Intuitionistic Descriptive Set Theory

Transcription

1 Intuitionistic Descriptive Set Theory Master Thesis Mathematics Peter t Hart supervised by Wim Veldman July 2014

2 Peter t Hart BSc Intuitionistic Descriptive Set Theory Master Thesis Mathematics Radboud Universiteit Nijmegen July 31, 2014 Supervisor: Dr. W.H.M. Veldman 1

3 Contents: Introduction Symbols and notations Introduction to Descriptive Set Theory Approaching Descriptive Set Theory Intuitionistically Open and closed subsets of N Sets from Σ 0 1 Π 0 1 which are decidable 22 Decidable subsets of N are in Σ 0 1 Π D 2 (A 1 ) is not closed 25 4 Restricted Borel sets Main theorem The complexity of D 2 (A 1 ) 5 Enumerable sets The complexity of Q D The complexity of {δ n n N} 39 6 Some elements of Π and further 31 and further 36 and further 7 Four elements of Π 0 3 and Σ The complexity of P 3 44 and further The complexity of C 3 46 The complexity of S 3 47 The complexity of D 3 48 and further 8 Two difficult Borel sets The complexity of P3 The complexity of S3 9 More primitive sets and their problems and further 52 and further Sources Index

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5 Introduction Intuitionistic mathematics is the mathematics that results from insights brought forward by Brouwer. Characteristics are: a constructive way of reasoning, a view on infinite objects as ongoing unfinished projects and a critical attitude towards results gained from formalistic reasoning. It is constructive: logical statements have a different meaning in intuitionistic mathematics than in classical mathematics. A statement A B is proven once we have a proof of A or we have a proof of B. A statement A means: the assumption A leads to a contradiction. Infinite objects are ongoing unfinished projects: in general we only know a finite initial part of an infinite sequence. It is possible though to choose a method to generate the infinite object. Intuitionistic mathematics is critical towards formalistic mathematics: meaningful mathematics is more than a game according to rules that we follow without completely understanding the results. Some subjects in classical mathematics are difficult for the intuitionistic mathematician, he does not see how to treat them in an intuitionistic way. Other subjects are more attractive and an intuitionistic treatment can be very rewarding. It leads to surprising and refined results. One of these subjects is Descriptive Set Theory. This is the study of the hierarchy that arises when starting with open and closed sets from (in this thesis) Baire space and using countable unions and countable intersections. In the first chapter we study elementary notions of descriptive set theory ([2]), like Baire space N and the class of Borel sets B. We conclude with three basic theorems of the subject, without proof. This chapter is written entirely from classical point of view. In the second chapter we give a brief impression of intuitionistic mathematics and reconsider the first chapter; we correct results and comment on the results that do not hold intuitionistically. In the third chapter we prove some theorems about open (Σ 0 1) or closed (Π 0 1) sets. We present a set ([1]) that is classically closed, but intuitionistically not so. The intuitionistic axioms we use are devided over the second and third chapter. The fourth chapter contains what I like to call the Main Theorem of this thesis. We introduce a subclass B of B. Many of the classification problems can be solved by deciding whether a subset of N is in B or not. The key here is: the class B is downwards closed with respect to (Wadge-)reducibility. This is obvious in classical mathematics, because there B = B, but intuitionistically B B. The fifth chapter concerns the subsets of N that are finite, image-finite, enumerable or countable. These are Borel sets too: they are Σ 0 2. We locate them in the hierarchy as specifically as possible. The sixth chapter is both a completion of the fifth chapter and a study of Π 0 2, which gives results so far similar to the classical results. In the seventh and eighth chapter we study six sets that are strictly Σ 0 3 or strictly Π 0 3. Two sets turned out to be more difficult than the other four and are not introduced until chapter eight. This is the chapter with the largest proof. Chapter nine closes this thesis with six other sets which remind us of N, E 2 and A 3, but intuitionistically are not even equivalent with these sets. This might give the reader yet a new perspective on the differences between Intuitionistic Descriptive Set Theory and Classical Descriptive Set Theory. We hope the reader is surprised by the intuitionistic answers of the classification-problems, enjoys the pictures illustrating the results of each chapter and finds inspiration in this unfamiliar kind of mathematics called Intuitionism. 4

6 Symbols and notations k, m, n, k 0, k 1, K, M, N Natural numbers. i, i 0 Index numbers, hence natural numbers. k 99 A symbol created to tease classical mathematicians. N The set of all natural numbers. =, <, Relations on N, the usual meaning. We also use them in combination with k 99. α, β, γ n Infinite sequences, we write: α = (α(i)) i=0. These also can be seen as functions. The constant sequence (n) i=0, for example 0. N The set of all infinite sequences with i [α(i) N], called Baire space. C The set of all infinite sequences with i [α(i) {0, 1}], called Cantor space. α = β This is actually: i [α(i) = β(i)]. m n This is just: m < n m > n. α β This is: i [α(i) = β(i)]. α#β This is a constructive variant of α β: i [α(i) β(i)]. We say: α is separated from β. Classically: α β α#β, intuitionistically α#β is stronger. X, Y, Z, A, B, P Sets, in principle subsets of N or N. δ P (i) A Kronecker delta. For P N (possibly depending on other variables) δ P (i) is 1 if P (i), and 0 if not. To let this work intuitionistically, P should be decidable. R A relation, thus a subset of some A B. We could write arb for (a, b) R. U An open subset of Baire space. F X, Y, Z, U, F A closed set subset of Baire space. Generators (somehow) of X, Y, Z N resp. U, F. Generators are decidable subsets of N, hence they could be seen as elements of Cantor space: (δ i X ) i=0. A set could have many generators. Intuitionistically, one can not demand uniqueness. X = Y This is actually: X Y Y X (X Y x X [x Y ]). x X, X Y The same as: x X and X = Y. f, g, h (and d, q, r, H) Functions, for example N N. This means, for x in the domain of f, we can uniquely determine an object f(x) in its codomain. With the First Axiom of Continuous Choice, we could always see functions as elements of Baire space. s, t Finite sequences. These also can be seen as functions: N <n N for a certain n. A l The set of all finite sequences with entries in A, hence A = n=0 An. To let this work intuitionistically, we let A be a decidable subset of N (chapter 2). A function N N, taking the length of the elements of N. We have: s N [s = (s(i)) l(s) 1 i=0 ]. Short for: (α(i)) n 1 i=0. Hence l(ᾱn) = n and ᾱn N. We have l(s) n and sn = (s(i)) n 1 ᾱn or α(n) sn s - i=0. We have l(s) > 0 and s - = s (l(s) 1). s t Short for: l(s) l(t) t (l(s)) = s. s t or s α Short for: s t s t resp. ᾱ (l(s)) = s. s t s concatenated with t, in fact: ( δ i<l(s) s(i) + δ i l(s) t(i l(s)) ) l(s)+l(t) 1. i=0 s α In fact: ( δ i<l(s) s(i) + δ i l(s) α(i l(s)) ) i=0. A fixed bijection N N. We assume: s, t [s t (s) (t) ]. This means: = 0. We do not mind statements like 0, 1, 2 N, although 0, 1, 2 N. Then l,.,. -, and also work for N. n The unique m with s [ m s = n]. There is no 0. n The unique s with m [ m s = n]. There is no 0. µ n[p (n)] A symbol representing the minimal n satisfying P (n). This could be done, if n [P (n)] is provided. Intuitionistically, P needs to be decidable. µ n<m[p (n)] A symbol representing the maximal n < m satisfying P (n). If n < m [ P (n)], we define: µ n<m[p (n)] = m. Intuitionistically, P needs to be decidable. µ n m [P (n)] Short for: µ n[n m P (n)]; a minimum. α m The m-th subsequence of α: (α( m i)) i=0. In this way, we can look at α as a sequence of sequences. s m In fact: (s( m i)) ( µ k<l(s) [k 0 k =m]) i=0, or if k < l(s) [k 0 k m]. Inf n[p (n)] Short for N n N [P (n)]. Evt n[p (n)] Short for N n N [P (n)]. Reckless: Inf n[ P (n)] Evt n[p (n)]. Fin n[p (n)] Short for Evt n[ P (n)]. R: Inf n[p (n)] Fin n[p (n)], Fin n[ P (n)] Evt n[p (n)]. 5

7 1 Introduction to Descriptive Set Theory In this thesis, we will study Descriptive Set Theory as a part of intuitionistic mathematics. We first consider some notions of classical descriptive set theory ([2]). Baire space N is the set of all infinite sequences of natural numbers. In order to start descriptive set theory, we first introduce open subsets of N. Definition: A subset X of Baire space is a basic open set, if there exists s N (a finite sequence) such that: X = {α α N s α}. We will write N s for X. Definition: A subset X of Baire space is open, if it is a union of basic open sets, that is: there exists X N such that: X = N s. s X We call X a generator of the open set X. The class Σ 0 1 is defined to be the class of all open subsets of N. We give some alternative definitions of X N is open ( X Σ 0 1 ). (I) A subset X of Baire space is open, if and only if there exists X N such that: X = {α α N n [ᾱn X ]}. This is clear: for all α s X N s there exists s such that: ᾱ (l(s)) = s X and n [ᾱn X ]. For all α X: n [ᾱn X ] and: n [α Nᾱn s X N s]. So X = s X N s. Cantor space C is the set: {α α N i [α(i) {0, 1}]}. (II) A subset X of Baire space is open, if and only if there exists β C such that: X = {α α N n [β(ᾱn) = 1]}. (Here we make use of a bijection. : N N.) Let X be a generator of X. For β we can take: β = (δ i X ) i=0. Also, if β C is given, we can take: X = {s β(s) = 1} as a generator of X. (III) A subset X of Baire space is open, if and only if for every α X there exists n such that: α Nᾱn X. Let X be a generator of X. If α X we determine ᾱn X and observe: Nᾱn s X N s = X. Also, if α X n [Nᾱn X], we can take: X = {s N s X} as a generator of X. 6

8 Theorem: The class Σ 0 1 is closed under countable unions and finite intersections. Proof: Take a sequence U 0, U 1,... of open sets. With the relatively innocent Axiom of Countable Choice we determine for every n a generator U n of U n. Now: n=0 U n has n=0 U n as a generator; the countable union of open sets is open. Take a finite sequence U 0, U 1,..., U N 1 of open sets. Determine for each n < N a generator U n of U n. A generator of X = N 1 n=0 U n is: X = {s s N n < N m l(s) [ sm U n ]}. Suppose α N 1 n=0 U n. Determine t N N with: n < N [ᾱ (t(n)) U n ]. When we calculate M = max{t(n) n < N}, we obtain that ᾱm is an element of X, because for every n < N we see: ᾱm (t(n)) U n (here we use t(n) M). Suppose α has an M for which ᾱm is an element of X. Take n < N arbitrarily. Determine m M with ᾱmm U n. Then ᾱm U n, α U n and (n < N arbitrary): α N 1 n=0 U n. Definition: A subset X of Baire space is closed, if there exists U Σ 0 1 such that: X = N \ U. Whenever U is a generator of the open set U, we call X := N \ U a generator of the closed set X. The class Π 0 1 is defined to be the class of all closed subsets of N. Alternative definitions for X N is closed are: (I) A subset X of Baire space is closed, if and only if there exists X N such that: X = {α α N n [ᾱn X ]}. (II) A subset X of Baire space is closed, if and only if there exists β C such that: X = {α α N n [β(ᾱn) = 1]}. (III) A subset X of Baire space is closed, if and only if for every α N \ X there exists n such that: Nᾱn N \ X. This latter definition could be turned into a more familiar one. Definition: The closure of a set X N is the set: X := {α α N n β X [ βn = ᾱn]}. It is clear that X X for any X. We have: X = X if and only if X is closed. Suppose X = X. Then every α N \ X is not in X; there is an n with β X [ βn ᾱn], so β [ βn = ᾱn β X]. This means: Nᾱn N \ X. Suppose α N \ X n [Nᾱn N \ X]. Take α X arbitrarily. This α can not be an element of N \ X, so it is an element of X. That is: X X. Note: for every X: X = X, so X always is a closed set. 7

9 Theorem: The class Π 0 1 is closed under countable intersections and finite unions. Proof: Take a sequence F 0, F 1... of closed sets. Write, for every n, U n for N \ F n Σ 0 1. The intersection n=0 F n is equal to n=0 N \ U n = N \ n=0 U n, hence closed. Take a finite sequence F 0, F 1,..., F N 1 of closed sets. Write, for every n < N, U n for N \F n Σ 0 1. Then N 1 n=0 F n = N 1 n=0 N \ U n = N \ N 1 n=0 U n, hence closed. The next step in descriptive set theory is to introduce the classes Π 0 2 and Σ 0 2. A subset X of N belongs to the class Π 0 2 if and only if X is a countable intersection of open sets. Equivalently: X is Π 0 2 if there exists β C such that: X = {α α N m n [β m (ᾱn) = 1]}. (A sequence β always can be thought of a sequence of sequences β 0, β 1,..., where, for each m: β m : i β( m i).) A subset X of N belongs to the class Σ 0 2 if and only if X is a countable union of closed sets. Equivalently: X is Σ 0 2 if there exists β C such that: X = {α α N m n [β m (ᾱn) = 1]}. We use recursion now to define, for each n, the classes Σ 0 n and Π 0 n. For every n > 1: Σ 0 n := { X m m [X m Π 0 n 1]}, Π 0 n := { m=0 m=0 X m m [X m Σ 0 n 1]}. We obviously have: Σ 0 1 Π 0 2 Σ and Π 0 1 Σ 0 2 Π These chains are also intertwined. Theorem: n 1 [Σ 0 n Σ 0 n+1 Π 0 n Π 0 n+1]. Proof: For n = 1: Σ 0 1 Σ 0 2: Every open set is a countable union of closed sets. This is obvious, since basic open sets are closed. (A generator of N s as a closed set is {t t s s t}.) Π 0 1 Π 0 2: Take a closed set F arbitrarily and determine a generator F. Now F could be written as: F = {α α N ᾱm F} = s F N N m s Π 0 2. m=0 For n > 1: An element of Σ 0 n is m=0 X m, where X m Π 0 n 1 Π 0 n. So it is an element of Σ 0 n+1. An element of Π 0 n is m=0 X m, where X m Σ 0 n 1 Σ 0 n. So it is an element of Π 0 n+1. Note: for each n, Σ 0 n is closed under countable unions and Π 0 n under countable intersections. As a consequence, Σ 0 n is closed under finite intersections and Π 0 n under finite unions: ( ) ( ) X m Y k = ( ) ( ) (X m Y k ) and X m Y k = (X m Y k ). m=0 k=0 (m,k) N 2 m=0 k=0 (m,k) N 2 m=0 8

10 One may also prove by induction, using De Morgan s laws: n 1 [Π 0 n = {X N N \ X Σ 0 n}]. We could eventually go further and define the class B of all Borel sets: The basic open sets are in B. Whenever (X n ) n=0 is a sequence of Borel sets: n=0 X n is a Borel set. Whenever (X n ) n=0 is a sequence of Borel sets: n=0 X n is a Borel set. All sets in B appear in these ways. The definition of B is actually made by transfinite recursion. Hence, one may use the corresponding principle of transfinite induction. This means: whenever P B satisfies the following: The basic open sets are all in P, Countable unions and intersections of sets in P are in P, then: P = B. In each of the classes Σ 0 n and Π 0 n we select an element. E 1 := {α α N α#0} is open. It is generated by: {s s N s s (l(s) 1) 0}. A 1 := {0} is closed, it is N \ E 1. For n > 1: E n := {α α N m [α m A n 1 ]} Σ 0 n. We use the fact that: if A n 1 is Π 0 n 1, so is {α α m A n 1 } for arbitrary m. For n > 1: A n := {α α N m [α m E n 1 ]} Π 0 n. We use the fact that: if E n 1 is Σ 0 n 1, so is {α α m E n 1 } for arbitrary m. It is the case that: n 1 [A n = N \ E n ]. Now we are willing to think A n+1 is at least as complicated as E n etcetera. This idea of being at least as complicated is an elementary notion in descriptive set theory and can be caught by the definition of reducing. Definition: A function f : N N is continuous if and only if: α m n β Nᾱn [f(β)m = f(α)m]. Definition: For all Borel sets X and Y we define: X (Wadge-)reduces to Y, notation: X Y, if there exists a continuous f : N N (a reduction) for which: X = f -1 (Y ), that is: If f reduces X to Y, we write: α N [α X f(α) Y ]. f : X Y. The relation is reflexive: consider the identity-function. Also is transitive: the composition of continuous functions is again a continuous function. 9

11 Continuous functions have generators too. These are γ N with: α n [γ(ᾱn) f(α)]. Definition: If γ : N N has the properties: s, t [s t γ(s) γ(t)] and α m n [l (γ(ᾱn)) > m], we call γ a generator of a continuous function, namely it will generate: f(α)(i) := γ (α (µ n [l (γ(ᾱn)) > i])) (i). A generator of a continuous f : N N can be found by: γ(s) := µ t<s [ α N s [t f(α)]]. Theorem: X B n 1 [( Y Σ 0 n [X Y ] X Σ 0 n) ( Y Π 0 n [X Y ] X Π 0 n)]. Every Borel class is downwards closed w.r.t.. Proof: Take X B arbitrarily. We prove the theorem with induction to n. For n = 1: If X U for an open U, we have to prove: for every continuous f: f -1 (U) is open. Take α f -1 (U) arbitrarily. Determine U and m with f(α)m U. Determine n with: β Nᾱn [f(β)m = f(α)m]. Then f(β)m U and f(β) U; β f -1 (U) for all β Nᾱn. So: Nᾱn f -1 (U) and f -1 (U) is open by the last definition (III) of open. If X F for a closed F, we have to prove: for every continuous f: f -1 (F ) is closed. Determine U Σ 0 1 with F = N \ U. Then: f -1 (F ) = N \ f -1 (U) Π 0 1. For n > 1: Use the fact that: f -1 ( m=0 X m) = m=0 f -1 (X m ) and f -1 ( m=0 X m) = m=0 f -1 (X m ). Definition: An X B is said to be Σ 0 n-complete if X Σ 0 n Y Σ 0 n [Y X]. Equivalently (using the theorem): Y B [Y Σ 0 n Y X]. Likewise, X B is said to be Π 0 n-complete if Y B [Y Π 0 n Y X]. Theorem: n 1 [ X Σ 0 n [X E n ] Y Π 0 n [Y A n ] ]. The elements E n and A n are complete elements of their classes. Proof: For n = 1: Let X be an element of Σ 0 1. Determine X and define: Then for every α: f(α) = (δᾱi X ) i=0. α X i [ᾱi X ] i [δᾱi X 0] i [f(α)(i) 0] f(α)#0 f(α) E 1. 10

12 Let Y be an element of Π 0 1. Determine U with Y = N \ U and f : U E 1. In general: f : X Z α [α X f(α) Z] α [α N \ X f(α) N \ Z] f : N \ X N \ Z. In our case: f : Y A 1. For n > 1: Let X be an element of Σ 0 n. Determine, for each m, Y m Π 0 n 1 such that: X = m=0 Y m. Use the axiom of countable choice to determine: Define f by: h : N N N ; m [h(m) : Y m A n 1 ] f(α) = (δ i 0 (h(i )(α)) (i )) i=0. For all α, m and i: (f(α)) m (i) = (h(m)(α)) (i), so (f(α)) m = h(m)(α). Then: α X m [α Y m ] m [h(m)(α) A n 1 ] m [(f(α)) m A n 1 ] f(α) E n, that is: f : X E n. Also, each Y Π 0 n reduces to A n. The proof is similar. Corollary: n 1 [E n E n+1 A n E n+1 E n A n+1 A n A n+1 ]. Finally we mention three theorems characteristic for classical descriptive set theory, each having a too difficult proof to be treated here. Theorem: n 1 [E n A n A n E n ]. This means: the classes Σ 0 n and Π 0 n form a real hierarchy (theorem 22.4 in [2]). {N } Σ 0 1 Π 0 1 Σ 0 1 Σ 0 2 Π 0 2 Π 0 2 Σ 0 3 Π { } Π 0 1 Σ Theorem: n 1 X B [X A n E n X X E n A n X]. The proof is by means of Borel determinacy. (In [2] this theorem is theorem ) In particular: elements of Σ 0 n \ Π 0 n are Σ 0 n-complete and elements of Π 0 n \ Σ 0 n are Π 0 n-complete. Then entirely mimics in the structure of the Borel hierarchy (still leaving multiple possibilities for within the classes Σ 0 n Π 0 n): N X (Σ 0 1 Π 0 1)\{N, } X Σ 0 1 \Π 0 1 X (Σ 0 2 Π 0 2)\(Σ 0 1 Π 0 1) X Π 0 2 \Σ 0 2 X Π 0 1 \ Σ 0 1 X Σ 0 2 \ Π 0 2 Theorem: n 1 X Σ 0 n+1 Π 0 n+1 Y Σ 0 n+1 Π 0 n+1 [Y X]. This is a con- There are no complete elements in Σ 0 n Π 0 n for n > 1 (exercise in [2]). sequence of the Hausdorff-hierarchy Theorem. 11

13 We concern the following sets (most of them from Kechris chapter 23). They will be reintroduced in the thesis when we come to speak of them. Σ 0 2-complete ([2], p.179): Countable dense sets Q D. This implies the completeness of Q 2 = {α Evt n [α(n) = 0]}. (We use Evt n instead of.) Π 0 2-complete ([2], p.179): The complements of Q D, and N 2 = {α Inf n [α(n) = 0]} and N 2 = {α Inf n [α(n) = 0] Inf n [α(n) = 1]}. (We use Inf n instead of.) Π 0 3-complete ([2], p.179, 180): P 3 = {α m Evt n [α m (n) = 0]}, C 3 = {α M Evt n [α(n) M]} the sequences with infinite limit and P 3 = {α Inf m [α m = 0]}. Σ 0 3-complete ([2], p.179, 180): Their complements: S 3 = {α m Inf n [α m (n) 0]} (in [2] this is α m (n) = 0 ), D 3 = {α M Inf n [α(n) < M]} and S 3 = {α Evt m [α m #0]} (in [2] this is α m #1 ). Sets from [1]: Fin = {α Fin n [α(n) 0]} = Q 2. We will use the name Fin. Its complement Inf = {α Inf n [α(n) 0]} which is very similar to N 2. And D 2 (A 1 ) = {α α 0 = 0 α 1 = 0}. An extra countable set: Count = {δ n n N}, where δ n = (δ i=n ) i=0. A 3 E 3 P 3, C 3, P 3 S 3, D 3, S 3 E 2 A 2 Q D, Fin (Q D ) C, Inf, N 2, N 2. Count D 2 (A 1), A 1 E 1 N The classical picture of the first three classes of the Borel hierarchy. The open sets, Σ 0 1, are in the four positions below E 1. Also the closed sets appear in four positions. Every set in Σ 0 n \ Π 0 n necessarily is in the same position as E n. Likewise for Π 0 n \ Σ 0 n and A n. The rectangles represent (Σ 0 n+1 Π 0 n+1) \ (Σ 0 n Π 0 n). There are many positions here. 12

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15 2 Approaching Descriptive Set Theory Intuitionistically The previous chapter was written from the point of view of the classical mathematician. Lots of the mathematical facts mentioned turn out not to be true, when we interpret them, in a direct way, intuitionistically. In intuitionistic mathematics logical statements have a constructive meaning: x [P (x)] is considered to be true only if it is known which object x lets P (x) be true. In particular: k {0, 1} [P (k)], that is: P (0) P (1) is only true if one either has a proof of P (0) or has a proof of P (1). This means it is forbidden to assume P (0) P (0) in general. The statement (P (0) P (0)) is equivalent to the statement: P (0) P (0), hence always false. Thus we have: (P (0) P (0)). The step of crossing away a double negation for every proposition P : P is equivalent to P implies: for every P : P P. This means this step is wrong and double negations occur in intuitionistic mathematics. We have seen P (0) P (0) is never false; it will never lead to a contradiction. Although the assumption of its truth should not be made. Intuitionists say it is reckless (Dutch: vermetel ) to assume P (0) P (0) in general. If a statement is reckless, intuitionistic mathematicians are able to come up with a so called weak counterexample (Dutch: vermetelheidstegenvoorbeeld ). The term weak here warns the reader that the counterexample does not provide a proof of the negation of the statement. We will make a weak counterexample against: P (0) P (0). If one thinks about the decimal expansion of the real number π: d(0) = 3, d(1) = 1, d(2) = 4,... one may formulate many problems concerning d we can not solve. When looking at finite sequences appearing as a subsequence of d, we see they come in all kinds. We take a long sequence, such as the sequence of 99 nines. Do we obtain: If so, we can determine: n 99 i < 99 [d(n i) = 9]? µ n 99 [ i < 99 [d(n i) = 9]]. (Determine such an n. Search through the earlier entries of d to determine the smallest n.) We let: k 99 = N and N = k 99 be short notation for: n 99 i < 99 [d(n i) = 9] N = µ n 99 [ i < 99 [d(n i) = 9]]. 14

16 The point here is: the problem what is the value of k 99 is too hard to solve. We can not construct an n with n = k 99, so it is reckless to state: n [n = k 99 ]. It also is reckless to state that this sequence will never appear among the infinitely many sequences; n [n k 99 ] is reckless. So the following statement is reckless: n [n = k 99 ] n [n = k 99 ]. We could transfer this problem to sets. Define: X = {0 n [n = k 99 ]} N. It is reckless to state 0 X 0 X. Definition: Let P be a proposition. We say: P is decidable if and only if P P. A subset X of a set Y is a decidable subset of Y if and only if for each y Y : y X is decidable. The set {0 n [n = k 99 ]} is a weak counterexample against the statement: all subsets of N are decidable (subsets of N), or even against: all subsets of {0} are decidable. It is reckless to state: X = {0 n [n = k 99 ] n [n k 99 ]} is a decidable subset of N (although we know: 0 X). There are a lot of things we could state about k 99 without a problem: n [n < k 99 n k 99 ]. Fix n. We could calculate the first n decimals of π. If we have found 99 nines in this range, we know n k 99. Else: n < k 99. n [n k 99 n > k 99 ]. Of course, after fixing n we could as well calculate n 1 decimals. n [n = k 99 n k 99 ]. Fix n. Determine whether both: n k 99 and n k 99. If so, n = k 99. Else: n k 99. We go through the previous chapter and see what remains in intuitionistic mathematics. Baire space N is accepted in intuitionistic mathematics. Baire space is a prime example of a so-called spread. Brouwer introduced spreads in order to be able to speak meaningfully about sets like N. The basic open sets N s are certainly acceptable, they are even decidable subsets of N. The definition of open is of not much use if we do not demand a generator to be a decidable subset of N. So we adjust the definition by adding: X is a decidable subset of N. The definitions (I) and (II) of open are indeed equivalent with the definition given. (Because of definition (II) we see it is harmless to think of Σ 0 1 as a set, for it is the image under a function defined on Cantor space, which is a spread. We could even put: Σ 0 1 := C and use an equivalence relation as equality of open sets.) The first real problem arises in definition (III) of open. The generator {s N s X} is not necessarily decidable. This means definition (III) of open could be weaker than the other three. We will see it is not weaker for strictly analytic sets on page 25. For now we try to make a weak counterexample. The following set seems to be one: X = {α n [n k 99 ]}. Indeed α X n [Nᾱn X] (n := 0), but assuming X is a generator of X, for every s: s X is reckless. So X is expected to be empty, but n [n k 99 ] is reckless. It looks like X Σ 0 1 is reckless. In contrast: {α n [n = k 99 ]} is open. 15

17 (Using the doubtful Brouwer-Kripke-principle, X is thought to be open after all.) The theorem (p.7) is true, however only the second generator mentioned really is decidable. To define a generator for n=0 U n, we also first have to determine all the U n. Therefore we need the intuitionistic Second Axiom of Countable Choice: leads to: n α N [nrα], in our case: n β C [U n = {α n [β(ᾱn) = 1]}] α N n [nr(α n )], in our case: β n [U n = {α n [β n (ᾱn) = 1]}]. This axiom is plausible, because we think of infinite objects as objects which are never completely finished. This means the sequence which is to be constructed has to be constructed for an arbitrary long initial part. Indeed we can make an arbitrary number of choices. Then a generator for X = n=0 U n is given by: X = {s s N m, n l(s) [ sm U n ]}. Let α be an element of X. Determine n with α U n, and m with ᾱm U n. Let M := max(m, n) and s = ᾱm. Note: m, n l(s) and: sm U n. So: ᾱm = s X and: M [ᾱm X ]. Suppose ᾱm X. Determine m, n M with ᾱm U n. Clearly α is an element of U n, hence of n=0 U n = X. In chapter 1 the first definition of closed is by means of a complement. Intuitionistically, complements for as far as they exist are not really unique. Thus we have to be precise which kind of complement is meant. Definition: The weak complement of X N is the set: X := {α α N α X}. Now: X X = N is as strong as: X is a decidable subset of N. So in general, this is reckless. The other characteristic of complements, X X =, is still valid. Closed sets are thus initially defined as the X N for which: U Σ 0 1 [X = U ]. The next two definitions (I) and (II) of closed again are equivalent with the definition given, definition (III) is not. The statement: every X with X = X is closed is reckless. We mentioned it is reckless to state: {α n [n k 99 ]} Σ 0 1, likewise it is reckless to state: X = {α n [n k 99 ]} Π 0 1, although X = X. It is true that for every closed set X: X = X. Although the double closure X is just the closure X for every X, stating: all closures are closed is reckless. In the proof of the next theorem (p.8) about Π 0 1 being closed under countable intersections and finite unions, we use the laws of De Morgan. The rule: k [P (k)] k [ P (k)] is right. This means the countable intersection of closed sets is closed (using the Second Axiom of Countable Choice to determine the U n with F n = U n ). The rule: (P (0) P (1)) P (0) P (1) works only from right to left. This means: the union 16

18 of two closed sets does not have to be closed. A counterexample will be provided in the next chapter. The definition of the classes of Borel sets is certainly acceptable. We could even define all the classes to be Cantor space, and use an equivalence relation on Cantor space to describe equality in Σ 0 n and Π 0 n. Also the subsequent theorem of the inclusion of all the classes intuitionistically holds. Obviously Σ 0 n is closed under countable unions also if n > 1. Let X 0, X 1,... be a sequence of Σ 0 n-sets. Use the Second Axiom of Countable Choice to determine for each m a sequence X m,0, X m,1,... of Π 0 n 1-sets such that: X m = k=0 X m,k. Then: m=0 X m = (m,k) N N X m,k Σ 0 n. Likewise Π 0 n is closed under countable intersections. Indeed, Σ 0 n is closed under finite intersections also if n > 1. Let X and Y be Σ 0 n-sets. Determine sequences X 0, X 1,... and Y 0, Y 1,... of Π 0 n 1-sets such that: X = m=0 X m and Y = k=0 Y k. Then: X m Y k = (X m Y k ) Σ 0 n. m=0 k=0 (m,k) N N As mentioned: Π 0 n does not have to be closed under finite unions. We should not state: for every X Σ 0 n: X Π 0 n. In general it is reckless to state the weak complement of a Borel set is even a Borel set. This is why elements of the intuitionistic B are called positively Borel in [1]. In this article, an example is given of a (positively) Borel set with a weak complement that is not (positively) Borel. Intuitionists are tolerant towards transfinite recursive definitions and transfinite induction. So we accept B as a set. In intuitionistic mathematics all functions f : N N are continuous as a result of the axiom BCP introduced in the next chapter. This simplifies the definition of reducibility. Indeed (continuous) functions f : N N have generators γ N. Note that the properties mentioned (p.10) are sufficient to define f from γ. Now if we imagine any f : N N to be just an R N N with special properties, it is trickier to find a generator γ than described on page 10. The formula α N s [t f(α)] should be decidable to use this method. Instead we need another axiom of choice to provide the γ: The First Axiom of Continuous Choice: Let R N N. If R satisfies: α n [αrn], then one can construct a decidable X N N with α k, n [(ᾱk, n) X] and: α k, n [(ᾱk, n) X αrn]. The axiom actually implies the continuity principle. Then for every (continuous) f : N N we can construct a decidable X N N with α k, n [(ᾱk, n) X] and (in fact): [( ) ] m α k, n ( m α)k, n X β Nᾱn [f(β)m = f(α)m]. 17

19 In this case, ( 0, 0) X may be assumed. A generator γ of f can then be determined as follows: [ ( ) γ(s) := f(s 0) µ m<l(s) µ n [ k n [ ( m s 0)(k + 1), n ( ) ] ] X ( m s 0)(n + 1), k X] l(s). The theorem stating Borel classes are downwards closed with respect to does intuitionistically hold. However, we need a different proof for: f -1 (U) is open. Determine a generator γ of f. Define a generator X of X Σ 0 1: (Indeed X is decidable.) X = {s s N m l (γ(s)) [γ(s)m U]}. f -1 (U) X: Take α f -1 (U) arbitrarily. Determine m with f(α)m U and n with l (γ(ᾱn)) m. Then γ(ᾱn)m = f(α)m U and: ᾱn X ; α X. X f -1 (U): Take α X arbitrarily. Determine n with ᾱn X ; determine m l (γ(ᾱn)) with γ(ᾱn)m U. Then f(α)m = γ(ᾱn)m U and: f(α) U; α f -1 (U). f -1 (U) = X Σ 0 1 The theorem stating E n and A n are complete elements intuitionistically holds. In the proof we only need the direction: f : X Y f : X Y. The other direction: f : X Y f : X Y is reckless. A weak counterexample against this, seems to be: X = {α n [n = k 99 ] n [n k 99 ]} and Y = E 1. The statement X is open is reckless to make, although: X = A 1 = Y. Finally we comment on the last three theorems of the first chapter. The classes Σ 0 n and Π 0 n form a real hierarchy: n [E n A n A n E n ]. This is intuitionistically true. It is the main subject of [1] (together with the transfinite equivalent). [ Elements of Σ 0 n \ Π 0 n are Σ 0 n-complete: X Σ 0 n X Π 0 n Y Σ 0 n [Y X] ]. This is intuitionistically false. We will see an element that is strictly Σ 0 n, that is: in Σ 0 n \ Π 0 n, and is not Σ 0 n-complete. Furthermore, this will be an X A 1 with E 1 X. Also elements that are strictly Π 0 n do not have to be Π 0 n-complete, at least for n > 2. There are no complete elements in Σ 0 n Π 0 n for n > 1: n > 1 X Σ 0 n Π 0 n Y Σ 0 n Π 0 n [Y X]. I do not know if this is true in intuitionistic mathematics. It is not my aim now to examine this further. 18

20 A 3 E 3 E 2 A 2 A 1 E 1 The fact that Borel sets can be strictly Σ 0 n or strictly Π 0 n and not complete, makes the schematic classical picture of the Borel hierarchy useless for the intuitionistic Borel hierarchy. We need more space, as provided in this picture. The completeness of E n and A n makes us draw triangles for the classes Σ 0 n and Π 0 n and place E n and A n at the tops. 19

21 3 Open and Closed subsets of N Recall that we gave the following definitions for open and closed. Every X N is open if and only if there exists a decidable subset X of N such that: X = {α α N n [ᾱn X ]}. Every X N is closed if and only if there exists a decidable subset X of N such that: X = {α α N n [ᾱn X ]}. A closed set is the weak complement of an open set. It is reckless to state an open set is the weak complement of a closed set. (Hence: the weak complement of a closed set is open also is reckless.) We take E 1 as a weak counterexample. If, for certain F, E 1 = F, we can prove: F = A 1. Is E 1 = A 1 reckless? Indeed, taking α A 1 it is not clear how the construction of the n with α(n) 0 should be made. However, when one uses the generalized Markov Principle as described in [1], the statement is obviously valid. At the other hand, Brouwer wanted to create a weak counterexample against what we now formulate as E 1 = A 1. He had to use a doubtful axiom, now called Axiom of Brouwer-Kripke. The general axiom will lead to a contradiction when sufficient choice is allowed. Therefore, we will not introduce it. ([3]) For the moment we accept Brouwer s conclusion: there is a β 0 for which β#0 is reckless. We obtain: the statements A 1 = E 1 and U F [U = F ] both are reckless. In the case of Baire space we could also introduce a positive kind of complement. Definition: Let X be a subset of N. The strong complement of X is the set: X C = {α α N β X [β#α]}. The strong complement is a subset of the weak complement. This is a consequence of X X C =. Indeed, the assumption α X X C leads to α#α, a contradiction. We thus have: A C 1 A 1, but A 1 A C 1 is as reckless as A 1 = E 1 (because A C 1 = E 1 ). Lemma 3.1: U Σ 0 1 [U = U C ]. Weak complements and strong complements of open sets are the same. Proof: Let U N be open. Determine a generator U. Assume α is a sequence in U, this means: n [ᾱn U]. For arbitrary β U: n [ βn U], so: n [ βn ᾱn], that is: α#β. We thus see (β arbitrary): α U C. We obtain: U = U C. One could also expect (U ) C = U. Indeed U (U ) C, but for the other direction we can create a weak counterexample. Again take a sequence β 0 for which β#0 is reckless. Define U by: U = {s s N β (l(s)) 0}. 20

22 Then U =, so (U ) C U leads to: U = N. We now can determine n with 0n U, but this is reckless. The strong complement of a spread is open. Definition: A generator F of a closed set F is a spreadlaw if and only if: s F n [s n F] and s F [s s - F]. (Every closed set F has a generator with s F [s s - F], but not necessarily one with s F n [s n F].) A closed set is a spread (Dutch: spreiding ) if there exists a spreadlaw that generates the set. We introduce the following abbreviation: Spr(F ) for F is a spread. A useful characteristic of spreads is that they give rise to a retraction from Baire space to the spread. We will spell this out. We will write f n for functions with the properties: n [f n : N n N n ] and n > 0 s N n [f n (s) - = f n 1 (s - )]. When defining f n, we gain a (continuous) f : N N ; we let its generator be: γ : s f l(s) (s). Now let F be a spread, and let F be its spreadlaw. We define f by: f n (s) = s if s F, f n (s) = f n 1 (s - ) µ n [s n F] if s F. Always: f n (s) F, so always: f(α) F. Furthermore: α F [f(α) = α]. Having a retraction from Baire space to a spread of choice we can prove the announced. Lemma 3.2: U Σ 0 1 [Spr(U ) U = (U ) C ]. Spreads have an open strong complement. Proof: Let U have Spr(U ); define F = U and determine a spreadlaw F of F. Assume α is a sequence in F C. We have: α#f(α). Determine n with ᾱn f(α)n. Apparently: ᾱn F. So ᾱn N \ F and (F decidable): α U. Hence (α F C arbitrary): F C U. We already made the remark: U (U ) C. We obtain: F C = U. The earlier set U for which (U ) C U was reckless, was thus an example of an F = U for which: F is a spread is reckless. I was not able to find an open F for which F is a spread is reckless. This is important, considering the following theorem. 21

23 Theorem: X Σ 0 1 Π 0 1 [Spr(X) α [α X α X]]. Open spreads are decidable subsets of N. Proof: Take an open spread X arbitrarily. Determine the spreadlaw F that generates X and the generator U of X as an open set. Let f be a retraction from N onto X as described on the previous page. Take α N arbitrarily. Since f(α) X we can determine n with f n (ᾱn) = f(α)n U. Now we have: ᾱn = f n (ᾱn) ᾱn f n (ᾱn). The first part: ᾱn = f n (ᾱn) obviously leads to: α X. The second part: ᾱn f n (ᾱn) leads to ᾱn F. So then: α X. We have obtained: α X α X and X is a decidable subset of Baire space. Spreads are important, as the Continuity Principle a crucial axiom in intuitionistic mathematics generalizes to spreads. Brouwer s Continuity Principle (BCP) Whenever R N N has the property: α N n [αrn], we may conclude: α N m, n β Nᾱm [βrn]. An important consequence of BCP is: every f : N N is continuous (compare the definition of continuous with BCP, p.9). Theorem: Generalized Continuity Principle (GCP). Let F be a spread. Whenever R F N has the property: we may conclude: α F n [αrn], α F m, n β Nᾱm F [βrn]. Proof: Choose a spreadlaw F and use the retraction f described on the previous page. Let R F N have the property α F n [αrn]. We obtain: α n [f(α)rn]; α N m, n β Nᾱm [f(β)rn]. For α, β F we have f(α) = α, f(β) = β, which completes the proof. 22

24 Making use of BCP and GCP, one may turn weak counterexamples into proofs of contradiction. If, for a certain x X, P (x) is reckless, we hope with BCP and GCP for the result: x X [P (x)]. We already made the remark the First Axiom of Continuous Choice is a generalization of BCP. To obtain desired proofs of contradiction, often one needs the even stronger Second Axiom of Continuous Choice. 1 The Second Axiom of Continuous Choice: Let R N N. If R satisfies: α β [αrβ], then one can construct an f : N N with: α [αr (f(α))]. Then we obtain indeed: Not all propositions are decidable: β [β#0 β = 0] (BCP). Not all subsets of {0} are decidable: β [0 {0 β#0} 0 {0 β#0}] (BCP). Not all sets X N with α X n [Nᾱn X] are open: β [{α β = 0} Σ 0 1] (SACC). Not all sets X N with X = X are closed: β [{α β 0} Π 0 1] (SACC). Not all sets X B with X E 1 have X E 1 : β [{α β#0 β = 0} Σ 0 1] (SACC). Not all closed sets are spreads: F [F = F ] (BCP). We prove β [{α β#0 β = 0} Σ 0 1]. Assume β [{α β#0 β = 0} Σ 0 1], that is: β γ α [(β#0 β = 0) n [γ(ᾱn) = 1]]. Determine f : N N with: β α [(β#0 β = 0) n [f(β)(ᾱn) = 1]]. Using BCP, f is continuous. Put β := 0 and α := 0 to determine n with: f(0)( 0n) = 1. Determine m with: β N 0m [f(β)( 0n + 1) = f(0)( 0n + 1)]. Now β N 0m [(β#0 β = 0) n [f(β)( 0n) = 1]] turns into: β N 0m [β#0 β = 0]; β N 0m k [βrk]. Here βr0 stands for β#0 and βrk with k > 0 for β = 0. Apply GCP: M, k β N 0M [βrk]. 1 We have chosen a formulation of the axiom that does not imply the First Axiom of Continuous Choice. This does not matter, for we assume, when one is willing to use the Second Axiom of Continuous Choice, one is also willing to use the First Axiom of Continuous Choice. 23

25 Determine M and k. This k can not be 0: we can put β := 0. So k > 0. Put β := 0M 1. Then β = 0. Contradiction. We conclude: β [{α β#0 β = 0} Σ 0 1]. With the First Axiom of Continuous Choice we can place the decidable sets in the picture of the Borel hierarchy. (Compare the theorem on page 22.) Theorem: For every decidable subset X of N : X Σ 0 1 Π 0 1. There are no decidable subsets of N outside Σ 0 1 Π 0 1. Proof: Let X be a decidable subset of N. Define R: αrn α X if n = 0, αrn α X if n > 0. Note: α n [αrn]. Apply the First Axiom of Continuous Choice and determine a decidable subset Y of N N such that α k, n [(ᾱk, n) Y ] and: α k, n [(ᾱk, n) Y αrn]. Let the following decidable set generate an open set U: U = {s s N (s, 0) Y }. X U: Assume α is an element of X. Determine k and n with (ᾱk, n) Y. The assumption that n is positive leads to a contradiction. So (ᾱk, 0) Y and: ᾱk U; α U. U X: Assume α is an element of U. Determine k with (ᾱk, 0) Y. Then αr0, that is: α X. So X = U Σ 0 1. Furthermore, because X is decidable, so is X, and also: (X ) = X. Write X = U Σ 0 1, then: X = (X ) = U Π 0 1. We introduce D 2 (A 1 ). For each subset X of N, for each n, we define a set D n (X) N, the n-fold disjunction: We also define the n-fold conjunction: So D 2 (A 1 ) is the set: D n (X) = {α m < n [α m X]} = n 1 m=0 {α αm X}. C n (X) = {α m < n [α m X]} = n 1 m=0 {α αm X}. D 2 (A 1 ) = {α α N α 0 = 0 α 1 = 0}. Classically one does not doubt about D 2 (A 1 ) Π 0 1. principle: This is an application of the pigeonhole for every α: n [α 0 n 0 α 1 n 0] leads to: n [α 0 n 0] n [α 1 n 0]. 24

26 Therefore classically: D 2 (A 1 ) = D 2 (A 1 ). But intuitionistically the pigeonhole principle is reckless, and GCP leads us to D 2 (A 1 ) D 2 (A 1 ). Lemma 3.3: D 2 (A 1 ) D 2 (A 1 ) (Veldman, [1]). D 2 (A 1 ) is not closed. Proof: We need D 2 (A 1 ) to be a spread. Indeed, it is generated by the spreadlaw: {s s N s 0 0 s 1 0}. So we can apply GCP, when assuming D 2 (A 1 ) = D 2 (A 1 ): Determine k and M and consider: k, M α D 2 (A 1 ) N 0k [α M = 0]. α = (δ i k i 0 i =M ) i=0 N 0k. Note: α M = ( δ M i k ) i=0 #0; α D2 (A 1 ). Also note: for every m M: α m = 0. This means α is an element of D 2 (A 1 ). Contradiction. In the next chapter, it will be clear that D 2 (A 1 ) is strictly Σ 0 2. A more general notion than the notion of a spread is the notion of a strictly analytic set: Definition: A subset X of Baire space is strictly analytic, if we can construct a surjection g : N X. This means a (continuous) g : N N with both α [g(α) X] and x X α [g(α) = x]. All the sets E n and A n are strictly analytic. Let X N be strictly analytic. Assume (III): α X n [Nᾱn X]. We prove: X Σ 0 1. Determine the surjection g : N X. Construct a generator γ of g. β N m [N γ( βm) X] Using the First Axiom of Continuous Choice, determine a decidable subset Y of N N with β k, m [( βk, m) Y ] and: β N k, m [( βk, m) Y N γ( βm) X]. We define a decidable subset of N to generate an open U: U = {s t s m, n l(t) [( tn, m) Y γ(t) s]}. U X: Let α be an element of U. Determine k with ᾱk U and t ᾱk, m, n l(t) with ( tn, m) Y γ(t) ᾱk. We could take β = t 0: N γ( tm) = N γ( βm) X. 25

27 Note: tm t and: γ( tm) γ(t) ᾱk. So: α Nᾱk N γ( tm) X. X U: Let α be an element of X. Determine β with g(β) = α. Determine n and m with ( βn, m) Y, and define k = max(m, n) and t = βk. Let N be large enough to satisfy: t ᾱn and l (γ(t)) N. (For example: N = max (t, l (γ(t))).) Now γ(t) g(β)n = ᾱn and ᾱn U; α U. X = U Σ 0 1 A 3 E 3 E 2 A 2 D 2 (A 1 ) A 1 E 1 By the theorem on page 24 decidable subsets of N occur in the small triangle only. 26

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29 4 Restricted Borel sets In the previous chapter we have seen that intuitionistically the set D 2 (A 1 ) falls out of the class of closed sets, although it is a union of two closed sets. In this chapter, we shall prove the stronger result that D 2 (A 1 ) is not restricted Borel. We explain what we mean by that. Definition: Let Y be a subset of N and X 0, X 1,... a sequence of subsets of N. We call Y the limit of the sequence X 0, X 1,... and write: if and only if: n [X n X n+1 ] Y = n=0 X n. Y = Lim n X n Definition: The class B of restricted Borel sets is defined as follows (by transfinite recursion): The open sets are in B. Whenever (X n ) n=0 is a sequence of sets in B : n=0 X n is restricted Borel. Whenever (X n ) n=0 is an increasing sequence of sets in B : Lim n X n is restricted Borel. All sets in B appear in these ways. Clearly: B B. In classical mathematics: B = B. We now give the classical proof, by making use of transfinite induction (p.9) on both the classes B and B. Lemma 4.1: Classically, X, Y B [X Y B ]. Proof: We have to prove: for each X B, the set: P X = {Y Y B X Y B } is equal to B ; that is: is equal to B. P = {X X B P X = B } The open sets are in P : Let U be an open set. With transfinite induction we prove: P U = B. The open sets clearly are in P U. Let Y 0, Y 1,... be a sequence of sets in P U. U Y n = (U Y n ) B n=0 n=0 So n=0 Y n P U. Let Y 0, Y 1,... be an increasing sequence of sets in P U. U Lim n Y n = Lim n (U Y n ) B So Lim n Y n P U. With transfinite induction we conclude: P U = B, and: U P. Countable intersections of sets in P are in P : Let X 0, X 1,... be a sequence of sets in P. We prove: P = n=0 Xn B. Take Y B arbitrarily. Note: n [X n Y B ]. ( ) X n Y = (X n Y ) B n=0 n=0 28

30 We conclude (Y B arbitrary): P n=0 Xn = B, and: n=0 X n P. The limits of sets in P are in P : Let X 0, X 1,... be an increasing sequence of sets in P. prove: P LimnX n = B. Take Y B arbitrarily. Note: n [X n Y B ]. We (Lim n X n ) Y = Lim n (X n Y ) B We conclude (Y B arbitrary): P LimnX n = B, and: Lim n X n P. With transfinite induction over the class B we conclude: P = B ; all finite unions of restricted Borel sets are restricted Borel. Then using transfinite induction, we prove: B B. The basic open sets clearly are all in B. Let X 0, X 1,... be a sequence of restricted Borel sets, then clearly n=0 X n B. Let X 0, X 1,... be a sequence of restricted Borel sets, then: ( N ) X n = Lim N X n B. n=0 With transfinite induction over the class B we conclude: B B ; the restricted Borel sets are just the Borel sets. n=0 Intuitionistically, lemma 4.1 is false. We use the wrong step: ( ) (X n Y ) X n Y. n=0 By the end of this chapter we will be absolutely sure, that B B. First we will demonstrate how B is of great use to us. n=0 Theorem: X B [ Y B [X Y ] X B ]. The class of restricted Borel sets is downwards closed w.r.t.. Proof: With transfinite induction to Y over the class B. Thus we define: P := {Y Y B X B [X Y X B ]}. Let U be open. Every X B that reduces to U, is itself open (p.18), hence in B. So: the open sets are in P. Let Y 0, Y 1,... be a sequence of sets in P. Take X B with X n=0 Y n arbitrarily. Determine f : X n=0 Y n. ( ) X = f -1 Y n = f -1 (Y n ) n=0 For every n: f -1 (Y n ) is reducing to Y n, hence restricted Borel. So X is restricted Borel. n=0 29

31 Let Y 0, Y 1,... be an increasing sequence of sets in P. Take X B with X Lim n Y n arbitrarily. Determine f : X Lim n Y n. X = f -1 (Lim n Y n ) = Lim n f -1 (Y n ) (Note: n [f -1 (Y n ) f -1 (Y n+1 )].) Again for every n: f -1 (Y n ) is restricted Borel. So X is restricted Borel. Using transfinite induction we conclude: P = B. The same proof works for an alternative restriction B defined to contain the closed sets and be closed under countable unions and limit-intersections. However, B = B also intuitionistically, because the statement corresponding to lemma 4.1 that we need, about finite intersections, is similarly proved, and intuitionistically correct. For every α and k {0, 1}, define α k : α k := (δ i 0 i k α(i)) i=0 D2 (A 1 ). Now we will first tackle a small lemma, to prepare ourselves for a larger lemma, which has B B as a corollary. Lemma 4.2: α D 2 (A 1 ) [α#α 0 α = α 1 ]. Proof: Choose α D 2 (A 1 ) with α#α 0 and determine i with α 0 (i) α(i). This means both δ i 0 i 0 1 and α(i) 0. Now choose n; we want: α(n) = α 1 (n). This is clear for n with δ n 0 n 1 = 1. So suppose n 0 n = 1. Take N = max(i, n) + 1. Determine β D 2 (A 1 ) with ᾱn = βn. For this β: β(i) = α(i) 0. Use i 0 i = 0: β 0 (i ) 0 and β 0 #0. This means (β D 2 (A 1 )): β 1 = 0. Then α(n) = β(n) = β 1 (n ) = 0 = (α 1 ) 1 (n ) = α 1 (n) and (n arbitrary): α = α 1. Lemma 4.3: X B [D 2 (A 1 ) X D 2 (A 1 ) X]. Proof: Choose α D 2 (A 1 ). We prove: X B [(α 0 X α 1 X) α X]. Then an X in B with D 2 (A 1 ) X will certainly satisfy α 0 X α 1 X, so α X, and (α arbitrary): D 2 (A 1 ) X. The proof is again by transfinite induction over the class B. Let U be an open set with α 0 U α 1 U. Determine U and n with ᾱ 0 n U. If ᾱn = ᾱ 0 n: α U. Else, use the previous lemma: α = α 1 U. Let X 0, X 1,... be a sequence of restricted Borel sets with, for every n:(α 0 X n α 1 X n ) α X n. Assume: α 0 n=0 X n α 1 n=0 X n. Then we know: n [α 0 X n α 1 X n ], so: Indeed: α n=0 X n. n [α X n ]. 30

32 Let X 0, X 1,... be an increasing sequence of restricted Borel sets with, for every n: (α 0 X n α 1 X n ) α X n. Assume: α 0 Lim n X n α 1 Lim n X n. Determine n and m with α 0 X n and α 1 X m. Take N = max(m, n): α 0 X n X N α 1 X m X N, so: Indeed: α Lim n X n. α X N. We now know X B [(α 0 X α 1 X) α X] and for restricted Borel sets X the statement D 2 (A 1 ) X will lead to D 2 (A 1 ) X. Corollary: D 2 (A 1 ) is not an element of the class B of restricted Borel sets (lemma 3.3, p.25). Corollary: D 2 (A 1 ) does not reduce to any set in B (theorem, p.29). I would like to call the two corollaries combined with the earlier theorem the Main Theorem of my thesis. So what is new here? Veldman [1] proved already: D 2 (A 1) does not reduce to Fin, a certain element from which we will see it is restricted Borel. He ends up with an n having D 2 (A 1) N 0n D 2 (A 1) as a consequence of the assumption D 2 (A 1) Fin, which is the contradiction. For the proof of D 2 (A 1) B I had to try something new to handle the countable intersections. The idea of α 0 and α 1 worked perfectly. It is nice to know that the idea of introducing the restricted Borel sets was suggested by Veldman, after he read my theorem about D 2 (A 1). Can we also make a theorem generalizing which sets do not reduce to D 2 (A 1 )? (Such a theorem would be less special though, because D 2 (A 1 ) is quite low in the Borel hierarchy.) Proposition: X B [X D 2 (A 1 ) X (X ) ]. Proof: Assume X B reduces to D 2 (A 1 ). Determine f : X D 2 (A 1 ). One easily proves that: α X [f(α) D 2 (A 1 )]. But ([1]): D 2 (A 1 ) = (( D 2 (A 1 ) ) ). So: every α X has its image f(α) in (( D 2 (A 1 ) ) ). Note: f : X ( D 2 (A 1 ) ) and f : (X ) (( D 2 (A 1 ) ) ). So every α X is in (X ). Corollary: E 1 does not reduce to D 2 (A 1 ) ((E 1 ) does not contain 0). This means D 2 (A 1 ) is a Borel set for which D 2 (A 1 ) A 1 and E 1 D 2 (A 1 ). This contradicts the classical theorem: n 1 X B [X A n E n X]. We also realize: D 2 (A 1 ) is strictly Σ 0 2 (it does not reduce to A 2 B ) but not Σ 0 2-complete. Veldman [1] even proves: m, n [D m (A 1) D n (A 1) m n]. The incompleteness of D 2 (A 1) is a very special case of this, for n [D n (A 1) E 2]. When we have a notion of D, which is yet to be introduced (p.45), we obtain E 2 = D (A 1). This completes the illustration of this infinite chain of sets in Σ