The right choice? An intuitionistic exploration of Zermelo s Axiom. Dion Coumans. July 2008

Transcription

1 The right choice? An intuitionistic exploration of Zermelo s Axiom Dion Coumans July 2008 Supervisor: r. W.H.M. Velman Secon reaer: prof. r. A.C.M. van Rooij Faculty of Science Rabou University Nijmegen Stuent number:

2 Contents List of Symbols 4 1 Introuction History Overview Prerequisites Basic sets Special subsets of N Spreas Brouwer s Continuity Principle Axioms of Choice Fans Fan Theorem Unrestricte Fan Theorem Extene Fan Theorem The set of the real numbers Discrete spreas an fans Discrete spreas Discrete fans On open, close an enumerable subsets of N All kins of close an open Sequentially close subsets of N Finitely enumerable subsets of N Enumerable subsets of N On the existence an non-existence of certain choice functions Choice functions on collections of subsets of N Choice functions on collections of subsets of C Finitely enumerable subsets of C Sequentially close subsets of C Effectively open subsets of C

3 6 Sets of representatives for equivalence relations Equivalence relations on N The prisoners an their hats Equivalence relation on C Connection with the Vitali relation Equivalence relations an choice functions Equivalence relation given by the inhabitely enumerable subsets of N Hamel basis A Hamel inepenent function from C to R A Hamel complete function from C to R? Connection with the Vitali relation Constructing weakly Vitali inepenent functions Weakly Vitali inepenence of g h Incompleteness of weakly Hamel inepenent functions Further research References 94 3

4 List of Symbols AC 0,0 First Axiom of Countable Choice, 15 CE α subset of N co-enumerate by α, 11 D α subset of N ecie by α, 11 E α subset of N enumerate by α, 11 GAC 1,0 First Axiom of Continuous Choice, 16 GAC 1,1 Secon Axiom of Continuous Choice, 16 GCP Brouwer s Generalize Continuity Principle, 14 IE α subset of N inhabitely enumerate by α, 47 S successor function, 42 α n subsequence of α, 14 # apartness relation between real numbers an intervals, 69 on Baire space, 9 on the set of the real numbers, 23 relation on the collection of intervals, 23 concatenation function, 12 composition function, 42 bijective function from N to N, 12 N set of all finite sequences of natural numbers, 12 C Cantor space, 13 N Baire space, 9 A closure of the set A, 36 αm initial segment of α of length m, 12 incompatibility relation, 42 / a strongly not a relate, 64 a equivalence relation on C, 62 orering relation on N, 12 on the collection of intervals, 23 m infinite sequence with constant value m, 9 lg length function on N, 12 on the collection of intervals, 23 4

5 Chapter 1 Introuction Imagine a queue of infinitely many prisoners numbere 0, 1, 2,.... Ranomly, each of them is assigne a black or a white hat. Each prisoner can only see the hats of the fellow inmates in front of him (i.e. the hats of the inmates who have a higher number than he has). The guar asks each prisoner in turn to guess the colour of his hat, without the other prisoners being able to hear his reply. If the prisoner answers correctly, he will be release. If not, he has to stay in prison for the rest of his life. After being given the rules of the game, the prisoners get one hour to etermine their strategy. One of them, a classical mathematician accepting the Axiom of Choice, says I have a plan that ensures at most finitely many of us guess wrongly. The topic of this thesis is the Axiom of Choice, a statement which has le to a huge amount of controversy an iscussion since its formulation. The example above is just one of the many peculiar consequences of the Axiom of Choice. Banach an Tarksi explaine, for example, how one, using the Axiom, can cut a sphere into finitely many pieces that can be rearrange into two new spheres that are both of the same size as the original one. At first sight the Axiom of Choice souns quite plausible, as it states: for every set A, there exists a function f that assigns to every non-empty subset B of A an element f(b) of B. We will refer to this statement as AC. For instance, for the set A = {0, 1}, both an {0} 0, {1} 1, {0, 1} 0, {0} 0, {1} 1, {0, 1} 1, (AC) woul o as such a function. This is a very simple example. Notice, AC states for every set, which makes it a strong statement, in particular if one has to eal with an infinite set A. 5

6 In this thesis we stuy the Axiom of Choice from Brouwer s intuitionistic point of view. There is much confusion about the role of the Axiom of Choice in constructive mathematics. Some people seem to regar it as a non-constructive principle, while others efen it from a constructive point of view. This confusion is largely cause by the many ifferent, classically equivalent, formulations of the Axiom. At face value, some of these formulations seem plausible in a constructive context, while others seem outright false. We will show that some forms of the Axiom of Choice are acceptable to the intuitionistic mathematician, while others are highly ebatable. Before giving a more etaile escription of the content of this thesis, we briefly sketch the origin an evelopment of the Axiom of Choice. Some familiarity with classical set theory is assume, but such knowlege is not require for the rest of this essay. 1.1 History At the en of the 19 th century, Georg Cantor claims that, for all sets X an Y, either X can be embee into Y or Y can be embee into X. In orer to prove this claim he introuces the Well-Orering Principle, which states that every set can be well-orere. He proposes this principle as a law of thought, a statement that, accoring to him, is obvious from the way one shoul think about sets. Not everyone agrees. In 1904, searching for a proof of the Well-Orering Principle, Ernst Zermelo introuces the Axiom of Choice. 1 The formulation of AC by Zermelo makes people aware of the numerous implicit uses of this proposition in earlier arguments. It leas to vehement iscussions: shoul one accept the Axiom of Choice or shouln t one? In 1908 Zermelo publishes an axiomatization of set theory. One of his aims is to clarify the role an meaning of AC. Further stuy an ebate leas to a list of nine axioms an axiom schemes for set theory, calle the Axioms of Zermelo an Fraenkel, abbreviation ZF, which fins wie acceptance. Due to the work of Fraenkel (1922), Göel (1938), Mostowski (1945), an Cohen (1963) we now know that the Axiom of Choice is inepenent of ZF, that is, if ZF is consistent then both ZF+AC an ZF+ AC are consistent. Cohen use a new metho for his proof, calle forcing, which turne out to be a very fruitful metho to obtain inepenence results [1]. 1 The Axiom of Choice is in fact equivalent to the Well-Orering Principle. 6

7 1.2 Overview This thesis is written from an intuitionistic point of view. A basic introuction to intuitionism is given in Chapter 2. The statement AC oes not make immeiate sense to the intuitionistic mathematician. The concept for every set is far too general an we prefer to first consier some special sets A, like the set N of the natural numbers an Baire space N, to see what becomes of AC in those cases. Some forms of the Axiom of Choice are acceptable to us, as these follow from the way we think about the objects of Baire space. We will explain an justify those in Section 2.5 an we will use these forms in the rest of the thesis. In the intuitionistic stuy of sets, the equality relation eserves special treatment. On the set of the natural numbers, the equality relation is eciable, but on Baire space it is not. We have trie to fin a characterization of both spreas an fans (two typically intuitionistic notions, explaine in Chapter 2) with a eciable equality. The result is presente in Chapter 3. A secon ifficulty in AC is how to unerstan the phrase every non-empty subset of A. Again, the notion of an arbitrary subset of a given set is too general. In intuitionistic mathematics not all subsets are of the same kin. In the example at the beginning, we have only efine the function for the eciable subsets of {0, 1}. The statement becomes stronger if we consier the enumerable subsets of A or an even broaer notion of subset. We want it to be clear which collection of (sub)sets we are working with, an therefore we start from the classically equivalent statement: for every family F of non-empty sets, there exists a function f : F F such that, for each S F, f(s) S. Given a family F, a function satisfying the above conition is calle a choice function on F. Inspire by this formulation of the Axiom of Choice, we consier various families of non-empty sets an ask whether one can efine a choice function on them. This is the topic of Chapter 5. Before we go into the existence an non-existence of choice functions, we have to gain a better insight into some of the families of sets we wish to consier, in particular the collection of the close subsets of N an the collection of the open subsets of N. As we explain in Chapter 4, the intuitionistic mathematician has to istinguish between several efinitions of the notions open subset of N an close subset of N, which woul be equivalent from a classical point of view. In classical mathematics, the Axiom of Choice has a large number of equivalent formulations [2]. In the last chapters we stuy two of these an see what becomes of them if we try to interpret them intuitionistically. In Chapter 6 we start from the statement: every equivalence relation has a set of representatives. We try to fin out if, for several kins of equivalence relations on the natural numbers, there exists a set of representatives an in what sense. 7

8 We also return to the problem of the prisoners an their hats formulate at the beginning of this introuction an explain the classical strategy of the prisoner that claims to be able to get most of them free. We will see that he uses the fact that there exists a set of representatives for a certain equivalence relation on Cantor space C. This equivalence relation appears to be connecte with the famous equivalence relation efine by Vitali on the set R of the real numbers to construct a non-lebesgue measurable subset of R. Consiering the relation on C from an intuitionistic point of view, we prove some theorems to the effect that it fails to have an easily obtainable set of representatives. In Chapter 7, we consier the following classical equivalent of AC: every vector space has a basis. We stuy R as a vectorspace over Q an investigate the existence of a basis for R over Q, a so-calle Hamel basis. We show that, as in classical set theory, there is a close connection between Hamel bases an Vitali sets. Using this connection we prove some positive results showing there is no easily obtainable Hamel basis. 8

9 Chapter 2 Prerequisites This chapter provies a short introuction to intuitionistic mathematics. Some important basic concepts of intuitionism are explaine an I hope that by going through the examples the reaer gets some feeling for a number of typically intuitionistic principles. Definitions of more specific concepts use in this thesis will be given along the way. 2.1 Basic sets We stuy two basic sets. The first is N, the set of the natural numbers. N has a so-calle eciable equality. This means we can ecie for all m, n in N whether m = n or m n. This is not so for the secon funamental set, N, the collection of all infinite sequences of natural numbers, or, one coul say, the collection of all functions from N to N. N is calle Baire space. The elements of N are thought of as being constructe step by step. One efines an element α of N by giving, one by one, its values α(0), α(1), α(2),.... Such a sequence may be given by an explicit algorithm (for example pick 0 in each step ), but, even then, it might be imagine to be the result of choosing ranomly in each step. Definition 2.1. For each m N, we efine an element m of N by m(n) = m for all n in N. For example, 0 is the sequence consisting of only zeros. For α, β N, we efine α = β n [ α(n) = β(n) ]. Saying (α = β) is saying that the assumption n [ α(n) = β(n) ] leas to a contraiction. This is quite weak. We only know that α an β are not everywhere the same, but we may not be able to inicate a place where they iffer. Therefore, we also efine a stronger relation on N. Definition 2.2. We say: α is apart from β, notation α # β, if an only if there exists n N such that α(n) β(n). 9

10 In intuitionistic mathematics the phrase there exists is interprete in a constructive way. That is, for any property A of natural numbers, the statement n N [ A(n) ] means that we are able to explicitly fin a natural number n 0 satisfying A(n 0 ). We cannot ecie for each α, β N, α = β α # β. Let us first explain how we, following Brouwer, interpret the logical symbol. For any two mathematical statements A an B, A B means that we can pick one of the two statements (A or B) an give a proof of this statement. Now consier for instance α N efine by: Note that an α(n) = 0 if there is no block of 99 nines in the first n igits of the ecimal expansion of π, = 1 if there is a block of 99 nines in the first n igits of the ecimal expansion of π. n [ α(n) = 0 ] there is no block of 99 nines in the ecimal expansion of π, n [ α(n) 0 ] there is a block of 99 nines in the ecimal expansion of π. As, at this moment, we o not know whether or not there exists a block of 99 nines in the ecimal expansion of π, we cannot prove α = 0, nor α # 0. Hence we cannot ecie α = 0 α # 0. We will use the example of the 99 nines in the ecimal expansion of π more often, an therefore, we introuce some abbreviations. For each natural number n we efine n < k 99 = there is no block of 99 nines in the first n igits of the ecimal expansion of π. n = k 99 = n is the first position in the ecimal expansion of π where a block of 99 nines is complete. n > k 99 = there is a block of 99 nines in the first n 1 igits of the ecimal expansion of π. Note that we o not give a efinition of k 99 on its own. Nevertheless, we will occasionally talk about k 99 as if it exists in its own right. We coul, for example, say k 99 exists, when we mean: n [ n = k 99 ]. 10

11 2.2 Special subsets of N Let A be a subset of N an n a natural number. We cannot ecie in general n A n / A. (2.1) One coul efine, for example, A = {n N n = 0 the Riemann Hypothesis}. As we can neither prove nor reject the Riemann Hypothesis (at this moment), we cannot ecie 0 A 0 / A. A subset A of N for which we are able to ecie (2.1) for all natural numbers n is calle a eciable subset of N. In fact, we will use the following efinition of the notion eciable subset of N, which is slightly stronger. Definition 2.3. For any α N, we efine: D α = {n N α(n) = 1}. Let X N. We say that α ecies X if an only if X coincies with D α. X is a eciable subset of N if an only if some α N ecies X. Two other special collections of subsets of N are the collection of the enumerable subsets of N an the collection of the co-enumerable subsets of N. These are efine as follows. Definition 2.4. For any α N, we efine: E α = {n N k N [ n = α(k) 1 ]}, CE α = N \ E α = {n N (n E α ) }. Let X N. We say that α enumerates (resp. co-enumerates) X if an only if X coincies with E α (CE α ). X is a enumerable subset (resp. co-enumerable subset) of N if an only if some α N enumerates (co-enumerates) X. Not every enumerable subset of N is eciable. For instance, the set A = {n N n = 1 k 99 exists} is not eciable. As we o not know whether or not there exists a block of 99 nines in the ecimal expansion of π, we are not able to ecie whether the natural number 1 is in A or not. A is enumerable for we can efine α N by α(k) = 0 if k < k 99, = 2 if k k 99. Then A = E α. We will say some more about this topic in Section

12 2.3 Spreas A typical intuitionistic concept is the notion of a sprea. In orer to efine spreas we first introuce some other concepts. To start with, N is the set of all finite sequences of natural numbers. We efine a bijective function : N N by efining, for all k N, for all s 0,..., s k 1 N, s 0,..., s k 1 = p s ps k 2 k 2 ps k 1+1 k 1 1 (2.2) where p n is the n th prime number an by efinition the empty prouct is 1 (whence = 1 1 = 0). Using this function, we may view N an N as the same set. We efine a function lg : N N by lg( s 0,..., s k 1 ) = k. We also efine, for each finite sequence s an all i < lg(s), s 0,..., s k 1 (i) = s i. We efine two concatenation functions, both enote by. The first one is between finite sequences an the secon one is between finite sequences an infinite sequences. Definition 2.5. Let α N an r, s N, say r = r 0,... r m, s = s 0,..., s n. r s = r 0,..., r m, s 0,..., s n, r α = (r 0,..., r m, α(0), α(1), α(2),...). The collection of all finite sequences of natural numbers may be orere as follows. Definition 2.6. Let a, b N. We say a extens to b, notation a b, if an only if c N [ b = a c ]. We say a is a proper initial segment of b, notation a b, if an only if a b a b. Definition 2.7. For all α N, for all m N, α(m) = α(0),..., α(m 1). In case there is no confusion possible, we leave out the parentheses an write αm. We call αm an initial segment of α. Similary, for all s N, for all m lg(s), s(m) = s(0),..., s(m 1). 12

13 Now we are reay to efine the notion of a sprea. Definition 2.8. Let σ : N {0, 1}. We say σ is a sprea law if an only if: (L1) σ( ) = 0, (L2) for all s N, σ(s) = 0 n [ σ(s n ) = 0 ]. For all α N, we say: α is amitte by σ, notation α σ, if an only if for all n N, σ(αn) = 0. The set of all sequences that are amitte by a sprea law σ is calle a sprea. We write σ for both the sprea an the sprea law. One can think of N as a tree with the finite sequences of natural numbers as noes. In this tree there is a path from the finite sequence a to the finite sequence b if an only if a extens to b. The elements of N are all the infinite paths one may construct by going upwars in this tree. For a given sprea σ, the sprea law tells you, in each noe, which branches you can walk if you want to construct an infinite sequence that is an element of σ. Due to property L2 there are no ea ens. Property L1 guarantees that σ contains at least one element. Cantor space, C, is the collection of all infinite sequences of natural numbers assuming no other values than zero an one. This is a sprea, as we can efine the appropriate sprea law σ by: σ(s) = 0 i < lg(s) [ s(i) = 0 s(i) = 1 ] 2.4 Brouwer s Continuity Principle Let R N N be a relation such that, for all α in N, we can fin a natural number n satisfying R(α, n). Then Brouwer s Continuity Principle (CP) states that, for all α in N, we can alreay etermine a natural number n satisfying R(α, n) after knowing only a finite part of α, i.e. α N m N n N β N [ αm = βm R(β, n) ] This might soun improbable at first, but recall how we view elements of Baire space. Every infinite sequence of natural numbers may be thought of as being constructe step by step. If you state α N n N [ R(α, n) ], you claim that if someone (let us call that person your opponent) gives you an element α of N by giving you one by one its values α(0), α(1), α(2),..., at some point you are able to come up with a natural number n such that R(α, n). Suppose you o so after your opponent has given the first 100 values of α. Then for any β that agrees on the first 100 positions with α (i.e. α100 = β100), also R(β, n), as you o not yet know anything about the future values of α. 13

14 In Section 2.1 we have constructe a sequence α for which the statement α = 0 α # 0 is reckless. Using the Continuity Principle one can obtain a contraiction from the assumption α, β N [ α = β α # β ]. First note that we can coe a sequence of elements α 0, α 1,... of N by just one element α of N. To o so we use the map efine at (2.2) on page 12. Definition 2.9. Let α N. For each n N, we efine α n N by, for all m N, α n (m) = α( n, m ). Now suppose α, β N [ α = β α # β ]. (2.3) Define R N N by R(α, 0) α 0 = α 1 R(α, 1) α 0 # α 1. By (2.3), in particular α N [ α 0 = α 1 α 0 # α 1 ], so α N n N [ R(α, n) ]. Applying the Continuity Principle yiels: α N m N n N β N [ αm = βm R(β, n) ]. (2.4) Note that 0 0 = 0 = 0 1, so R(0, 0) an not R(0, 1). Hence, by applying (2.4) for α = 0, we can etermine m N such that: β N [ 0m = βm R(β, 0) ]. (2.5) Define β N by β(n) = 0 if n 1, m, = 1 if n = 1, m. Then, as 1, m > m, βm = 0m. Hence, by (2.5), R(β, 0). But β 0 = 0 β 1. Contraiction. So, using the Continuity Principle, we have proven α, β N [ α = β α # β ]. In Section 2.3 we have introuce spreas. The argument given to justify the Continuity Principle for N can be applie to any sprea. Definition Brouwer s Generalize Continuity Principle (GCP). Let σ N be a sprea an R σ N. If, for all α in σ, there exists a natural number n satisfying R(α, n), then α σ m N n N β σ [ αm = βm R(β, n) ]. The (Generalize) Continuity Principle is a powerful tool. Using GCP one can for example prove that every real function is continuous (the efinition of the real numbers will be given in Section 2.7). The Continuity Principle will play an important role in many proofs in this thesis. 14

15 2.5 Axioms of Choice In this thesis we will investigate what becomes of the various forms an equivalents of the Axiom of Choice if we try to interpret them intuitionistically. Some forms of the Axiom of Choice are acceptable for the intuitionistic mathematician, as these follow from the way we think about elements of Baire space. The first of these is the First Axiom of Countable Choice Definition First Axiom of Countable Choice (AC 0,0 ) Let P N N such that, for each natural number n, we can fin at least one m in N with P (n, m). Then there exists α N such that n N [ P (n, α(n)) ]. In classical set theory AC 0,0 is a theorem (i.e. a statement one can prove from ZF, the usual axiom system for set theory without the Axiom of Choice), as one may efine a function f : N N, by f(n) = µm [ P (n, m) ]. Intuitionistically, this formula oes not efine an element of N, as we may not be able to ecie, for a given n, which m N is the smallest natural number satisfying P (n, m). However, AC 0,0 is accepte as an axiom, because it follows from the way we view N. The elements of N are efine step by step. Hence for any subset P of N N with m N n N [ P (m, n) ], we can efine an element α as follows: in step n fin m N with P (n, m) an efine α(n) = m. This α satisfies the requirement. The First Axiom of Countable Choice is about relations on N. The next choice axiom we consier is about relations between N an N. Before we can formulate this axiom we nee to be clear on what we mean by a function from N to N. We immeiately efine a more general notion, namely the one of a function from a sprea to the natural numbers. Definition Let γ, α N. We efine γ : α n = γ assigns to α the natural number n, = k N [ γ(αk) = n + 1 m N [ m k γ(αm) = 0 ] ]. For any sprea σ an any γ N we efine γ : σ N = γ efines a function from σ to N, = α σ n N [ γ : α n ]. For all α σ, we write γ(α) for the unique natural number n such that γ : α n. 15

16 Why oes it make sense to efine functions from a sprea σ to N like this? Well, such a function sens each element of σ to a natural number. If someone is builing an element of the sprea by giving its values step by step, at some point the function shoul be able to supply the image of this sequence. This is exactly what happens with the γ efine above. For any α N, for all m N, as long as the initial segment αm of α oes not give enough information to etermine the image of α, γ(αm) = 0. However, at some point we know what the image of α is, say after knowing the first k values of α we know its image is n. Then γ(αk) = n + 1. Now we are reay to efine the next choice axiom, the First Axiom of Continuous Choice. Definition First Axiom of Continuous Choice (GAC 1,0 ) Let σ be a sprea an R σ N such that, for all α σ, there exists n N with R(α, n). Then there exists a function γ : σ N, such that, for all α N, R(α, γ(α)). This axiom may be efene as follows. Let σ be a sprea an R σ N such that α σ n N [ R(α, n) ]. We efine γ N step by step. For each a N ecie whether a gives enough information to fin n N with R(α, n) for a sequence α starting with (the finite sequence coe by) a. If this is the case an there is no b N with b a an γ(b) 0, we efine γ(a) = n + 1. Otherwise, γ(a) = 0. One may convince oneself that γ is a function from σ to N an, for all α σ, R(α, γ(α)). The last choice axiom we introuce is about relations on N. We first efine what we mean by a function from a sprea to N. Definition Let α, β, γ N. We efine: γ : α β = γ assigns to α the sequence β, = n N [ γ n : α β(n) ]. For any sprea σ an γ N we efine: γ : σ N = γ efines a function from σ to N, = α σ β N [ γ : α β ]. For all α σ, we write γ α for the unique sequence β such that γ : α β. Definition Secon Axiom of Continuous Choice (GAC 1,1 ) Let σ be a sprea an R σ N such that, for all α σ, there exists β N with R(α, β). Then there exists a function γ : σ N such that, for all α N, R(α, γ α). 16

17 It takes a bit more effort to efen this axiom. For the justification of this axiom an a thorough explanation of all choice axioms introuce in this section the reaer is referre to [4]. GAC 1,1 is the strongest axiom we have introuce in this section. Both AC 0,0 an GAC 1,0 are a consequence of GAC 1, Fans We have seen two examples of spreas up to now: N an C. There is an important ifference between these two spreas. When constructing an element of N you have infinitely many possible choices at each step (you can pick any natural number). On the other han, you only have two options (zero an one) for each position when you are constructing an element of C. A sprea in which you only have finitely many possible choices in each step is calle a fan. A precise efinition of the notion of a fan is the following. Definition Let σ : N {0, 1}. We say σ is a fan law if an only if σ is a sprea law, i.e. σ satisfies: (L1) σ( ) = 0, (L2) for all s N, σ(s) = 0 n [ σ(s n ) = 0 ], an in aition, (L3) for all s N, σ(s) = 0 k N n N [ σ(s n ) = 0 n < k ]. For any fan law σ the set {α N n N [ σ(αn) = 0 ]}, i.e. the set of all infinite sequences that are amitte by σ, is calle a fan. One may view a fan as a tree, where in each noe there are finitely many (but never zero) branches going up. The elements of a fan are constructe by walking up the tree an in each noe picking one of the finitely many possible branches to a following noe Fan Theorem The Fan Theorem is a powerful tool use in intuitionistic mathematics. Brouwer use this theorem to prove that every real-value (continuous) function efine on a close interval is uniformly continuous. First we introuce the notion of a bar. Definition Let σ be a fan an B a subset of N. We call B a bar in σ if an only if every α in σ has an initial segment αn belonging to B. 17

18 The Fan Theorem says every eciable bar has a finite subbar. What o we mean by this? Let σ be a fan an B a eciable subset of N that is a bar in σ. Then the Fan Theorem states we can fin a finite subset B of B, such that B is a bar in σ. One may formulate the Fan Theorem in the following equivalent way as well. Definition Fan Theorem For every fan σ an for every eciable subset B of N that is a bar in σ, there exists a natural number M such that, for all α σ, there exists a natural number n M with αn B. Classically, the Fan Theorem is equivalent to its contraposition, which is known as König s Lemma: every infinite finitely branching tree has an infinite path. There are some expressions in the statement above that perhaps have to be explaine. So let us start with that. Definition Let T N. T is a tree if an only if for all s N, for all n N, if s n belongs to T, then s belongs to T. A tree T is finitely-branching if an only if for all elements s of T, there are only finitely many natural numbers n with s n T. A tree T is infinite if an only if for all n N, there exists s T with lg(s) = n. An element α of N is an infinite path of T if an only if for all n N, αn T. The classical mathematican may prove König s Lemma as follows. Let T be an infinite finitely-branching tree. The classical mathematician now constructs an infinite path α of T step by step. In the first step he notes that T is finitely-branching, hence there are only finitely many n N with n T, say n 0,..., n k. As any element of T starts with either n 0 or n 1 or... or n k an T is infinite, at least for one of the n i the set A i = {s N s(0) = n i } is infinite. α(0) is efine as the smallest such n i. Continuing this way, the classical mathematician efines, for each n N, α(n) = µk [ {s T αn k s} is infinite ] He proves by inuction that such k exists for each n, hence α is a well-efine infinite path in T. So classically König s Lemma hols an thereby also the Fan Theorem hols. This proof of König s Lemma is not acceptable for the intuitionistic mathematician. Have another look at how α(0) is efine. Although the union of the sets A 0,..., A k is infinite, we may not be able to ecie which of the sets 18

19 is infinite. We only know that they are not all finite. The same problem arises when efining α(1), α(2),.... Having seen this, we nee not be suprise that König s Lemma oes not hol intuitionistically. One coul efine a tree T by s T n N [ s = 0n (n < k 99 (n k 99 k 99 is even) ] n N [ s = 1n (n < k 99 (n k 99 k 99 is o) ]. Then T is an infinite finitely-branching tree, but it is reckless to say T has an infinite path. For suppose T has an infinite path, say α. Then α(0) = 0 n N [ n = k 99 n is even ], α(0) = 1 n N [ n = k 99 n is o ]. As we are able to ecie whether α(0) = 0 or α(0) = 1, it follows from the equivalencies above that we can ecie n N [ n = k 99 n is even ] n N [ n = k 99 n is o ]. But this is reckless. Brouwer prove the Fan Theorem using his Principle of Bar Inuction. For a full explanation of his argument the reaer coul consult [3] or [4] Unrestricte Fan Theorem Up to now we have only spoken about eciable bars. Leaving out the ajective eciable in the formulation of the Fan Theorem gives us a more general form of this theorem. Note that this oes not change anything for the classical mathematician. He assumes the law of exclue mile by which every bar is automatically a eciable bar. Definition Unrestricte Fan Theorem Let σ be a fan an let B N be a bar in σ. Then we can fin N N such that α σ n N [ n N (α, n) B ]. Theorem Assuming GAC 1,0, the Unrestricte Fan Theorem can be erive from the Fan Theorem. Proof. Let σ be a fan an B N a bar in σ. We will show that B has a finite subbar B. As B is a bar in σ, α σ n N [ αn B ]. So, by GAC 1,0, there exists a function γ : σ N such that: α σ [ α(γ(α)) B ]. Define of subset C of N by C = {s N γ(s) 0}. 19

20 First note that C is a bar in σ. Let α σ. γ : σ N, so by efinition there exists n N with γ(αn) 0. So α σ n N [ αn C ]. C is a eciable subset of N. Applying the Fan Theorem yiels that C has a finite subbar C, say C = {s 0,..., s n }. The iea for the efinition of B is the following: For 0 i n: Consier γ(s i ) an istinguish two cases: 1. γ(s i ) 1 lg(s i ). Make sure s i (γ(s i ) 1) is in B. 2. γ(s i ) 1 > lg(s i ). Make sure all finite sequences of length γ(s i ) 1 that start with s i an are in σ are in B. To accomplish this we efine B N by: s B i n [ s i (γ(s i ) 1) = s ] i n [ s i s lg(s) = γ(s i ) 1 σ(s) = 0 ]. Note that B is a eciable subset of N. We will prove that B is a finite subbar of B. B B Let s B. It follows from the efinition of B that there are two possibilities: 1. i n [ s i (γ(s i ) 1) = s ]. By efinition of γ this yiels s B. 2. i n [ s i s lg(s) = γ(s i ) 1 σ(s) = 0 ]. Define α N by α(n) = s(n) if n < lg(s), = µi [ σ(αn i ) = 0 ] if n lg(s). α σ, so α(γ(α)) B. (2.6) As s i s = α(lg(s)) an γ(s i ) 0, γ(α) = γ(s i ) 1. (2.7) 20

21 Combining (2.6) an (2.7) we conclue, α(γ(s i ) 1) B. As γ(s i ) 1 = lg(s), So B B. s = α(lg(s)) = α(γ(s i ) 1) B. B is a bar in σ. Let α σ. Determine i N such that α(lg(s i )) = s i (C is a bar in σ). Then α(γ(s i ) 1) B. B is finite. For every 0 i n there are two possibilities: 1. γ(s i ) 1 lg(s i ) s i causes the aition of one extra element to B. 2. γ(s i ) 1 > lg(s i ) σ is a fan, so for each t N with σ(t) = 0 there are only finitely many natural numbers k such that σ(t k ) = 0. So there are only finitely many t σ of length γ(s i ) 1 that start with s i. Any element in B is in there because of either reason 1 or 2. So B only has finitely many elements. Hence B is a finite subbar of B, as require Extene Fan Theorem We often use an extension of the Fan Theorem which states: if, for every element in a certain fan, we can fin a natural number with a esire property, then we can fin N N such that, for each element of the fan, we can fin a natural number with the esire property below N. Definition Extene Fan Theorem Let σ be a fan an R σ N such that for each α σ there exists n N such that R(α, n). Then we can fin N N such that α σ n N [ R(α, n) ]. Theorem Assuming GAC 1,0, the Extene Fan Theorem can be erive from the Fan Theorem. Proof. Let σ be a fan an R σ N with the property α σ n N [ R(α, n) ]. 21

22 By GAC 1,0, there exists a function γ : σ N satisfying α σ [ R(α, γ(α)) ]. Define a subset B of N by, for all s N, s B γ(s) 0. B is eciable an, as γ is a function on σ, B is a bar in σ. So we can apply the Fan Theorem an fin M N such that α σ n M [ γ(αn) 0 ]. Define N = max( { γ(s) lg(s) M } ). Then α σ n N [ R(α, n) ], as require. The Extene Fan Theorem is much stronger than the Fan Theorem an, in contrast to the Unrestricte Fan Theorem, it oes not hol classically. One coul efine for example R C N by, for all α C, for all n N, R(α, n) (α = 0 n = 0) α(n) 0. The classical mathematician woul say this relation satisfies: α C n N [ R(α, n) ]. (2.8) To him, any element α of C is either equal to 0 (in which case R(α, 0)) or is not equal to 0 (in that case he is convince there exists a position n with α(n) 0). Clearly there is no natural number N, such that for all α in C we can fin n N with R(α, n). Hence the Extene Fan Theorem oes not hol classically. The intuitionistic mathematician oes not agree with (2.8), however. Hence the relation R oes not satisfy the requirements of the Extene Fan Theorem. 2.7 The set of the real numbers Starting from the two basic structures we have introuce, N an N, we want to construct more complex structures. To begin with, we efine the integers by Z = (N N)/ 0, where 0 is the equivalence relation: m, n 0 p, q m + q = n + p. Aition, multiplication an orer on Z are efine as usual. The rational numbers are efine by Q = (Z N >0 )/( 1 ), where 1 is the equivalence relation: m, n 1 p, q m q = n p. 22

23 Aition, multiplication an orer on Q are efine as usual. Note that the equality on both Z an Q is eciable. To efine the real numbers we first introuce the set S of all (coe numbers of) rational intervals. Let ρ be a bijective map from N to Q. We say m N coes a rational interval if an only if there exist m 0, m 1 N such that 1. m = m 0, m 1, 2. ρ (m 0 ) ρ (m 1 ). For each m = m 0, m 1 in S we efine m = ρ(m 0 ) an m = ρ(m 1 ), lg(m) = m m. So m an m are the left an the right enpoint of the interval coe by m an lg(m) is the length of the interval coe by m. For rational numbers p, q, we will usually just write [p, q] for the rational interval ρ 1 (p), ρ 1 (q). For all a, b S we efine a b = b a a b (a is containe in b), a b = a b b a (a touches b). Notice we also efine a function lg an a relation on the collection of finite sequences of natural numbers. By stating which collection we are working with it will be clear which notion we mean. Definition Let x N. We say x is a real number, i.e. x R, if an only if x is a shrinking an shriveling sequence of rational intervals. That is, (i). for all n N, x(n) S, (ii). for all n N, x(n + 1) x(n) (shrinking) (iii). for all n N, there exists m in N such that lg(x(n)) 2 m (shriveling) For all x, y R, we efine x y = n N [ x(n) y(n) ], x # y = n N [ ( x(n) y(n)) ], x < y = n N [ x(n) < y(n), ], x y = n N [ x(n) y(n) ]. 23

24 Note that equality on the reals is not eciable. We coul, for example, efine a real number ψ by ψ(n) = [ 1 n, 1 n ] if n k 99, As one can show = [ 1 k, 1 k ] if k = k 99 an k n. ψ 0 k 99 exists, ψ # 0 k 99 oes not exist, we cannot ecie ψ 0 ψ # 0. Far more can be sai about the construction of the real numbers, but as this is not of particular importance for this thesis we o not go into this subject any further. 24

25 Chapter 3 Discrete spreas an fans In the previous chapter we have seen that equality on N is not eciable. One coul, however, stuy subsets of N that o have a eciable equality, i.e. subsets A of N with α, β A [ α = β α # β ]. (3.1) A subset A of N satisfying (3.1) is calle iscrete. In this chapter we stuy iscrete spreas an fans. We will see that these can be characterize in a surprising an nice way. The characterization of iscrete fans leas us to an equivalent of the Fan Theorem, which we prove in the last section of this chapter. 3.1 Discrete spreas In section 2.3 we have introuce spreas. When working with spreas we have a powerful tool, the (Generalize) Continuity Principle (Definition 2.10). Using GCP we will show that a sprea is iscrete if an only if it is enumerable. Definition 3.1. Let A N. A is enumerable iff there exist α 0, α 1,... in N such that A = {α 0, α 1,...}. We start with a lemma an show: for all α an β in a iscrete sprea, we can alreay ecie whether α = β or α # β after knowing finitely many values of α an β. Lemma 3.2. Let σ be a iscrete sprea. Then α σ m N β σ [ αm = βm α = β ]. Proof. Let σ be a iscrete sprea. We efine for each α N α o = (α(1), α(3), α(5),...) α even = (α(0), α(2), α(4),...). 25

26 Similarly, s o an s even are efine for s N. Define a sprea law τ by, for all s N, τ(s) = 0 σ(s o ) = σ(s even ) = 0. As, for all γ τ, both γ o an γ even are in the iscrete sprea σ, γ τ [ γ o = γ even γ o # γ even ]. (3.2) Define a relation R τ N by γ R 0 γ o = γ even, γ R 1 γ o # γ even. Then, by (3.2), for all γ in τ, γ R 0 or γ R 1. So using GCP, γ τ n, m N δ τ [ γm = δm δ R n ]. (3.3) Let α σ. Define γ τ such that γ o = α an γ even = α. Then: γ R 0 an (γ R 1) (as γ o = α = γ even ). Determine m N such that δ τ [ γm = δm δ R 0 ] (see (3.3)). Then δ τ [ γm = δm δ o = δ even ]. (3.4) Now let β τ with αm = βm. Define δ τ by δ o = α an δ even = β. Then γm = δm, so by (3.4), α = δ o = δ even = β. An we conclue, β σ [ αm = βm α = β ]. What oes this lemma tell us? Let us view a sprea as a tree again. Then the lemma states: if you are walking up a tree representing a iscrete sprea, from some point onwars there is only one way up (in each noe there is only one irection to choose). Using this lemma, we prove that every iscrete sprea is enumerable. To efine such an enumeration we first enumerate the set of all finite sequences of natural numbers. For each amitte finite sequence (i.e. each noe in the tree), we construct an element of the sprea by starting at the given finite sequence 26

27 an choosing, at each step, the leftmost branch. We a the resulting infinite sequence to our list. For every element α of the iscrete sprea, we can fin a natural number m such that any sequence in the sprea starting with αm is equal to α. Hence, if you start in the noe αm an take, in each step, the leftmost branch up, you construct α (as α is the only path in the sprea starting with αm), so α is in our enumeration. We make this iea precise in the proof of the following theorem. Theorem 3.3. For every sprea σ, σ is iscrete if an only if σ is enumerable. Proof. Let σ be a sprea. ) Suppose σ is iscrete. Let s 0, s 1,... be an enumeration of N. First efine α by, for all n N, α(n) = µk [ σ(αn k ) = 0 ]. α is well-efine, as σ is a sprea an α σ. We efine a sequence α 0, α 1,... as follows: Let p N an consier s p. We istinguish two cases: 1. σ(s p ) = 1. Then s p is not amitte by σ. Define α p = α. 2. σ(s p ) = 0. Define α p by: α p (n) = s p (n) if n < lg(s p ), = µk [ σ(α p n k ) = 0 ] if n lg(s p ). Note that this sequence is well-efine an that for each p, α p σ. So {α 0, α 1,...} σ. Let β σ. Determine, using Lemma 3.2, m N such that γ σ [ βm = γm β = γ ]. (3.5) Determine p N such that βm = s p. It follows from the efinition of α p that α p m = s p = βm. As α p σ, by (3.5), β = α p {α 0, α 1,...}. So σ {α 0, α 1,...}. 27

28 Hence σ = {α 0, α 1,...} an therefore, σ is enumerable. ) Suppose σ is enumerable. Let {α 0, α 1,...} be an enumeration of σ. Then, α σ n N [ α = α n ]. Applying GCP yiels: α σ m, n N γ σ [ αm = γm γ = α n ]. Let α, β σ. Determine m, n such that γ σ [ αm = γm γ = α n ]. In particular, as αm = αm, α = α n. Hence γ σ [ αm = γm γ = α ]. This means that, by consiering αm an βm, we are able to ecie α = β α # β. So σ is iscrete. 3.2 Discrete fans Every fan is in particular a sprea, so Theorem 3.3 also hols for iscrete fans. However, using the Fan Theorem, we can prove an even stronger statement for fans, namely every iscrete fan is a finite set. (3.6) In the context of elementary intuitionistic analysis, this statement is equivalent to the Fan Theorem. Theorem 3.4. The statement every iscrete fan is a finite set is equivalent to the Fan Theorem Proof. ) Assume: every iscrete fan is a finite set. Let σ be a fan an B N a eciable bar in σ. We efine a function σ : N {0, 1} by: σ (s) = 0 ( s σ n lg(s) [ sn / B ] ) ( s σ n lg(s) [ sn B m < n [ sm / B ] i N [ n < i < lg(s) s(i) = µj [ σ(si j ) = 0 ] ] ] ). 28

29 The iea of σ is that in the beginning, you put all elements of σ also in σ, but once you are in B, you only take the leftmost way up in the fan σ. σ is a fan law, as: L1) The statement on the right-han sie of the - sign is eciable, so σ is a well-efine function from N to {0, 1}. Clearly σ ( ) = 0. L2) For any s N with σ (s) = 0, So σ (s µj [σ( s j ) = 0] ) = 0. s N [ σ (s) = 0 n N [ σ (s n ) = 0 ]. The other irection ( ) follows immeiately from the way σ is efine. L3) From the fact that σ σ an σ is a fan, it follows that s N [ σ (s) = 0 k N n N [ σ (s n ) = 0 n k ]. Furthermore, σ is a iscrete fan. We prove this as follows. Let α, β σ. Determine k N such that αk B. It follows from the efinition of σ that: β σ [ βk = αk β = α ]. Hence, by consiering βk, we can ecie: α = β α # β. As σ is a iscrete fan, we can apply the assumption an conclue that σ is finite. Say σ = {α 0,..., α p }. Define, for 0 i n, n i = µj [ α i j B ]. An efine: B = {α 0 n 0,..., α 0 n p }. Then clearly B B. 29

30 The only thing left to show is that B is a bar in σ. Let α σ. B is a eciable bar in σ, so we can efine n = µi [ αi B ]. Define β N by: β(k) = α(k) if k n, = µj [ βk j σ ] if k > n. Then β σ, so we can etermine i N such that β = α i. As α i n = βn = αn, it follows from the efinition of n that n = n i. An we conclue αn = α i n = α i n i B. So, for all α in σ, there exists a natural number n such that αn is in B. Hence B is a finite subbar of B. ) Assume the Fan Theorem. Let σ be a iscrete fan. Accoring to Lemma 3.2, α σ m N β σ [ αm = βm α = β ]. Using the Unrestricte Fan Theorem we can fin M N such that So α σ m M β σ [ αm = βm α = β ]. α, β σ [ αm = βm α = β ]. (3.7) As σ is a fan, there are only finitely many s N with lg(s) = M an σ(s) = 0. Combining this with (3.7) gives that σ is finite, as require. 30

31 Chapter 4 On open, close an enumerable subsets of N In Chapter 5 we will examine whether there exist choice functions on various collections of subsets of N. The collection of the close subsets of N an the collection of the open subsets of N are two of these collections. Classically, there are a number of equivalent ways to efine the notions of close an open. Intuitionistically, not all those efinitions are equivalent. Before we can go into the problem of the existence of choice functions we have to be clear on what collections we are talking about. In this chapter we stuy the ifferent efinitions an their epenencies. 4.1 All kins of close an open In classical mathematics the most commonly use efinition of open is the one that states: a subset A of N is open iff every element of A has a neighbourhoo that is containe in A. Consiering N with the metric given by (α, β) = n=0 sg(α(n) β(n)) 2 n, where, for every natural number m, sg(m) = 0 m = 0 an sg(m) = 1 m 0, this notion of open correspons with the intuitionistic notion of weakly open. One coul also efine an open subset of N as a union of basic open subsets of N, where a subset A of N is basic open iff there exists s N with A = {α N α(lg(s)) = s }. This efinition correspons with the intuitionistic notion of effectively open, provie we start from a eciable collection of basic open sets, i.e. A N is effectively open iff there exists a eciable subset B of N such that A = {α N α(lg(s)) = s}. s B 31

32 A precise efinition of these two notions of open is the following. Definition 4.1. Let G N. We say G is weakly open if an only if for every α G, there exist n N, such that every γ N passing through αn belongs to G. We say G is effectively open if an only if there exists β N such that, for all α N, α belongs to G if an only if there exists n N with αn D β (i.e. β(αn) = 1). In the efinition of effectively open we coul just as well replace D β by E β. To see this, efine for each β N an element γ β of N by: Then: γ β (n) = 1 if k n [ β(k) 0 β(k) 1 n ], = 0 otherwise. {α N n [ αn E β ]} = {α N n, k [ β(k) = αn + 1 ]}, = {α N m [ γ β (αm) = 1 ]}, = {α N m [ αm D γβ ]}. One can easily show that every effectively open subset of N is also weakly open. The converse is not true intuitionistically. We will show this using the Secon Axiom of Continuous Choice (GAC 1,1 ). Theorem 4.2. Not: every weakly open subset of N is effectively open. Proof. Suppose every weakly open subset of N is effectively open. Define, for every α N, a subset G α of N by G α = {γ N n [ α(n) = 0 ]}. G α is weakly open for each α N (just pick n = 1 for every γ G α ). So from the assumption it follows that, for each α in N, G α is effectively open, i.e.: α N β N γ N [ γ G α n [ β(γn) = 1 ] ]. Accoring to GAC 1,1, there exists a function δ : N N such that α N γ N [ γ G α n [ (δ α)(γn) = 1 ] ]. Consier α = 0. G 0 = N, so in particular 1 G 0. Hence, n [ (δ 0)(1n) = 1 ]. Determine n such that (δ 0)(1n) = 1 an efine m = 1n. Determine k such that δ m (0k) 0. 32

33 By efinition, (δ 0)(m) = δ m (0). So, for all α N with αk = 0k, δ m (α) = δ m (0) = 1. Hence, α N [ αk = 0k 1 G α ]. (4.1) Define α N by: α(n) = 0 if n k, = 1 if n > k. Then αk = 0k an by (4.1), 1 G α. But α(k + 1) = 1, so n [ α(n) = 0 ] an G α =. Contraiction. We conclue: not every weakly open subset of N is effectively open. Let us move on to the close subsets of N. To begin with, we give a efinition of the notion of the complement of a subset of N. Definition 4.3. Let A N. The complement of A, notation A, is efine by A = {α N α / A}. So A is the set of all α in N for which the assumption α belongs to A leas to a contraiction. Note that not for every subset A of N, (A ) = A. Having two ifferent notions of open, the iea: a set is close if an only if it is the complement of an open set, also leas to two notions of close subset of N. Note that we say it is the complement of an open set an not its complement is open. Classically these two statement are equivalent, but intuitionistically they are not. We also introuce a thir notion of close, namely sequentially close. Definition 4.4. Let F N. The set F is weakly close if an only if there exists a weakly open subset G of N such that F is the complement of G. We say F is effectively close if an only if F is the complement of an effectively open set. Or, equivalently, F is effectively close if an only if there exists β N such that, for all α N, α belongs to F iff for all n N, αn D β. We call F sequentially close if an only if, for every γ N, if, for every n N, there exists α F passing through γn, then γ itself belongs to F. We will now investigate how those ifferent notions of being close are relate. 33

34 Theorem 4.5. Every weakly close subset of N is sequentially close. Proof. Let F be a weakly close subset of N. Suppose γ N an n N α F [ γn = αn ]. We have to prove: γ F. As F is weakly close, we can fin a weakly open set G such that F = G. Note that: γ F γ / G, ( n β [ γn = βn β G ] ), n ( β [ γn = βn β G ] ). Let n N. By assumption, there exists α F, such that γn = αn. As α F, ( β [ αn = βn β G ] ). Hence, using γn = αn, ( β [ γn = βn β G ] ). This hols for all n N, so γ F an F is sequentially close. The converse is not true constructively: not every sequentially close subset is weakly close. Let P be an unsolve mathematical problem an efine F = { α N P P }. F is sequentially close, but not weakly close. For suppose F is weakly close. Then we can fin a weakly open set G such that F = G an, for all α N, (α F ) (α / G). As (α / G) is equivalent to α / G, for all α N, (α F ) α G = F. So if (P P ), then P P. But (P P ) is true intuitionistically an P P may be reckless [5]. It is clear that every effectively close subset of N is also weakly close. However, as in the case of the open subsets of N, the converse oes not hol. The proof of this theorem resembles the proof of Theorem 4.2. Theorem 4.6. Not: every weakly close subset of N is effectively close. Proof. Just as in the proof of Theorem 4.2, efine, for every α N, G α = {γ N n [ α(n) = 0 ]}. G α is weakly open, an therefore, F α = (G α ) is weakly close. Suppose F α is effectively close for each α, then: α β γ [ γ F α n [ β(γn) = 1 ] ]. 34