aa r X i v : . [ m a t h . L O ] F e b INTUITIONISM: AN INSPIRATION?
WIM VELDMAN τοιγὰρ ἐγώ τοι, ξεῖνε, μάλ᾿ ἀτρεκέως ἀγορεύσω
Well then, let me you, stranger, precisely explain these matters
Od. α Introduction
Topology and foundations.
We want to introduce the reader to the in-tuitionistic view of mathematics proposed, developed and defended by the Dutchmathematician L.E.J. Brouwer (1881-1966) .Brouwer obtained his fame as a mathematician by a number of very importantresults in topology such as the Dimension Theorem :For all positive integers m, n ,if there is a homeomorphism from [0 , m to [0 , n , then m = n ,and the closely related Brouwer Fixed-point Theorem :For each positive integer n , for every continuous function f from[0 , n to [0 , n , there exists p in [0 , n such that f ( p ) = p .These results were obtained in 1911.Brouwer had started thinking on the foundations of mathematics earlier thanthat. In his dissertation from 1907, see [9], he wrestled already with the concept ofthe continuum. In a famous paper from 1908, see [10], he attacked the principle ofthe excluded third X ∨ ¬ X .1.2. Against the formalists.
In his inaugural lecture at the University of Ams-terdam in 1912, entitled: ‘
Intuitionism and Formalism ’, see [11], Brouwer declareshimself an opponent of the formalists . The formalists, in Brouwer’s words, explainthe exactness and precision of the statements of mathematics by describing math-ematics as a game with meaningless strings of symbols according to very strict andprecise rules. The game may turn out to be useful for certain purposes, and thenmay give ‘ a vague sensation of delight ’, but these other purposes are not the con-cern of the mathematician. In the formalist’s ‘game’, there is no place for truth ,only for correctness .Brouwer most emphatically did not want to explain away the meaningfulness ofmathematical statements and the experience of truth. In his intuitionistic view, thesource of mathematical exactness is not to be found ‘ on paper ’ but ‘ in the humanintellect ’.1.3.
I. Kant.
Brouwer was inspired by the philosopher I. Kant (1724-1804) whodescribed the simple theorem ‘7+5=12’ as the result of a construction in ‘ pureintuition ’.Pure intuition has to be distinguished from observation by the senses. In Kant’sterms, it is the ‘form’ of all observations by the senses, and, as such, independentof all individual observations by the senses, i.e. a priori . The title of the paper is taken from an exclamation by the philosopher L. Wittgenstein, see[64]. t is very important that one has to do something in order to come to the insightof the truth of ‘7+5=12’: counting 1, 2, 3, . . . , analytic , i.e. a matter oflogic, but synthetic , see [37, § Languageless constructions.
For the intuitionistic mathematician, everymathematical theorem is, like ‘7+5=12’, the result of a construction in pure in-tuition . The successful completion of this construction, the proof of the theorem ,gives joy and a sense of beauty, and one wants to tell one’s friends.A construction in pure intuition is languageless . Language comes in if I want topreserve the memory of the construction for myself, or if I want to make an attemptto explain to my fellow mathematician what I did, in the hope she will be able todo something similar. Language is not trustworty: it always may fail to do whatone expects it to do, as we all know from our experience as unsuccessful teachersand forgetful researchers.There is no such thing as an exact linguistic description of a mathematical con-struction. The linguistic description only is the accompaniment of an act of un-derstanding, either between me and one of my former selves, or between me and afellow mathematician.Nevertheless, it is our task to study the language of mathematics carefully andto find out where it perhaps may be improved. Although it is impossible we expressourselves in a perfect way, we always have to try to do better than we did untilnow.1.5.
Intuitionistic mathematics is constructive mathematics.
Intuitionisticmathematics is not a new, alternative kind of mathematics. Mathematics is intu-itionistic mathematics. Brouwer’s revolution is a call for reflection . A sharpenedawareness of what it is we do when doing mathematics makes us more careful andmore precise and, thereby, hopefully, better mathematicians.Intuitionistic mathematics is constructive mathematics. The reason is that math-ematics itself essentially is constructive although this fact has been obscured by ourmistaken trust in (classical) logic.Thinking of Brouwer’s ability in topology one may reflect that a continuousfunction f from [0 , m to [0 , n has the property that, for each p in [0 , m , onemay effectively find approximations of the function value f ( p ) from approximationsof the argument p . Continuous functions are constructive functions: there is a deepconnection between Brouwer’s foundational interests and his topological concerns.1.6.
The contents of the paper.
The paper is divided into 21 Sections. Section2 treats the basic intuition giving rise to the nonnegative integers. Section 3 is ashort reflection on theorems and their proofs. In Sections 4-6, we consider threefamous results, that usually are taken to be negative statements although theyare not negative at all. In Section 4, we have a look at Euclid’s theorem thatthere are infinitely many primes. In Section 5, we consider the theorem that √ reals oscillating around
0: if x is such a number, one is unable to make out if x < x = 0 or x > X ∨ ¬ X . In Section 9 we studya basic result of real analysis: the Intermediate Value Theorem . The theorem isan existential statement that fails when taken at its constructive face value, but, if uitably reformulated and adapted, leads to several true and useful results. Findingsuch reconstructions of a constructively invalid result is part of the intuitionisticprogram .Intuitionistic mathematicians hold that every real function is (pointwise) con-tinuous. In Section 10, we treat the preliminary observation that a real functioncan not have a positive discontinuity. In Section 11, we go into the intuitionisticmeaning of negation . In Section 12, we introduce the Continuity Principle , an ax-iom used by Brouwer and we show that this axiom leads to the positive result thatreal functions are (pointwise) continuous.In Section 13, we introduce the
Fan Theorem , a theorem that perhaps, like theContinuity Principle, should be called an axiom: we discover Brouwer’s ‘proof’:most of us would call it a philosophical argument. One should note, however, that,as long as we do not organize our mathematical results in a formal way, it is difficultto distinguish axioms from theorems. As we shall see, the Fan Theorem implies thata real function from [0 ,
1] to R is not only pointwise but even uniformly continuous.As an aside, we prove that the Fan Theorem fails to be true in computable analysis .In Section 14, we go into the rˆole of the Fan Theorem in game theory.In Section 15, we consider Brouwer’s Thesis on Bars in N = ω ω , formulated asa Principle of Bar Induction . The Thesis on Bars in N implies the Fan Theorembut is a far stronger statement. We show how the argument for the Fan Theoremextends to the Thesis on Bars in N . As an application we prove, in Section 16, the Principle of Open Induction in [0 , stumps : inductively generatedsets of finite sequences of natural numbers, whose rˆole may be compared to the rˆoleof countable ordinals in classical, non-intuitionistic analysis. As an application wegive, in Section 18, an intuitionistic formulation and proof of the Clopen RamseyTheorem , itself an impressive extension of Ramsey’s result from 1928. The proof ofthis intuitionistic theorem has not been published before. In Section 19 we studyBorel sets. Because of the failure of De Morgan’s Law ¬∀ n ¬ P ( n )] → ∃ n [ P ( n )], thesubject of descriptive set theory needs a complete reconstruction. In Section 20 wereflect on the notion of a finite subset of the set N of the non-negative integers.This mathematical notion may be made intuitionistically precise in uncountablymany ways. It is a perfect illustration of the subtlety and expressivity the languageof mathematics has obtained from Brouwer’s intervention.In Section 21, we briefly describe how E. Bishop, who founded his own schoolof constructive analysis in the 1960’s, treated Brouwer’s legacy, and how also P.Martin-L¨of’s view on constructive mathematics was influenced by Brouwer.Many Sections may be read independently from other Sections.2. The basic intuition
If you don’t become like little children . . . . Mathematics is a child’s game and each of us, from his tender days, is familiarwith the infinite sequence: 0 , , , , . . . The sequence of these natural numbers or non-negative integers has no end .No end? Can we always go on? Yes, yes, we can always go on!We never die, or get tired of the game. Please, do not plague us with such sillyobjections.We courageously stick to the fascinating perspective of the never finished infinite. Matthew 18:3 .2. Induction and recursion.
The function S produces, given any non-negativeinteger n , the next one, S ( n ): S (0) = 1 , S (1) = 2 , S (2) = 3 , . . . Let P be a property of non-negative integers such that one has a proof of P (0),and, for each n , a proof of P (cid:0) S ( n ) (cid:1) from P ( n ), i.e. a proof of P (1) from P (0), aproof of P (2) from P (1), a proof of P (3) from P (2), and so on. One then proves,step by step, first P (1), then P (2), then P (3), and so on, that is, one proves, foreach n , P ( n ). We see this happening although the process is never finished.The principle of complete induction on the non-negative integers thus is obvious.Suppose one defines: for each m , m +0 = m and for each n , m + S ( n ) = S ( m + n ).One then calculates, first m + 1 = m + S (0), then m + 2 = m + S (1), then m + 3 = m + S (2), and so on, and, in this way, one finds, for each n , the meaningof m + n .This also is obvious.2.3. The ‘logicist’ attempt to ‘prove’ induction.
In the nineteenth century,G. Frege and R. Dedekind tried to define the non-negative integers, logically orset-theoretically, and then to prove the principle of complete induction and thepossibility of defining functions by recursion.Brouwer considered such attempts futile and misleading.2.4.
Awakening consciousness.
Brouwer spent a lot of thought on finding the origin of the idea of the non-negative integers, see for instance [14].
A first thing, then a next thing, then another thing, and again another thing . . .
Mark the beginning: a first thing, then a next thing . Brouwer called it the basic intuition of two-ity . He saw it happening in an awakening of slumberingconsciousness, a shift of attention, a moment’s reflection, or a move of time : hereI am, and now I see: ‘here I am’, and that is two things. Iterating this shift ofattention, I obtain three things, four things, and so on.One is reminded of R. Dedekind’s proof of [23, Theorem 66]:
There exist infi-nite sets , for instance, the ‘ world of my thoughts, meine Gedankenwelt ’. Dedekindinvokes a function s s ′ that associates to every thought s of mine the thought s ′ : ‘I am thinking of s ’ and concludes, from the fact that this function should beone-to-one and non-surjective, the latter because ‘ I myself, mein eigenes Ich ’ am athought of mine not of the form s ′ , that I have infinitely many thoughts.As we mentioned already, Brouwer refers to Kant. Kant held that the conceptsof ‘time’ and ‘space’, the ‘forms’ of ‘ inner sense ’ and ‘ outer sense ’, respectively,are apriori concepts, that is, independent from every observation by the senses.These apriori concepts are studied by mathematics, in its two divisions: ‘arithmetic’and ‘geometry’. Brouwer wants to maintain Kant’s conviction about the aprioricharacter of ‘time’ but rejects the apriori character of ‘space’. This is partly becauseKant believed space to be Euclidean, and, after Kant, one had the non-euclideanrevolution. Also, geometry may be ‘arithmetized’ and be robbed of its status as anindependent discipline.2.5. The ‘set’ N . We use N to denote the set of the nonnegative integers. This ‘set’ is a well-understood totality but an ongoing project, in no sense finished orcomplete. We use m, n, . . . , s, t, . . . as variables over N .3. Theorems and their proofs
Mere announcements.
One does not understand a mathematical theoremimmediately upon hearing its statement. The statement of the theorem is only a reliminary announcement, a partial communication, a promise. Its full meaningwill unfold itself in the proof , and understanding this full meaning is reserved forsomeone who takes the trouble to go through the proof and to reconstruct it forherself.3.2. No to ‘Platonism’.
The statement of a theorem tells us that some mentalconstruction has been succesfully completed. It is not reporting a fact one might‘observe’ in a mathematical reality lying outside us.The latter view has been strongly defended by Plato (427-347 B.C.). He believedthat the contemplation of the timeless mathematical reality, that one approacheswith the mind and not with the senses and that one should not touch with one’sfingers , would prepare us for the contemplation of what is the true interest of thephilosopher: the timeless reality of the ‘ Ideas ’.The intuitionistic mathematician rejects this view. In the ancient debate betweenSpeusippus (405-338 B.C.) and Menaechmus (380-320 B.C.) as reported by Proclus(410-485), she would definitely have sided with Menaechmus. Speusippus heldthat every ‘problem’ essentially is a ‘theorem’, i.e. a truth to behold, whereasMenaechmus defended the view that every ‘theorem’ essentially is a ‘problem’, i.e.a construction to find and carry out. Proclus, in his wisdom, judged that bothviews contain some truth , and, indeed, one might say, both views try to explain apart of our mathematical experience.In our days, many mathematicians still feel they are discovering hard facts not oftheir own making. They prefer a so-called realistic view, reminiscent of Plato’s,while admitting it is difficult to give a convincing account of the mathematicalreality see [35, p. 123].As to Brouwer’s formalist adversaries, some of them develop their formalismswith an eye to ‘reality’ while others, like Brouwer himself, keep far from it. A famousdefender of the realistic point of view is K. G¨odel (1906-1972). The formula CH inthe language of set theory that is taken to represent Cantor’s famous continuumhypothesis has been shown, by G¨odel himself in 1938, and by P. Cohen in 1963,respectively, not to be refuted and not to be proven by the rules of first-orderpredicate logic from the formulas accepted as representing ‘the’ axioms of set theory,provided the latter do not give a contradiction. G¨odel observed, even before Cohenproved his result, that this fact does not solve the problem: we still have to findout if CH holds in the mathematical universe, see [30] .The intuitionistic mathematician discards the realistic view: he believes it iswrong and harmful for the practice of mathematics.3.3. Propositions (as yet) without proof.
From our intuitionistic perspective,it seems impossible to fully understand a proposition that has (as yet) no proof.We like to say, however, that we also ‘ understand ’ the proposition ‘0 = 1’, in thesense that we see this proposition never will have a proof.Besides, propositions that are only partially understood in the sense that we haveas yet no proof and also do not see that we never will have one, have an importantpart to play in mathematics. Let us consider an example. See Plato,
Politeia (the Republic) VII The 48 propositions of the First Book of Euclid’s
Elements (300 B.C.) are divided into 34‘theorems’ and 14 ‘problems’. ‘Both parties are right’, see [45, Prologue, Part Two, § In an intuitionistic context, the classical continuum hypothesis has no immediate meaning.One may however study meaningful statements that seem to come close tot it, see [29] and [65,Sections 7 and 8]. “Of course it is happening inside your head, Harry, but why on earth should that mean thatit is not real?” Dumbledore in: J.K. Rowling, Harry Potter and the Deathly Hallows tudying Euclid’s proof that there are infinitely many primes , one comes to un-derstand the statement: ‘ there are infinitely many primes ’ and to enjoy its beautyand depth. One now may ask: ‘ Are there also infinitely many twin primes, i.e.prime numbers p such that also p + 2 is prime? ’. By analogy , we have some under-standing of the latter proposition. We can imagine what a proof of this propositionmight be and, in this sense, catch its meaning, although, up to now, no proof orrefutation has been found.By his exemplary proof, Euclid suggests us how to (partially) understand otherpropositions of the form ‘ there are infinitely many numbers n such that P ( n )’.3.4. Negative theorems.
There is a large supply of theorems whose statementis negative. Such theorems report us that a construction of a kind we hoped foris impossible. We obtain such conclusions imagining ourselves possessed of a con-struction of the required kind and seeing that we then could prove 0 = 1. Wehastily retreat, admitting we were imagining the impossible.Indeed, when constructing proofs, we often use our imagination . We make tem-porary extra assumptions for the sake of the argument , that later perhaps will haveto be given up. G. Gentzen (1909-1945), the inventor of ‘
Natural Deduction ’, putgreat emphasis on this fact. Gentzen was inspired by Brouwer, see [27, Section II, § second-rank . They tellus about a lost hope, a failure. Fortunately, many mathematical theorems are unnecessarily negative . Their negative formulation is caused by laziness or lack ofattention. The intuitionistic mathematician likes to find the positive content ofseemingly negative statements.In Sections 4, 5 and 6, trying to make this clear, we give some examples.4. Infinitely many prime numbers
The following theorem is due to Euclid, see [25, Book IX, Proposition 20]. It isimportant and rewarding to study the proof in the original.
Theorem 1.
The prime numbers are more than every quantity of prime numbersgiven beforehand.Proof.
Let the prime numbers given beforehand be
A, B, C . I say that there aremore prime numbers than
A, B, C . Let the least number measured by A, B, C betaken, and let it be DE , and let the unit DF be added to it; we thus obtain EF . EF is prime or EF is not prime . First, let it be prime. Then the prime numbers A, B, C, EF are more than
A, B, C . Next, let EF be not prime. Then there is aprime number that measures EF . Let it be G . I say G is not the same with anyof A, B, C . For, if possible, let it be so. Now
A, B, C measure DE and thus also G measures DE . But the number G also measures EF and thus the number G measures the remainder DF which is a unit , and that is absurd. So G is not thesame with any of A, B, C . Thus prime numbers
A, B, C, G are found which aremore than than the quantity of prime numbers
A, B, C given beforehand. (cid:3)
We now paraphrase Euclid’s proof in contemporary terms. Euclid explains us,how, given any finite list q , q , . . . , q n − of prime numbers, one produces a primenumber q that is not on the list: consider c := the least common multiple of q , q , . . . , q n − . If c is prime, take q := c . If c is not prime, determine the leastnumber q > c . Then q is prime, and, for every i < n , q = q i . See Section 4. the least common multiple of Euclid makes this case distinction as he did not consider a number to be a divisor of itself. ne may seriously wonder if this paraphrase is better than the original.Note that Euclid only treats the case: n = 3. He of course expects the reader topick up the general idea from this exemplary case. The reader must be willing tounderstand him . This is a conditio sine qua non for every written proof.Also note that Euclid provides an algorithm : he demonstrates how, given anyfinite list of prime numbers, one calculates a prime number that still is missing.Most importantly, as we observed already in the previous Section, Euclid ina way lays down the meaning of the word ‘infinite’ . His proof suggests how to understand a proposition of the form: ‘ there are infinitely many numbers satisfying A ’. One may prove such a thing by doing something like he does in the case of theprime numbers.Sometimes, a description of Euclid’s proof is given that does no justice to theargument. According to this description, one is invited to assume : there is a finitelist containing all the primes. Euclid then reduces this assumption to absurdity andforces one to admit that the initial assumption is false: there is no such finite list, which is to say: there are not finitely many primes.G.H. Hardy, a great mathematician and a convinced platonist, has this view, see[35, Section 12]. In the context of Euclid’s theorem, he calls reductio ad absurdum ‘ one of a mathematician’s finest weapons ’, but adds a footnote stating: ‘The proofcan be arranged so as to avoid a reductio and logicians of some schools would preferthat it should be.’ √ is irrational The theorem that √ ‘those who know aboutthe odd and the even’ . The theorem uses the following Lemma 2.
For all positive integers n , if n is odd, then n is odd.Proof. Find p such that n = 2 p + 1. Then n = (2 p + 1)(2 p + 1) = 4 p + 4 p + 1 =2(2 p + 2 p ) + 1. (cid:3) Theorem 3.
There are no positive integers m, n such that √ mn .Proof. Suppose there are. Find such m, n and assume: at least one of m, n is odd.Then √ mn , so 2 n = m , so m is even and, by Lemma 2, also m is even and4 divides m = 2 n , so n is even and, again by Lemma 2, n is even. We thus see:both m, n are even. Contradiction. (cid:3) Already in ancient times, some people worried about this proof, in particularabout the sentence: ‘
Suppose there are. ’. How is it possible to suppose that √ Suppose ’ must mean something like ‘
Imagine ’. How can one make aclear mental picture of a fact that will turn out to be false? How to imagine theimpossible?In particular, Parmenides (fl. 475 B.C.) and Zeno (fl. 450 B.C.), leading mem-bers of the
Eleatic school of philosophy , felt this difficulty. The Dutch mathemati-cian G.F.C. Griss (1898-1953) shared their concern and suggested to Brouwer toremove the proposition without proof, negation and the empty set from the dis-course of mathematics. He started a redevelopment of intuitionistic mathematics,see, for instance, [33]. Brouwer did not want to go into his suggestion, and offeredthe somewhat surprising argument that, following it, one would impoverish thesubject, see [13]. Nevertheless, in intuitionistic, and, more generally, in construc-tive mathematics, one avoids negation as much as possible, also for the reason thatnegative results lack constructive content . he negatively formulated Theorem 3 may be replaced by a positive and affir-mative result, as follows. Definition 1.
A real number x is positively irrational if and only if, given anyrational number q = mn one may effectively find a positive integer p such that | x − mn | ≥ p . Theorem 4. √ is positively irrational.Proof. Let m, n be given. Note 2 n = m , and therefore: | n − m | ≥ | − m n | ≥ n , that is |√ mn | · |√ − mn | ≥ n . If mn >
2, then mn − √ > and,if mn ≤
2, then |√ mn | < |√ − mn | > n . (cid:3) Uncountably many reals
We sketch how to treat integers and rationals and then define the notion of a real number , essentially like G. Cantor (1845-1918) did it, and Brouwer after him.6.1.
Integers and rationals.
One first introduces, on the set N of the naturalnumbers, the operations + , · , exp and the relations <, ≤ by recursive definitions, assuggested in Section 2, and proves, by induction , their well-known nice properties.One then defines a pairing function from N × N to N , for instance the function( m, n ) = 2 m (2 n + 1) −
1, with inverse functions
K, L : N → N . We will write m ′ = K ( m ) and m ′′ = L ( m ), so, for each m , m = ( m ′ , m ′′ ).Then an equivalence relation = Z on N is introduced by: m = Z n ↔ m ′ + n ′′ = m ′′ + n ′ . Integers, for us, are just natural numbers and not,as in the usual treatment, the equivalence classes of the relation = Z . Functions andrelations on Z are functions and relations on N that respect the relation = Z . Thefunctions + Z , · Z , the relations < Z , ≤ Z ,and the special elements 0 Z , Z are defined inthe way one would expect.One then defines Q := { m | ¬ ( m ′′ = Z Z ) } . An equivalence relation = Q isintroduced on Q by: m = Q n ↔ m ′ · Z n ′′ = Z m ′′ · Z n ′ . Rationals, for us, arejust natural numbers, but, whenever we consider natural numbers as rationals, werespect = Q . The functions + Q , · Q , the relations < Q , ≤ Q , and the special elements0 Q , Q are defined straightforwardly.In general, we omit the subscripts ‘ Z ’, ‘ Q ’ unless we fear for confusion,It is important to realize that all functions and relations mentioned so far arealgorithmic. We learned the algorithms at elementary school and know, for instance,how to find out, given rationals p, q , if p = Q q or not.6.2. Real numbers.
We now are ready to introduce the reals.
Definition 2.
A real number is an approximation process , an infinite sequence x = x (0) , x (1) , . . . of pairs x ( n ) = (cid:0) x ′ ( n ) , x ′′ ( n ) (cid:1) of rationals such that (i) x is shrinking: for all n , x ′ ( n ) ≤ x ′ ( n + 1) ≤ x ′′ ( n + 1) ≤ x ′′ ( n ) and (ii) x is dwindling: for all m , there exists n such that x ′′ ( n ) − x ′ ( n ) ≤ m . R is the set of all real numbers. We use x, y, . . . as variables over R . Brouwer gave Cantor’s notion his own colouring.Every real number x = x (0) , x (1) , . . . is an infinite sequence of successive approxi-mations , and, like the infinite sequence 0 , , , . . . itself, never finished and alwaysunder construction. Even if an algorithm has been given determing the successivevalues of x , the finding of these successive values keeps us busy forever. The number is the process of approximation itself: it makes no sense to ask for the limit point that the successive approximations are striving for.For all reals x , for all rationals p, q , for all n , if p < Q x ′ ( n ) < Q x ′′ ( n ) < Q q , wesay that ( p, q ) is an approximation of x of precision q − Q p . efinition 3. For all real numbers x, y , we define: x < R y ( x is smaller than y ) if and only if ∃ n [ x ′′ ( n ) < Q y ′ ( n )] , and: x ≤ R y ( x is not greater than y ) if and only if ∀ n [ x ′ ( n ) ≤ Q y ′′ ( n )] , and: x R y ( x is apart from y ) if and only if either x < R y or y < R x , and: x = R y ( x coincides with y ) if and only if x ≤ R y and y ≤ R x . We suppress the subscript ‘ R ’ if we expect that, doing so, we will not confusethe reader.Traditionally, our ‘ real numbers ’ would be called ‘ number generators ’.The name ‘ real number ’ would then be reserved for equivalence classes of the form[ x ] = { y ∈ R | y = R x } .Also Brouwer does so, using the terms ‘ point ’ and ‘ point kernel ’.We will not make this distinction but will only consider operations and relationson R that respect the equivalence relation = R .The relations < R and R are positive relations. From a constructive point ofview, they are more important than the relations ≤ R and = R . The latter tworelations are negative , and, moreover, expressible in terms of the positive relations,as: x ≤ R y ↔ ¬ ( y < R x ) and: x = R y ↔ ¬ ( x R y ). Given real numbers x, y onemay prove : ‘ x = R y ’ by assuming: ‘ x R y ’ and obtaining a contradiction.It is important that the positive relations < R and = R are co-transitive , i.e. x < R y → ( x < R z ∨ z < R y ) and x R y → ( x R z ∨ z R y ).It suffices to prove this for the first relation.Given x, y, z such that x < R y , find n such that z ′′ ( n ) − z ′ ( n ) < Q y ′ ( n ) − x ′′ ( n )and decide: either x ′′ ( n ) < Q z ′ ( n ) and x < R z or: z ′′ ( n ) < Q y ′ ( n ) and z < R y .6.3. Cantor’s proof.
In 1873, Cantor proves the following, see [19, § Theorem 5.
For every infinite sequence x , x , . . . of reals, there exists a real x apart from every real in the sequence, that is: x R x , x R x , . . . Proof.
Let an infinite sequence x , x , x , . . . of reals be given.Define x = x (0) , x (1) , . . . , as follows, step by step.First, define x (0) = (0 , x ( n ) has been defined and x ′′ ( n ) − x ′ ( n ) = Q n .Find m := the least m such that x ′′ n ( m ) − x ′ n ( m ) < Q n +1 .Note: either x ′ ( n ) + Q x ′′ ( n ) < Q x ′ n ( m ) or x ′′ n ( m ) < Q x ′ ( n ) + Q x ′′ ( n ). If x ′ ( n ) + Q x ′′ ( n ) < Q x ′ n ( m ), define: x ( n + 1) := (cid:0) x ′ ( n ) , x ′ ( n ) + x ′′ ( n ) (cid:1) and, if not , define x ( n + 1) := (cid:0) x ′ ( n ) + x ′′ ( n ) , x ′′ ( n ) (cid:1) .By this choice of x ( n + 1), we ensure x ′′ ( n + 1) − x ′ ( n + 1) = Q n +1 and x R x n . (cid:3) Our constructive argument differs but little from Cantor’s. Cantor makes itperfectly clear how, given any countable list x , x , . . . , of reals, one produces areal positively different from every real on the list.We thus see that the statement ‘ R is uncountable ’, essentially, is a positive state-ment.Note that, by his exemplary proof, Cantor suggest to us how we should under-stand and prove a proposition of the form ‘ A ⊆ R is uncountable ’.One defines operations + R and · R on R by:for all n , ( x + R y )( n ) := (cid:0) x ′ ( n ) + Q y ′ ( n ) , x ′′ ( n ) + Q y ′′ ( n ) (cid:1) and( x · R y )( n ) = (cid:0) min Q ( x ′ ( n ) · Q y ′ ( n ) , x ′ ( n ) · Q y ′′ ( n ) , x ′′ ( n ) · Q y ′ ( n ) , x ′′ ( n ) · Q y ′′ ( n )) , max Q ( x ′ ( n ) · Q y ′ ( n ) , x ′ ( n ) · Q y ′′ ( n ) , x ′′ ( n ) · Q y ′ ( n ) , x ′′ ( n ) · Q y ′′ ( n )) (cid:1) .Also subtraction and an absolute value function x
7→ | x | may be defined. or every rational q , we define q R in R such that, for all n , q R ( n ) = ( q − Q n , q + Q n ).We omit subscripts ‘ R ’ where we think it does no harm to do so.7. Fugitive integers and oscillating reals
The following definition lies at the basis of many counterexamples in Brouwer’sstyle to various mathematical results that do not stand a constructive reading.We consider the decimal expansion of π , evaluating it step by step, hunting forthe first block of 99 consecutive 9’s. We would like to define k as the least number n such that at the places n, n + 1 , . . . n + 98 in the decimal expansion of π we findthe value 9, but we have to be careful as we do not know if such a number n exists. Definition 4 (The fugitive ‘number’ k ) . Let d = d (0) , d (1) , . . . be the decimalexpansion of π , that is: d is the function from N to { , , . . . , } such that π = 3 + P ∞ n =0 d ( n ) · − n − .For each n , we define: k ≤ n if and only if ∃ j ≤ n ∀ i < d ( j + i ) = 9] , and: n < k if and only if ∀ j ≤ n ∃ i < d ( i + j ) = 9] , and: n = k if and only if k ≤ n and ∀ j < n [ j < k ] . Note that we do not define a natural number k but only the meaning of anexpression like ‘ k ≤ n ’.Also note that, for each n , one may find out, for each of the propositions ‘ k ≤ n ’,‘ n < k ’, and ‘ n = k ’, if they are true or not, by simply calculating the first n values of the decimal expansion of π .The statement ‘ P := ∃ n [ n = k ]’ is a prime example of an undecided proposition ,i.e. we do not have a proof of P , but we also do not have a proof that we neverwill find one, and we have no procedure to find either one of these proofs in finitelymany steps.The problem of the 99 9’s in the decimal expansion of π is not important initself. It only is a pedagogical example , showing that we have no method to solve this kind of problems . Should someone, by historical accident, find the 99 9‘s, thenone easily formulates a similar proposition that is still undecided. Definition 5 (The ‘oscillating’ reals ρ , ρ , ρ ) . We define real numbers R , ρ , ρ and ρ , as follows. R is the real number such that, for all n , R ( n ) = ( − n , n ) . ρ = ( ) k is the real number such that, for all n < k , ρ ( n ) := ( − n , n ) ,and, for all n ≥ k , ρ ( n ) := ( k , k ) . ρ = ( − ) k is the real number such that, for all n < k , ρ ( n ) = ( − n , n ) ,and, for all n ≥ k , if k is even, then ρ ( n ) := ( k , k ) , and, if k is odd,then ρ ( n ) := ( − k , − k ) .Finally, ρ := ρ + R ρ . Now note the following:(1) 0 R = R ρ ↔ ∀ n [ n < k ], and we have no proof of ‘ ∀ n [ n < k ]’.Also: ¬ (0 R = R ρ ) ↔ ¬∀ n [ n < k ], and we have no proof of ‘ ¬∀ n [ n < k ]’.Finally: 0 R < R ρ ↔ ∃ n [ n = k ], and we have no proof of ‘ ∃ n [ n = k ]’.We thus see that, in general, given real numbers x, y , we have no means ofproving one of the propositions ‘ x = R y ’, ‘ ¬ ( x = R y )’ , ‘ x < R y ’.(2) Note: 0 R ≤ R ρ , but, as we saw, we have no proof of ‘0 R = R ρ ’, nor of‘0 R < R ρ ’. e thus see that, in general, given real numbers x, y such that x ≤ R y , we haveno means of proving either one of the propositions ‘ x = R y ’ or ‘ x < R y ’.(3) Note: 0 R ≤ R ρ ↔ ∀ n [ n = k → ∃ m [ n = 2 m ]], and we have no proof of:‘ ∀ n [ n = k → ∃ m [ n = 2 m ]]’.Note: ρ ≤ R R ↔ ∀ n [ n = k → ∃ m [ n = 2 m + 1]], and we have no proof of:‘ ∀ n [ n = k → ∃ m [ n = 2 m + 1]]’.We thus see that, in general, given real numbers x, y , we have no means ofproving either one of the propositions ‘ x ≤ R y ’, ‘ y ≤ R x ’.(4) Note: ρ = R · ρ ↔ ∀ n [ n = k → ∃ m [ n = 2 m ]] and ρ = R R ↔ ∀ n [ n = k → ∃ m [ n = 2 m + 1]]. We have no means of proving eitherone of the propositions ‘ ρ = R · ρ ’, ‘ ρ = R R ’.We will use use the number ρ in Section 9. ρ , ρ , ρ are examples of real numbers oscillating around R . ρ oscillates above R and ρ oscillates up and down around R . ρ = ρ + ρ oscillates between R and · ρ .8. Rejecting P ∨ ¬ P Mathematics and Logic.
Brouwer describes mathematics as being devel-oped by the mind having come to awareness and playfully exploring its possibili-ties, see, for instance, [14]. Mathematics is not dependent on any evidence from thesenses. All other science is applied mathematics, as mathematics recognizes andenforces patterns on the data acquired by observation.Logic in particular is not a foundational science, prescribing the way one shouldthink, but an observational science. Mathematics is not reigned by logic. Math-ematics is a languageless activity of the mind . As we observed already in Section1, only when we want to communicate about mathematics, language comes in.We communicate with other people, trying to induce them to make mathematicalconstructions like the ones we ourselves are making, and we communicate withourselves, understanding the weakness of our memory, and hoping to be able toremind ourselves tomorrow of what we did today. Observing ourselves and otherswhen we are busy communicating about mathematics, we take notes of the soundsand signs that are used. We discover patterns and regularities, and then promotesuch regularities to laws of logic . There is no guarantee, however, that someone,even myself, who is making sounds or giving other signs in accordance with theselaws is actually succeeding in making successful mathematical constructions. Inthis sense, the ‘ laws of logic ’ are unreliable .Learning logic is part of learning the language of mathematics. Learning themeaning of the connectives ‘ . . . or . . . ’, ‘ . . . and . . . ’, ‘ . . . if . . . ’, and ‘ not : . . . ’, andthe quantifiers ‘ ∃ x ∈ V [ . . . x . . . ]’ and ‘ ∀ x ∈ V [ . . . x . . . ]’, is like learning the meaningof the expressions ‘infinite’ or ‘uncountable’ in the case of Theorems 1 and 5.A connective ∗ is a general method to obtain a new proposition, P ∗ Q , from anytwo given propositions, P and Q . Understanding P and Q means having some ideaabout what counts as a proof of P, Q , respectively. If we are able to point out ingeneral terms what we should consider as a proof of P ∗ Q , given our understandingof proofs of P, Q , we will have explained the meaning of the connective ∗ .8.2. Disjunction.
Let us start with the case of disjunction, ∨ .If someone says to us: ‘ I have a proof of P ∨ Q , P or Q ’, we expect , from ourexperience with earlier situations in which the word ‘or’ occurred, that he will comeup with a proof of P or with a proof of Q . We therefore adopt the rule that a proofof P ∨ Q must consist either in a proof of P or in a proof of Q . ne might bring forward that disjunction is not always taken in this construc-tively strong sense. There were other situations too! Some mathematicians mighteven say: ‘ When I make a statement P ∨ Q , I merely want to remind my readerthat ¬ P and ¬ Q are not both true ’.The intuitionistic mathematician then answers: ‘ Sometimes one uses the strongerinterpretation and at other times one uses the weaker one. Let us end this confusionand make the first and stronger interpretation the canonical one and always watchwhat we want to say. ’Our decision implies that the statement: ∃ n [ k ≤ n ] ∨ ∀ n [ n < k ], i.e. ∃ n [ k ≤ n ] ∨ ¬∃ n [ k ≤ n ].is not a true statement.Why?Until now, we did not find a place in the decimal expansion of π where there isan uninterrupted sequence of 99 digits 9: we have no proof of ‘ ∃ n [ k ≤ n ]’.Until now, we also did not find a clever argument showing that such an un-interrupted sequence of 99 digits 9 never will be found: we have no proof of‘ ¬∃ n [ k ≤ n ]’.Until now, therefore, we have no proof of ‘ ∃ n [ k ≤ n ] ∨ ¬∃ n [ k ≤ n ]’.Note that there is an asymmetry in the two horns of the dilemma. The (possible)truth of ∃ n [ k ≤ n ] will become clear to anyone who just patiently calculates thesuccessive values of the decimal expansion of π , but the discovery of the (possible)truth of ∀ n [ n < k ] requires mathematical ingenuity.The example shows that the principle If ∀ n [ P ( n ) ∨ ¬ P ( n )], then ∃ n [ P ( n )] ∨ ¬∃ n [ P ( n )]is unreliable.How did we come to trust it?We unthinkingly generalized our experience with handling finite sets and bounded quantifiers. After all, the following holds, for any given m , even for m = 10 : If ∀ n ≤ m [ P ( n ) ∨ ¬ P ( n )], then ∃ n ≤ m [ P ( n )] ∨ ¬∃ n ≤ m [ P ( n )],because, at least in principle, we can check each of the numbers 0 , , . . . , m and findout if one of them has the property P .The case { , , . . . , m } does not extend to the case { , , , . . . } .Checking each of the infinitely many numbers 0 , , , . . . is not feasible. If we thinkwe can do it we must imagine ourselves to be angels rather than human beings.8.3. ‘Reckless’ or ‘hardy’ statements. When in a playful mood, we call thestatement: ∃ n [ k = n ] ∨ ∀ n [ n < k ]a reckless or hardy statement. Actually, it is foul play. We call the mathematicianupholding the above statement as true ‘reckless’ because we take him to read thestatement constructively. Probably, he will protest.For us, the terms ‘reckless’ and ‘hardy’ indicate that the statement called sohas no constructive proof, although the non-intuitionistic mathematician, using hisown reading, sees no objection.Other examples of reckless statements are: ¬∀ n [ n < k ] ∨ ∀ n [ n < k ] ∃ n [ n = k ] → ∃ n [ n = k ] ∀ n [2 n = k ] ∨ ∀ n [2 n + 1 = k ] ∃ n [ n = k ] → ( ∀ n [2 n = k ] ∨ ∀ n [2 n + 1 = k ]) n = k ’ here stands for: ‘ n is the least number m such that at the places m, m + 1 , . . . , m + 87 in the decimal expansion of π we find the value 8’.Theorems implying a reckless statement themselves will be considered reckless.8.4. Brouwer’s example and the halting problem.
The above example of anundecided proposition, ‘ ∃ n [ n = k ]’, is simple and very important. A first suchexample was given by Brouwer in [10].The example is related to the discovery, in 1936, of the ‘ halting problem ’ byA.M. Turing (1912-1954). We briefly describe this problem using the approachdeveloped by S.C. Kleene (1909-1994), see [40, Chapter XI].Natural numbers are used for coding algorithms for (partial) functions from N to N . There is a computable predicate on triples of natural numbers called T . Forall e, n, z , T ( e, n, z ) if and only if z codes a successful computation according to thealgorithm coded by e at the argument n . The algorithm coded by e halts at n ifand only if ∃ z [ T ( e, n, z )]. There is a function U from N to N that extracts fromevery z coding a computation the outcome U ( z ) of the computation. A function α from N to N is computable if and only if there exists e such that ∀ n ∃ z [ T ( e, n, z ) ∧ U ( z ) = α ( n )]. A number e with this property is called an index of the function α .Every number e determines a partial function ϕ e from N to N :for all n , ϕ e ( n ) = U ( z ), where z is the least number w such that T ( e, n, w ). ϕ e is only defined at n if ∃ w [ T ( e, n, w )].The (self-)halting problem is the question if there exists a computable function α from N to N such that, for each e , α ( e ) = 0 ↔ ∃ z [ T ( e, e, z )] ↔ ϕ e is defined at e . Such a computable function α would be called a solution to the halting problem .Turing’s theorem is that there is no solution to the halting problem.One may prove, without much difficulty, that there exists e such that ∃ n [ k ≤ n ] ↔ ∃ z [ T ( e, e, z )]: the example we considered is one of the problemsin the scope of the halting problem. Therefore, a solution to the halting problemwould also solve the problem: ‘ ∃ n [ k ≤ n ]’. This fact gives some apriori probabilityto Turing’s result.8.5. The other logical ‘constants’.
The so-called proof-theoretical interpretation of the disjunction we sketched in Subsection 8.2 is extended to the other logical‘constants’ as follows.A proof of ‘ P and Q ’ is a proof of ‘ P ’ together with a proof of ‘ Q ’.A proof of ‘ if P , then Q ’ isa proof of ‘ Q ’ possibly using ‘ P ’ as an extra assumption.A proof of ‘ not P ’ is a proof of ‘if P , then 0=1’.For explaining the quantifiers, we need the notion of a propositional function . Let V be a set and let v P ( v ) a function that associates to every v in V a proposition P ( v ). We then may introduce propositions ∃ x ∈ V [ P ( x )] and ∀ x ∈ V [ P ( x )] bystipulating:A proof of ‘ ∃ x ∈ V [ P ( x )]’ consists of an element v of V together with a proof ofthe proposition ‘ P ( v )’.A proof of ∀ x ∈ V [ P ( x )] is a function v p ( v ) associating to every v in V aproof p ( v ) of the proposition ‘ P ( v )’.These explanations are not to be considered as exact definitions. They onlyindicate a provisional intention as to how we want to use our words. . A test case: the Intermediate Value Theorem
The theorem. A function from X ⊆ R to R is an effective method f , that,given any input x from X , will produce a well-defined outcome f ( x ) in R , of courserespecting the fundamental relation of real coincidence , i.e. for all x, y in X , if x = R y , then f ( x ) = R f ( y ).We take the notion of a function as a primitive notion, i.e. we do not explain whata function is by using notions that have been introduced before. Definition 6.
Let f be a function from X ⊆ R to R and let x in X be given. f is continuous at x if and only if ∀ p ∃ m ∀ y ∈ X [ | y − x | < m → | f ( y ) − f ( x ) | < p ] . f is continuous if and only if f is continuous at every x in X . The
Intermediate Value Theorem states the following:
Let f be a continuous function from [0 , to R .Then ∀ y ∈ R [ f (0) ≤ y ≤ f (1) → ∃ x ∈ [0 , f ( x ) = y ]] . The first version of this Theorem dates from 1817 and is due to B. Bolzano(1781-1848), see [5].Note that the intuitionistic mathematician reads the statement of the Theoremdifferently from his non-intuitionistic colleague: he understands the logical con-stants constructively, as sketched in Section 8.9.2.
The proof as we learnt it.
One may use the fruitful method of successivebisection , as follows.Let y be given such that f (0) ≤ y ≤ f (1).Define x in [0 , x (0) := (0 , n be given such that x ( n ) has been defined already.Consider m := x ′ ( n )+ x ′′ ( n )2 , the midpoint of the rational interval (cid:0) x ′ ( n ) , x ′′ ( n ) (cid:1) .If f ( m ) ≤ y , define x ( n + 1) = (cid:0) m, x ′′ ( n ) (cid:1) .If y < f ( m ), define x ( n + 1) = (cid:0) x ′ ( n ) , m (cid:1) .This completes the definition of x . x is a well-defined real number, and, as one verifies by induction,for every n , x ′′ ( n ) − x ′ ( n ) = n and f (cid:0) x ′ ( n ) (cid:1) ≤ y ≤ f (cid:0) x ′′ ( n ) (cid:1) .We claim: f ( x ) = y , and we prove this claim as follows.Assume y < R f ( x ). Using the continuity of f at x , find r such that ∀ z ∈ [0 , | x − z | < r → y < R f ( z )].Note: | x − x ′ ( r + 1) | < r , and, therefore, y < R f (cid:0) x ′ ( r + 1) (cid:1) ], but also f (cid:0) x ′ ( r + 1) (cid:1) < R y .Contradiction.Conclude: ¬ (cid:0) y < R f ( x ) (cid:1) , i.e. f ( x ) ≤ R y .By a similar argument, conclude: y ≤ R f ( x ) and: f ( x ) = R y .9.3. An objection to the ‘proof ’.
The construction of the point x , in the aboveargument, can not be carried out, as, in general, we are unable to decide: f ( m ) ≤ R y or y < R f ( m ).9.4. Two counterexamples to the theorem.
We give two examples showingthat the statement of the theorem, if one reads it constructively, sometimes fails tobe true.
First example .Let ρ = ( − ) k be the number oscillating up and down around 0 R introducedin Definition 5. Define a function f from [0 ,
1] to R such that i) f (0) = 0 and f (1) = 1 and f ( ) = f ( ) = + ρ , and(ii) f is linear on the interval [0 , ], on the interval [ , ] and on the interval[ , < ρ ↔ ∀ x ∈ [ , f ( x ) > ] and ρ < ↔ ∀ x ∈ [0 , ][ f ( x ) < ].Assume we find x such that f ( x ) = .Find p such that x ′′ ( p ) − x ′ ( p ) < and note: either < x ′ ( p ) or x ′′ ( p ) < .If < x ′ ( p ), then < x and ¬ (0 < ρ ), that is: ρ ≤
0, and,if x ′′ ( p ) < Q , then x < and ¬ ( ρ < ≤ ρ .Conclude: either 0 ≤ ρ or ρ ≤ reckless or hardy conclusion. Second example .Let ρ = ( ) k the number oscillating above 0 R introduced in Definition 5 andlet ρ = ρ + ρ = ( ) k + ( − ) k be the number oscillating between 0 R and 2 · ρ ,also introduced in Definition 5.Define a function f from [0 ,
1] to R such that(i) f (0) = 0 and f ( ) = ρ and f (1) = ρ , and(ii) f is linear on the interval [0 , ] and on the interval [ , f (0) ≤ · ρ ≤ f (1).If ∃ n [2 n = k ],then 0 < ρ ∧ ρ = 2 · ρ and ∀ x ∈ [0 , f ( x ) = · ρ → x = ], and,if ∃ n [2 n + 1 = k ],then 0 < ρ ∧ ρ = 0 and ∀ x ∈ [0 , f ( x ) = · ρ → ( x = 0 ∨ x = 1)].Suppose we find x in [0 ,
1] such that f ( x ) = · ρ .Find p such that x ′′ ( p ) − x ′ ( p ) < and note: either x ′′ ( p ) < or < x ′ ( p ) or < x ′ ( p ) and x ′′ ( p ) < x ′′ ( p ) < or < x ′ ( p ), then ¬∃ n [2 n = k ].If 0 < x ′ ( p ) and x ′′ ( p ) <
1, then ¬∃ n [2 n + 1 = k ].We thus see: ¬∃ n [2 n = k ] or ¬∃ n [2 n + 1 = k ].This is a reckless or hardy conclusion.Note that the above counterexamples pose a serious problem to the mathemati-cian. If the Intermediate Value Theorem does not stand a straightforward construc-tive reading, what then is its meaning? The defender of the theorem must mumblesomething like: ‘Well, I did not mean that you really could find a point where f assumes the intermediate value, but . . . ’Mathematical theorems without ‘numerical content’ are a shame and an em-barassment.We should not, because of these examples, put aside the Intermediate ValueTheorem as nonsense, but, arguing carefully, try and find related statements thatare constructively true.9.5. An approximate version.
It helps to weaken the conclusion of the Inter-mediate Value Theorem by requiring only that, for every given accuracy p , thereexists a point where the function assumes a value closer to the given intermediatevalue than p .Brouwer, who knew his own famous fixed-point theorem to be constructivelyfalse, formulated and proved such an approximate fixed-point theorem, see [15] . The 1-dimensional case of Brouwer’s fixed-point theorem is closely related to the Intermediatevalue Theorem. ote that the proof of the new theorem still is using the method of successivebisection. Theorem 6 (The Approximate Intermediate Value Theorem) . Let f be a continuous function from [0 , to R .Then ∀ p ∀ y ∈ R [ f (0) ≤ y ≤ f (1) → ∃ x ∈ [0 , y − p < f ( x ) < y + p ] .Proof. Let p be given and let y be given such that f (0) ≤ y ≤ f (1).Define x in [0 , x (0) := (0 ,
1) and note: f (cid:0) x ′ (0) (cid:1) − p +1 < y < f (cid:0) x ′′ (0) (cid:1) + p +1 .Now let n be given such that x ( n ) has been defined already.Consider m := x ′ ( n )+ x ′′ ( n )2 and z := f ( m ).Find l := the least k such that both z ′′ ( k ) − z ′ ( k ) < p +1 and y ′′ ( k ) − y ′ ( k ) < p +1 and define s := z ( l ).1. If s ′′ < y ′ ( l ) + p +1 , define x ( n + 1) = ( m, x ′′ ( n )).Then: f ( m ) ≤ s ′′ < y ′ ( l ) + p +1 ≤ y + p +1 and: f ( m ) − p +1 < y .2. If not s ′′ < y ′ ( l ) + p +1 , define x ( n + 1) = ( x ′ ( n ) , m ).Then: y ′′ ( l ) < y ′ ( l ) + p +1 ≤ s ′′ < s ′ + p +1 and: y < f ( m ) + p +1 .This completes the definition of x . x is a well-defined real number, and,for every n , x ′′ ( n ) − x ′ ( n ) = n and f (cid:0) x ′ ( n ) (cid:1) − p +1 < y < f (cid:0) x ′′ ( n ) (cid:1) + p +1 .Using the fact that f is continuous at x , find r such that ∀ z ∈ [0 , x − r < z < x + r → f ( x ) − p +1 < f ( z ) < f ( x ) + p +1 ].1. Note: x − r < x ′ ( r + 1) ≤ x , and: f ( x ) < f ( x ′ ( r + 1)) + p +1 < y + p +1 + p +1 = y + p .2. Note: x ≤ x ′′ ( r + 1) < x + r , and: y < f (cid:0) x ′′ ( r + 1) (cid:1) + p +1 < f ( x ) + p +1 + p +1 = f ( x ) + p .Conclude: y − p < f ( x ) < y + p . (cid:3) Locally non-constant functions.
One may also try to restrict the class offunctions to which the Intermediate Value Theorem applies. Observe that the twofunctions mentioned as counterexamples in Subsection 9.4 have the property that,over a whole interval, one does not know if they change their value. Let us try toforbid such behaviour. We require constructive evidence that, for every y , for everyinterval ( p, q ), the function does not have, on the interval ( p, q ), the constant value y . Definition 7.
A function f from [0 , to R is locally non-constant if and only if ∀ y ∈ R∀ p ∈ Q ∀ q ∈ Q [0 < p < q < → ∃ x ∈ [0 , p < x < q ∧ f ( x ) R y ]] . The following Lemma says that, if, for a given intermediate value y , the function f assumes, in any given interval, a value positively different from y , then there is apoint where f assumes the value y . Lemma 7.
Let f be a continuous function from [0 , to R .Let y be given such that f (0) ≤ y ≤ f (1) and ∀ p ∈ Q ∀ q ∈ Q [0 < p < q < → ∃ x ∈ [0 , p < x < q ∧ f ( x ) R y ]] .Then ∃ x ∈ [0 , f ( x ) = y ] .Proof. Let y be given such that f (0) ≤ y ≤ f (1).Define x in [0 , x (0) := (0 , n be given such that s := x ( n ) has been defined. sing the continuity of f , find a rational number q such that s ′ + Q s ′′ < q < s ′ + Q s ′′ and f ( q ) R y .If f ( q ) < R y , define x ( n + 1) := (cid:0) x ′ ( n ) , q (cid:1) , and,if y < R f ( q ), define x ( n + 1) := (cid:0) q, x ′′ ( n ) (cid:1) .This completes the definition of x . x is a well-defined real number, as, for each n , x ′ ( n ) ≤ x ′ ( n + 1) ≤ x ′′ ( n + 1) ≤ x ′′ ( n ) and x ′′ ( n + 1) − x ′ ( n + 1) ≤ (cid:0) x ′′ ( n ) − x ′ ( n ) (cid:1) .Moreover, for each n , f (cid:0) x ′ ( n ) (cid:1) ≤ R y ≤ R f (cid:0) x ′′ ( n ) (cid:1) .Assume: f ( x ) < R y .Find p such that p < y − R f ( x ), and find q such that ∀ z ∈ [0 , | z − x | < q → | f ( z ) − f ( x ) | < R p ].Then ∀ z ∈ [0 , | z − x | < q → f ( z ) < R y . Find r such that ( ) r < ) q .Conclude: f ( x ′′ ( r )) < R y . Contradiction.We thus see: ¬ ( f ( x ) < R y ), that is: f ( x ) ≥ R y .For similar reasons: f ( x ) ≤ R y . Conclude: f ( x ) = R y . (cid:3) Theorem 8 (The Intermediate Value Theorem for locally non-constant functions) . Let f be a continuous and locally non-constant function from [0 , to R . Then ∀ y [ f (0) ≤ y ≤ f (1) → ∃ x ∈ [0 , f ( x ) = y ]] .Proof. Use Lemma 7. (cid:3)
Many real functions may be proven to be locally non-constant and Theorem 8is widely applicable .From a constructive point of view, the next result, Theorem 9 is not a nice result.The result nevertheless is instructive. The intuitionistic mathematician may say:‘ Yes, this explains the classical mathematician’s belief that the Intermediate ValueTheorem is true. He means no more than that it is impossible to have constructiveevidence, for every point x in the domain of the function, that the value assumed at x is apart from the given intermediate value. Unfortunately, even if this impossible,it may also be impossible to find a point where the intermediate value is assumed. ’ Theorem 9 (The Negative Intermediate Value Theorem) . Let f be a continuous function from [0 , to R .Then ∀ y [ f (0) ≤ y ≤ f (1) → ¬∀ x ∈ [0 , f ( x ) R y ]] .Proof. Use Lemma 7. (cid:3)
At most countably many exceptions.
One may also observe that theIntermediate Value Theorem goes through for ‘most’ intermediate values.
Theorem 10 (The Intermediate Value Theorem holds with at most countablymany exceptions ) . Let f be a continuous function from [0 , to R .There exists an infinite sequence y , y , . . . of elements of [0 , such that ∀ y [ (cid:0) f (0) ≤ y ≤ f (1) ∧ ∀ n [ y R y n ] (cid:1) → ∃ x ∈ [0 , f ( x ) = y ]] . Proof.
Let q , q , q . . . be an enumeration of all rational numbers q in [0 , n , define y n := f ( q n ).Let y in [0 ,
1] be given such that ∀ n [ y R y n ].Define x in [0 , x (0) := (0 ,
1) and note: f (0) < R y < R f (1). Theorems 6 and 8 may be found in [8, Theorem 2.5]. In [4], one finds various theorems ‘ with countably many exceptions ’, for instance, Theorem4.9: if f ; [0 , → R is uniformly continuous, then for all but countably many y ≥ inf( f ) the set { x ∈ [0 , | f ( x ) ≤ y } is (constructively) compact. ow let n be given such that x ( n ) is defined and f (cid:0) x ′ ( n ) (cid:1) < R y < R f (cid:0) x ′′ ( n ) (cid:1) .Consider m := x ′ ( n )+ x ′′ ( n )2 . If f ( m ) < R y , define: x ( n + 1) := ( m, x ′′ ( n )), and,if y < R f ( m ), define: x ( n + 1) := ( x ′ ( n ) , m ). Then f (cid:0) x ′ ( n + 1) (cid:1) < R y < R f ( (cid:0) x ′′ ( n + 1) (cid:1) .This completes the definition of x .Using the argument given in the last part of the proof of Lemma 7, one proves: f ( x ) = R y . (cid:3) Perhaps, perhaps, perhaps, . . . . Suppose we change the conclusion of the Intermediate Value Theorem into( ∗ ) ∃ x [ f ( x ) R y → ∃ x [ f ( x ) = R y ]], Perhaps , f assumes the value y . One may think of ( ∗ ) as follows. You are offered x where f should assume thevalue y . If you discover x clearly does not work, you may go back to the shop andask for a better number and will be offered x and x certainly will work.One may reconsider the two counterexamples given in Subsection 9.4, the func-tions f and f , in the light of this definition.We have seen that the statement ‘ f assumes the value ’ is reckless, but, if f ( ) R , then ρ R either < R ρ and f ( · · ρ ) = , or ρ < R f ( − · ρ − · ρ ) = . We thus see: perhaps, f assumes the value . We also have seen that the statement ‘ f assumes the value · ρ ’ is reckless,but, if f (0) R · ρ , then ρ R R and ρ = R ρ and f ( ) = R · ρ . We thussee: perhaps, f assumes the value · ρ .The funny thing is that the operation Perhaps may be repeated. One mayconsider the statement: ∃ x [ f ( x ) R y → ∃ x [ f ( x ) R y → ∃ x [ f ( x ) = R y ]]], Perhaps, perhaps, f assumes the value y .We give an example showing that this statement indeed may be weaker than theprevious one, with only one ‘perhaps’. Another example .Consider ρ = ( − ) k but also ρ := ( − ) k .Define a function f from [0 ,
1] to R such that(i) f (0) = 0 and f (1) = 1 and f ( ) = f ( ) = + ρ , and f ( ) = f ( ) = + ρ , and,(ii) for each i ≤ f is linear on the interval [ i , i +15 ].Let us first observe: if f ( ) R and also f ( ) R , then ρ R ρ R ∃ x [ f ( x ) = R ]. Conclude: Perhaps, perhaps, f assumes the value .Now assume: Perhaps, f assumes the value .Find x such that, if f ( x ) R , then ∃ x [ f ( x ) = R ].Distinguish two cases. Case (1) . x < R . Assume: ρ < R x := sup( x , ) and note f ( x ) = f ( ) + ( x − ) · ρ − ρ ) = + ρ + ( x − ) · ρ − ρ ) ≤ R + ρ + ( − ) · ρ − ρ ) = + ρ + ρ .Conclude: f ( x ) ≤ R f ( x ) < R or 0 < R ρ . A similar use of the expression ‘
Perhaps ’ is made in Section 20. n both cases, one may conclude: ∃ x ∈ [0 , f ( x ) = R ] and: 0 ≤ R ρ ∨ ρ ≤ R ρ < R
0, then 0 ≤ R ρ ∨ ρ ≤ R
0, i.e.if ∃ n [2 n + 1 = k ], then ∀ n [2 n + 1 = k ] ∨ ∀ n [2 n = k ],a reckless or hardy conclusion. Case (2) . < R x . By a similar argumnet, we obtain the result:if 0 < R ρ , then 0 ≤ R ρ ∨ ρ ≤ R
0, i.e.if ∃ n [2 n = k ], then ∀ n [2 n + 1 = k ] ∨ ∀ n [2 n = k ],a reckless or hardy conclusion.We thus see that the statement‘ Perhaps, f assumes the value ’has reckless consequences.One now may define: P erhaps ( f, y ) if and only if ∃ x [ f ( x ) = y ]and, for each n , P erhaps n +1 ( f, y ) if and only if ∃ x [ f ( x ) R y → P erhaps n ( f, y )]and also P erhaps ω ( f, y ) if and only if ∃ n [ P erhaps n ( f, y )],and P erhaps ω +1 ( f, y ) if and only if ∃ x [ f ( x ) R y → P erhaps ω ( f, y )].and prove that in general, alle these statements have different meanings. Goingthis path further, increasing the number of ‘ perhapses ’ further into the transfi-nite, one finds uncountably many intuitionistically different versions of the notion:‘ ∃ x [ f ( x ) = R y ]’, see [56].10. A function from R to R is nowhere positively discontinuous In the next three Sections we study the intuitionistic claim that every real func-tion is continuous. The first theorem, Theorem 11, is weak and negative.10.1.
No positive discontinuity.Definition 8.
Let f be a function from X ⊆ R to R and let x in X be given. f is positively discontinuous at x if and only if ∃ p ∀ m ∃ y ∈ X [ | y − x | < m ∧ | f ( y ) − f ( x ) | ≥ p ] . Theorem 11. Let f be a function from X ⊆ R to R . Assume there exists x in X such that f is positively discontinuous at x . Then there exists z in R such thatwe can not calculate f ( z ) , so z can not belong to X .Proof. Let X , f, x satisfy the conditions of the theorem.Without loss of generality, we may assume: x = 0 = f (0).Find p such that ∀ m ∃ y ∈ X [ | y | < m ∧ | f ( y ) | > p ] . Find an infinite sequence y , y , . . . of reals such that ∀ m [ | y m | < m ∧ | f ( y m ) | > p ] . We give two arguments establishing the conclusion of the theorem.
First argument .I start constructing a real number z in an infinite sequence of steps, definingsuccessively z (0) , z (1) , z (2) , . . . .I promise to take care that for each n , z ′ ( n ) ≤ z ′ ( n + 1) ≤ z ′′ ( n + 1) ≤ z ′′ ( n ) and Cf. [17, Theorem 1]. ′′ ( n ) − z ′ ( n ) = n − . I am free, within these bounds, to choose each z ( n ) as I likeit. I have in mind to perhaps ‘freeze’ the free development of z , at some step n , bymaking a decision that determines all values from that step n on.I define z (0) = ( − ,
1) and do not freeze z at step 0. For each n >
0, if z has notyet been frozen at one of the earlier steps, either I do not freeze z at this step anddefine z ( n ) = ( − n , n ), or I freeze z by defining, for each i , z ( n + i ) = y n − ( q + i ),where q is the least k such that − n − < y ′ n − ( k ) ≤ y ′′ n − ( k ) < n − . z is a well-defined real number.Note: if I never make use of my freedom to freeze z , then z = R f ( z ) = R n >
0, then z = R y n − and | f ( z ) | = R | f ( y n − ) | > R p .But I am free and do not know myself if I ever will make use of the possibilityto freeze z or not.I thus am unable to find an approximation of f ( z ) smaller than p .We must conclude: f is not defined at the argument z and z does not belong tothe domain X of f . Second argument .Define z such that, for each n ,if n < k , then z ( n ) = ( − n , n ), and, if k ≤ n , then z ( n ) = y k ( q + n ),where q is the least k such that − k < y ′ k ( k ) ≤ y ′′ k ( k ) < k .Note: if ∀ n [ n < k ], then f ( z ) = R f (0 R ) = R
0, and,if ∃ n [ n = k ], then | f ( z ) | = R | f ( y k | > R p .Assume we may calculate f ( z ). By finding an approximation of f ( z ) smallerthan p we will be able to decide: f ( z ) R | f ( z ) | < R p . Case (a) . f ( z ) R
0, and, therefore: ¬∀ n [ n < k ]. Case (b) . | f ( z ) | < p , and, therefore: ∀ n [ n < k ].Conclude: ¬∀ n [ n < k ] ∨ ∀ n [ n < k ].This is a reckless conclusion.We must admit that, at this point of time, f is not defined at z and z does not belong to the domain X of f . (cid:3) Clearly, a function from
X ⊆ R with a positive discontinuity can not be calcu-lated at every point of R . In Brouwer’s terminology: such a function is not a full function. 11. The meaning of ‘not’
A weak and a strong interpretation.
May we conclude, from Theorem 11:
A function f from R to R is not positively discontinuous at anypoint x in R ?We have to be careful. There is some ambiguity in the use of the word ‘ not ’.When saying: ‘ not: ( ∃ n [ n = k ] ∨ ∀ n [ n < k ])’we mean no more than:‘ As yet, we have no proof of ‘ ∃ n [ n = k ∨ ∀ n [ n < k ] ’ .Most of the time, however, we take a proposition ‘not P ’, ‘ ¬ P ′ , in the strongersense indicated in Subsection 8.5: ‘ Assuming P , one is led to a contradiction ’. Thisstrong interpretation of ‘ not ’ is the canonical one. It follows from reading ‘ not - P ’as ‘ P → P → Q ’ as: ‘ Assuming P , onemay prove Q ’. ollowing this strong and canonical interpretation, one may prove, for any propo-sition P , ¬¬ ( P ∨ ¬ P ),as follows: Assume: ¬ ( P ∨ ¬ P ), i.e. P ∨ ¬ P leads to a contradiction. Thenboth P and ¬ P will lead to a contradiction. We thus see: ¬ P and: ¬ P leads to acontradiction. We therefore have a contradiction.Conclude: ¬ ( P ∨ ¬ P ) leads to a contradiction, i.e. ¬¬ ( P ∨ ¬ P ).In particular, we find: ¬¬ ( ∃ n [ n = k ] ∨ ∀ n [ n < k ]).11.2. Some useful remarks.
Defining Q := ∃ n [ n = k ] ∨ ¬∃ n [ n = k ], we see that there are propositions Q such that ¬¬ Q is true but Q itself is reckless or hardy. On the other hand, forevery proposition Q , if Q , that is: Q has a proof, then ¬ Q , (the statement that Q leads to a contradiction), leads indeed to a contradiction, i.e. if Q , then ¬¬ Q .Later in this paper, we need the following law of contraposition : if P → Q , then ¬ Q → ¬ P , and its consequence: if P → Q , then ¬¬ P → ¬¬ Q .This law is easily seen to be valid for the proof-theoretical interpretation,Using it, one sees that, as Q → ¬¬ Q is always true, also ¬¬¬ Q → ¬ Q is alwaystrue. (The scheme ¬¬¬ Q → Q was discovered by Brouwer and may be called Brouwer’s first law of logic ).Also note:if ( P ∨ ¬ P ) → ¬ Q , then, as ¬¬ ( P ∨ ¬ P ), one has: ¬¬¬ Q , and, therefore: ¬ Q .This fact may be useful in practice: if it is our aim to prove a negative statement ¬ Q it does no harm to make an extra ssumption P ∨ ¬ P .One should keep in mind: ¬ P means: P is reducible to absurdity . Therefore,the constructive mathematician has nothing to criticize if someone proves ¬ P byreducing P to absurdity. If one does so, one does what one has to do. Shefeels unhappy, however, if one proves P itself by reducing ¬ P to absurdity. Theconclusion of the latter procedure is ¬¬ P only, a statement that, most of the time,is weaker than P .11.3. More fugitive numbers.
The following Definition extends Definition 4.
Definition 9 (The fugitive numbers k α ) . Let an infinite sequence α = α (0) , α (1) , α (2) , . . . of natural numbers be given. For each n we define : k α ≤ n if and only if ∃ j ≤ n [ α ( j ) = 0] , and: n < k α if and only if ∀ j ≤ n [ α ( j ) = 0] , and: n = k α if and only if k α ≤ n and ∀ j < n [ j < k α ] , i.e. n is the least j such that α ( j ) = 0 . LPO , the limited principle of omniscience , is the statement ∀ α [ ∃ n [ k α ≤ n ] ∨ ∀ n [ n < k α ]] . WLPO , the weak limited principle of omniscience is the statement ∀ α [ ¬∀ n [ n < k α ] ∨ ∀ n [ n < k α ]] . Hardy’s finest weapon is not bad in itself but can only be used for proving negative state-ments. Constructively, negative statements are not so useful. There are better statements andfiner weapons. Note that we do not define a natural number k α bot only the meaning of an expression like‘ k α ≤ n ’. This statement got its name from the American analyst E. Bishop, who, fifty years afterBrouwer, founded his own school of constructive analysis , see [8, page 3]. The name comes fromcalling the Principle of the Excluded Third P ∨ ¬ P the Principle of Omniscience , a perhapsinfelicitous decision, as the name suggests mathematics is a matter of ‘ knowing facts ’. sing the second argument in the proof of Theorem 11 one may see: If a real function from R to R has a positive discontinuity, then WLPO . For an informal constructive mathematician like Brouwer,
LPO , and also
WLPO , represent the absurd , the contradiction , a statement one feels sure neverto be able to prove. For her, the argument for Theorem 11 establishes, for every‘full’ function f from R to R , ¬ (there exists x in R such that f is positively discontinuous at x ).In Section 12, we introduce Brouwer’s Continuity Principle , an axiomatic as-sumption one perhaps finds plausible after having seen and accepted the first argu-ment in the proof of Theorem 11. Brouwer’s Continuity Principle proves: ¬ WLPO ,see Theorem 12. 12.
The Continuity Principle
Let f be a (‘full’) function from R to R . The statement that there is no pointwhere f is positively discontinuous is, constructively, a weak statement and not avery useful one. The positive result that f is continuous at every point is obtainedfrom an axiomatic assumption first used by Brouwer.12.1. Understanding Brouwer’s Continuity Principle.
We first have to agreeupon some notation.In the following definition, we use k, n , n , . . . , s, t, u, . . . as variables over theset N . Definition 10 (Coding finite sequences of natural numbers by natural numbers) . Let p : N → N be the function enumerating the primes: p (0) = 2 , p (1) = 3 , p (2) =5 , . . . .For each k , for all n , n , . . . , n k − , h n , n , . . . , n k − i := p ( k − · Q i BCP ) . For every R ⊆ N × N ,if ∀ α ∈ N ∃ n [ αRn ] , then ∀ α ∈ N ∃ m ∃ n ∀ β ∈ N [ αm ⊏ β → βRn ] . How may we convince each other that thia axiom is ‘reasonable’ and that weshould accept it as a rule for our common mathematical game?We should ask what we mean by the ‘set’ N of all infinite sequences of naturalnumbers. This ‘set’ is a kind of framework in which all kinds of infinite sequenceswill grow, although only a very few of them have been realized up to now.Cantor’s suggestion that a set is the result of taking together its (earlier con-structed?) elements is a notion that does not make sense to us. n infinite sequence α may be given by a finite description or an algorithm thatenables us to find, one by one, the successive values α (0) , α (1) , . . . of α . Brouwercalls such infinite sequences lawlike sequences .As we saw in the first argument in the proof of Theorem 11, there are also infinitesequences α that are not governed by a law. The successive values α (0) , α (1) , α (2) , . . . of α then are freely chosen, step by step . At any given moment, only a finite initialpart (cid:0) α (0) , α (1) , . . . , α ( n − (cid:1) of α is known. One may ask for further values, but,having received the answers, one still only knows a finite initial part of α , albeit alonger one.There are intermediate possibilities. Having begun with choosing freely, step bystep, α (0) , α (1) , . . . α ( n − 1) one may decide, to ‘freeze’ α and to determine theremaining values by means of some rule or algorithm.There is even a possibility I make α dependent on my future mathematical ex-perience , by saying, for instance, α ( n ) = 0 if at the moments 0 , , . . . , n , I did notfind a proof of the abc -conjecture, and α ( n ) = 1, if I did.If we think of the set N as a framework making room for lawlike sequences aswell as for free-choice-sequences and sequences that are of some intermediate kind,we are thinking of, what Brouwer calls, the full continuum .Now assume R ⊆ N × N is given such that ∀ α ∃ n [ αRn ]. Every infinite sequence α , also a lawlike sequence α , may be thought of as being produced, step by step, by a black box . At the moment we produce a number n such that αRn , only finitely manyvalues of α , say α (0) , α (1) , . . . , α ( m − β (0) = α (0), β (1) = α (1) , . . . and β ( m − 1) = α ( m − βRn .Adopting Axiom 1 as a starting point for our further mathematical discourse,we lay down a canonical meaning for expressions of the form ‘ ∀ α ∃ n [ αRn ]’. Earlier,we did so for expressions of the form ‘infinite’, ‘uncountable’ and ‘or’ . Theorem 12. BCP ⇒ ¬ WLPO .Proof. Assume WLPO i.e. ∀ α [ ¬∀ n [ n < k α ] ∨ ∀ n [ n < k α ]], i.e. ∀ α [ ¬∀ n [ α ( n ) = 0] ∨ ∀ n [ α ( n ) = 0]].Consider α = 0 and, applying BCP , find m such that either ∀ β [0 m ⊏ β → ¬∀ n [ β ( n ) = 0]] or ∀ β [0 m ⊏ β → ∀ n [ β ( n ) = 0]].The first of these two statements is not true, consider: β = 0, and the second onealso fails, consider: β = 0 m ∗ (cid:3) The (pointwise) continuity of functions from R to R . In the next Definition, we introduce a function α u α associating to every α in N a real number u α . Definition 12. We let ( r ′ , r ′′ ) , ( r ′ , r ′′ ) , ( r ′ , r ′′ ) , . . . be a fixed enumeration of allpairs of rationals.We define a mapping associating to each s = 0 a pair ( u ′ s , u ′′ s ) of rationals, asfollows.For each n , if 0 < r ′′ n − r ′ n ≤ , then ( u ′h n i , u ′′h n i ) = ( r ′ n , r ′′ n ) , and, if not , then ( u ′h n i , u ′′h n i ) = (0 , .For each s = 0 , for each n , if u ′ s < r ′ n < r ′′ n < u ′′ s and < r ′′ n − r ′ n ≤ ( u ′′ s − u ′ s ) , then ( u ′ s ∗h n i , u ′′ s ∗h n i ) = ( r ′ n , r ′′ n ) , and, if not , then ( u ′ s ∗h n i , u ′′ s ∗h n i ) = ( u ′ s + u ′′ s , u ′ s + u ′′ s ) .For each α in N we let u α be the infinite sequence of pairs of rationals such that,for all n , u α ( n ) = ( u ′ α ( n +1) , u ′′ α ( n +1) ) . heorem 13. (i) For every α , u α is a real number. (ii) For every real number x , there exists α such that x = R u α . (iii) BCP ⇒ Every function from R to R is continuous at every point x .Proof. The proof of (i) and (ii) is left to the reader.(iii) Let f be a real function and let a real x be given. We shall prove that f iscontinuous at x .Let p be given. We have to find m such that, for every real y , if | y − x | < m ,then | f ( y ) − f ( x ) | < p .Note: ∀ α ∃ n [ r ′ n < f ( u α ) < r ′′ n ] and find α such that x = R u α .Applying BCP , find q, n such that ∀ β [ αq ⊏ β → r ′ n < f ( u β ) < r ′′ n ].Find i, j such that u α ( n − 1) = ( r ′ i , r ′′ i ) and u α ( n ) = ( r ′ j , r ′′ j ).Note: r ′ i < Q r ′ j < Q r ′′ j < Q r ′′ i .Find m such that m < min Q ( r ′ j − Q r ′ i , r ′′ i − Q r ′′ j ).Note: for every real y , if | y − x | < m , then r ′ i < y < r ′′ i and there exists β suchthat αn ⊏ β and y = R u β and, therefore: r ′ n < f ( u β ) < r ′′ n and r ′ n < f ( y ) < r ′′ n .Conclude: for every real y , if | y − x | < m then | f ( y ) − f ( x ) | < r ′′ n − r ′ n < p . (cid:3) The Fan Theorem Proving the Fan Theorem. The intuitionistic mathematician holds notonly that every function from [0 , 1] tot R , like every function from R to R , seeTheorem 13(iii) is continuous at every point, but also that every function from[0 , 1] to R is continuous uniformly on [0 , Fan Theorem .We need some definitions in order to formulate the Fan Theorem. Definition 13. Let X ⊆ N and B ⊆ N . B is a bar in X , notation: Bar X ( B ) , if and only if: ∀ α ∈ X ∃ s ∈ B [ s ⊏ α ] . C := 2 ω := { α | ∀ n [ α ( n ) < } . C is sometimes called Cantor space . Bin := 2 <ω := { s | ∀ i < length ( s )[ s ( i ) < } .For each n , Bin n := { s ∈ Bin | length ( s ) = n } = { s ∈ ω n | ∀ j < n [ s ( j ) < } . Definition 14. For each X ⊆ N , for each s , we define: X ∩ s := { α ∈ X | s ⊏ α } . The Fan Theorem is the statement that every bar in C has a finite subbar.Brouwer actually proved the more general but equivalent statement that every barin a so-called fan has a finite subbar. Here, B ⊆ N is called finite if and only if ∃ u ∀ s [ s ∈ B ↔ ∃ i < length ( u )[ s = u ( i )]].We avoid defining the notion of a fan . For understanding the name of theTheorem, it suffices to know every fan is a subset of N and that C is the primeexample of a fan.The proof of the Fan Theorem turns on a philosophical claim . Theorem 14 (Brouwer’s Fan Theorem) . For every B ⊆ N , if Bar C ( B ) , then thereexists u such that ∀ i < length ( u )[ u ( i ) ∈ B ] and ∀ α ∈ C∃ i < length ( u )[ u ( i ) ⊏ α ] . Proof. Let B ⊆ N be given such that Bar C ( B ).Define, for every s in 2 <ω , s is B -safe , or simply: safe , if and only if Bar C∩ s ( B ). See [48], [52, § 2] and [65, § In [12] and [16], the Fan Theorem is obtained as a corollary of the ‘Bar Theorem’, in thispaper Theorem 19. We give an argument for the Fan Theorem inspired by Brouwer’s proof of theBar Theorem, like A. Heyting did in [36, § s Bar C ( B ), h i is safe.We now ask the probably somewhat surprising question: ‘ How is it possible weknow this? ’ and, after some reflection, we come forward with the claim : there mustbe a canonical proof of the fact: ‘ h i is safe.’What do we mean by: ‘a canonical proof’ ?The canonical proof is an arrangement of statements of the form: ‘ s is safe ’.The conclusion of the canonical proof is the statement: ‘ h i is safe.’The starting points of the canonical proof have the form: s ∈ <ω and s ∈ B . Therefore: s is safe.There are only two kinds of reasoning steps: forward reasoning steps and back-ward reasoning steps .A forward reasoning step has the form: s ∈ <ω , and s ∗ h i is safe, and s ∗ h i is safe. Therefore: s is safe.A backward reasoning step either has the form: s ∈ <ω and s is safe. Therefore: s ∗ h i is safe.or it has the form s ∈ <ω and s is safe. Therefore: s ∗ h i is safe.Clearly, the canonical proof may be visualized as a finite tree, with the statement‘ h i is safe’ at its bottom node and statements ‘ s ∈ B ’ at its top nodes. Each nodethat is not a top node has either one or two upstairs neighbours.One easily sees that all resoning steps are sound and that a canonical proof of‘ h i is safe’ is indeed a proof of ‘ h i is safe’.Our claim gives expression to the feeling that, if we know: ‘ Bar C ( B )’, thisknowledge must be based upon an orderly organization of elementary pieces ofinformation of the form: ‘ s ∈ B ’.Now, trusting our claim, let us take a canonical proof of: ‘ h i is safe.’ We aregoing to use it in order to build another proof.Define, for every s in 2 <ω , s is supersafe if and only if there exists u such that ∀ i < length ( u )[ u ( i ) ∈ B ] and ∀ α ∈ C ∩ s ∃ i < length ( u )[ u ( i ) ⊏ α ].Note that the conclusion we are after is: ‘ h i is supersafe.’Replace, in the canonical proof of ‘ h i is safe’, every statement ‘ s is safe’ by thestatement ‘ s is supersafe’.The result will be another valid proof .Why?First, the new starting points are sound: if s ∈ <ω and s ∈ B , define: u := h s i and note: u (0) ∈ B and ∀ α ∈ C ∩ s [ u (0) ⊏ α ].Then, the new forward reasoning steps are sound.For assume: s ∈ <ω and both s ∗ h i and s ∗ h i are supersafe.Find u , u such that: for each j < ∀ i ≤ length ( u j )[ u j ( i ) ∈ B ] and ∀ α ∈ C ∩ s ∗ h j i∃ i < length ( u j )[ u j ( i ) ⊏ α ] . efine u := u ∗ u and note ∀ i < length ( u )[ u ( i ) ∈ B ] and ∀ α ∈ F ∩ s ∃ i < length ( u )[ u ( i ) ⊏ α ] , that is: s is supersafe.Also, the new backward reasoning steps are sound.For assume: s ∈ <ω and s is supersafe.Find u such that ∀ i < length ( u )[ u ( i ) ∈ B ] and ∀ α ∈ C ∩ s ∃ i < length ( u )[ u ( i ) ⊏ α ] . Note: for both j < C ∩ s ∗ h j i ⊆ C ∩ s and s ∗ h j i is supersafe.Conclude: our new proof is sound and its conclusion is true, i.e. h i is supersafe. (cid:3) Generalizing the proof of Theorem 14 we obtain the following conclusion: Theorem 15 (Bar Induction in C ) . Let B ⊆ N be a bar in C . (i) Assume B ⊆ C ⊆ <ω and, for all s in <ω , s ∈ C if and only if ∀ j < s ∗ h j i ∈ C ] . Then h i ∈ C . (ii) Assume B ⊆ C ⊆ <ω and, for all s in <ω ,if ∀ j < s ∗ h j i ∈ C ] , then s ∈ C . Then h i ∈ C .Proof. (i) Define, for each s in 2 <ω , s is safe if and only if Bar C∩ s ( B ). In thecanonical proof of: ‘ h i is safe’, replace every statement ‘ s is safe’ by ‘ s ∈ C ’. Theresult will be a proof of: ‘ h i ∈ C ’.(ii) Define C ∗ := { s ∈ <ω [ ∃ t ⊑ s [ t ∈ C ] } .Note: for all s in 2 <ω , if s ∈ C ∗ , then ∀ j < s ∗ h j i ∈ C ∗ ].Now assume: s ∈ <ω and ∀ j < s ∗ h j i ∈ C ∗ ]. We may distinguish two cases. Case 1. ∀ j < s ∗ h j i ∈ C ]. Then s ∈ C ⊆ C ∗ . Case 2. ∃ t ⊑ s [ t ∈ C ]. Then s ∈ C ∗ .We thus see: for all s in 2 <ω , s ∈ C ∗ if and only if ∀ j < σ ( s ∗ h j i ∈ C ∗ ].Applying (i), we conclude: h i ∈ C ∗ , and: h i ∈ C . (cid:3) Theorem 15(ii) shows that, in the canonical proof of Bar C ( B ), introduced in theproof of Theorem 14, the backward steps might have been left out. They may notbe left out in the canonical proof of the Bar Theorem , the infinitary analog of theFan Theorem, see Subsection 15.2.13.2. Every function from [0 , to R is uniformly continuous. Brouwer’sContinuity Principle implies that every function from [0 , 1] to R , like every functionfrom R to R , see Theorem 13(iii), is continuous at every point. Using the FanTheorem, we now prove that every function from [0 , 1] to R is even continuous uniformly on [0 , α c α associating to every α in C a real number c α in [0 , Definition 15. We define a mapping associating to every s in <ω a pair ( a s , b s ) of rationals such that ( a h i , b h i ) = (0 , and for each s in <ω , ( a s ∗h i , b s ∗h i ) = ( a s , a s + b s ) and ( a s ∗h i , b s ∗h i ) = ( a s + b s , b s ) .For each α in C we let c α be the real x such that, for each n , x ( n ) = ( a αn , b αn ) . Theorem 16 (Uniform-Continuity Theorem) . Every function f from [0 , to R iscontinuous uniformly on [0 , , i.e. ∀ p ∈ N ∃ m ∈ N ∀ x ∈ [0 , ∀ y ∈ [0 , | x − y | < m → | f ( x ) − f ( y | < p ] . roof. Let f be a function from [0 , 1] to R and let p be given. We want to find m such that ∀ x ∈ [0 , ∀ y ∈ [0 , | x − y | < m → | f ( x ) − f ( y | < p ] . Let B be the set of all s in 2 <ω such that, for all x, y in [ a s , b s ], | f ( x ) − f ( y ) | < p .We first prove that B is a bar in C , i.e. Bar C ( B ).Let α in C be given. As f is continuous at c α , see Theorem 13(iii), find m such that ∀ y ∈ [0 , | c α − y | < m → | f ( c α ) − f ( y | < p +1 ]. Find n such that b αn − a αn < m and note: αn ∈ B .We thus see that every α in C has an initial part in B .Now let C be the set of all s in 2 <ω such that, for some m ,for all x, y in [ a s , b s ], if | x − y | < m , then | f ( x ) − f ( y ) | < p .Note: B ⊆ C .Let s in 2 <ω be given such that ∀ i < s ∗ h i i ∈ C ]. We are going to prove: s ∈ C .First find m such that, for all x, y in ( a s ∗h i , b s ∗h i ) = ( a s , a s + b s ), if | x − y | < m , then | f ( x ) − f ( y ) | < p . Then find m such that, for all x, y in ( a s ∗h i , b s ∗h i ) =( a s + b s , b s ), if | x − y | < m , then | f ( x ) − f ( y ) | < p .Now find n := length ( s ).Note: b s − a s = ( ) n , and, for all x, y in [ a s , b s ], if | x − y | < ( ) n , then either x, y both are in [ a s , a s + b s ] or x, y both are in [ a s + b s , b s ].Find m such that m ≥ m and m ≥ m and m < ( ) n and note: for all x, y in [ a s , b s ], if | x − y | < m , then | f ( x ) − f ( y ) | < p . We thus see: s ∈ C .Clearly, for all s in Bin , if ∀ i < s ∗ h i i ∈ C ], then s ∈ C .Using Theorem 15(ii), conclude: h i ∈ C , that is: ∃ m ∀ x ∈ [0 , ∀ y ∈ [0 , | x − y | < m → | f ( x ) − f ( y ) | < n ] . (cid:3) The failure of the Fan Theorem in computable analysis. In the nextTheorem, we use some notations introduced in Subsection 8.4. Theorem 17 (Kleene’s Alternative ) . There exists B ∗ ⊆ N such that (i) Every computable α in C has an initial part in B , i.e. ∃ s ∈ B ∗ [ s ⊏ α ] . (ii) Every finite subset of B positively fails to be a bar in the set { α ∈ C | α is computable } , i.e.for all u, p , if length ( u ) = p > and ∀ i < p [ u ( i ) ∈ B ∗ ] ,then there exists a computable α in C such that ∀ i < p [ u ( i ) ⊥ α ] . (iii) B ∗ is an algorithmically decidable subset of N .Proof. First, let B be the set of all s in Bin such that, for some e , length ( s ) = e + 1and ∀ i ≤ e ∃ z [ T ( e, i, z ) ∧ U ( z ) = s ( i )] . h i / ∈ B and, for every e , there is at most one s in B such that length ( s ) = e + 1.Let α be a computable element of C . Find e such that α = ϕ e .Note: α ( e + 1) = ϕ e ( e + 1) ∈ B .Let u, p be given such that length ( u ) = p > ∀ n < p [ u ( n ) ∈ B ].We may assume: 0 < length (cid:0) u (0) (cid:1) < length (cid:0) u (1) (cid:1) < . . . < length (cid:0) u ( p − (cid:1) and, therefore, ∀ n < p [ length (cid:0) u ( n ) (cid:1) > n ]. Define α such that, for each n < p , α ( n ) = 1 − (cid:0) u ( n ) (cid:1) ( n ), and, for each n ≥ p , α ( n ) = 0. Note: α is computable and ∀ n < p [ u ( n ) ⊥ α ].Unfortunately, B is not an algorithmically decidable subset of N .We therefore introduce another subset of N .Let B ∗ be the set of all s in Bin such that, for some e < length ( s ), ∀ i ≤ e ∃ z < length ( s )[ T ( e, i, z ) ∧ U ( j ) = s ( i )], and, therefore, s ( e + 1) ∈ B .Note that B ∗ is an algorithmically decidable subset of N . Kleene discovered this theorem in 1950, see [39] and [41, Lemma 9.8]. et α be a computable element of C . Find e such that α = ϕ e . Find n > e suchthat ∀ i ≤ e ∃ z < n [ T ( e, i, z ) ∧ U ( z ) = s ( i )] and conclude: αn ∈ B ∗ .We thus see that every computable α in C has an initial part in B ∗ .Let u, p be given such that length ( u ) = p > ∀ n < p [ u ( n ) ∈ B ∗ ]. Find v such that length ( v ) = p and ∀ n < p [ v ( n ) ⊑ u ( n ) ∧ v ( n ) ∈ B ]. Find a computable α in C such that ∀ n < p [ v ( n ) ⊥ α ] and conclude ∀ n < p [ u ( n ) ⊥ α ].We thus see that every finite subset of B ∗ positively fails to be a bar in { α ∈ C | α is computable } . (cid:3) The subject of computable analysis starts from the assumption that every α in N is given by an algorithm, i.e., if one uses Church’s Thesis , by a Turing-algorithm.One may study this subject from an intuitionistic point of view. As appears fromTheorem 17, computable analysis is dramatically different from intuitionistic anal-ysis. For instance, there exists a real function from [0 , 1] to R that is everywherecontinuous but positively unbounded. Many more such results may be found in[60].14. The Determinacy Theorem as an equivalent of the Fan Theorem The Uniform-Continuity Theorem, Theorem 16, is the first application of theFan Theorem, and may be called the goal for which it was devised. In this Sectionwe want to introduce to the reader to a second and more recent application, anapplication in the theory of games.In the first two Subsections we consider two kinds of two-move-games for Players I, II and define when such games are constructively determinate from the viewpointof Player I . Games from the first Subsection may fail to have this property but wewill see that the Fan Theorem proves that the games from the second Subsectionalways have it. In the third Subsection we prove that, conversely, this result impliesan important case of the Fan Theorem. In the fourth Subsection we introduce theintuitionistic Determinacy Theorem.14.1. Games in × ω . For every C ⊆ × ω , we introduce a game G × ω ( C ).There are two players, Player I and Player II . Every play goes a follows. Player I chooses i in { , } , thereafter Player II chooses n in ω and the play is finished.Player I wins the play if and only if h i, n i ∈ C . Player II tries to prevent thatPlayer I wins the play.Clearly, Player I has a winning first move if and only if ∃ i ∀ n [ h i, n i ∈ C ]. A strategy for Player II , on the other hand, is a pair of numbers h p , p i , and the strategy is winning if ∀ i < h i, p i i / ∈ C ].The game G × ω ( C ) is called determinate if and only if either Player I has awinning first move or Player II has a winning strategy. As one may conjecture, itmay happen that this decision can not be taken in a constructive way. We will seethis in a moment.We now define: the game G × ω ( C ) is determinate from the viewpoint of Player I if and only if: If Player II does not have a winning strategy in the constructivelystrong sense: ∀ p ∀ p [ h , p i ∈ C ∨ h , p i ∈ C ] , then Player I hasa winning first move, i.e. ∃ i < ∀ n [ h i, n i ∈ C ]and ask ourselves the question: is the game G ω × ( C ) always determinate from theviewpoint of Player I ?The following example shows that, sometimes, it is not.Define C := {h i, n i | k ≤ n → ∃ l [ k = 2 l + i ] } .Let p , p be given. If k ≤ p , one may find i < l such that k = 2 l + i and h i, p i i ∈ C . If p < k , then h , p i ∈ C . We thus see Player I has an answer o any given strategy of Player II .On the other hand, if Player I has a winning first move, then ∃ i < ∀ n [ h i, n i ∈ C ]and ∃ i < ∃ j [ j = k ] → ∃ n [2 n + i = k ]], i.e. ¬∃ n [2 n = k ] ∨ ¬∃ n [2 n +1 = k ],a reckless or hardy statement.Note that the statement: ‘the game G × ω ( C ) is determinate’ is also reckless, as,for every C ⊆ × ω , if G × ω ( C ) is determinate, then G × ω ( C ) is determinate fromthe viewpoint of Player II .14.2. Games in ω × . For every C ⊆ ω × 2, we introduce a game G ω × ( C ).There are again two players, Player I and Player II . Every play goes a follows.Player I chooses n in ω , thereafter Player II chooses i in { , } and the play isfinished. Player I wins the play if and only if h n, i i ∈ C . Player II tries to preventthat Player I wins the play. Clearly, Player I has a winning first move if and only if ∃ n ∀ i < h n, i i ∈ C ]. A strategy for Player II is an element τ of C and the strategy τ is winning if and only if ∀ n [ h n, τ ( n ) i / ∈ C ].The game G ω × ( C ) is called determinate if and only if either Player I has awinning first move or Player II has a winning strategy.Considering the example C := {h n, i i ∈ ω × | n = k ], we see that it mayhappen that the statement: ‘ G ω × ( C ) is determinate’ is a reckless one.We now define: the game G ω × ( C ) is determinate from the viewpoint of Player I if and only if If Player II does not have a winning strategy in the strong sense: ∀ τ ∈ C∃ n [ h n, τ ( n ) i ∈ C ] , then Player I has a winning first move,i.e. ∃ n ∀ i < h n, i i ∈ C ] . The Fan Theorem proves that, for every C ⊆ ω × 2, the game G ω × ( C ) isdeterminate from the viewpoint of Player I .For assume ∀ τ ∈ C∃ n [ h n, τ ( n ) i ∈ C ].Define B := S m { t ∈ Bin m | ∃ n < m )[ h n, t ( n ) i ∈ C ] } and note: B is a bar in C . Applying the Fan Theorem, find u such that ∀ i < length ( u )[ u ( i ) ∈ B ] and ∀ τ ∈ C∃ i < length ( u )[ u ( i ) ⊏ τ ]. Find m such that ∀ i < m [ length ( u ( i )) < m ] andconclude: ∀ t ∈ Bin m ∃ n < m [ h n, t ( n ) i ∈ C ].We now prove, by backwards induction:for each j ≤ m , either ∀ t ∈ Bin j ∃ n < j [ h n, t ( n ) i ∈ C ] or ∃ n ∀ i < h n, i i ∈ C ].This clearly holds if j = m .Now assume j + 1 ≤ m and ∀ t ∈ Bin j +1 ∃ n < j + 1[ h n, t ( n ) i ∈ C ]. For every t in Bin j , one may consider t ∗ h i and conclude: ∃ n < j [ h n, t ( n ) i ∈ C ] or h j, i ∈ C .Therefore: either h j, i ∈ C or ∀ t ∈ Bin j ∃ n < j [ h n, t ( n ) i ∈ C ]. For similarreasons, either h j, i ∈ C or ∀ t ∈ Bin j ∃ n < j [ h n, t ( n ) i ∈ C ]. Conclude: either: ∀ i < h j, i i ∈ C ] and ∃ n ∀ i < h n, i i ∈ C ] or: ∀ t ∈ Bin j ∃ n < j [ h n, t ( n ) i ∈ C ].Repeating this step m times we find the conclusion: ∃ n ∀ i < h n, i i ∈ C ].14.3. Recovering the Fan Theorem. Let us consider the result of the last sub-section: ( C ⊆ ω × ∀ τ ∈ C∃ n [ h n, τ ( n ) i ∈ C ], then ∃ n [ h n, i ∈ C ∧ h n, i ∈ C ].We prove that this statement implies an important case of the Fan Theorem. TheFan Theorem is the statement that every B ⊆ N that is a bar in C has a finitesubset that is a bar in C . We prove that B is a decidable subset of N .Assume ( B ⊆ N be given such that Bar C ( B ) and one may decide,for each n , n ∈ B or n / ∈ B . For all n , we define: B n := { s ∈ B | s < n } . See [49, § e want to prove the statement QED := ∃ n [ Bar C ( B n )] .We let C be the set of all pairs h s, i i such that either : QED or : Bar C∩ s ∗h i i ( B n )and not Bar C∩ s ∗h − i i ( B n ). Note that, for each s in Bin , if both s ∗ h i and s ∗ h i are in C , then QED . Also note that C , like B , is a decidable subset of N .Let τ in C be given. Find α in C such that ∀ i [ α ( i ) = τ ( αi )]. Find n such that αn ∈ B and define m := αn + 1. Note: Bar C ∩ αn ( B m ).Note that, for each i , if i +1 ≤ n and Bar C∩ α ( i +1) ( B m ), then either : Bar C∩ α ( i ) ( B m ) or : ¬ Bar C∩ α ( i ) ∗h − α ( i ) i ( B m ) and h αi, α ( i ) i ∈ C . Using backwards induction,starting from the fact: Bar C∩ αn ( B m ), we prove:for each i ≤ n , either Bar C∩ αi ( B m ) or ∃ j < n [ h αj, α ( j ) i ∈ C ].Taking i = 0, we find: either Bar C ( B m ) or ∃ j [ h αj, α ( j ) i ∈ C ], i.e. either QED or ∃ j [ h αj, τ ( αj ) i ∈ C ]. In both cases, one may conclude: ∃ s [ h s, τ ( s ) i ∈ C ].We thus see: ∀ τ ∈ C∃ s [ h s, τ ( s ) i ∈ C ]. Using ( ∃ s [ s ∗ h i ∈ C ∧ s ∗ h i ∈ C ], and, therefore: QED , i.e. ∃ n [ Bar C ( B n )].This shows that every decidable subset B of N that is a bar in C has a finitesubset that is a bar in C .14.4. The Intuitionistic Determinacy Theorem. We now study games with infinitely many moves rather than two moves only. Definition 16. ( ω × ω := { α | ∀ n [ α (2 n + 1) < } .For each σ , for each α ∈ ( ω × ω , α ∈ I σ := ∀ n [ α (2 n ) = σ (cid:0) α (2 n ) (cid:1) ] For each τ in C , for each α in ( ω × ω , α ∈ II τ := ∀ n [ α (2 n +1) = τ (cid:0) α (2 n +1) (cid:1) ] . For every X ⊆ ( ω × ω , we introduce a game G ( ω × ω ( X ). There are two players,Player I and Player II . Every play takes infinitely many moves and goes as follows.Player I chooses n in ω , thereafter Player II chooses i in 2 = { , } , then Player I chooses n in ω and, thereafter, Player II chooses i in 2, and so on. Player I wins the play if the infinite sequence n , i , n , i , . . . is in X . Player II tries toprevent that Player I wins the game. Theorem 18 (Intuitionistic Determinacy Theorem) . For all X ⊆ ( ω × ω , if ∀ τ ∈ C∃ α ∈ ( ω × ω [ α ∈ II τ ∧ α ∈ X ] , then ∃ σ ∀ α ∈ ( ω × ω [ α ∈ I σ → α ∈ X ] . The Theorem is good news for Player I . Suppose X ⊆ ( ω × ω is such thatPlayer I is able, once Player II has told her the strategy τ she intends to follow,to find a play following τ that is in X . She then may devise a strategy σ , such thatevery play following σ is in X . Using σ , Player I can do without information aboutthe strategy followed by Player II .We will not give the proof of Theorem 18 . As in Subsection 14.2, the FanTheorem plays a key rˆole in the argument. From Subsection 14.1, one may see it iscrucial for Theorem 18 that, in ( ω × ω , Player II has, at each one of her moves,only 2 choices. The theorem extends to the case that Player II has, at each one ofher moves, finitely many choices.15. Brouwer’s Thesis The Principle of Bar Induction. The key point in the proof of the nextTheorem is a philosophical assumption one might call Brouwer’s Thesis (on barsin N ) . We read QED not in its usual sense: ‘ quod erat demonstrandum, what had to be proven ’but as: ‘ quod est demonstrandum, what has to be proven ’. See [47, Chapter 16], [59] and [61, Section 9]. The theorem better might be called an axiom. For a discussion, see [57]. heorem 19 (Bar Induction) . Let B ⊆ N be given such that Bar N ( B ) . Let C ⊆ N be given such that B ⊆ C and, for all s , s ∈ C if and only if ∀ n [ s ∗ h n i ∈ C ] .Then h i ∈ C .Proof. Let B ⊆ N be given such that Bar N ( B ).Define, for every s , s is B-safe , or simply: safe , if and only if Bar N ∩ s ( B ).Note: h i is safe.There must exist a canonical proof of the fact: h i is safe.The conclusion of the canonical proof is: ‘ h i is safe.’The starting points of the canonical proof have the form: s ∈ B . Therefore: s is safe.There are two kinds of reasoning steps, forward steps and backward steps .A forward reasoning step has infinitely many premises and is of the followingform: s ∗ h i is safe, s ∗ h i is safe, s ∗ h i is safe, . . . Therefore: s is safe.A backward reasoning step has the form: s is safe. Therefore: s ∗ h n i is safe.The canonical proof is not a finite tree but an infinite one. The canonical proofcan not be written out explicitly as a finite text.Let C ⊆ N be given such that B ⊆ C and, for all s , s ∈ C if and only if ∀ n [ s ∗ h n i ∈ C ].Now replace in the canonical proof every statement ‘ s is safe’ by the statement‘ s ∈ C ’. The result will be another valid proof .Why?As B ⊆ C , the new starting points are sound.As, for every s , if ∀ n [ s ∗ h n i ∈ C ], then s ∈ C , the new forward reasoning stepsare sound.As, for every s , for every n , if s ∈ C , then s ∗ h n i ∈ C , the new backwardreasoning steps are sound.We must conclude: the new conclusion is true, that is: h i ∈ C . (cid:3) One needs backward steps.Theorem 20 (Kleene’s example ) . There exist B ⊆ C ⊆ N such that Bar N ( B ) and, for all s , if ∀ n [ s ∗ h n i ∈ C ] , then s ∈ C , while the statement ‘ h i ∈ C ’ is areckless one. C ⊆ N with this property is called inductive . C ⊆ N with this property is called monotone . See [41, § roof. Let B consist of all s such that either ∃ n < k [ s = h n i ] or s = h i and ∃ n [ n = k ] ∨ ∀ n [ n < k ].Let α be given. Either α (0) < k and α ∈ B , or α (0) ≥ k and h i = α ∈ B .We thus see: Bar N ( B ).Let C coincide with B .For all s , if, for all n , s ∗ h n i ∈ C , then s = h i and ∀ n [ n < k ] and h i ∈ C .The statement ‘ h i ∈ C ’ is equivalent to ‘ ∃ n [ n = k ] ∨ ∀ n [ n < k ]’, and isreckless. (cid:3) Consider the set B mentioned in the proof of Theorem 20. Note that everycanonical proof of ‘ h i is (B-)safe ’ must use backward steps. The last step in acanonical proof of ‘ h i is (B-)safe ’ must be a forward step: h i is safe because: h i is safe, h i is safe, h i is safe, . . . . For each n < k the conclusion: h n i is safe follows by a basic step from: h n i ∈ B . For each n ≥ k the conclusion: h n i is safe follows by a backward step from: h i is safe and that follows in its turn by a basicstep from: h i ∈ B (as n ≥ k ).We thus see that, in this particular case, the backward steps can not be missed.In the case of the Fan Theorem, Theorem 14, as we learned from Theorem 15, onemight have claimed there is a canonical proof using only forward reasoning steps.16. Open Induction in [0 , Definition 17. We let ( r ′ , r ′′ ) , ( r ′ , r ′′ ) , ( r ′ , r ′′ ) , . . . be a fixed enumeration of allpairs of rationals.For each α , G α := { x ∈ R|∃ n [ r ′ α ( n ) < x < r ′′ α ( n ) ] } . For each a , G a := { x ∈ R | ∃ n < length ( a )[ r ′ a ( n ) < x < r ′′ a ( n ) ] } . G ⊆ R is open if and only if, for some α , G = G α .For all x, y in R such that x < y , we define [ x, y ) := { z ∈ R | x ≤ R z < R y } . G ⊆ R is progressive in [0 , if and only if ∀ x ∈ [0 , , x ) ⊆ G → x ∈ G ] . Theorem 21 (Principle of Open Induction in [0 , . For every open G ⊆ R , if G is progressive in [0 , , then [0 , ⊆ G . Although the theorem may be new to the reader, ´E. Borel in fact introducedand used it when he proved, in 1895, see [6], what is now called the Heine-BorelTheorem: for each α , if [0 , ⊆ G α , then ∃ n [[0 , ⊆ G αn ] . Borel’s argument may be described as follows. Let α be given such that [0 , ⊆ G α .Define H := { y ∈ R | ∃ n [[0 , y ] ⊆ G αn ] } . Note that H is open and progressive in[0 , , ∈ H and ∃ n [[0 , ⊆ G αn ].How does the classical mathematician convince herself of the validity of thePrinciple of Open Induction in [0 , G ⊆ R that is progressive in [0 , x , x , . . . of reals. She defines x := 0 and notes x ∈ G . She then finds x > x such that [0 , x ) ⊆ G . If x ≥ 1, she is done and defines, for each n ≥ x n = 1. If not, she observes: x ∈ G and finds x > x such that [0 , x ) ⊆ G . If x ≥ n ≥ x n = 1. If not, she observes: x ∈ G and finds x > x such that [0 , x ) ⊆ G . And so on. She thus finds an infinitesequence 0 ≤ x ≤ x ≤ x ≤ . . . of reals such that, for each n , [0 , x n ) ⊆ G and: x n +1 = x n if and only if x n = 1. The infinite non-decreasing sequence x , x , . . . is ounded by 1 and thus converges, to, say, x ω +0 = x ω . Then: x ω ∈ G as [ x, x ω ) ⊆ G .If x ω ≥ 1, she is done and defines, for each n ≥ x ω + n = 1. If not, she startsagain and finds suitable x ω +1 , x ω +2 , . . . such that, for each n , [0 , x ω + n ) ⊆ G and x ω + n ≤ x ω + n +1 and: x ω + n +1 = x ω + n if and only if x ω + n ≥ 1. She then considers x ω · = x ω + ω := lim n x ω + n . She has inexhaustible energy and, if needed, finds x ω · x ω · , . . . x ω · ω , . . . . She thus continues her infinite sequence through the countableordinals.She believes this procedure will come to an end, that is: there must exist acountable ordinal β such that x β = 1. She argues a follows. For each k > k countable ordinals α such that x α + k < x α +1 . Therefore, there areonly countably many countable ordinals such that x α < x α +1 . Find a countableordinal β that lies beyond every countable ordinal α such that x α < x α +1 . Then x β = x β +1 and: x β = 1.Borel argued in this way, proudly using the countable ordinals introduced byCantor some 20 years earlier, see [34].This argument, from a constructive point of view, is quite fantastic. We partedcompany with the classical mathematician already at the point where she believedto find x ω . It is not true, constructively, that every non-decreasing bounded infinitesequence of reals has a limit: consider the sequence defined by: x n = 0 if n < k and x n = 1 if k ≤ n . The classical mathematician may propose a second line of thought. She might use a classical mathematician’s finest weapon , as follows. Assume G 6 = [0 , y := inf([0 , \ G ). As G is open, y / ∈ G , but also [0 , y ) ⊆ G . Contradiction.Again, we can not partake in her joy. It is not clear that we may construct y .And, of course, it is not clear that, having proved: ¬∃ x ∈ [0 , x / ∈ G ], we mightconclude: ∀ x ∈ [0 , x ∈ G ].In order to make the problem more pictorial, let us define: Achilles arrives at x if and only if [0 , x ) ⊆ G .How might we, constructive mathematicians, convince ourselves that Achillesarrives at 1? We should first prove that Achilles will arrive at , but it does notseem easier to prove that Achilles will arrive at than to prove that he will arriveat 1. Proof. Let G ⊆ R be open and progressive in [0 , α such that G = G α .We define a mapping s ( c s , d s ) that associates to every s in N a pair of rationals.The elements s of N are thought of as coding finite sequences of natural numbers,see Subsection 12.1, Definition 10.(i) ( c h i , d h i ) := (0 , s , for each n , find out if ∀ x ∈ [0 , c ( s )+ d ( s )2 ] ∃ i < n [ r ′ α ( i ) < x < r ′′ α ( i ) ]. If so , define ( c s ∗h n i , d s ∗h n i ) = ( c s + d s , d s ), and if not , define ( c s ∗h n i , d s ∗h n i ) = ( c s , c s + d s ).Note: for each s , ( c s ∗h i , d s ∗h i ) = ( c s , c s + d s ).Also note: for each s , ∃ n ∀ x ∈ [0 , c s ) ∃ i < n [ r ′ α ( i ) < x < r ′′ α ( i ) ].One proves this by induction on the length of s . Assume c := lim n x n exists. If c > 0, then ∃ n [ n = k ] and, if c < 1, then ∀ n [ n < k ]. See Section 4. See [52, p. 338]. The theorem was found and proven by Thierry Coquand, in 1997. We were discussing thenthe problem if positive contrapositions of the minimal bad sequence arguments used in [43] mightbe intuitionistically true, see [54, § Note that this a decidable proposition. or each β in N , we let x β be the real number such that, for each n , x β ( n ) = ( c βn , d βn ). Note: for each β , [0 , x β ) ⊆ G and thus x β ∈ G .Let B be the set of all s such that ∃ n ∀ x ∈ [0 , d s ] ∃ i < n [ r ′ α ( i ) < x < r ′′ α ( i ) ].Let β be given. Using the fact: x β ∈ G , find q such that r ′ α ( q ) < x β < r ′′ α ( q ) .Find m such that r ′ α ( q ) < c βm < d βm < r ′′ α ( q ) .Find p such that ∀ x ∈ [0 , c βm ] ∃ i < p [ r ′ α ( i ) < x < r ′′ α ( i ) ].Define n := max( p, q + 1).Conclude: ∀ x ∈ [0 , d βm ] ∃ i < p ][ r ′ α ( i ) < x < r ′′ α ( i ) ]] and: βm ∈ B .We thus see: Bar N ( B ), i.e.: B is a bar in N .Note: B is monotone, i.e.: for all s , if s ∈ B , then ∀ n [ s ∗ h n i ∈ B ].We now prove that B is also inductive.Let s be given such that ∀ n [ s ∗ h n i ∈ B ]. Then, in particular: s ∗ h i ∈ B .Note d s ∗h i = c s + d s and find n such that ∀ x ∈ [0 , c s + d s ] ∃ i < n [ r ′ α ( i ) < x < r ′′ α ( i ) ].Conclude: ( c s ∗h n i , d s ∗h n i ) = ( c s + d s , d s ). Using the fact: s ∗ h n i ∈ B , find m suchthat ∀ x ∈ [0 , d s ∗h n i ] ∃ i < m [ r ′ α ( i ) < x < r ′′ α ( i ) ].Note: d s ∗h n i = d s and conclude: s ∈ B .We thus see: if ∀ n [ s ∗ h n i ∈ B ], then s ∈ B .As Bar N ( B ) and ∀ s [ s ∈ B ↔ ∀ n [ s ∗ h n i ∈ B ]], we may conclude, using Theorem19: h i ∈ B , that is: ∃ n ∀ x ∈ [0 , ∃ i < n [ r ′ α ( i ) < x < r ′′ α ( i ) ] and: [0 , ⊆ G . (cid:3) The Principle of Open Induction in [0 , 1] plays a large rˆole in [62].17. Brouwer’s Thesis, again Using stumps. In intuitionistic analysis, stumps play the rˆole fulfilled bycountable ordinals in classical analysis. Definition 18. For every s in N , for every A ⊆ N , s ∗ A := { s ∗ t | t ∈ A } . Stp , a collection of subsets of N , called stumps , is defined as follows. (i) ∅ ∈ Stp , and (ii) for every infinite sequence S , S , . . . of elements of Stp , the set S := {h i} ∪ S n ∈ N h n i ∗ S n is again an element of Stp , and (iii) nothing more: every element of Stp is obtained by starting from ∅ and apply-ing the operation mentioned in (ii) repeatedly. Note that, for every stump S , either S = ∅ or h i ∈ S . Definition 19. For every A ⊆ N , for every n , A ↾ h n i := { s | h n i ∗ s ∈ A } .For every non-empty stump S , for every n , S ↾ h n i is again a stump. S ↾ h n i is called the n -th immediate substump of S . Axiom 2 (Principle of Induction on Stp ) . Let P ⊆ Stp be given. If ∅ ∈ P , andfor every non-empty stump S , if ∀ n [ S ↾ h n i ∈ P ] , then S ∈ P , then Stp = P . Theorem 22 (Brouwer’s Thesis on bars in N ) . For every B ⊆ N such that Bar N ( B ) , there exists a stump S such that Bar N ( S ∩ B ) .Proof. Let B ⊆ N be given such that Bar N ( B ).Let C be the set of all s such that, for some stump S , ∀ α ∃ n [ αn ∈ S ∧ ∃ t ⊑ s ∗ αn [ t ∈ B ]] . Let s be given such that s ∈ B . Define S := {h i} and note: ∀ α [ α ∈ S ∧ s ∗ α ∈ B ] . See [47, § e thus see: B ⊆ C .We now prove that C is inductive.Let s be given such that ∀ m [ s ∗ h m i ∈ C ].Find an infinite sequence S , S , . . . of stumps such that ∀ m ∀ α ∃ n [ αn ∈ S m ∧ ∃ t ⊑ s ∗ h m i ∗ αn [ t ∈ B ]] . Define a non-empty stump S suchthat, for each m , S ↾ h m i = S m and conclude: ∀ α ∃ n [ αn ∈ S ∧ ∃ t ⊑ s ∗ αn [ t ∈ B ]and: s ∈ C .We thus see: for each s , if ∀ m [ s ∗ h m i ∈ C ], then s ∈ C .Finally, we show that C is monotone.Let s in C be given.Find a stump S such that ∀ α ∃ n [ αn ∈ S ∧ ∃ t ⊑ s ∗ αn ∈ B ]].Let m be given and distinguish two cases: Case (1) . S ↾ h m i 6 = ∅ . Then ∀ α ∃ n [ αn ∈ S ↾ h m i ∧ ∃ t ⊑ s ∗ h m i ∗ αn ∈ B ]],and: s ∗ h m i ∈ C . Case (2) . S ↾ h m i = ∅ . Define T := {h i} and note: ∀ α [ α ∈ T ∧ ∃ t ⊑ s ∗ h m i ∗ α t ∈ B ]], and: s ∗ h m i ∈ C .We thus see: for all s , for all m , if s ∈ C , then s ∗ h m i ∈ C .Using Theorem 19, we conclude: h i ∈ C , that is: there exists a stump S suchthat ∀ α ∃ n [ αn ∈ S ∧ αn ∈ B ], that is: Bar N ( S ∩ B ). (cid:3) In the next Section, we show that Theorem 22 is a useful conclusion from The-orem 19. 18. Ramseyan Theorems Dickson’s Lemma.Definition 20. For all α, β in N , α ◦ β is the element of N such that, for each n , α ◦ β ( n ) = α (cid:0) β ( n ) (cid:1) . [ ω ] ω := { ζ ∈ N | ∀ i [ ζ ( i ) < ζ ( i + 1)] } . The following Lemma, a humble member of a family of results called RamseyanTheorems , plays a rˆole in Computer Algebra. The Lemma lies at the basis ofBuchberger’s algorithm. Theorem 23 (Dickson’s Lemma) . For all p > , for all α , α , . . . α p − in N ,there exist i, j such that i < j and ∀ k < p [ α k ( i ) ≤ α k ( j )] . We first sketch a proof along traditional lines that fails to be constructive.The proof is by induction to p .First consider the case p = 1. Let α be given and consider α (0). Note: ∃ i ≤ α (0)[ α ( i ) ≤ α ( i + 1)].Now let p be given such that the case p of the Theorem has been established.We want to prove the case p + 1.Let α , α , . . . , α p be given.( ♭ ) Find ζ in [ ω ] ω such that ∀ i [ α p ◦ ζ ( i ) ≤ α p ◦ ζ ( i + 1)].Applying the induction hypothesis, determine i, j such that, for all k < p , α ◦ ζ ( i ) ≤ α ◦ ζ ( j ) and conclude: for all k ≤ p + 1, α ◦ ζ ( i ) ≤ α ◦ ζ ( j ).This completes the proof of the induction step.Unfortunately, there are difficulties with ( ♭ ). It is, in general, not possible tofind ζ in [ ω ] ω such that ∀ i [ α p ◦ ζ ( i ) ≤ α p ◦ ζ ( i + 1)]. The following example makesthis clear. See [22, Theorem 5]. efine α such that, for all n , if n < k , then α ( n ) = 1, and, if k ≤ n , then α ( n ) = 0. Suppose we find ζ in [ ω ] ω such that ∀ i [ α ◦ ζ ( i ) ≤ α ◦ ζ ( i + 1)].If α ◦ ζ (0) = 0, then k ≤ ζ (0), and, if α ◦ ζ (0) = 1, then ∀ n [ n < k ].The statement ‘ ∃ ζ ∈ [ ω ] ω ∀ i [ α ◦ ζ ( i ) ≤ α ◦ ζ ( i + 1)]’ thus is reckless or hardy .We now give a constructive proof of Dickson’s Lemma. Proof. The proof is by induction on p . The case p = 1 is easy and is done asin the sketch preceding this proof. Now let p given such that the case p has beenestablished. We prove the case p + 1.Let α , α , . . . , α p be given.Define QED := ∃ i ∃ j [ i < j ∧ ∀ k ≤ p [ α k ( i ) ≤ α k ( j )].We first prove:( ∗ ) for each m , ∀ n ∃ j ≥ n [ m ≤ α p ( j ) ∨ QED ].This again is done by induction. Note that the case m = 0 is trivial.Now assume m is given and the case m has been established.We prove the case m + 1.Let n be given.Using the induction hypothesis, find ζ in [ ω ] ω such that ∀ i [ α p ◦ ζ ( i ) ≥ m ∨ QED ].Find i, j such that n < i < j and ∀ k < p [ α k ◦ ζ ( i ) ≤ α k ◦ ζ ( j )].Note: either m + 1 ≤ α p ◦ ζ ( i ) or m + 1 ≤ α p ◦ ζ ( j ) or QED .We thus see: ∀ n ∃ j ≥ n [ m + 1 ≤ α p ( j ) ∨ QED ]. This completes the proof of ( ∗ ).Using ( ∗ ) we find ζ in [ ω ] ω such that ∀ n [ α p ◦ ζ ( n ) ≤ α p ◦ ζ ( n + 1) ∨ QED ].Find i, j such that i < j and ∀ k < p [ α k ◦ ζ ( i ) ≤ α k ◦ ζ ( j )].Note: α p ◦ ζ ( i ) ≤ α p ◦ ζ ( j ) ∨ QED and conclude: QED .This concludes the proof of the case p + 1.The theorem thus is proven by induction. (cid:3) How to formulate and extend Ramsey’s Theorem? The following Definition extends Definition 10 in Subsection 12.1. Definition 21. For all α , for all n , for all s in ω n , α ◦ s is the element of ω n suchthat, for each i < n , α ◦ s ( i ) = α (cid:0) s ( i ) (cid:1) . For all m , for all s in ω m , for all n , for all t in ω n such that ∀ i < n [ t ( i ) < m ] , s ◦ t is the element of ω n such that for each i < n , s ◦ t ( i ) = s (cid:0) t ( i ) (cid:1) . [ ω ] n := { s ∈ ω n | ∀ i [ i + 1 < n → s ( i ) < s ( i + 1)] } . [ ω ] <ω := S n [ ω ] n . Definition 22. A ⊆ N is almost-full if and only if ∀ ζ ∈ [ ω ] ω ∃ s ∈ [ ω ] <ω [ ζ ◦ s ∈ A ] . Let k > k -dimensional case of Ramsey’s Theorem is the followingstatement: IRT ( k ): For all A, B ⊆ [ ω ] k ,if A, B are almost-full, then A ∩ B is almost-full. As, for all A ⊆ [ ω ] k , A ∩ ([ ω ] k \ A ) = ∅ is not almost-full, the classical mathe-matician draws the following conclusion from IRT ( k ): IRT ( k ) class : For all A ⊆ [ ω ] k ,either A is not almost-full or [ ω ] k \ A is not almost-full , See [54, Theorem 1.1]. QED (again) stands for: quod est demonstrandum, ‘what we (still) have to prove’ . This definition improves the definition as used in [54, Section 4.2]. The proof of [54, Theorem4.4] is not correct as was pointed out to me by M. Bickford. The argument may be saved if onerestricts attention to strictly increasing sequences as we do here. .e. either ∃ ζ ∈ [ ω ] ω ∀ s ∈ [ ω ] k [ ζ ◦ s ∈ A ] or ∃ ζ ∈ [ ω ] ω ∀ s ∈ [ ω ] k [ ζ ◦ s / ∈ A ] . This is the formulation of the theorem that will be familiar to the classical reader.She also will be able to conclude (her reading of) IRT ( k ) from IRT ( k ) class . F.P. Ramsey proved: ∀ k > IRT ( k ) class ], see [46].There is an extension of Ramsey’s result into the transfinite that is called theClopen Ramsey Theorem .We need the following definition. Definition 23. For every stump U , we define U := { s | s / ∈ U ∧ ∀ t ⊏ s [ t ∈ U ] } .The set U is called the border of the stump U . Let a stump U be given. We introduce the following statement: IRT ( U ): For all A, B ⊆ U ,if A, B are almost-full, then A ∩ B is almost-full. The Clopen Ramsey Theorem , CRT , is the statement: for every stump U , IRT ( U ) . The Infinite Ramsey Theorem , IRT , is the statement: for all k , IRT ( k ) . IRT class is the (constructively wrong) statement: for all k , IRT ( k ) class .18.3. The proof of the Clopen Ramsey Theorem. The following Definition extends Definition 21 in Subsection 18.2. Definition 24. For each n, for each k , [ n ] k := { s ∈ [ ω ] k | ∀ i < k [ s ( i ) < n ] } .For each n , for each k , [ n ] For each A ⊆ N , if A is almost-full, then there exists a stump S securing A .Proof. Let A ⊆ N be almost-full.Define B := S n { u ∈ ω n | u / ∈ [ ω ] n ∨ ∃ s ∈ [ n ] 1) + 1.Note: ζ ∈ [ ω ] ω .Find n such that ∃ s ∈ [ n ] Let a stump U be given and assume A ⊆ U and u ∈ U ∪ U . u U - secures A if and only if either: u ∈ A or: u ∈ U and ∀ t ∈ U [ u ⊏ t → t ∈ A ] .Let also a stump S be given. S U -secures A if and only if ∃ m ∀ ζ ∈ [ ω ] ω [ ζ (0) > m → ∃ n [ ζn ∈ S ∧ ∃ s ∈ [ n ] IRT (1) class is a reckless or hardy statement:consider A := {h n i | n < k } . See [21] and [26]. See [66, Theorem 7.3] The Theorem has been announced in [66, § § 6] but, until now, no proof appearedin print. orollary 25. For all stumps U , for each A ⊆ U , if A is almost-full, then thereexists a stump S U -securing A .Proof. Note: if S secures A , then S U -secures A and use Corollary 24. (cid:3) Theorem 26 (Intuitionistic Ramsey Theorem) . (i) CRT : For all stumps U , for all A, B ⊆ U ,if A, B are both almost-full, then A ∩ B is almost-full. (ii) IRT : For all k , for all A, B ⊆ [ ω ] k ,if A, B are both almost-full, then A ∩ B is almost-full.Proof. (i) According to Corollaries 24 and 25, it suffices to prove:for all stumps U , P ( U ),where P ( U ) is the statement: for all stumps S , Q ( U, S ),where Q ( U, S ) is the statement: for all stumps T , R ( U, S, T ),where R ( U, S, T ) is the statement:for all A, B ⊆ U , if S U -secures A , and T U -secures B , then A ∩ B is almost-full.We use Axiom 2 from Subsection 17.1, the principle of induction on the set Stp of stumps. Our proof is a proof by triple induction .1. Note: ∅ = {h i} .For all A, B ⊆ {h i} , if A, B are almost-full, then A = B = A ∩ B = {h i} .We thus see that P ( ∅ ) is true.2. Let a non-empty stump U be given such that, for all n , P ( U ↾ h n i ).We intend to prove P ( U ), that is: for all stumps S , Q ( U, S ).We use again use Axiom 2.2.1. Note that Q ( U, ∅ ) holds a trivial reason: no subset of N is secured by ∅ .2.2. Let a non-empty stump S be given such that, for all n , Q ( U, S ↾ h n i ).We intend to prove: Q ( U, S ), that is, for all stumps T , R ( U, S, T ).Again, we use Axiom 2.2.2.1. Note that R ( U, S, ∅ ) holds for a trivial reason: no subset of N is securedby ∅ .2.2.2. Let a non-empty stump T be given such that, for all n , R ( U, S, T ↾ h n i ).We intend to prove: R ( U, S, T ).Let A, B ⊆ U be given such that S U -secures A and T U -secures B .We have to prove that A ∩ B is almost-full, i.e.( ∗ ) ∀ ζ ∈ [ ω ] ω ∃ s ∈ [ ω ] <ω [ ζ ◦ s ∈ A ∩ B ].It is useful, however, to first prove a slightly weaker statement:( ∗∗ ) ∀ ζ ∈ [ ω ] ω ∃ s ∈ [ ω ] <ω [ ζ ◦ s ∈ A ∩ B ∨ ( s = 0 ∧ s (0) = 0 ∧ ζ ◦ s ∈ A )].Let ζ in [ ω ] ω be given.We distinguish two cases. Case (a) . S ↾ h ζ (0) i = ∅ . Note: {h i} U -secures A , and A = U and A ∩ B = B is almost-full. One may conclude: ( ∗ ), and, in particular, ∃ s ∈ [ ω ] <ω [ ζ ◦ s ∈ A ∩ B ]. Case (b) . S ↾ h ζ (0) i 6 = ∅ . Then h i ∈ S ↾ h ζ (0) i .Define a statement QED := ∃ s ∈ [ ω ] <ω \ { } [ s (0) = 0 ∧ ζ ◦ s ∈ A ].Define A ζ := { u ∈ U | ¬∃ t [ u = ζ ◦ t ] ∨ u ∈ A ∨ QED } .Note that, for all u in U ∪ U , if ¬∃ t [ u = ζ ◦ t ], then u U -secures A and also A ζ .We now want to prove: QED (again) stands for: quod est demonstrandum, ‘what we (still) have to prove’ . ↾ h ζ (0) i U -secures A ζ .First, using the fact that S U -secures A , find m such that ∀ η ∈ [ ω ] ω [ η (0) > m → ∃ n [ ηn ∈ S ∧ ∃ s ∈ [ n ] 0. Find u such that length ( u ) = length ( s ) and ∀ i < length ( u )[ u ( i ) = s ( i ) − ρ ( n − ◦ u = ηn ◦ s U -secures A and also A ζ . Case (bb) . s (0) = 0. Find u such that s = h i ∗ u . Note: s (0) = η (0) = ζ (0).If ¬∃ t [ ρn ◦ u = ζ ◦ t ], then also ¬∃ t [ ρ ( n − ◦ u = ζ ◦ t ] and ρ ( n − ◦ u U -secures A and also A ζ .If ∃ t [ ρ ( n − ◦ u = ζ ◦ t ], find η in [ ω ] ω such that ζ ◦ s ⊏ η and ∀ i ∃ j [ η ( i ) = ζ ( j )],so η is a subsequence of ζ . Find n such that ηn ∈ U . As ζ ◦ s U -secures A , we mayconclude: ηn ∈ A . Taking t := ηn , we see: ∃ t [ t = 0 ∧ t (0) = 0 ∧ ζ ◦ t ∈ A ], i.e. QED and we may conclude: ρ ( n − ◦ h i = h i U -secures A ζ .We thus see: for all ρ in [ ω ] ω , if ρ (0) > q , then ∃ n [ ρn ∈ S ↾ h ζ (0) i ∧ ∃ t ∈ [ ω ] <ω [ ρn ◦ t U -secures A ζ ]], that is: S ↾ h ζ (0) i U -secures A ζ .Also: T U -secures B .Using the assumption R ( U, S ↾ h ζ (0) i , T ), we conclude: A ζ ∩ B is almost-full.In particular, we may find s in [ ω ] <ω such that ζ ◦ s ∈ A ζ ∩ B , and therefore,either ζ ◦ s ∈ A ∩ B or QED .We thus see: ∀ ζ ∈ [ ω ] ω ∃ s ∈ [ ω ] <ω [ ζ ◦ s ∈ A ∩ B ∨ ( s = 0 ∧ s (0) = 0 ∧ ζ ◦ s ∈ A )].Now let ζ in [ ω ] ω be given.Define a statement QED := ∃ s ∈ [ ω ] <ω [ ζ ◦ s ∈ A ∩ B ].Also define C := { u ∈ U ↾ h ζ (0) i | ¬∃ t ∈ [ ω ] <ω ][ u = ζ ◦ t ] ∨ h ζ (0) i ∗ u ∈ A ∨ QED } .We want to prove that C is almost-full.First, note that, for each η in [ ω ] ω , if η (0) = ζ (0) and η is a subsequence of ζ , i.e.: ∀ i ∃ j [ η ( i ) = ζ ( j )], then either ∃ s ∈ [ ω ] <ω [ η ◦ s ∈ A ∩ B ], and, therefore, QED , or ∃ s ∈ [ ω ] <ω \{ } [ s (0) = 0 ∧ η ◦ s ∈ A ], and, therefore, ∃ s ∈ [ ω ] <ω [ h ζ (0) i∗ ( η ◦ s ) ∈ A ].We may even conclude: for every η in [ ω ] ω that is a subsequence of ζ , either QED , or ∃ s ∈ [ ω ] <ω [ h ζ (0) i ∗ ( η ◦ s ) ∈ A ], for, given η such that η (0) = ζ (0) wemay apply the previous result to the sequence η + := h ζ (0) i ∗ η .We thus see that, for every η in [ ω ] ω that is a subsequence of ζ , there exists s in[ ω ] <ω such that η ◦ s ∈ C .Now let η in [ ω ] ω be given. Define η ζ in [ ω ] ω as follows.If ∃ i [ η (0) = ζ ( i )], define η ζ (0) = η (0), and, if not, define η ζ (0) = ζ (0). or each n , if ∃ t ∈ [ ω ] n +2 [ η ( n + 2) = ζ ◦ t ], then η ζ ( n + 1) = η ( n + 1), and, if not,then η ζ ( n + 1) = ζ ( i + 1), where i satisfies η ζ ( n ) = ζ ( i ).As η ζ is a subsequence of ζ , find s in [ ω ] <ω such that η ζ ◦ s ∈ C .Either η ◦ s = η ζ ◦ s ∈ C or η ◦ s = η ζ ◦ s . In the latter case, let n be the least n such that ¬∃ i [ η ( n ) = ζ ( i )]. Define ρ in [ ω ] ω such that ∀ n [ ρ ( n ) = n + n ] and find m such that η ◦ ρ ( m ) ∈ U ↾ h ζ (0) i . Define s := ρm and note: η ◦ ρ ( m ) = η ◦ s and ¬∃ u ∈ [ ω ] <ω [ η ◦ s = ζ ◦ u ] and η ◦ s ∈ C .We thus see that C is almost-full.We obtained this result using the assumption: Q ( U, S ↾ h ζ (0) i ).Now define D := { u ∈ U ↾ h ζ (0) i | ¬∃ t ∈ [ ω ] <ω ][ u = ζ ◦ t ] ∨ h ζ (0) i ∗ u ∈ B ∨ QED } .Using the assumption: R ( U, S, T ↾ h ζ (0) i ), one may prove: D is almost-full, byan argument similar to the one that established that C is almost-full.Using the assumption P ( U ↾ h ζ (0) i ), one then concludes: C ∩ D is almost-full.Find s in [ ω ] <ω such that ζ ◦ s ∈ C ∩ D or h ζ (0) i ∗ ζ ◦ s ∈ A ∩ B , so, in any case, QED .Using the assumptions: for all n , P ( U ↾ h n i ) (2), and: for all n , Q ( U, S ↾ h n i )(2.2) and: for all n , R ( U, S, T ↾ h n i ) (2.2.2), one thus proves: R ( U, S, T ) , i.e.: forall A, B ⊆ U , if S, T U -secure A, B , respectively, then A ∩ B is almost-full, i.e.: ∀ ζ ∈ ω ω ∃ s ∈ [ ω ] <ω [ ζ ◦ s ∈ A ∩ B ].Axiom 2 and Corollary 25 then guarantee the conclusion: CRT .(ii) This immediately follows from (i), as, for all k , ω For all k > , for all r > , for all subsets A , A , . . . , A r − of [ ω ] k ,if, for each j < r , A j is almost-full, then T j The (extended) Finite Ramsey Theorem andthe ‘compactness argument’. The following Definition extends Definition 24 in Subsection 18.3. Note that, de-viating from the usual treatment of Ramsey theorems, we consider finite increasingsequences of nonnegative integers rather than finite sets of nonnegative integers. Definition 26. For all r > , r <ω := { s | ∀ i < length ( s )[ s ( i ) < r ] } .For all positive integers c, m, k, r , c : [ m ] k → r , ( c is an r -colouring of [ m ] k ) , if and only if c ∈ r <ω and ∀ s ∈ [ m ] k [ s < length ( c )] .If c : [ m ] k → r , one may consider c as an r -colouring of the k -element subsetsof m = { , , . . . , m − } .For all positive integers k, r, n, M , M → ( n ) kr if and only if,for every c : [ M ] k → r , there exists t in [ M ] n such that ∃ j < r ∀ u ∈ [ n ] k [ c ( t ◦ u ) = j ] , i.e. t is c -monochromatic .If M → ( n ) kr , then, for every r -colouring c of the k -element subsets of M = { , , . . . , M − } there exists an n -element subset A = { t (0) , t (1) , . . . , t ( n − } of M such that all k -element subsets of A obtain, from c , one and the same colour.For all positive integers k, r, n, M , M → ∗ ( n ) kr if and only if,for every c : [ M ] k → r , there exist t, p such that p ≥ n and t ∈ [ M ] p and p = t (0) and ∃ j < r ∀ u ∈ [ p ] k [ c ( t ◦ u ) = j ] . IRT is proven in [66, Theorem 7.3] from what is, in this paper, Theorem 19, see also [65, § IRT (2) is obtained in [54, § 4] and [59, Theorem 9] from what is, in this paper Theorem 22. he meaning of this statement is similar to that of the previous one but now the p -element subset A = { t (0) , t (1) , . . . t ( p − } of M is required to satisfy the conditions n ≤ p and p = min( A ) = t (0) . A finite set A with the property A ) ≥ min( A ) issometimes called relatively large , see [44] . FRT , the Finite Ramsey Theorem , is the statement: ∀ k ∀ r ∀ n ∃ M [ M → ( n ) kr ] . FRT P H , the Extended Finite Ramsey Theorem , is the statement: ∀ k ∀ r ∀ n ∃ M [ M → ∗ ( n ) kr ] . Like IRT class , FRT was proven in 1928 by F.P. Ramsey, see [46].Ramsey thought proving IRT class was a useful training for someone who wantedto prove FRT . FRT was given an independent proof.Around 1950, it was seen that a so-called compactness argument proves FRT from IRT class , see [18] and [32, § § FRT P H was formulated by J. Paris and L. Har-rington. Also FRT P H may be proven (classically) from IRT class by a compactnessargument, see [44] but FRT P H , although it may be formulated in the language of Peano Arithmetic , can not be proven from the axioms of P A . Paris and Harring-ton thus found a mathematical Incompleteness Theorem, thereby creating a stir inthe logic community.The proof of the next Theorem is an intuitionistic version of the compactnessargument deriving FRT P H from IRT class . Definition 27. For each r > , r ω := { χ | ∀ i [ χ ( i ) < r ] } . Theorem 28. IRT → FRT P H .Proof. Let n, k, r be given positive integers. Let χ be an element of r ω . We wantto prove: QED := ∃ p ≥ n ∃ t ∈ [ ω ] p [ t (0) = p ∧ ∃ j < r ∀ u ∈ [ p ] k [ χ ( t ◦ u ) = j ]] . For each j < r , define: A j := { t ∈ [ ω ] k | χ ( t ) = j ∨ QED } .Let j < r and ζ in [ ω ] ω be given. Find p := max (cid:0) n, ζ (0) (cid:1) and define t := ζp .Either: ∀ u ∈ [ p ] k [ χ ( t ◦ u ) = j ] and QED , or: ∃ u ∈ [ p ] k [ χ ( t ◦ u ) = j ].We thus see: for each j < r , A j is almost-full.Using Corollary 27, conclude: T j Bar r ω ( B ) and: r ω is a fan. Using the Fan Theorem, Theorem 14, find z, m such that length ( z ) = m and ∀ i < m [ z ( i ) ∈ B ] and ∀ χ ∈ r ω ∃ i < m [ z ( i ) ⊏ χ ].Define N := max i Some preparations. We introduce the important notion of a spread . Definition 28. Let β be given. β is a spread-law if and only if ∀ s [ β ( s ) = 0 ↔ ∃ n [ β ( s ∗ h n i ) = 0]] . X ⊆ N is a spread if and only if there exists a spread-law β such that X = F β := { α | ∀ n [ β ( αn ) = 0] } . Note that the spread-law β in a sense governs the set F β . The law makesclear which steps are allowed during the step-by-step construction of an element α = α (0) , α (1) , . . . of F β . In his early publications, Brouwer used the term ‘ set,Menge ’ for what he later called spreads. This is because the notion expresses hisview in what sense a totality like for instance R might be called a set .Brouwer’s Continuity Principle, Axiom 1, extends to spreads: Theorem 29 (The extended Continuity Principle) . Let β be a spread-law and let R be a subset of F β × N .If ∀ α ∈ F β ∃ n [ αRn ] , then ∀ α ∈ F β ∃ m ∃ n ∀ β ∈ F β [ αm ⊏ β → βRn ] .Proof. Let β be a spread-law such that β ( h i ) = 0.Define ϕ : N → N such that, for each α , for each n ,if β ( α (cid:0) n + 1) (cid:1) = 0, then ( ϕ | α )( n ) = α ( n ), and,if not, then ( ϕ | α )( n ) = the least p such that β (cid:0) ( ϕ | α ) n ∗ h p i (cid:1) = 0].Note: ∀ α [ ϕ | α ∈ F β ] and ∀ α ∈ F β [ ϕ | α = α ].The function ϕ is called a retraction of N onto F β .Now assume: ∀ α ∈ F β ∃ n [ αRn ]. Then ∀ α ∃ n [( ϕ | α ) Rn ]. Using Axiom 1 conclude: ∀ α ∃ m ∀ β [ αm ⊏ β → ( ϕ | β ) Rn ] and: ∀ α ∈ F β ∃ m ∃ n ∀ β ∈ F β [ αm ⊏ β → βRn ]. (cid:3) Definition 29. For each α in N , for each n , we define α n in N by: ∀ m [ α n ( m ) = α (cid:0) m (2 n + 1) − (cid:1) ] . α n is called the n -th subsequence of α .For all α , for all n, j , we define: α n,j := ( α n ) j . The next definition explains how continuous functions fom N to N and contin-uous functions from N to N are coded by elements of N . Definition 30. For each ϕ , we define: ϕ : N → N if and only if ∀ α ∃ n [ ϕ ( αn ) = 0] .For each ϕ such that ϕ : N → N , for each α , we let ϕ ( α ) be the number p suchthat ∃ n [ ϕ ( αn ) = p + 1] ∧ ∀ i < n [ ϕ ( αi ) = 0]] .For each ϕ , we define: ϕ : N → N if and only if ∀ n [ ϕ n : N → N ] .For each ϕ such that ϕ : N → N , for each α , we let ϕ | α be the element β of N such that ∀ n [ β ( n ) = ϕ n ( α )] . The following notion plays a key rˆole in intuitionistic descriptive set theory , thetheory of Borel and projective sets. Definition 31 (Reducibility) . For all X , Y ⊆ N , for all ϕ : N → N , we define: ϕ reduces X to Y , if and onlyif ∀ α [ α ∈ X ↔ ϕ | α ∈ Y ] .For all X , Y ⊆ N , we define: X (cid:22) Y , X reduces to Y , if and only if there exists ϕ : N → N reducing X to Y . If ϕ reduces X to Y , then, for each α , the question In classical descriptive set theory, the reducibility notion introduced here is known as Wadge-reducibility , see [38, Section 21.E]. One may compare this notion to the notion of many-onereducibility from recursion theory. oes α belong to X ?is reduced to the question: Does ϕ | α belong tot Y ?19.2. Borel sets of finite rank. We first introduce open subsets and closed sub-sets of N = ω ω . Definition 32. For each β , we define G β := { α | ∃ n [ β ( αn ) = 0] } and F β := N \ G β = { α | ∀ n [ β ( αn ) = 0] } .We define E := { α | ∃ n [ α ( n ) = 0] } and A := N \ E = { α | ∀ n [ α ( n ) = 0] } = { } . X ⊆ N is open , or: Σ , if and only if ∃ β [ X = G β ] ,and closed , or: Π , if and only if ∃ β [ X = F β ] . Note that spreads, as introduced in Definition 28, are closed subsets of N . It isnot true that every closed subset of N is a spread. A closed subset F of N is aspread if and only if it is located , i.e. ∃ β ∀ s [ ∃ α ∈ F [ s ⊏ α ] ↔ β ( s ) = 0].Note that, if G ⊆ N is open, then its complement N \ G is a closed subset of N .It is not true that the complement of a closed subset of N always is an open subsetof N .We now introduce Borel sets of finite rank. Note that we avoid the operation oftaking the complement of a given Borel set. Definition 33. For each n > , for each β , we define, inductively, G n +1 β = S m F nβ m , and F n +1 β = T m F nβ m .For each n > , we define, inductively, E n +1 := { α | ∃ m [ α m ∈ A n ] } and A n +1 := { α | ∀ m [ α m ∈ E n ] } .For each n > , X ⊆ N is Σ n if and only if ∃ β [ X = G nβ ] and X is Π n if and only if ∃ β [ X = F nβ ] . X ⊆ N belongs to Σ n +1 if and only if there exists an infinite sequence Y , Y , . . . of elements of Π n such that X = S m Y m . X ⊆ N belongs to Π n +1 if and only if there exists an infinite sequence Y , Y , . . . of elements of Σ n such that X = T m Y m .19.3. The fine structure of the Borel hierarchy. One may prove that theunion of two opens sets is open again but it is not true that the union of two closedsets is a closed sets itself. It is useful to introduce the following operation. Definition 34. For each n > , for each X ⊆ N , we define D n ( X ) = { α | ∃ i < n [ α i ∈ X ] } . The next Theorem, Theorem 30, offers an example of a union of two closedsubsets of N that is not an intersection of countably many open subsets of N and,therefore, certainly not a closed subset of N , see item (iii). Theorem 30(iv) makesit clear that there are unions of three closed sets not coinciding with any unionof two closed sets, and unions of four closed sets not coinciding with any union ofthree closed sets, and so on. Theorem 30. (i) For all X ⊆ N , X is Π if and only if X (cid:22) A . (ii) For all X ⊆ N , for all n > , X (cid:22) D n ( A ) if and only if there exists Π sets F , F , . . . , F n − such that X = S i 2, ( α i ) i = 0 and, for all s , if ¬∃ p [ s = h i, p i ], then α i ( n ) = α ( n ).Note: for both i < α i ∈ D ( A ). Also note: ∀ n [ αn = α n ∨ αn = α n ].Find n , n such that, for both i < β m ( α i n i ) = 0. Note: either α n ⊏ α or α n ⊏ α , and, in both cases, ∃ n [ β m ( αn ) = 0].We thus see: ∀ α ∈ T [ α ∈ D ( A )]. Using Theorem 29, find m, i such that i < ∀ α [0 m ⊏ α → α i = 0]. This gives a contradiction as one may find α in T suchthat 0 m ⊏ α and α i = 0.We may conclude: D ( A ) is not Π .(iv) Again, consider T := { α | ∀ m ∀ n [ (cid:0) α ( m ) = 0 ∧ α ( n ) = 0 (cid:1) → m = n ] } .Let n > D n +1 ( A ) (cid:22) D n ( A ).Find β such that D n +1 ( A ) = { α | ∃ i < n ∀ j [ β i ( αj ) = 0] } .For each i , define B i := { α | α i = 0 } . Note that each B i is a spread.Also note: D n +1 ( A ) = S i The next Theorem, Theorem 31, makes astart with establishing the Borel hierarchy itself. Theorem 31. (i) For all n > , for all X ⊆ N , X is Σ n if and only if X (cid:22) E n ,and X is Π n if and only if X (cid:22) A n . (ii) For every ϕ : N → N , there exists α such that,for all m , α m ∈ E ↔ ( ϕ | α ) m ∈ E , and, therefore, α ∈ A ↔ ϕ | α ∈ A and α ∈ E ↔ ϕ | α ∈ E . (iii) For every ϕ : N → N ,if ∀ α [ α ∈ E → ϕ | α ∈ A ] , then ∃ α [ α ∈ A ∧ ϕ | α ∈ A ] . (iv) For every ϕ : N → N , if ∀ α [ α ∈ A → ϕ | α ∈ E ] , then ∃ α [ α ∈ E ∧ ϕ | α ∈ E ] .Proof. (i) The proof is left to the reader.(ii) Let ϕ : N → N be given. Define α , inductively, as follows.Let p be given such that α (0) , α (1) , . . . , α ( p − 1) have been defined already.Find m, n such that such that p = 2 n (2 m + 1) − α ( p ) = α m ( n ) := 1 if ∃ q < p [ ϕ m ( αq ) > ∧ ∀ i < q [ ϕ m ( αi ) = 0]],and α ( p ) := 0 if not .Note that, for all m , ∃ n [ α m ( n ) = 0] ↔ ∃ n [( ϕ | α ) m ( n ) = 0], and, therefore, α m ∈ E ↔ ( ϕ | α ) m ∈ E and α m ∈ A ↔ ( ϕ | α ) m ∈ A .(iii) Let ϕ : N → N be given such that ∀ α [ α ∈ E → ϕ | α ∈ A ].Using (i), find α such that ∀ m [ α m ∈ E ↔ ( ϕ | α ) m ∈ E ].Let m be given. efine α such that ( α ) m = 0 and, for all n = m , ( α ) n = α n .Note: α ∈ E , and, therefore, ϕ | α ∈ A and ( ϕ | α ) m ∈ E .Find n such that ( ϕ | α ) m ( n ) = 0. Either ( ϕ | α ) m ( n ) = 0 and ( ϕ | α ) m ∈ E and also α m ∈ E , or ( ϕ | α ) m ( n ) = 0 and ϕ | α ⊥ ϕ | α and α ⊥ α and α m ⊥ α m ∈ E .We thus see: ∀ m [ α m ∈ E ]. Therefore, both α and ϕ | α are in A .(iv) The following observation is easy but crucial: ∀ α [ ∃ σ ∀ m [ α m (cid:0) σ ( m ) (cid:1) = 0] → α ∈ A ].For each α , for each σ , we let σ ⋊⋉ α be the element of N such that, for each m ,( σ ⋊⋉ α ) m (cid:0) σ ( m ) (cid:1) = max (cid:0) , α m (cid:0) σ ( m )) (cid:1) and, for all n = σ ( m ),( σ ⋊⋉ α ) m ( n ) = α m ( n )]. σ ⋊⋉ α may be thought of as ‘ α -corrected-by- σ ’.Note: ∀ α [ ∃ σ [ α = σ ⋊⋉ α ] → α ∈ A ].Now let ϕ : N → N be given such that ∀ α [ α ∈ A → ϕ | α ∈ E ].Conclude: ∀ α ∃ n [ (cid:0) ( ϕ | ( α ⋊⋉ α ) (cid:1) n = 0].Using Axiom 1, find m, n such that ∀ α [0 m ⊏ α → (cid:0) ( ϕ | ( α ⋊⋉ α ) (cid:1) n = 0].Define β in C such that ∀ i ∀ j [ β i ( j ) = 1 ↔ ( i < m ∧ j = 0)].Let j be given. Find p such that ϕ n,j ( βp ) = 0 and ∀ i < p [ ϕ n,j ( βi ) = 0].Find α such that 0 m ⊏ α and βp ⊏ α ⋊⋉ α .Conclude: ( ϕ | β ) n ( j ) = (cid:0) ϕ | ( α ⋊⋉ α ) (cid:1) n ( j ) = 0 and ϕ n,j ( βp ) = 1.We thus see: ∀ j [( ϕ | β ) n ( j ) = 0] and ( ϕ | β ) n = 0 and ϕ | β ∈ E .We thus found β such that both β and ϕ | β are in E . (cid:3) Note that the proofs of Theorem 31(ii) and (iii) are elementary in the sensethat they do not use intuitionistic axioms like the Continuity Principle or the FanTheorem. The constructive argument for Theorem 31(ii) extends to a constructiveargument establishing For each n > , for each ϕ : N → N , there exists α such that α ∈ E n ↔ ϕ | α ∈ E n and α ∈ A n ↔ ϕ | α ∈ A n . Unfortunately, one can’t conclude from this, constructively, that A n does not reduceto E n . Assuming that ϕ : N → N reduces A n to E n , one finds, using Theorem31(ii), α such that α ∈ A n ↔ α ∈ E n . If n > 1, one can’t derive a contradictionfrom this statement.The intervention of the Continuity Principle in the proof of Theorem 31(iv) iscrucial. One may extend this argument to a proof of: For each n > , for each ϕ : N → N ,if ∀ α [ α ∈ E n → ϕ | α ∈ A n ] , then ∃ α [ α ∈ A n ∧ ϕ | α ∈ A n ] , and,if ∀ α [ α ∈ A n → ϕ | α ∈ E n ] , then ∃ α [ α ∈ E n ∧ ϕ | α ∈ E n ] . This shows that E n positively refuses to reduce to A n , and A n positively refusesto reduce to E n .The theorem extends into the transfinite and then may be called the IntuitionisticBorel Hierarchy Theorem , see [58, Theorems 7.9 and 7.10].20. Notions of finiteness Some examples. We study decidable subsets of the set N .Such sets may be called ‘finite’ in various constructively different ways. Definition 35. For every α , we define D α := { n | α ( n ) = 0 } . D α is called the subset of N decided by α . Fin := { α | ∃ n ∀ m > n [ α ( m ) = 0] } . D α is finite if and only if α ∈ Fin . ote: α ∈ Fin if and only if we can calculate the number of elements of D α . Definition 36. For every X ⊆ N , we define: X + := { α | ∃ n ∀ m > n ∀ n [ α ( m ) = 0 → α ∈ X ] } . D α is perhaps-finite if and only if α ∈ Fin + , D α is perhaps-perhaps-finite if and only if α ∈ Fin ++ = ( Fin + ) + , Some examples are useful.1. Define α such that ∀ n [ α ( n ) = 0 ↔ n = k ]. Then { k } = { n | n = k } is a decidable subset of N . The statement ‘ α ∈ Fin ’ is equivalent to ‘ ∃ n [ n = k ]or ∀ n [ n < k ]’ and thus is reckless or hardy . We do know D α has at most oneelement, but we do not know if the number of elements of D α is 0 or 1.On the other hand, α ∈ Fin + . For assume we find m such that α ( m ) = 0, Then m = k and D α = { m } is finite. Therefore: ∀ m [ α ( m ) = 0 → α ∈ Fin ] and: α ∈ Fin + .2. Define α such that ∀ n [ α ( n ) = 0 ↔ k ≤ n < · k ]. Again, the statement‘ α ∈ Fin ’ is reckless, and again, D α is perhaps-finite, although, unlike in the caseof example 1, we are unable to give an upper bound for the number of elements of D α .3. Let γ, δ be given and define α such that ∀ n [ α ( n ) = 0 ↔ ( n = k γ ∨ n = k δ )], so D α = { k γ , k δ } . The reader may find out herself that D α is perhaps-perhaps-finiteand that the statement: ‘ D α is perhaps-finite’ may be reckless .20.2. Extension into the transfinite. The process of taking so-called ‘ perhapsiveextensions ’ of the set Fin may be continued into the transfinite. Definition 37. We define a collection E of subsets of N by means of the followinginductive definition. (i) Fin ∈ E . (ii) For every X in E , also X + ∈ E . (iii) For every infinite sequence X , X , . . . of elements of E , such that, for each n , ( X n ) + ⊆ X n +1 , also S n X n ∈ E . (iv) Clauses (i), (ii), (iii) produce all elements of E . Definition 38. For every X ⊆ N , X ¬ := N \ X := { α | ¬ ( α ∈ X ) } and X ¬¬ = ( X ¬ ) ¬ . Theorem 32. (i) For all X in E , Fin ⊆ X ⊆ Fin ¬¬ . (ii) For all X in E , for all α , ∀ s [ α ∈ X ↔ s ∗ α ∈ X ] . (iii) For all X in E , X ( X + .Proof. (i) The proof is by induction, following the definition of the class E .1. Obviously, Fin ⊆ Fin ¬¬ , see Subsection 11.2.2. Assume Fin ⊆ X ⊆ Fin ¬¬ . Assume α ∈ X + .Find m such that ∀ n > m [ α ( n ) = 0 → α ∈ X and distinguish two cases: Case (a) . ∃ n > m [ α ( n ) = 0]. Then α ∈ X and, therefore: α ∈ Fin ¬¬ . Case (b) . ¬∃ n > m [ α ( n ) = 0]. Then ∀ n > m [ α ( n ) = 0], so α ∈ Fin ⊆ Fin ¬¬ .We thus see: if ∃ n > m [ α ( n ) = 0] ∨ ¬∃ n > m [ α ( n ) = 0], then α ∈ Fin ¬¬ .We may conclude: α ∈ Fin ¬¬ , see Subsection 11.2.We thus see: ∀ α ∈ X + [ α ∈ Fin ¬¬ ], and conclude: Fin ⊆ X ⊆ X + ⊆ Fin ¬¬ .3. Assume: for all n , Fin ⊆ X n ⊆ Fin ¬¬ . Clearly, then Fin ⊆ S n X n ⊆ Fin ¬¬ .(ii) We again use induction on the definition of the class E . Cf. Subsection 9.8. See [50], [51] and [55]. . Note: ∀ α ∀ s [ α ∈ Fin ↔ s ∗ α ∈ Fin ].2. Let X in E be given such that ∀ α ∀ s [ α ∈ X ↔ s ∗ α ∈ X ]. Let s, α be given.If s ∗ α ∈ X + , find m such that ∀ n > m [ s ∗ α ( n ) = 0 → s ∗ α ∈ X ] and conclude:for all n > m , if α ( n ) = 0 then length ( s ) + n > m and s ∗ α ∈ X and α ∈ X , i.e. α ∈ X + . Conversely, if α ∈ X + , find m such that ∀ n > m [ α ( n ) = 0 → α ∈ X ] andconclude: for all n > m + length ( s ), if s ∗ α ( n ) = 0, then α (cid:0) n − length ( s ) (cid:1) = 0 and n − length ( s ) > m and α ∈ X and s ∗ α ∈ X , i.e. s ∗ α ∈ X + .3. Assume: for all n , ∀ α ∀ s [ α ∈ X n ↔ s ∗ α ∈ X n ].Clearly, then ∀ α ∀ s [ α ∈ S n X n ↔ s ∗ α ∈ S n X n .(iii) We shall prove: for all X in E there exists a spread F such that F ⊆ X + and not F ⊆ X . The proof is by induction, following the definition of the class E .1. Consider T = { α | ∀ m ∀ n [ (cid:0) α ( m ) = 0 ∧ α ( n ) = 0 (cid:1) → m = n ] } .Note that T is a spread. Note that, for every α in T , for every n , if α ( n ) = 0, then ∀ m > n [ α ( m ) = 0] and α ∈ Fin . We thus see: T ⊆ Fin + Now assume T ⊆ Fin . Then ∀ α ∈ T ∃ n ∀ m > n [ α ( m ) = 0]].Using Theorem 29, find p, n such that ∀ α ∈ T [0 p ⊏ α → ∀ m > n [ α ( m ) = 0]].Conclude: n + 1 ≥ p and consider α := 0( n + 1) ∗ h i ∗ α ∈ T , 0 p ⊏ α and α ( n + 1) = 0.Clearly, we reached a contradiction.2. Let X in E be given and let F be a spread such that F ⊆ X + and not F ⊆ X .Define F ∗ := { α | ∀ n ∀ p ∀ β [ α = 0 n ∗ h p + 1 i ∗ β → β ∈ F ] } .Note: F ∗ is a spread and 0 ∈ F ∗ .Note that, for every α in F ∗ , for every n , if 0 n ⊏ α and α ( n ) = 0, then there exists β in F such that α = α ( n + 1) ∗ β , and by (ii), also α ∈ F and α ∈ X + .We thus see: F ∗ ⊆ X ++ .Now assume F ∗ ⊆ X + .Then ∀ α ∈ F ∗ ∃ n ∀ m > n [ α ( m ) = 0 → α ∈ X ]. Using Theorem 29,find p, n such that ∀ α ∈ F ∗ [0 p ⊏ α → ∀ m > n [ α ( m ) = 0 → α ∈ X ]].We may assume n + 1 > p . Note that, for each β in F , 0 n ∗ h i ∗ β ∈ F ∗ and:0 n ∗ h i ∗ β ∈ X , and by (ii), β ∈ X .We thus see: F ⊆ X and we know that this leads to a contradiction.Conclude: not: F ⊆ X + .3. Let X , X , . . . be an infinite sequence of elements of E such that, for each n ,( X n ) + ⊆ X n +1 , and let F , F , . . . be an infinite sequence of spreads such that, foreach n , F n ⊆ ( X n ) + ⊆ X n +1 and not F n ⊆ X n .Define F ∗ := { α | ∀ n ∀ p ∀ β [ α = 0 n ∗ h p + 1 i ∗ β → β ∈ F n } .Note: F ∗ is a spread and 0 ∈ F ∗ .Note that, for every α in F ∗ , for every n , if 0 n ⊏ α en α ( n ) = 0, then there exists β in F n ⊆ X n +1 such that α = α ( n + 1) ∗ β , and by (ii), also α ∈ X n +1 ⊆ S i X i .We thus see: F ∗ ⊆ ( S i X i ) + .Now assume: F ∗ ⊆ S i X i . Using Theorem 29, find p, i such that ∀ α ∈ F ∗ [0 p ⊏ α → α ∈ X i ]. Define n := max( p, i ) and note: for all β in F n ,0 n ∗ h i ∗ β ∈ F ∗ , so 0 n ∗ h i ∗ β ∈ X i , and, by (ii), also β ∈ X i .We thus see: F n ⊆ X i ⊆ X n . Contradiction. (cid:3) Theorem 32 shows the expressive power of the language of intuitionistic mathe-matics. In Theorem 32(i) the set Fin ¬¬ might be replaced by the set AlmostFin := { α | ∀ ζ ∈ [ ω ] ω ∃ n [ α ◦ ζ ( n ) = 0] } ,as one may prove: S E ⊆ AlmostFin .Using Theorem 22, one may prove: AlmostFin ⊆ S E ⊆ Fin ¬¬ . he set AlmostFin is an example of a ‘simple’ Π or co-analytic set that failsto be positively Borel, see [55, § Avoiding Brouwer’s ‘axioms’. Bishop-style constructivism. E. Bishop (1928-1983) started his own schoolof constructive mathematics in the 1960s, see [3] and [4]. Although Bishop admiredBrouwer for his criticism of the non-constructive nature of much of mathematics,and for his heroic attempt to do something about it, he also had strong hesitationsabout Brouwer’s work. He did not accept Brouwer’s adoption of the ContinuityPrinciple and the Fan Theorem. He thought the arguments in favour of these prin-ciples bizarre, metaphysical and mystical and judged that Brouwer, by bringingthem in, had spoiled his own good cause. He declared Brouwer’s intuitionism to bedead and wanted a new beginning.Nevertheless, developing his constructive mathematics, he felt the need for someof the consequences of Brouwer’s principles.Bishop’s course was the following. He declared the notion of ‘ pointwise continu-ity ’ of real functions to be ‘ irrelevant ’. There is no need then for arguments as givenin Section 12. He defined a function from R to R to be continuous if and only ifit is uniformly continuous on every closed and bounded subinterval of R . He thusbuys for nothing what was for Brouwer the main consequence of the Fan Theorem,Theorem 16. There is no need then for arguments ‘proving’ the Fan Theorem asgiven in Section 13. Bishop’s attitude might perhaps be called ‘ pragmatic ’ in view of Brouwer’smore deeply going analysis of mathematical thinking. The principles proposedby Brouwer, even if one does not want to subscribe to the way he defends them,deserve to be discussed as possible starting points for our common mathematicaldiscourse. Not doing so, Bishop denied himself the possibility of a further goingperspective.Bishop calls his own attitude realistic , contrasting it with the attitude of the‘classical’ mathematician that uses non-constructive arguments. Unlike Brouwer,he does not fiercely attack the classical mathematician but calls him ‘ idealistic ’. Inview of Subsection 3.2, where the platonists were dubbed ‘realists’, this terminologyis funny.21.2. Martin-L¨of ’s ‘constructive mathematics’. P. Martin-L¨of (1942- ) madea serious study of Brouwer’s work and came to the following view in [42]. Everyobject in constructive mathematics should be given by a ‘ a finite configuration ofsigns ’. Therefore, an infinite sequence α of natural numbers always should be givenby a (finitely given) algorithm. Creating α by choosing its values α (0) , α (1) , . . . one-by-one in an infinite process of free choices, as Brouwer wanted to do, is out of thequestion. As we saw in Subsection 13.3, the Fan Theorem fails in such a context.Nevertheless, Martin-L¨of formulates and upholds a ‘Fan Theorem’. His strategyis similar to Bishop’s as sketched in the previous Subsection. He redefines themeaning of the statement ‘ ∀ α ∈ C∃ n [ αn ∈ B ]’. Presenting his definition somewhatfreely, one might say: he proposes to define ‘ ∀ α ∈ C∃ s ∈ B [ s ⊏ α ]’ as: ‘there existsa canonical proof of ‘ h i is B -secure’ as we explained this in Section 13. Martin-L¨ofextends this strategy to the Bar Theorem, see the proof of Theorem 19.21.3. An application. The Fan Theorem and Brouwer’s Continuity Principle areused for a reconstruction of Cantor’s Uniqueness Theorem in [63]. Bishop’s definition is questionable. F. 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