# Polish spaces and the Effros Borel structure *Mathematical exposition by Claude Opus 5.5 (Anthropic), self-checked by the writing AI. CC0 1.0. Independent review of this edition is not asserted.* This chapter develops Polish subspaces, Borel refinements, and the standard Borel structures of closed sets and closed linear subspaces. ## Conventions and prerequisites - \(\mathbb K\) is \(\mathbb R\) or \(\mathbb C\), and Banach spaces are over \(\mathbb K\). Hilbert spaces are complex, inner products are linear in the first variable, and the zero space is allowed. "Separable" includes finite-dimensional. - For a Banach space \(E\), \(E^*\) is its dual with the weak\* topology, \(E^*_1\) is the closed unit ball of \(E^*\), and \(d(x,A)=\inf\{\|x-a\|:a\in A\}\). For \(S\subseteq E\), \(S^\perp=\{f\in E^*:f=0\text{ on }S\}\). For \(T\subseteq E^*\), \(T_\perp=\{x\in E:f(x)=0\text{ for all }f\in T\}\). - \((\Gamma,\Sigma)\) is a measurable space, and *measurable* means \(\Sigma\)-measurable. A map into a topological space is measurable if the preimage of every Borel set lies in \(\Sigma\). No standardness of \(\Gamma\) is needed, and no measure is needed either. - **(D1)** (*Hahn–Banach theorem*) Let \(q\) be a seminorm on a vector space \(W\) over \(\mathbb K\), \(V\subseteq W\) a subspace and \(g:V\to\mathbb K\) linear with \(|g|\leq q\) on \(V\). Then \(g\) extends to a linear functional on \(W\) with \(|g|\leq q\). Over \(\mathbb R\) it is enough that \(g\leq q\), and the extension satisfies \(g\leq q\). Proved in [Hahn–Banach, Baire and the basic theorems on Banach spaces, Section 2](course:foundations-of-von-neumann-algebras/hahn-banach-baire-and-the-basic-theorems-on-banach-spaces#OA-FND-HB-02). - **(D2)** (*Annihilators*) For \(S\subseteq E\), \((S^\perp)_\perp\) is the norm-closed linear span of \(S\). For \(T\subseteq E^*\), \((T_\perp)^\perp\) is the weak\*-closed linear span of \(T\). Proved in [Weak topologies, Tychonoff, Banach–Alaoglu, Mazur, bipolars, Krein–Milman and Eberlein–Šmulian, Section 5](course:foundations-of-von-neumann-algebras/weak-topologies-tychonoff-banach-alaoglu-mazur-bipolars-krein-milman-and-eberlein-smulian#OA-FND-WT-05). - **(D3)** (*Banach–Alaoglu theorem*) \(E^*_1\) is weak\*-compact. Proved in Section 3 of the same lesson. ## 1. Borel spaces and Polish spaces A reference for this section and the next is [Tserunyan, Parts 1 and 3]. **Definitions 1.1.** - A *Borel space* consists of a set \(X\) and a \(\sigma\)-algebra \(\mathcal B\) of subsets of \(X\), whose members are called *Borel sets*. A subset \(Y\subseteq X\) carries the *relative* Borel structure \(\{B\cap Y:B\in\mathcal B\}\). - A map \(f:X\to Y\) of Borel spaces is *Borel* if \(f^{-1}(B)\) is Borel for every Borel \(B\subseteq Y\). A *Borel isomorphism* is a bijection \(f\) such that \(f\) and \(f^{-1}\) are Borel. - \(\sigma(\mathcal S)\) is the \(\sigma\)-algebra generated by a family \(\mathcal S\) of sets. For maps \(f_i:X\to Y_i\) into Borel spaces, \(\sigma(f_i:i\in I)\) is the smallest \(\sigma\)-algebra that makes every \(f_i\) Borel. A family of Borel sets is *generating* if it generates \(\mathcal B\), and *separating* if for any two distinct points some member contains one of them and not the other. \(X\) is *countably generated* or *countably separated* if it has a countable family of that kind. A stronger condition asks for a single countable family that both generates and separates. [Proposition 3.2(5)](#oa-fnd-ef-03) supplies such a family, so its conclusion holds in either sense. - The Borel structure of a topological space is the \(\sigma\)-algebra generated by its open sets. - A topological space is *Polish* when its topology comes from a complete metric and it has a countable dense subset. A Borel space is *standard* when some Polish space, with its Borel sets, is Borel isomorphic to it. - A *measure* on \((X,\mathcal B)\) is a countably additive map \(\mu:\mathcal B\to[0,\infty]\). It is *standard* if \(X\setminus N\), with its relative Borel structure, is standard for some Borel set \(N\) with \(\mu(N)=0\). **Example 1.2.** A Borel map need not carry Borel sets to Borel sets. The identity map from \(\mathbb R\) with its Borel sets to \(\mathbb R\) with the \(\sigma\)-algebra of countable and co-countable sets is Borel, since every set of the second kind is Borel. It carries the Borel set \([0,1]\) to a set that is neither countable nor co-countable. **Lemma 1.3.** 1. (*Testing on generators*) Suppose \(\mathcal B=\sigma(f_i:i\in I)\). A map \(g:\Gamma\to X\) is measurable if and only if every \(f_i\circ g\) is measurable. 2. (*Countable bases*) If a topological space has a countable base \(\mathcal U\), every open set is a countable union of members of \(\mathcal U\), so the Borel sets are \(\sigma(\mathcal U)\). The Borel structure of a product of countably many second countable spaces is generated by the coordinate maps. In particular, for a countable set \(D\), the Borel structure of \(\mathbb R^D\) is generated by the coordinates. 3. (*A stock of Polish spaces*) Closed subsets and countable products of Polish spaces are Polish, and so is the disjoint union of two Polish spaces, each piece open in the union. The space \(\mathbb N^{\mathbb N}\) is Polish. A separable metrizable space, and hence every standard Borel space, has at most \(2^{\aleph_0}\) points. 4. (*Pieces*) Let \(\Gamma\) be the union of countably many sets \(\Gamma_d\in\Sigma\). A map \(g\) on \(\Gamma\) whose restriction to every \(\Gamma_d\) is measurable for the relative \(\sigma\)-algebra is measurable. **Proof.** (1) The sets \(B\subseteq X\) with \(g^{-1}(B)\in\Sigma\) form a \(\sigma\)-algebra. If every \(f_i\circ g\) is measurable, it contains every \(f_i^{-1}(C)\) with \(C\) Borel, so it contains \(\mathcal B\). The converse is clear. (2) Let \(V\) be open. For each \(x\in V\) pick \(U_x\in\mathcal U\) with \(x\in U_x\subseteq V\). Only countably many distinct sets \(U_x\) occur, and their union is \(V\). For a product \(\prod_kX_k\) with countable bases \(\mathcal U_k\), the finite intersections of sets \(\pi_k^{-1}(U)\), \(U\in\mathcal U_k\), form a countable base. Each of them lies in \(\sigma(\pi_k:k)\), and the coordinates are continuous, so this \(\sigma\)-algebra is the Borel one. (3) A closed subset of a complete metric space is complete, and a subset of a separable metric space is separable. For Polish spaces \(X_k\), choose complete compatible metrics \(d_k\leq1\) (replace \(d_k\) by \(\min(d_k,1)\)). Then \(\rho(x,y)=\sum_k2^{-k}d_k(x_k,y_k)\) is a metric for the product topology. A \(\rho\)-Cauchy sequence is Cauchy in every coordinate, so it converges in every coordinate, and then in \(\rho\), because the tails of the series are uniformly small. The points that agree with a fixed point outside finitely many coordinates, and lie in fixed countable dense sets inside them, form a countable dense set. For two Polish spaces, use their metrics bounded by \(1\) inside each piece and distance \(1\) across. \(\mathbb N^{\mathbb N}\) is a countable product of copies of the complete separable discrete space \(\mathbb N\). Finally, if \(Q\) is a countable dense set in a separable metrizable space, sending each point to a sequence in \(Q\) that converges to it is injective, and there are at most \(|Q^{\mathbb N}|\leq2^{\aleph_0}\) such sequences. (4) For Borel \(B\), \(g^{-1}(B)=\bigcup_d(g|_{\Gamma_d})^{-1}(B)\), and each term is a subset of \(\Gamma_d\) lying in the relative \(\sigma\)-algebra, hence in \(\Sigma\). \(\square\) ## 2. Polish subspaces and Borel subsets **Theorem 2.1** (*Polish subspaces*). Let \(X\) be a Polish space. 1. Every \(G_\delta\) subset of \(X\) is Polish in the relative topology. 2. Conversely, a subset of \(X\) that is Polish in the relative topology is a \(G_\delta\) subset of \(X\). 3. Every Polish space is homeomorphic to some \(G_\delta\) subset of \([0,1]^{\mathbb N}\), and every such subset is Polish. **Proof.** Fix a complete compatible metric \(d\) on \(X\). (1) Let \(Y=\bigcap_nG_n\) with \(G_n\) open. Drop every \(G_n\) equal to \(X\); if none is left, \(Y=X\). For \(y\in Y\) put \(h_n(y)=1/d(y,X\setminus G_n)\), a positive continuous function on \(G_n\), and \[ d_Y(y,y')=d(y,y')+\sum_n2^{-n}\min\big(1,|h_n(y)-h_n(y')|\big). \] This is a metric on \(Y\), and it defines the relative topology: \(d_Y\geq d\), and if \(d(y_k,y)\to0\) inside \(Y\), then \(h_n(y_k)\to h_n(y)\) for each \(n\), so the series tends to \(0\) term by term under the bounds \(2^{-n}\). Let \((y_k)\) be \(d_Y\)-Cauchy. It is \(d\)-Cauchy, so it converges in \(X\) to some \(x\). For each \(n\), \((h_n(y_k))_k\) is Cauchy, hence bounded by some \(c_n\). Then \(d(y_k,X\setminus G_n)\geq1/c_n\) for all \(k\), so \(d(x,X\setminus G_n)\geq1/c_n>0\) and \(x\in G_n\). Hence \(x\in Y\) and \(d_Y(y_k,x)\to0\). Finally \(Y\) is separable, being a subset of a separable metric space. (2) Let \(\rho\) be a complete metric on \(Y\) compatible with its relative topology. For \(n\geq1\), let \(Y_n\) be the set of points \(x\) of the closure \(\overline Y\) that have an open neighbourhood \(U\) in \(X\) with \(\rho\)-diameter of \(U\cap Y\) at most \(1/n\). Each \(Y_n\) is relatively open in \(\overline Y\), because a witness \(U\) for \(x\) is a witness for every point of \(U\cap\overline Y\). Also \(Y\subseteq Y_n\): for \(y\in Y\), the ball \(\{\rho(\cdot,y)<1/(3n)\}\) is relatively open in \(Y\), so it equals \(U\cap Y\) for some open \(U\ni y\) in \(X\). Conversely, let \(x\in\bigcap_nY_n\), with witnesses \(U_n\). The open sets \(W_n=U_1\cap\cdots\cap U_n\cap\{d(\cdot,x)<1/n\}\) contain \(x\in\overline Y\), so we can pick \(y_n\in W_n\cap Y\). Then \(y_n\to x\) in \(X\), and \(\rho(y_m,y_n)\leq1/n\) for \(m\geq n\), since both lie in \(U_n\cap Y\). So \((y_n)\) converges in \((Y,\rho)\) to some \(y\in Y\), hence to \(y\) in \(X\), and \(y=x\). Thus \(Y=\bigcap_nY_n\). Write \(Y_n=G_n\cap\overline Y\) with \(G_n\) open in \(X\). Since \(\overline Y=\bigcap_m\{d(\cdot,\overline Y)<1/m\}\) is a \(G_\delta\), so is \(Y\). (3) \([0,1]^{\mathbb N}\) is a compact metrizable space, hence Polish, so its \(G_\delta\) subsets are Polish by (1). Conversely, let \(X\) be Polish, with compatible metric \(d\) and dense sequence \((a_n)\), and put \(\varphi(x)=(\min(1,d(a_n,x)))_n\). The map \(\varphi\) is continuous. It is injective: for \(x\neq y\) put \(\delta=\min(1,d(x,y))\) and choose \(a_n\) with \(d(a_n,x)<\delta/3\); then the \(n\)-th coordinate of \(\varphi(x)\) is below \(\delta/3\) and that of \(\varphi(y)\) is at least \(2\delta/3\). Its inverse is continuous on \(\varphi(X)\): if \(\varphi(x_k)\to\varphi(x)\) and \(0<\varepsilon<1/2\), choose \(a_n\) with \(d(a_n,x)<\varepsilon\); then eventually \(d(a_n,x_k)<\varepsilon\), so \(d(x_k,x)<2\varepsilon\). So \(\varphi(X)\) is homeomorphic to \(X\), hence Polish, hence a \(G_\delta\) subset of \([0,1]^{\mathbb N}\) by (2). \(\square\) **Theorem 2.2** (*Borel subsets of standard spaces*). Let \((X,\tau)\) be a Polish space and \(B\subseteq X\) a Borel set. There is a Polish topology \(\tau'\supseteq\tau\) on \(X\), with the same Borel sets as \(\tau\), in which \(B\) is open and closed. Consequently \(B\), with its relative Borel structure, is a standard Borel space. The same holds for a Borel subset of any standard Borel space. The proof uses only part (1) of Theorem 2.1. **Proof.** Call a topology \(\tau'\) on \(X\) *admissible* if it is Polish, contains \(\tau\), and has the same Borel sets as \(\tau\). *(a) Open sets.* Let \(U\) be \(\tau\)-open, and let \(\tau_U\) be the topology generated by \(\tau\) together with the set \(X\setminus U\). Its open sets are the unions of sets \(V\) and \(V\cap(X\setminus U)\) with \(V\in\tau\). So \((X,\tau_U)\) is the disjoint union of the open subspace \(U\) and the closed subspace \(X\setminus U\) of \((X,\tau)\), each open in \(\tau_U\). Both pieces are Polish: \(U\) by Theorem 2.1(1), since an open set is a \(G_\delta\), and \(X\setminus U\) because closed subsets of Polish spaces are Polish ([Lemma 1.3(3)](#oa-fnd-ef-01)). By the same lemma, their disjoint union is Polish. The new open sets are \(\tau\)-Borel. So \(\tau_U\) is admissible, and \(U\) is \(\tau_U\)-clopen. *(b) Countable joins.* Let \(\tau_1,\tau_2,\ldots\) be admissible, and let \(\tau_\infty\) be the topology generated by their union. The diagonal map \(x\mapsto(x,x,\ldots)\) from \((X,\tau_\infty)\) to the Polish space \(\prod_n(X,\tau_n)\) is a homeomorphism onto the diagonal \(\Delta\), since the preimage of \(\pi_n^{-1}(V)\) is \(V\) for \(V\in\tau_n\). The diagonal is closed: if \(y\notin\Delta\), then \(y_m\neq y_n\) for some \(m,n\); since \(\tau\) is Hausdorff, there are disjoint \(\tau\)-open sets \(A\ni y_m\) and \(C\ni y_n\), and \(\pi_m^{-1}(A)\cap\pi_n^{-1}(C)\) is a neighbourhood of \(y\) that misses \(\Delta\). So \(\tau_\infty\) is Polish, because a closed subset of a Polish space is Polish. It has a countable base consisting of finite intersections of members of countable bases of the \(\tau_n\). These are \(\tau\)-Borel, and a countable base generates the Borel sets ([Lemma 1.3(2)](#oa-fnd-ef-01)). So \(\tau_\infty\) has the same Borel sets as \(\tau\), and it is admissible. *(c) All Borel sets.* Let \(\mathcal A\) be the family of sets that are clopen for some admissible topology. It contains \(\tau\) by (a), and it is closed under complements. If \(B_n\in\mathcal A\) is clopen for an admissible \(\tau_n\), then \(\bigcup_nB_n\) is open for the \(\tau_\infty\) of (b); applying (a) to \((X,\tau_\infty)\) makes it clopen for a topology that is admissible. So \(\mathcal A\) is a \(\sigma\)-algebra containing \(\tau\), and it contains every Borel set. Now let \(B\) be clopen for an admissible \(\tau'\). Then \(B\) is \(\tau'\)-closed, hence Polish in the relative \(\tau'\)-topology ([Lemma 1.3(3)](#oa-fnd-ef-01)). The Borel sets of a subspace are the traces of the Borel sets of the whole space: the traces form a \(\sigma\)-algebra on \(B\) containing the relatively open sets, and the sets \(C\) whose trace is Borel in \(B\) form a \(\sigma\)-algebra containing the open sets. The \(\tau'\)-Borel sets are the \(\tau\)-Borel sets, so the relative Borel structure of \(B\) is the Borel structure of a Polish space. For a standard Borel space, transport \(B\) to a Polish space by a Borel isomorphism. \(\square\) ## 3. The Effros Borel structure Throughout, \(E\) is a separable Banach space, and \(\mathfrak W(E^*)\) is the set of weak\*-closed linear subspaces of \(E^*\). For \(F\in\mathfrak W(E^*)\) put \(F_1=F\cap E^*_1\) and \[ p_F(x)=\sup\{|f(x)|:f\in F_1\},\qquad x\in E. \tag{3.1} \] This is the norm of \(x\) as a functional on \(F\). **Definition 3.1.** The *Effros Borel structure* on \(\mathfrak W(E^*)\) is \(\sigma(F\mapsto p_F(x):x\in E)\). **Proposition 3.2.** Let \(F\in\mathfrak W(E^*)\), and let \(D\subseteq E\) be dense. 1. \(p_F\) is a seminorm with \(p_F\leq\|\cdot\|\). Hence \(|p_F(x)-p_F(y)|\leq\|x-y\|\). Its kernel is \(F_\perp\). 2. (*Distance formula*) \(p_F(x)=d(x,F_\perp)\) for every \(x\in E\). 3. (*Domination*) A linear functional \(f\) on \(E\), not assumed continuous, lies in \(F_1\) if and only if \(|f(x)|\leq p_F(x)\) for every \(x\in E\). For \(\mathbb K=\mathbb R\) this is the same as \(f\leq p_F\). 4. (*Subspaces of \(E\)*) \(F\mapsto F_\perp\) is a bijection of \(\mathfrak W(E^*)\) onto the set \(\mathcal S(E)\) of closed linear subspaces of \(E\), with inverse \(N\mapsto N^\perp\). It carries \(p_F\) to \(d(\cdot,F_\perp)\). 5. \(F\) is determined by the values \(p_F(x)\), \(x\in D\), and the Effros structure equals \(\sigma(F\mapsto p_F(x):x\in D)\). If \(D\) is countable, the sets \(\{F:p_F(x) (W) for every \(x\in D\) and every rational \(\varepsilon>0\) there is \(y\in D\) with \(c(y)<\varepsilon\) and \(\|x-y\|\|x-y\|-\varepsilon\}\) is open in \(\mathbb R^D\). So the set of \(c\in S\) that satisfy (W) is a \(G_\delta\) subset of \(\mathbb R^D\). *Claim: for \(c=p|_D\in S\), (W) holds if and only if \(p(x)=d(x,\ker p)\) for all \(x\in E\).* First, \(p(x)=p(x-z)\leq\|x-z\|\) for \(z\in\ker p\), so \(p\leq d(\cdot,\ker p)\) always. If \(p=d(\cdot,\ker p)\), let \(x\in D\) and \(\varepsilon>0\). Pick \(z\in\ker p\) with \(\|x-z\|0\) be rational. Applying (W) first to \(x\) and \(\varepsilon/2\), and then to each \(y_j\) and \(2^{-j-1}\varepsilon\), choose \(y_1,y_2,\ldots\in D\) with \[p(y_1)<\varepsilon/2,\quad\|x-y_1\|0\), a limit point \(z\) with \(f(z)=0\) and \(d(x,z)