{d(x,a):aA}d(x, A) = \inf \{d(x, a): a \in A\}, and for α>0\alpha \gt 0 define A α={xX:d(x,A)<α}A_\alpha = \{x \in X: d(x, A) \lt \alpha\} and α=\emptyset_\alpha = \emptyset. Define the Prokhorov metric d Pd_P by

    d P(μ,ν)=inf{α>0: BorelAμ(A)<ν(A α)+αandν(A)<μ(A α)+α}.d_P(\mu, \nu) = \inf \{\alpha \gt 0: \forall_{Borel\; A} \mu(A) \lt \nu(A_\alpha) + \alpha \; and \; \nu(A) \lt \mu(A_\alpha) + \alpha\}.

    Then the map u X:XP(X)u_X: X \to P(X) sending xx to the Dirac measure δ x\delta_x is continuous. If x nx_n is a countable dense set of XX, then the set of rational convex combinations i=1 nq nδ x n\sum_{i=1}^n q_n \delta_{x_n} (with 0q n10 \leq q_n \leq 1 and iq i=1\sum_i q_i = 1) is a countable dense subset of P(X)P(X). More at Giry monad.

  • Spaces of structures and models (in the model theory sense), and spaces of nn-types? (again in the model theory sense), quite often provide examples of Polish spaces. For example, if LL is a countable language (a countable signature), then the collection of possible LL-structures MM on the countable universe \mathbb{N}, topologized by taking as basic opens

    U ϕ={MStruct(L):Mϕ}U_\phi = \{M \in Struct(L): M \models \phi\}

    where ϕ\phi is a quantifier-free sentence, is a Polish space homeomorphic to the product space

    relationsR2 arity(R)× functionsf arity(f)\prod_{relations\; R} \mathbf{2}^{\mathbb{N}^{arity(R)}} \times \prod_{functions\; f} \mathbb{N}^{\mathbb{N}^{arity(f)}}

    (taking constants to be functions of arity 00 in the signature).

As an example of the last principle, we have a kind of continuum hypothesis for substructures:

Proposition

Let XX be a countable structure of a language; then the number of substructures of XX is either countable or the continuum.

Proof

A subset of XX is specified by its characteristic function χ2 X\chi \in 2^X, where 2 X2^X is regarded as a Polish space. In order for the subset not to support a substructure, then must be some function symbol ff of the language, say of arity nn, and elements a 1,,a na_1, \ldots, a_n such that χ(a i)=1\chi(a_i) = 1 for all ii and χ(f X(a 1,,a n))=0\chi(f_X(a_1, \ldots, a_n)) = 0. The basic open

U f;a 1,,a n{ω2 X:ω(a 1)=1,,ω(a n)=1,ω(f X(a 1,,a n))=0}U_{f; a_1, \ldots, a_n} \coloneqq \{\omega \in 2^X: \; \omega(a_1) = 1, \ldots, \omega(a_n) = 1, \; \omega(f_X(a_1, \ldots, a_n)) = 0\}

would thus contain χ\chi and also exclude any substructure; we conclude that the collection of substructures forms a closed subset CC of 2 X2^X. Then the Cantor-Bendixson theorem (i.e., Corollary ) shows that |C|{|C|} is either countable or the continuum.

Polish spaces of mappings

In his study of wild knots, Kulnikov studied the space of embeddings of S 1S^1 in S 3S^3, and implicit in his work, though perhaps not explicitly proved, is the fact that this space is a Polish space. More generally, we can show that the space of embeddings of a compact Polish space XX in a Polish space YY is again a Polish space. For this we first give the space Y XY^X the metric

d(f,g)=sup xXd(f(x),g(x)) d(f,g) = \sup_{x \in X} d(f(x),g(x))

where the supremum is well-defined because XX is compact. The topology coming from this metric is called the ‘’uniform topology’’.

Proposition

If XX is a compact Polish space and YY is a Polish space, then Y XY^X with the above metric is a Polish space, and its uniform topology is the compact-open topology.

Proof

Given any compact metric space XX and metric space YY, the uniform topology on Y XY^X agrees with the compact-open topology. If XX is any locally compact Polish space and YY is a Polish space, Y XY^X with its compact-open topology is a Polish space, by A.10 of Kerr and Li.

We then have:

Proposition

Let XX be a compact Polish space and YY a Polish space. Let Emb(X,Y)Y X\mathrm{Emb}(X,Y) \subseteq Y^X be the space of embeddings of XX in YY, with its subspace topology. Then Emb(X,Y)\mathrm{Emb}(X,Y) is a Polish space.

Proof

Since XX is compact and YY is Hausdorff, a map f:XYf : X \to Y is an embedding, i.e. ff is a homeomorphism from XX to its image, if and only if ff is injective. Thus we have

Emb(X,Y)={fY X:x,yXf(x)f(y)} \mathrm{Emb}(X,Y) = \{ f \in Y^X: \forall x, y \in X \; f(x) \ne f(y) \}

To show Emb(X,Y)\mathrm{Emb}(X,Y) is a Polish space it is sufficient to show that is a G δG_\delta, i.e. a countable intersection of open subsets of Y XY^X (Marker, Theorem 1.33). For this we consider the sets

O m,n={fY X|x,yXd(x,y)1/md(f(x),d(y))>1/n} O_{m,n} = \{ f \in Y^X \; \vert \; \forall x, y \in X \; d(x,y) \ge 1/m \implies d(f(x),d(y)) \gt 1/n \}

If each set O m,nO_{m,n} is open then Emb(X,Y)\mathrm{Emb}(X,Y) is a G δG_\delta, since

Emb(X,Y)= m=1 n=1 O m,n \mathrm{Emb}(X,Y) = \bigcap_{m = 1}^\infty \bigcup_{n = 1}^\infty O_{m,n}

To complete the proof we show O m,nO_{m,n} is open. Note that for each m,nm,n and x,yXx,y \in X the set

O m,n,x,y={fY X|d(x,y)1/md(f(x),d(y))>1/n} O_{m,n,x,y} = \{ f \in Y^X \; \vert \; d(x,y) \ge 1/m \implies d(f(x),d(y)) \gt 1/n \}

is open. Then we use a lemma: if KK is a compact topological space and AA is an arbitrary topological space, and UK×AU \subseteq K \times A is open, then

{aA|kK(k,a)U} \{ a \in A \; \vert \; \forall k \in K \;\; (k,a) \in U \}

is open. Thanks to this and the fact that K 2K^2 is compact, the set

O m,n={fC(X,Y)|(x,y)X 2fO m,n,x,y} O_{m,n} = \{ f \in C(X,Y) \; \vert \; \forall (x,y) \in X^2 \;\; f \in O_{m,n,x,y} \}

is indeed open.

References

  • David Kerr, Hanfeng Li, Ergodic Theory. Springer Monographs in Mathematics (2016). doi:10.1007/978-3-319-49847-8
  • Polish spaces, blog discussion, nn-Category Café, 2008.

  • David Marker, Descriptive Set Theory, UIC Course Notes (Fall 2002) (pdf)

  • S.K. Berberian, Borel spaces, (pdf)
  • A.S. Kechris, Classical descriptive set theory, Springer-Verlag (1994).

Internal groupoids in Polish spaces are considered in

  • Martino Lupini, Polish groupoids and functorial complexity. (arxiv)

The space of embeddings of S 1S^1 in S 3S^3 was studied using Polish spaces in

  • Vadim Kulikov, A non-classification result for wild knots, Transactions AMS, 369 (2017), 5829-5853. (arXiv).

Last revised on May 28, 2024 at 21:39:32. See the history of this page for a list of all contributions to it.