THE VERY BASICS of DESCRIPTIVE SET THEORY 1. Standard Borel Spaces Definition 1.1. (A) a Metric Space Px, Dq Is Called Polish If

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THE VERY BASICS OF DESCRIPTIVE SET THEORY

PHILIPP SCHLICHT

Abstract. This note presents elementary proofs that any two uncountable Polish metric spaces are Borel isomorphic, a function is Borel measurable if and only if its graph is a Borel set, and the auxiliary result that disjoint analytic sets can be separated by Borel sets. The proofs are much shorter than those in the literature.

1. Standard Borel spaces Deﬁnition 1.1. (a) A metric space pX, dq is called Polish if it is countably based and complete. (b) A topological space pX, τq is called Polish if there is a Polish metric d on X that induces τ. (c) A pair pX, σq is called a standard Borel space if σ is the Borel σ-algebra generated by a Polish topology on X. Our notation is pX, dq, pY, dq, pZ, dq for Polish metric spaces and the metric d is often omitted. We further write dpAq for the diameter supx,yPA dpx, yq of a subset A of X, and Bpxq for the open ball of radius around x.

Example 1.2. Rn, Baire space NN, Cantor space 2N, compact metric spaces, countably based locally compact metric spaces.

N N 1 Here N caries the product topology. The standard metric on N is deﬁned by dpx, yq “ 2n for the least n with xpnq ‰ ypnq, where x ‰ y. The basic open subsets of NN (clopen balls) are ă denoted Nt “ tx P NN | t Ď xu for t P N N. The following combinatorial presentation of NN via trees is useful for constructions of continuous maps. A tree on N is a subset of NăN that is closed under initial segments. It is easy to see that the closed subsets of NN are exactly the sets of the form rT s “ tx P NN | @n xæn P T u for trees T on N. Let |t| denote the length of a sequence t P NăN. Let xæn denote the restriction of a sequence x P NN to t0, . . . , n ´ 1u. A tree is called pruned if it has no end nodes. Let LevnpT q denote the n-th level of a tree T , so the elements of LevnpT q are sequences of length n.

Note that Polish spaces are closed under countable products. To see this, take a sequence ~ X “ xXn | n P Ny of Polish spaces and assume that each metric d is bounded by replacing d with maxtd, 1u. Then form the weighted sum over all coordinates with factors 2´n. It is easy to show that this metric on the product is complete and induces the product topology.

We next show that any two uncountable Polish spaces are Borel isomorphic in the sense of a bijection that preserves Borel sets both ways. Proposition 1.3. Any two uncountable Polish spaces are Borel isomorphic. This will follow from the next two lemmas. To state them, we ﬁx some notation. An -cover ~ of a subset A of X is a sequence A “ xAi | i ă Ny with N P N Y tNu whose union equals A with dpAiq ă and Ai Ď A for all i P N. An Fσ-set is a countable union of closed sets. Note that all open sets are Fσ. Lemma 1.4. Let ą 0. (1) Any open set has an -cover consisting of open sets. (2) Any Fσ-set has a disjoint -cover consisting of Fσ-sets.

Date: November 15, 2019. 1 2 PHILIPP SCHLICHT

Proof. (1) Suppose that A is an open subset of X. Fix a countable dense subset Q of X and let ~ A “ xAi | i P Ny list all open balls Bqpxq with x P Q, q P p0, 2 q X Q and Bqpxq Ď A. It suffices to show that A “ iPN Ai. To see this, take any x P A and find some q P Q with B2qpxq Ď A. Since Q is dense, there is some y P Bqpxq X Q. We then also have x P Bqpyq and thus it suffices to show Ť that Bqpyq Ď A. This holds since Bqpyq Ď B2qpxq Ď A. (2) Suppose that A is an Fσ subset of X. Take a sequence A~ “ xAi | i P Ny whose union equals A with Ai closed. Then A is the disjoint union of the sets Bi “ AizpA0 Y ¨ ¨ ¨ Y Ai´1q. Since Bi is of the form C X D with C open and D closed, it suffices to prove the claim for sets of this form. To see this, note that by (a), there is an -cover C~ “ xCi | i P Ky of C consisting of open sets. Let ~ ~ D “ xDi | i P Ky with Di “ pCizpC0 Y ¨ ¨ ¨ Y Ci´1qq X D. Then D is a disjoint -cover of C X D consisting of Fσ-sets. The existence of Borel isomorphisms between uncountable Polish spaces follows from the second part of the next lemma by noting that countable sets can be discarded. We call a subset of X co-countable if its complement is countable. Lemma 1.5. 1 (1) There is a closed subset A of NN and a continuous Borel isomorphism f : A Ñ X. (2) If X is uncountable, then there is a continuous Borel isomorphism f : NN Ñ X onto a co-countable subset of X.

` Proof. (1) Let ~ “ xi | i P Ny be a sequence in R converging to 0. By iterative applications ă ~ of Lemma 1.4, we obtain a tree T Ď N N without end nodes and a family A “ xAt | t P T y of nonempty Fσ-subsets of X with A0 “ X that satisﬁes the following properties for all t P T : (a)

At “ tiPT Ati is a disjoint union, (b) Ati Ď At for ti P T , and (c) dpAtq ă n if |t| “ n. We then deﬁne f : rT s Ñ X by choosing fpxq as the unique element of Ax n. The latter nPN æ is nonemptyŤ by (b), (c) and completeness of X, and has a unique element by (c). Ş Claim. f is bijective. Proof. It follows from (a) that f is injective. To see that f is surjective, take any x P X. By (a), there is for each n P N a unique tn P T of length n with x P Atn . By (b), ti Ď tj for i ď j. Hence fpyq “ x for y “ ti. iPN Claim. f is continuous.Ť

Proof. If U is open and fpxq P U, then Axæn Ď U for suﬃciently large n by (c). Claim. f is a Borel isomorphism. Proof. Since f is continuous, it suﬃces that the images of Borel sets are Borel. This holds since f maps each basic open set Nt onto At and f is bijective. Thus f is as required.

(2) A slight modiﬁcation of the construction in (a) works. We will ensure that At is uncountable for all t P T . The change is that we remove a set Ati if it is countable. Note that one can easily ensure that Ati is uncountable for inﬁnitely many i P N by partitioning the sets further. Therefore one can assume that ti P T for all i P N and thus T “ NăN. During the construction only a countable set is removed in total, yielding the claim. Note that in (2), for any X with isolated points, one cannot replace the co-countable set by X.

2. Analytic sets Let p: X ˆ Y Ñ X denote the ﬁrst projection. Deﬁnition 2.1. A subset A of X is called analytic if A “ ppBq for some Borel subset B of X ˆ Y for some standard Borel space Y .

1In fact, any continuous bijection is a Borel isomorphism, but we don’t prove this here. BASICS OF DESCRIPTIVE SET THEORY 3

Clearly this class contains all Borel sets. By Proposition 1.3, one can take Y to be a fixed uncountable Polish space. Moreover, the analytic subsets of an uncountable standard Borel space X are precisely the images of analytic subsets of NN under Borel isomorphisms. Hence we can work in NN to study their properties. To do this, we will go through a slight detour. We call a subset A of NN ˚analytic if A “ prCs for some closed subset C of NN ˆ NN, equivalently A “ prT s “ tx | Dy px, yq P rT su for some subtree T of pN ˆ NqăN. Lemma 2.2. (1) The class of ˚analytic sets is closed under countable unions and intersections. (2) Every analytic subset of NN is ˚analytic. Proof. (1) Closure under countable unions is easily seen by forming unions of disjointified trees. Closure under countable intersections is proved by merging trees. We do only the case of two ă trees, since the general case is similar. So suppose that T0 and T1 are subtrees of pN ˆ Nq N. We identify elements of pN ˆ NqăN with pairs of sequences. For t P NăN, define t0, t1 P NăN by letting t0piq “ tp2iq if 2i ă k and t1piq “ tp2i ` 1q if 2i ` 1 ă k. Let U be the subtree of pN ˆ NqăN k 0 1 that consists of all ps, tq P pN ˆ Nq with ps, t q P S and ps, t q P T1. One can now check that prSs X prT s “ prUs. (2) The collection of subsets generated from basic open sets by countable unions and intersections forms a σ-algebra and thus equals the Borel σ-algebra. Hence all Borel sets are ˚analytic. The claim follows by composing two projections.

The previous result also holds for ﬁnite products of NN, by observing that the proof still works or noting that they are homeomorphic to NN. In the next proof, subsets A, B of X are called Borel separated if there is a Borel set C with A Ď C and B Ď XzC.

Proposition 2.3. Any two disjoint analytic subsets A, B of X are Borel separated.

Proof. It suﬃces to show this for X “ NN, and by Problem 2.2 A and B are then *analytic. It follows that there are continuous surjections f : rSs Ñ A and g : rT s Ñ B, where S and T are pruned subtrees of NăN. It is an easy to see that there is a continuous surjection from NN onto rSs, in fact one that ﬁxes all elements of rSs. We can thus assume that rSs “ rT s “ NN. Suppose that A, B are not Borel separated. We then construct increasing sequences ~s “ xsi | ~ ăN i P Ny and t “ xti | i P Ny in N such that for all i P N, fpNsi q and gpNti q are not Borel separated as follows.

Let s0 “ t0 “H. Given si and ti, there are k, m P N such that fpNsikq and gpNtimq are not Borel separated. If each of these pairs were Borel separated by some B , then C B k,m k “ timPT k,m

separates fpNsikq and gpNti q for each k P N, and hence C “ Ck separates fpNsi q and sikPS Ş gpNt q, contradicting our assumption. Let si 1 “ sik and ti 1 “ tim. i ` ` Ť Let x “ si P rSs and y “ ti P rT s. Since A and B are disjoint, fpxq ‰ gpyq. Let iPN iPN further U, V be disjoint open subsets of X with fpxq P U and gpyq P V . By continuity of f and Ť Ť g, fpNxænq Ď U and gpNyænq Ď V for suﬃciently large n P N, so fpNxænq and gpNyænq are Borel separated. However, they are not Borel separated by the construction.

Next is a very useful characterization of analytic sets.

Lemma 2.4. Every nonempty analytic subset of X is the range of some continuous f : NN Ñ X.

Proof. Take an analytic subset A of X. By Lemma 1.5, there are a closed subset Y of NN and a continuous Borel isomorphism f : Y Ñ X. It suﬃces to ﬁnd a continuous bijection g : NN Ñ f ´1pAq. Then A equals the range of fg : NN Ñ X. Since analytic sets are preserved under Borel isomorphisms, f ´1pAq is an analytic subset of Y and thus of NN. Let B be a Borel subset of NN ˆ NN with f ´1pAq “ ppBq. Since Borel subsets of NN ˆ NN are *analytic by Problem 2.2 (1), there is a closed subset C of pNNq3 with f ´1pAq “ ppCq. Since pNNq3 and NN are homeomorphic, it is easy to see that there is a continuous surjection h: NN Ñ C. Then g “ fph is as required. 4 PHILIPP SCHLICHT

3. Borel measurable functions A function f : X Ñ Y between standard Borel spaces is called Borel measurable if f-preimages of Borel sets are Borel. Lemma 3.1. The following are equivalent for a function f : X Ñ Y : (a) f is Borel measurable. (b) The graph Gpfq of f is a Borel subset of X ˆ Y . ´1 ´1 Proof. (a) ñ (b): fpxq “ y ðñ x P f pNy kq ðñ px, yq P k f pNtq ˆ Nt. kPN æ kPN tPN (b) ñ (a): It suﬃces to show that f ´1pUq is Borel for any open subset of Y . By Proposition 2.3, it is suﬃcient to show that both Şf ´1pUq and its complement inŞX areŤ analytic. These claims follows from the equivalences x P f ´1pUq ðñ Dy P Y px, yq P Gpfq ðñ @y P Y ppx, yq P Gpfq ñ y P Uq.

Philipp Schlicht, School of Mathematics, University of Bristol, Fry Building, Woodland Road, Bristol, BS8 1UG, UK Email address: [email protected]