Lecture Notes on Descriptive Set Theory Jan Reimann∗ Department of Mathematics Pennsylvania State University Notation U"(x) Ball of radius " about x U Topological closure of U 2<N Set of finite binary strings σ, τ, . Finite binary strings 2N Cantor space, set of all infinite binary sequences NN Baire space, set of all infinite sequences of natural numbers α n Length n initial segment of sequence α, α(0) ... α(n 1) j − Nσ Open cylinder defined by σ LipL(X ) Set of L-Lipschitz functions on X diam(X ) Diameter of a set X in a metric space, diam(X ) = sup d(x, y): x, y X f 2 g ∗[email protected] Lecture 1: Perfect Subsets of the Real Line Descriptive set theory nowadays is understood as the study of definable subsets of Polish Spaces. Many of its problems and techniques arose out of efforts to answer basic questions about the real numbers. A prominent example is the Continuum Hypothesis (CH): If A R is uncountable, does there exist a bijection between A and ⊆ R? That is, is every uncountable subset of R of the same cardinality as R? [Cantor, 1890’s] Early approaches to this problem tried to show that CH holds for a number of sets with an easy topological structure. It is a standard exercise of analysis to show that every open set satisfies CH. (An open set contains an interval, which maps bijectively to R.) For closed sets, the situation is less clear. Given a set A R, we call x R an accumulation point of A if ⊆ 2 ε > 0 z A [z = x & z U"(x)], 8 9 2 6 2 1 where U"(x) denotes the standard "-neighborhood of x in R . Call a non-empty set P R perfect if it is closed and every point of P is an accumulation point. In other⊆ words, a perfect set is a closed set that has no isolated points. It is not hard to see that for a perfect set P, every neighborhood of a point p P contains infinitely many points. 2 Obviously, R itself is perfect, as is any closed interval in R. There are totally disconnected perfect sets, such as the middle-third Cantor set in [0, 1] Theorem 1.1: A perfect subset of R has the same cardinality as R. Proof. Let P R be perfect. We construct an injection from the set 2N of all ⊆ infinite binary sequences into P. An infinite binary sequence ξ = ξ0ξ1ξ2 ... can be identified with a real number [0, 1] via the mapping 2 X i 1 ξ ξi2− − . 7! i 0 ≥ Note that this mapping is onto. Hence the cardinality of P is at least as large as the cardinality of [0, 1]. The Cantor-Schröder-Bernstein Theorem (for a proof 0 see e.g. (author?) [Jec03]) implies that P = 2@ . j j 1There are some divergences in terminology. Some authors call an accumulation point a limit point. We reserve the latter term for any point that is the limit of a sequence of points from a given set. Hence every member of a set is a limit point of that set. In particular, isolated members of a set are limit points. 1 – 1 Choose x P, and let 1 20. Since P is perfect, P U x . Let x x "0 = = "0 ( ) 0 = 1 be two points2 in P U x , distinct from x. Let be\ such that 6 1 2, "0 ( ) "1 "1 = U x , U x U\ x , and U x U x , where U denotes≤ the "1 ( 0) "1 ( 1) "0 ( ) "1 ( 0) "1 ( 1) = closure of U. ⊆ \ ; We can iterate this procedure recursively with smaller and smaller diameters, using the fact that P is perfect. This gives rise to a so-called Cantor scheme, a family of open balls (Uσ). Here the index σ is a finite binary sequence, also called a string. The scheme has the following properties. σ C1) diam(Uσ) 2−| j, where σ denotes the length of σ. ≤ j j C2) If τ is a proper extension of σ, then Uτ Uσ. ⊂ C3) If τ and σ are incompatible (i.e. neither extends the other), then Uτ Uσ = . \ ; C4) The center of each Uσ, call it xσ, is in P. U U0 U1 U00 U01 U10 U11 Figure 1: Cantor Scheme Let ξ be an infinite binary sequence. Given n 0, we denote by ξ n the string formed by the first n bits of ξ, i.e. ≥ j ξ n= ξ0ξ1 ... ξn 1. j − 1 – 2 The finite initial segments give rise to a sequence x of centers. By (C1) and ξ n (C2), this is a Cauchy sequence. By (C4), the sequencej lies in P. Since P is closed, the limit xξ is in P. By (C3), the mapping ξ xξ is well-defined and injective. 7! Theorem 1.2: Every uncountable closed subset of R contains a perfect subset. Proof. Let C R be uncountable and closed. We say z R is a condensation point of C if ⊆ 2 " > 0 [U"(z) C uncountable]. 8 \ Let D be the set of all condensation points of C. Note that D C, since every condensation point is clearly an accumulation point and C is closed.⊆ Furthermore, we claim that D is perfect. Clearly D is closed. Suppose z D and " > 0. Then 2 U"(z) C is uncountable. We would like to conclude that U"(z) D is uncountable, \ \ too, since this would mean in particular that U"(z) D is infinite. The conclusion holds if C D is countable. To show that C D is countable,\ we use the fact that n n every open interval in R is the union of countably many open intervals with rational endpoints. Note that there are only countably many such intervals. If y C D, then for some δ > 0, Uδ(y) C is countable. y is contained in some 2 n \ subinterval Uy Uδ(y) with rational endpoints. Thus, we have ⊆ [ C D Uy C, n ⊆ y C D \ 2 n and the right hand side is a countable union of countable sets, hence countable. We will later encounter an alternative (more constructive) proof that gives additional information about the complexity of the closed set C. For now we conclude with the fact we started out to prove. Corollary 1.3: Every closed subset of R is either countable or of the cardinality of the continuum. 1 – 3 Lecture 2: Polish Spaces The proofs in the previous lecture are quite general, that is, they make little use of specific properties of R. If we scan the arguments carefully, we see that we can replace R by any metric space that is complete and contains a countable basis of the topology. Review of some concepts from topology Let (X , O) be a topological space. A family B O of subsets if X is a basis for the topology if every open set from O is the⊆ union of elements of B. For example, the open intervals with rational endpoints form a basis of the standard topology of R. (We used this fact in Lecture1.) S O is a subbasis if the set of finite intersections of sets in S is a basis for the topology.⊆ Finally, if S is any family of subset of X , the topology generated by S is the smallest topology on X containing S. It consists of all unions of finite intersections of sets in S X , . [f ;g A set D X is dense if for open U = there exists z D U. If a topological space (X⊂, O) has a countable dense6 subset,; the space is2 called\ separable. If (Xi)i I is a family of topological spaces, one defines the product topology on 2 1 Πi I Xi to be the topology generated by the sets π−i (U), where i I, U Xi is 2 2 ⊆ open, and πi : Πi I Xi Xi is the ith projection. 2 ! Now suppose (X , d) is a metric space. With each point x X and every " > 0 we associate an "-neighborhood or "-ball 2 U"(x) = y X : d(x, y) < " . f 2 g The "-neighborhoods form the basis of a topology, called the topology of the metric space (X , d). If this topology agrees with a given topology O on X , we say the metric d is compatible with the topology O. If for a topological space 2 (X , O) there exists a compatible metric, (X , O) is called metrizable . If a topological space (X , O) is separable and metrizable, then the balls with center in a countable dense subset D and rational radius form a countable base of the topology. 2Note that a compatible metric is not necessarily unique. 2 – 1 Polish spaces Definition 2.1: A Polish space is a separable topological space X for which exists a compatible metric d such that (X , d) is a complete metric space. As mentioned before, 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, R, the set of real numbers. One can obtain other Polish spaces using the following basic observations. Proposition 2.2: 1) A closed subset of a Polish space is Polish. 2) The product of a countable (in particular, finite) sequence of Polish spaces is Polish. n n Hence we can conclude that R , C, C , the unit interval [0, 1], the unit circle T = z C: z = 1 , and the infinite dimensional spaces RN and [0, 1]N (the Hilbertf 2 cube)j arej Polishg spaces.