ramsey's theorem and the pigeonhole …s theorem and the pigeonhole principle in intuitionistic...

RAMSEY'S THEOREM AND THE PIGEONHOLE PRINCIPLEIN INTUITIONISTIC MATHEMATICS

WIM VELDMAN AND MARC BEZEM

ABSTRACT

At first sight, the argument which F. P. Ramsey gave for (the infinite case of) his famous theorem from1927, is hopelessly unconstructive. If suitably reformulated, the theorem is true intuitionistically as wellas classically: we offer a proof which should convince both the classical and the intuitionistic reader.

1. Introduction

1.1 In 1927, Ramsey [12] proved the following theorem.

For every binary relation R on N there exists an infinite subset A of N which isR-homogeneous, that is, such that

either Vm eAVne A[m < n -*• R(m,«)] or Vm e A Vn 6 A[m <«->-> R{m,«)].

(Note that we write 'i?(m,«)' for \m,n)eR\)

What is the constructive content of this theorem? This question may be treated indifferent ways. Some kind of an answer has been given by Specker [13] and also byC. G. Jockusch [6]. They exhibited recursive binary relations RonN, such that thereis no recursive infinite i?-homogeneous subset of N, and thereby showed Ramsey'sTheorem to be false in that particular branch of constructive mathematics which onecalls recursive mathematics (cf. [2 or 14], for a survey of possible positions withinconstructive mathematics). One may be surprised that classical logic is not generallyavoided in recursion theory. It does not seem proper to use nonconstructivearguments when treating algorithmic objects.

We study Ramsey's Theorem from an intuitionistic point of view, and accordinglytake serious the 'either ... or . . . ' which occurs in its above formulation. The theoremdoes not stand this reading. The following one-dimensional case and easy consequenceof the theorem, which sometimes goes by the name of pigeonhole principle, alreadyfails to be true.

For every subset R of M there exists an infinite subset A of N such that

either Vm € A[m E R] or Vm e A[m $ R].

We refute this principle by a Brouwerian counterexample.

Let p: N -> {0,1,.. . , 9} be the decimal development of the real number n. Consider

R := {ne N 131 ^ riii < 99[p{l+ i) = 9]}. Whoever claims that this set R is (intuition-

Received 15 November 1990; revised 15 April 1991.

1991 Mathematics Subject Classification 03F55.

J. London Math. Soc. (2) 47 (1993) 193-211

7 JLM47

194 WIM VELDMAN AND MARC BEZEM

istically) infinite, implies, probably recklessly, that there exists an unbroken sequenceof 99 nines in the decimal development of n; whoever claims that N\/? is infinite,implies, probably recklessly, that there is no such sequence.

Observe that the set R constructed in this example is recursive, and that from aclassical point of view, either R itself or its complement is recursive and infinite. Theredo exist classical recursion-theoretic counterexamples to the pigeonhole principle,that is, subsets R of N such that neither R nor its complement contains an infiniterecursive subset. Such sets must be nonrecursive and are called bi-immune subsets ofN. It is interesting to note that Jockusch first studied bi-immune subsets of N andthen found his recursive counterexample to Ramsey's Theorem.

We shall see in Section 3 of this paper, that the pigeonhole principle plays a keyrole in some classical proofs of Ramsey's Theorem. We shall see in Section 8 of thispaper that the construction of a recursive counterexample to Ramsey's Theorem isrelated to the construction of a bi-immune subset of N.

1.2 Whenever a theorem from classical mathematics proves to be false intuition-istically, one may try, pondering the classical arguments and, while preserving itsclassical meaning, reformulating the theorem in many different ways, to findintuitionistically valid versions of it. We did so for Ramsey's Theorem. TheIntuitionistic Ramsey Theorem that we present in Section 6 of this paper is anegationless statement of intuitionistic analysis, classically equivalent to Ramsey'sTheorem itself. In the proof of this theorem we obey the laws of intuitionistic logic,but we use no more than one intuitionistic axiom, namely, the principle of inductionon monotone bars. As the principle is not, like the famous continuity principles,contrary to classical assumptions, but admits of a not too difficult classicaljustification, the proof is acceptable classically as well as intuitionistically. Beforeproving the Intuitionistic Ramsey Theorem, we first establish, in Section 5 of thispaper, a similar Intuitionistic Pigeonhole Principle.

1.3 The paper is organized as follows. In Section 2 we recapitulate the principles ofintuitionistic analysis, and we give some notation. In Section 3 we formulate aclassical proof of Ramsey's Theorem. In Section 4 we develop an intuitionisticargument from this classical proof. In Section 5 we study the (infinite) pigeonholeprinciple. In Section 6 we treat the Intuitionistic Ramsey Theorem. In Section 7 wegeneralize our theorem from binary to ternary, and further to «-ary relations on 1 .In Section 8 we discuss the recursion-theoretic counterexamples mentioned in 1.1from an intuitionistic point of view. In Section 9 we show how a recent result ofCoquand's easily follows from our main theorem. Section 10 contains someconcluding remarks.

2. Intuitionistic analysis

2.1 Our discussion concerns the set N of natural numbers and the set Jf of functionsfrom N to N. We use m,n,p,q,... as variables over the set N and a,/?,... as variablesover the set Jf. We use intuitionistic logic, as we interpret connectives and quantifiersconstructively. This means, in particular, that a proof of a disjunction A\i B shouldcontain either a proof of A or a proof of B (and we are able to decide which one), andthat a proof of an existential statement 3JC[^(A:)] should contain an effective methodto find an object x with the property A(x).

RAMSEY'S THEOREM AND INTUITIONISTIC MATHEMATICS 195

2.2 We introduce an axiom of countable choice.

AC0 0: For every binary relation C on foJ:If Vm 3n[C(m,«)], then 3a Vm[C(m, a(m))].

We accept this axiom, as, in our view, an element a of JV may be constructed step-by-step, first a(0), then a(l), and so on. We feel no obligation to describe a in finitelymany words. Observe that C need not be a decidable relation on N, therefore it isnot possible, in general, to define a by: for all mGN:<x(m) is the least neN such thatC(m, n).

We shall not use in our main argument the stronger intuitionistic axiom ofcountable choice AC0 15 nor any continuity principle.

2.3 fol* = | J n 6 ^N n is the set of all finite sequences of natural numbers. A finitesequence a = <a(0),a(l), ...,a(n— 1)> may be thought of as a function from theset {0,1,...,« — 1} to N. <> is the empty sequence, the only sequence of length 0.*:M*xN*->l\l* is the concatenation function, that is, for all a,beN*, a*bdenotes the finite sequence which results from putting b behind a.

2.4 We introduce the principle of induction on monotone bars.

BIM: Let P, Q be subsets of N* such that(i) \/aeN*Vm[P(a)-+P(a*(m})] (P is monotone),(ii) Va3/i[P«a(0), a(l),. . . , a(w- 1)»] (P is a bar in Jf),(iii) Va e N *[P{a) -> Q(a)] (P £ Q),(iv) Va e N*[V«[0(« * <«»] -* Q(a)] (Q is inductive).Then:<2«».

(We write lP(a)' for ' a e P ' and similarly in similar cases.)

This principle, which seems to be implicit in Brouwer's argument for his bartheorem, was made an axiom of intuitionistic analysis by Kleene. On this much-discussed axiom, the reader may consult [8 or 14].

2.5 S := {ae N* | V/i[n + 1 < length (a) -• a(n) < a(n + 1]} is the set of all finite strictlyincreasing sequences of natural numbers.

y := {a e Jf \ V«[a(«) < <x(n + 1)]} is the set of all strictly increasing functions fromN to N.

In elementary intuitionistic analysis, the following principle is an easy consequenceof BIM itself.

BI^,: Let P, Q be subsets of S such that(i) Va G S[P(a) -> Vm[5(a * <m» -> P(a * <m»]],(ii)(iii)(iv) VaG S[Vn[S(a * </i» -> Q(a * <«»] -+ Q(a)].Then:<2«».

7-2


3. A classical proof of Ramsey's Theorem

Let R be a binary relation on N. We sketch a classical argument by which oneobtains an infinite /^-homogeneous subset of N.

3.1 One defines, for each finite sequence a = <<z(0), «(1), ...,a(n—1)> of naturalnumbers: a is R-homeogeneous if and only if a is strictly increasing and for allm,p,q <n such that m< p and m < q.R{a{m),a{p))*-+R(a(m),a(q)). Observe thatevery strictly increasing finite sequence of length ^ 2 is /?-homeogeneous.

3.2 One uses the following fact. For all aeN*, and i,j,keN such that i <j < k: Ifa * </>, a * <y> and a * </c> are all three i?-homeogeneous, then at least one of the threesequences a * </,./>, a * </, A;> and a * (j, k) is iMiomeogeneous.

Proof Let a = <a(0),a(l), ...,a(n-1)> and suppose: «*</>, a*(j) and a*(k)are i?-homeogeneous, and a*(ij} is not iMiomeogeneous.

Then: -• (R(a(n — 1), a(i)) <-> R(a(n — 1), a(J)). Therefore either

R(a(n- \),a(i))~R(a(n- \),a(k)) or R(a(n- 1),a(j

that is, either a * </, k} is .R-homeogeneous, or a * (J, k) is i?-homeogeneous.

3.3 One defines, for each finite sequence a = <a(0), ...,a(n— 1)> of natural numbers: ais safe for R := Vp3q[p < q A a * (q) is /?-homeogeneous], that is, there are infinitelymany immediate extensions of a which are iMiomeogeneous.

Hence each finite sequence aeN*, which is safe for R, is also /?-homeogeneous.

3.4 One makes the following important remark.

For each aeS: If a is safe for R, then 3q[a* (q) is safe for R].

Proof. Let a e S be safe for R. Then A:={qsN\a* <#> is iMiomeogeneous} is aninfinite set of natural numbers. We distinguish two cases.

Case (i): VpeAVqeA[p < q^>a*(p,q) is iMiomeogeneous]; thenV^6^[fl*<^f> is safe for R].

Case (ii): 3peA3qeA[p < q A a*(p,q) is not /Miomeogeneous]. Let po,qo

be members of A such that p0 < q0 and a* (po,qo} is not i?-homeogeneous. Then(cf. 3.2), for each keA such that qo<k: either a*(po,ky is i?-homeogeneous,or a*(qo,k} is i?-homeogeneous; therefore: either the set {keA \a* (pQ,k} isi?-homeogeneous} is infinite, or the set {k e A \ a * <#0, k)> is i?-homeogeneous} is infinite,that is, either a*(p0} is safe for R or a* < ro> is safe for R.

3.5 The empty sequence < > is of course safe for R. Using Remark 3.4, one buildsan infinite strictly increasing sequence a e ^ such that Vw[<a(0),a(l), ...,a(« —1)>is safe for R]. One then observes: V«[<a(0), a(l), ...,a(«— 1)> is /?-homeogeneous].Finally, one considers the two sets B0:={a.(n)\neN \R(ot.(ri),a(n+1))} andB1 := {a(«) | n e N \ -> R(tx(n), a(« + 1))} and one concludes: these two sets arei?-homogeneous (see 1.1), and at least one of them is infinite.


4. A first attempt to make sense of the classical proof

4.1 From now on, we reason intuitionistically. First, we translate the classicalargument from Section 3 into intuitionistic language, using double negations, as inthe Godel-Gentzen translation of classical arithmetic into intuitionistic arithmetic.This approach to Ramsey's Theorem is due to the second author. In later sections,we shall see other ways of understanding the classical argument.

In 3.4 and 3.5, one applied the following classical truth:

For all subsets A,B of N:If A is infinite and B £= A, then either B is infinite or A\B is infinite.

We saw in 1.1 that this so-called pigeonhole principle is intuitionistically false. Thereis, however, a weak version of the principle which is intuitionistically provable.

4.2 Reasoning with negative statements is an art with its own peculiarities. Wemention two of them.

Firstly, it occurs, when we are busy proving a negative statement, that we reinforceour assumptions, leaving out some double negation and replacing an assumption'-i ->P' by ' P \ We may do so, as, in intuitionistic logic, ->-iP-*->Q follows fromP -> -i Q. In the sequel, we shall indicate such reinforcements by the words ' Supposeeven more\ Secondly, we may, again when we are aiming for a negative conclusion,use some statements of the form ' P V -> P ' as an extra assumption. (This is a specialcase of the first procedure, as -> -> (P V -> P) is an intuitionistic truth). We thenmay continue the proof by distinction of cases; 'Case (i): Suppose P . . . , Case (ii):Suppose ->P...\ In the sequel, whenever we apply this device, we use the words:'Distinguish two cases'.

4.3 We define, for each subset A of N:

A is weakly infinite := V/? -i -- 3q[p < q A qeA].

4.4 LEMMA. For all subsets A,B of N:If A is weakly infinite and B c A,then -i -i (B is weakly infinite V A\B is weakly infinite).

Proof. Suppose B ^ A ^ N and: A is weakly infinite, and B is not weaklyinfinite. Then: -> -i3p\/q[p < q^q^B]. Suppose even more: 3pVq[p < q -*• q$B].Calculate poeN such that Vq[p0 < q ^>q$B]. A is weakly infinite, thereforeV/7 -i -> 3q[q >pAq>p0AqeA] and: V/> -> -• 3q[q > p A qeA A q$B], that is, A\Bis weakly infinite.

4.5 We define, for each binary relation R and N and each finite sequencea = (a(0),a(\), ...,a(n—\)} of natural numbers:

a is weakly safe for R:=Vp-i-i 3q[p < q A a * <#> is i?-homeogeneous].

4.6 LEMMA. For all binary relations R on N, for all aeN*:If a is weakly safe for R, then -> -> 3q[a* <#> is weakly safe for R].


Proof. Suppose a is weakly safe for R. Define A:={qeN\a*<#> isiMiomeogeneous}. Then A is weakly infinite. Observe that our goal is a negativeconclusion, and distinguish two cases:Case (i): VpeAVqeA[p < q ->->-'(«*</>, #> is iMiomeogeneous)].Then: VpeA[a*(p) is weakly safe for R].Case (ii): ~>VpeA\/qeA[p<q-*->->(a*(p,q) is iMiomeogeneous)], that is-<-i3peA3qeA[p < q A -•(#*</?,q} is iMiomeogeneous)]. Suppose even more,and let po,q0 be such that p0 < q0 and a*(p0}, a*(q0) are iMiomeogeneous, anda*(po,qQ} is not iMiomeogeneous. Then, for each keA such that qo<k:-• -< (a • (p0,k) is iMiomeogeneous v a*(q0,k) is iMiomeogeneous). (This followsby the argument given in 3.2). Using Lemma 4.4, we conclude: ->->({k€A \a*(po,k)is iMiomeogeneous} is weakly infinite v {k€A\a*(qQ,k} is iMiomeogeneous} isweakly infinite), that is ->-'(a*(pQ) is weakly safe for R v a*(q0) is weakly safefor R), therefore: -<-> 3p[a * </?> is weakly safe for R].

In view of a later application, in Lemma 4.9, we rephrase Lemma 4.6.

4.7 COROLLARY. For all binary relations R on M,for all aeN*:IfVg[a*(q) is not weakly safe for R], then a is not weakly safe for R.

Proof. Obvious.

4.8 The difficulty now is that we do not see how to iterate Lemma 4.6 countably manytimes. (This is done in the classical proof, cf. 3.5). The following conclusion, let us callif (*), is surely out of reach:

(*) For all binary relations R on N:3ae«^V«[<a(0),a(l), ...,cn{n—1)> is iMiomeogeneous].

Assume (•). Given a subset A of fJ we may apply (*) to the binary relation R on Nwhich is defined by: VwVAj[i?(w,n)<r+A(n)]. Thus, (*) is seen to imply:

(•*) For every subset A of N: 3a e y V«[^(a(0)) <-• A(<x(n))].

(**) is refuted by the Brouwerian example mentioned in 1.1. One is tempted to try thefollowing weakening of (*):

For all binary relations R on N:a(0),a(l), ...,a(«—1)> is iMiomeogeneous],

or its corollary:

For every subset AofN:^^ 3a G ^ V/t[i4(a(0))<->i4(a(/i))].

In 8.6.2, however, we shall see, by a metamathematical argument, that theseconclusions cannot be obtained from the usual axioms of intuitionistic analysis. Inview of these impossibilities, we are happy with the following lemma, first establishedby the second author.


4.9 LEMMA. For all binary relations R on N:

->Vae«9'?3«[<(a(0), a(l), ...,a(«—1)> is not R-homeogeneous].

Proof. Let R be a binary relation on N such that: Vae^3«[<a(0),<x(l),...,a(« —1)> is not i?-homeogeneous]. We shall use the principle of induction on mono-tone bars, BI^, (cf. 2.5). We define subsets P and Q of S as follows. For each aeS:P{a) := a is not iMiomeogeneous, and Q(a) := a is not weakly safe for R. Observe thatP is a monotone bar in Sf, that P £ Q, and that, according to Corollary 4.7, Qis inductive. Using BI^,, we conclude: g « » , that is: the empty sequence < > is notweakly safe for R. But: < > is weakly safe for R, as Vq[(q} is /Miomeogeneous], andso we have a contradiction.

4.10 Lemma 4.9 says that, for every binary relation RonN, the assumption that therange of every strictly increasing sequence of natural numbers is, in a strong sense, not/Miomeogeneous, leads to a contradiction. In this very weak sense, the classicaltheorem that there exists an infinite i?-homeogeneous subset of N, is true. In 6.2, wewill establish a similar weak version of Ramsey's Theorem itself, that is the theoremthat, for every binary relation RonN, there exists an infinite R-homogeneous (see 1.1)subset of N. To obtain this result, we need one further step which we discuss in thenext section.

5. Almost full subsets of N

5.1 Containing as many double negations as it does, Lemma 4.4 is not a verywelcome substitute for the classical pigeonhole principle. In this section, we want toapproximate the principle by a lemma and a theorem, namely 5.2 and 5.4, which,having a more positive appearance, are less like ghosts from the lost classicalparadise.

5.2 LEMMA. For all subsets AofN:IfVyeSrin[A(y(n))], then 3yeSfVn[A(y(n))].

Proof. We use AC0 0, the axiom of countable choice introduced in 2.2. Let A bea subset of N such that Vy e 9* 3n[A(y(n))]. Then Vra 3n[n > m A A(n)]. (In order to seethis, one considers, for each me N, the function yeS? defined by V«[y(«) = m + n+ 1]).Applying AC0 0 we find OLEJV such that Vw[a(w) > m A A(a(m))]. Define ysjV by:y(0) = a(0) and V/i[y(«+ 1) = a(y(«))]. Then y e ^ and Vn[A(y(n))].

5.3 Considering Lemma 5.2, one might cherish some hopes for the following versionof Ramsey's Theorem:

For all binary relations R on N:If Vy G & 3m 3n[m < n A R(y(m), y(n))], then 3y g Sf Vm V«[w < n -»R(y(m), y(«))].

Such hopes are idle, as we show by an example, due to Mervyn Jansen.Let p:M ->{0,1, ...,9} be the decimal development of the number n, letA:={neN\3l*:n\/i<99[p(l+i) = 9}} and let R:={(m,n)eN*\A(m)~A(n)}. We


claim that Vye^3m3«|>w < n A R{y{m), y{n))]. (Let yeS and consider y(l).Either: ->A(y(\)), and therefore: R(y(0),y(\)), or: A(y(\)), and therefore:i?(y(l),y(2))). Suppose now: y e ^ and VmV«[m < n -> i?(y(m), y(n))]. Either ^4(y(0))and V«[y4(y(n))], a reckless conclusion, or ->A(y(0)) and V«[-> /4(y(«))], again areckless conclusion.

5.4 THEOREM (Intuitionistic Pigeonhole Principle).For all subsets A,BofN:!fVyeSfln[A(y(n))] and Vy e <f 3n[B(y(n))), then Vy e Snn[A(y(n)) A B(y(n))).

First proof. (Using AC0 0.) Let A, B be subsets of N which fulfil the requirementsof the theorem. Let yeSf. Observe that: \/Se^3n[A(yoS(n))].

Apply Lemma 5.2 and construct deSf such that Vn[A(yod(n))]. Calculate n0 suchthat B(yod(n0)), let p := <5(«0) and observe: A(y(p)) A B(y(p)).

Second proof. (Using BI^,.) Let A, B be subsets of N which fulfil the requirementsof the theorem. Define a subset P of the set S of strictly increasing finite sequencesby: For all a = <a(0),a(l), ...,a(n- \)}eS:

P(a) := 3/ < «3y < n[i <y A y4(fl(i)) A ^ 0 ) ) ] -

We claim that P is a monotone bar in S?.(Proof of this claim. Let yeSf. Calculate /oeN such that A(y(i0)). Define <5ey

by: V«[<5(«) = y(/0 + «)]- Calculate A:o such that B(S(kQ)). Lety0 := /o + o an<^ observe:/0 ^y 0 and A(y(iQ)) A B(y(JQ)). This proves the claim.)

Define a subset Q of the set S of strictly increasing sequences of natural numbersby: For all a e S: Q(a) := Vy 6 <f 3n[P(a * <y(«)»]. Observe that P c 0 , We claim thatg is inductive, that is: VaeSTV^Sfa * <#» -> g(a * <^r»] -• Q(a)].

{Proof of this claim. Let a e S" be such that V^Sfa • <#» -»• g(« * <^»]- Let y e 5^.Calculate «0 := fip[S(a * <y(/?)»].

Remark: G(a* (y(wo)))' an(^ calculate Wj such that P(a* <y(«0), y(«x)», observe:either P(«*<y(«0)» or P(fl*<y(/ix)» or A(y(n0)) A 5(y(«i))-

Remark: 2(a*<y(«1)», and calculate n2 such that P(a* <y(«1),y(«2)))' observe:either P(a*<y(«1)» or P(fl*<y(«2)» or ^(y(«x)) A 5(y(«2)).

We conclude: P(a*<y(«0)» V P(a* <y(w1)>) V P(a*<y(«2)», and:

Therefore: VyeS3«[P(a* <y(«)»], that is: Q(a). This proves the claim.)Using BI^ (cf. 2.5), we conclude g « » , that is, Vy6^3rt[i>«y(«)>)], therefore

Vye<?3n[A(y(n))AB(y(n))].

5.5. We define, for each subset A of N:

A is almost full := Vy 6 Sf 3n[A(y(n))].

Observe that, for each subset A of N : if 3« Vm[m > n -> A(m)], then V y e ^ 3«[y4(y(n))].Intuitionistically, the converse is not true: let p:N ->{0,1, . . . ,9} be the decimaldevelopment of n and consider

A:={neN\->Vi< 99[p(n + i) = 9] V 3/ < rNi < 99[p(/+ /) = 9]}.


Observe that A is a decidable subset of N, that is, Vn[nsA v ->(neA)] andA n$A) -• m = n], that is, N\A has at most one element. Therefore:

V A(y(\))].Whoever claims: 3n \/m[m > n -> >4(/w)], declares himself able to decide:3m[-> A(m)] V Vm[y4(ra)], and is probably reckless.(This example is related to the one given in [15, Section 1]).

It is evident from Theorem 5.4 that the class of all almost full subsets of N is afilter; N itself is of course almost full. For all subsets A, B of N: if A is almost full andA g B, then B is almost full, and: if both A and B are almost full, then A [\ B isalmost full.

Observe that Theorem 5.4 is classically equivalent to the pigeonhole principlewhich says that for every subset A of N, either A or its complement is infinite.Theorem 5.4 implies this principle, as, for any subset A of N, if neither A nor itscomplement have an infinite subset, both N\A and A are almost full, which iscontradictory, according to Theorem 5.4. Theorem 5.4 is implied by this principle, as,for all subsets A,B of N, if both A and B are almost full, every infinite subset of Ncontains an infinite subset of A and therefore a member of A n B.

6. The intuitionistic Ramsey Theorem

We start with an application of the Intuitionistic Pigeonhole Principle.

6.1 LEMMA. For all binary relations R,Ton N:IfV<xey3m3n[m < n A R(tx(m),<x(n))] and\/ixe^3m3n[m < n A T{<x{m),<x(n))],

then V<xe&'3m3n3p[m < n A m <p A R(a(m), a(«)) A T(<x{m), cn{p))].

Proof. Let R, T be binary relations on fJ which fulfil the requirementsof the theorem. Let oceS. Define subsets Aa,Ba of N by:Aa:={meN\3n[m < n A R(a(m),a(«))]}, and Ba:={meN\3n[m < n A T(<x(m),oc(«))]}.Both Aa and Ba are almost full subsets of N. (Observe that:V/?ey 3m3n[m < n A i?(aoj?(m),aoj?(«))], therefore: V/?e^3m[Aa{fi{m))}.)Using Theorem 5.4, we conclude: Aa (] Ba is almost full, in particular,3m[Aa(m) A Ba(m)], that is, 3m 3n3p[m < n A m <p A R{<x{m),on{n)) A T(<x(m), (x(p))].

6.2 COROLLARY. For all binary relations R on N:

^(xe9>3m3n3p3q\m < n A R(tx(m),tx(n)) A p < q A ->R(a(p),<x(q))].

Proof. Let i? be a binary relation on fJ such that

V(X€Sf3m3n3p3q[m < n A R(a(m),(x(n)) A p < q A -> R(oc(p), cc(q))].

Applying Lemma 6.1, we find:

Va e £f 3m 3n 3p[m < n A m <p A R(<x(m),0L(ri)) A -> R(oc(m),a(p))].

Therefore V a e y 3«[<a(0), a(l), ...,a(«— 1)> is not i?-homeogeneous]. This leads to acontradiction, according to Lemma 4.9.


Corollary 6.2 is the result we announced in 4.10. The second author, when he firstproved Corollary 6.2 for decidable binary relations, called it the intuitionistic RamseyTheorem, as it is equivalent to Ramsey's Theorem when one reads it classically. Wenow prefer to reserve this name for the stronger theorem, Theorem 6.5.

In the proof of this stronger theorem, we apply the so-called finite RamseyTheorem, which we state here as another lemma. In most classical proofs of theinfinite Ramsey Theorem, one does not use the finite Ramsey Theorem. One evengoes the other way around, proving the Ramsey Theorem from the infinite one by aso-called compactness argument, (cf. [5, pp. 13-17]). (In Section 9 we shall reconstructthis compactness argument intuitionistically, using the Fan Theorem.) Ramseyhimself however, gave the proof of the infinite theorem only to prepare the reader forthe finite theorem, which was the one he needed.

6.3 LEMMA (Finite Ramsey Theorem).For all natural numbers n, k there exists a natural number N such that for every

function g from {0,1,.. . , N— 1} x{0,1, . . . , JV—1} to {0,1, ...,k— 1} there exists a subsetA of{0,1, ...,N— 1} of at least n elements which is g-homogeneous, that is, such that forallp,q,r,seA such that p < q and r < s:g(p,q) = g(r,s).

We do not give the proof. Ramsey himself, indicating how, given the numbers n,k, the number N may be calculated, already argued constructively.

In the sequel, we frequently follow the set theoretic convention of identifying anatural number n with the set of its predecessors {0,1,...,«— 1}.

6.4 LEMMA. For all binary relations R,Ton N:IfV(xeyimln[m < n A R(<x(m),a(w))] and V<xeyimln[m < « A T(<x(m),a(/i))],

then 3 w 3 « [ m < « A R(m,n) A T(m,«)].

Proof Let R, T be binary relations on N which fulfil the requirements of thelemma. We define a subset P of the set S1 of strictly increasing finite sequences ofnatural numbers in the following way. For each a = <a(0),a(l), ...,a(n—\)}eS:

P(a) := 3/ < n3j < n3k < n[i <j A i < k A R(a(i), a(J)) A T(a(i), a(k))].

Observe that, in consequence of Lemma 6.1, P is a monotone bar in S. We define aproposition QED, as follows: QED := 3m 3«[m < n A R(m, ri) A T(m,«)]. We alsodefine a subset Q of the set S of strictly increasing finite sequences of natural numbers:for each a e S: Q(a) :=VyeSf 3n[P(a * <y(«))> V QED]. Observe that P £ Q. We claimthat Q is inductive, that is: VaeS[Vq[S(a*(qy) -> Q(a* <#»] -> Q(a)].

Proof of this claim. Let a = <a(0), a(\),..., a(n —1)> be such that Vq[S(a * <<?»

Using Lemma 6.3, the finite Ramsey Theorem, we calculate NeN such that,for each function g from NxN to n there exists ij,k < N such that i <j < k andg(i,j) = g(i,k) = g(j,k). Now let ye2>. Calculate n0 := fip[S(a* (y(p)y)]. AsQ(a * <y(«o)>). w e calculate B ^ N such that P{a*(y(no),y{n^) V QED. AsQ(a*(y(n0)}) and Q{a*(y{n^)>), we calculate, using Theorem 5.4, the IntuitionisticPigeonhole Principle, n2eN, such that both P(a*<j{n^),y{n^)>) V QED andP(a * <y(«i), y{n2)}) V QED. As Vz' < 3[Q(a * <y(»i)»], we calculate, using Theorem


5.4, «3e fol such that Vi < 3[P(a * (y(nt), y(«3)») V QED]. We continue this procedurefor N steps. In the end, we have defined natural numbers no,nlt ...,«iv-i s u c n that:Vi < NVj < N[i <./-> (P(a * <y(/i<), y(«,)) V £ED)]. We distinguish two cases.

Case(i): QED.Case (ii): Vi < NV/ < JV[i <y-» P(tf * <y(«<), y(«,)»]- We define, for each i,j < N

such that i<j: aij:=a*<j{n^,y{n^)'y. We build, in finitely many steps,three functions, / , g, h, from Nx.N to n + 2 such that, for all i j < N such thati<>: MJ)<g(i,J) and f(i,j) < h(i,j) and R(aiJ(f(i,j)),aiMi>J))) andT{aij{f{i,j)),aij{h{Uj))). We distinguish three subcases.

Case (ii)a: 3/ < N3y < JV[i <y A g(i,f) = h{i,j)]. Let ij < N be such numbers. Lets•= au{f{hj)) and t:=au{g{i,j)). Observe: R((s,t)) A 7T(<M», therefore: QJE:D.

Case (ii)b: 3/ < Nlj < N[i <j A (g(i,/) < n v A(i,/) < »)]. Let ij < N be suchnumbers. We distinguish two cases, namely: Either g{i,j) = n + \ or /i(/,/) = n + 1 .Then: P(a* (y(n}))). Or g(i,y) < n and / I ( / J ) ^ n. Then: P(a* (yC^))). In both cases:lp[P(a*<y(p)))l

Case (ii)c: Vi< iVV; < ^ [ / <j^{g(i,j),h(i,j)} = {«,/i+l}]. Then Vi < NVj <N[i <j ->/(/,/) < «]. In view of our choice of N, let i,j, k < N be such that i <j <kand f{i,j)=f(hk)=f{j,k). Now, let s:=a{f(i,j% r:=7(«{), w:=H«i) andy:=y(W;t). Observe: (i?(s, f) v R(s,u)) A (/?(>, r) V R(s,v)) A (R(s,u) V i?(.s,v)).

Spelling this out, we find two sequences from {<s, f>, <5, «>, <s, y>} that belong toi?. In a similar way, we find two sequences from {<j, />, <j, M>, <5, y>} that belong to7. Combining, we find a sequence from {<s, / > , < J , M>,<5,y>} that belongs to i? n 71.We conclude: 3p[P(a* (y(p)})] and 0£Z).

Thus in all cases, we have: 3p[P(a * (y(p)}) v QED]. This ends the proof of ourclaim, that Q is inductive.

Using the principle of induction on monotone bars, BI^, (cf. 2.5), we conclude:g « » , that is, Vyey 3«[i>«y(n)» V QED], therefore QED, that is: 3m3n[m < n AR{m,n) A T(m,n)].

6.5 THEOREM. (Intuitionistic Ramsey Theorem).For all binary relations R,T on N:If Vaeyimlnlm < n A R(<x(m),<x(n)] and V(xe^3m3n[m < n A T(<x(m),a(/i»],

< « A R(a(m), a(«)) A T(a(m), a(n))].

Let /?, r be binary relations on N which fulfil the requirementsof the theorem. Let a e ^ . Define binary relations Ra,Ta on N by: for allm,neN: RJ^m,n):= R(oc(m),a(«)) and 7;(/w,«) = T(<x(m),a(«)). Observe thatVpetf>lm3n[m < n A RJfi(m),P(n))] and V^e^3m3n[m < « A 7a0?(m),/?(«))].Use Lemma 6.4 and conclude: 3w3«[m < « A R{(x.(m),(x.{n)) A T(a.(m),<x(n))].

6.6 Let /? be a binary relation on N. We define

i? is almost full := Va e y 3m 3n[m < « A R(<x(m),«(»))].

Theorem 6.5 says that the class of all almost full binary relations on N is a filter.Observe that Theorem 6.5 is classically equivalent to Ramsey's Theorem. // impliesRamsey's Theorem, as, for any binary relation R on N, if there is no infinite.^-homogeneous subset of N, both R and its complement are almost full, which iscontradictory, according to Theorem 6.5. It is implied by Ramsey's Theorem as, for


all binary relations R,T on N, if both R and T are almost full, every infinite subsetA of N contains an infinite subset B such that VmeBVneB[m < n-> R(m,n)], andthere are m,neB such that m < n and T(m,n).

1. Generalization of the Intuitionistic Ramsey Theorem

7.1 We consider finite sequences of natural numbers as functions a whose domain,Dom (a), is a natural number n = {0,1,. . . , n — 1}.

For each b = (b(0),b(l), ...,b(m-\))eN* and each a = <a(0),a(l) , . . . ,a(«- l)>G N * we define boaeN* by: Dom(boa) = {keN \k < n A V/ ^ &[a(/) < m]} andVA:GDom(boa)[boa(k) = b(a(k))]. Then boa is called the composition of b and a.

For each txejV and each a = <(0), ...,a(n — \))eN* we may form the finitesequence <xoa:= <a(a(0)),a(a(l)), ...,a(fl(n —1))>. The finite sequence a o a is calledthe composition of a and a.

For each A:eN, 5 t will denote the set of strictly increasing finite sequences ofnatural numbers of length k (equivalently, with k as domain).

7.2 Let keN and let R be a fc-ary relation on N. We define

R is almost full :=^<xe^3ae Sk[R(a o a)].

7.3 THEOREM (Generalized Ramsey Theorem).For each keN, k > 0: If R and T are almost full k-ary relations on N, then

R 0 T is an almost full k-ary relation on N.

Proof The proof is by induction.The cases k = 1, k — 2 have been treated in Theorems 5.4 and 6.5 respectively.

Assume that keN and that the theorem has been proved for the cases \,...,k. Wesketch the proof for the case k+\. The proof is in three steps. Let R, The almost full(k+ l)-ary relations on N. Without loss of generality, we may assume that both R andT are subsets of Sk+1.

Step one (cf. Lemma 6.1): Define A>ary relations R~ and T~ on N by: for allaeN": R~{a) := 3n[R{a* <«»] and: T~(a) := 3n[T(a* <«»].

Observe that R~ and T~ are almost full. Using the induction hypothesis weconclude that R~ 0 T~ is almost full, in particular: 3aeSk[R~(a) A T~(a)].

We generalize this result as follows: let y e Sf and define /r-ary relations R~ and T~on N by: for all aeNk: R^(a) := 3n[R(yo(a* (n)))] and T~{a) := 3n[T(yo(a*<»»)].Again, R~ and T~, and therefore: R~ n T~ are almost full, in particular:3a e Sk[R~(a) A T~(a)].

Step two (cf. Lemma 6.4): Define a subset P of ftJ* by: for all aeN*:P(a) := 3b e Sk 3m 3n[R(a o (b * <m») A T(a o (b * <«»)].

In Step one, we have seen: P is a monotone bar in £f. We define a propositionQED by: QED:=3aeSk+1[R(a) A T(a)]. We define a subset 0 of N* by: for allaeN*: Q{a):=Vye2>3n[P{a*(y{n)y) V QED].

Observe that P £ Q. Arguing very much like we did in the proof of Lemma 6.4, wemay establish, applying once more the Finite Ramsey Theorem (Lemma 6.3), that Qis inductive.

Using BI^,, the principle of induction on monotone bars (cf. 2.5), we conclude:), and therefore: QED, that is: 3aeSk+1[R(a) A T(a)].


Step three (cf. Theorem 6.5): Let ye9* and consider the subsets Ry and Ty of Sk+1

denned by: for all a e Sk+1: Ry(a) := R(y o a) and 7J(a) := T(y o a).Observe that Ry and 7 are almost full (k+ l)-ary relations on ftl. Using the result of

Step two, we conclude: 3aeSk+1[Ry(a) A Ty(a)], that is: 3aeSk+1[R(yoa) A T(yoa)].Therefore: VyeS/'3aeSk+1[R{yoa) A T(yoa)], that is: i? n Tis almost full.

8. .Some remarks on the recursion-theoretic counterexamples

In this section, we examine the construction, due to Jockusch [6], of a recursivebinary relation R on N such that every infinite recursive subset W of N containsnumbers m,n such that m <n and R(m,n) and numbers p,q such that p < q and-> R(p, q). (The earlier counterexample to Ramsey's Theorem found by Specker [13] isa bit more involved, as it is based on the existence of two recursively enumerable setsof incomparable degrees of unsolvability.) We shall see that this construction isclosely related to the construction of a A°-subset A of N such that every infiniterecursive subset W of N is not a subset of A and not a subset of N\A. (A subset Aof N with the latter property is sometimes called a bi-immune set.)

We want to explore the intuitionistic meaning of these constructions. Moreover,we shall draw some information from them concerning the formal strength of ouraxioms. In doing so, we reinforce a remark by the second author, that the doublenegation of the classical Ramsey Theorem, that is, the statement that, for all binaryrelations R on N

-> -- 3a e Sf [Vra V«[w < n -> R(oc(m), a(«))] V Vm V«[m < n -> -> R(a(m), a(«))]]

is not derivable in any of the usual formalizations of intuitionistic analysis.

8.1 The question as to the existence and complexity of bi-immune subsets of N isconnected with the following more general problem:

Given some sequence Vo, Vx, V2,... of subsets of N, find subsets A,BofN such thatA (1 B = 0 and: V«[Fn is infinite -> (Vn n A # 0 A Vn n B * 0)].

Classically, one easily forms such sets A,B by going through the sequence Vo, Vx, V2,...,disregarding the finite sets among them, and choosing from each infinite set twoelements which are both different from every element chosen from one of the earliermentioned sets. One lets A consist of the first members of all pairs thus found, andB of the second members. Intuitionistically, the main difficulty with this procedure isthat we cannot decide in general whether a given set is infinite or not. In some specialcases, however, this difficulty is absent.

8.2 Recall from 7.1, that, for each neN, Sn is the set of all strictly increasing finitesequences of natural numbers of length n. We call a subset X of fJ finite if and onlyif these exist neN and aeSn such that X = {a(0),a{ 1),...,a(n — 1)}.

We also define, for each subset Xof N and each neN

# X ^ n (X has at least n elements) := 3ae Sn V/ < n[a(j) eX].

Observe that, for each finite subset X of N and each neN, we may decide:

V


8.3 LEMMA. There exists a functional F which assigns to each function/from N to theset of finite subsets ofNa one-to-one sequence a = F(f)ejV such that:

Vw[#/(«) ^ 2n + 2 -> (a(2/i)e/(/i) A a(2/i+ l)e/(/i))].

Proof Let /= / (0) , / ( l ) , / (2) , . . . be a sequence of finite subsets of N. Wedefine a = F{f) by recursion: for each neH: if #/(«) ^ In + 2, thena(2») :=npef{n)\ij < 2n[p # «(/)]) and a(2/i+1) :=///> e/(w)[Vy < 2/2 +1[/> ^ a(y)]],and, if -. (#/(«) ^ 2H + 2), then a(2«) :=pipeM[ij < 2n[p # a(/)]] anda(2« +1) := upe N[V/ < In + \[p ^ a(j)]]. Note that, if -. (#/(0) ^ 2), then a(0) = 0a n d a ( l ) = 1.

8.4 We call a subset V of N enumerable if and only if there exists a sequence Vo, Vv

Vz,... of finite subsets of N such that V = LLeN Vm. In the following, we restrictattention to recursively enumerable subsets of N.

We know from recursion theory that there exists a (recursive) universal double-sequence (W iTO)n>meN of finite subsets of N, such that VnVm[Wn m £ W^m+1] andsuch that, if we define, for each neN, Wn:= \JmeN Wn m, then Wo, Wv W2,... is acomplete list of the recursively enumerable subsets of N. We use such a universaldouble-sequence in the statement and the proof of the next theorem.

8.5 THEOREM.

(i) There exist recursive binary relations R, T on N such that:(ia) Rf] T=0,(ib) V/i[# Wn > 2w + 2 3m Vq nBr e Wn 3s e Wn[R(r, q) A T(s, q)]],(ic) V«[# Wn > In + 2 -* - - 3/-e Wn 35e Wn 3w V^ > m[R(r, q) A 71(5, q)]],(id) V/» -i -i 3m[V^ ^ m[R(p, q)} W ^ m[7Xp, g)]].

(ii) There exist ^jj'Subsets A,B of N such that:(iia) A n 5 = 0 ,(iib) Vn[# PFn ^ 2« + 2 -• -. -«3re PKn356 Wn[reA A 5G5]] ,

(iic) V/?[-. -. (p6 A V p € B)].

Proof Let {Wn m)n mGN be the universal double sequence that we mentionedin 8.4. Define, for each m e N , a sequence fm of finite subsets of N by: for all «e N:/m(n) := Wn m. Let Fbe the functional defined in the proof of Lemma 8.3. Define, foreachmGN:a m :=F( / J .

Define binary relations ^,7* on N by: R:={(am(2ri),m}\m,neN} andT:= {<am(2« +1), m) \ m, n e N}. We show that R, T fulfil the requirements.

(ia) According to Lemma 8.3, each am is injective, therefore: R f] T= 0.(ib) Assume neN and #Wn^2n + 2. Calculate me M such that #Wn m^ 2n + 2.

Observe that, for each q ^ m, # Wn q^2n + 2, and therefore: <xg(2n) e Wn anda,(2«+l)G Wn and R(ag(2n),q) and 7^(2/2 + 1 ),<?).

(ic) Assume neN and # Wn ^ 2« + 2. We have to prove a negative conclusion,namely: -. -. 3re W 3^e H n 3w V# ^ /w[.R(/-, ^) A T(s, q)].

Therefore, as we saw in 4.2, we assume without risk: V/< «[# W^ 2j+2 v-i (# W, ^ 2y + 2)]. Calculate ^ e N such that Mj ^n[#W}^ 2j+2~# WjtP> 2j+2].Calculate / E N such that V/ < /JVJCG W P[X ^ i\. We now make another harmlessassumption: VJC /Vy ^ / i ^ ^ V ->(^G W )]. Calculate meN such that: Vx ^ Nj ^


n[xG W} -> xe W, m]. Observe that, for each j ^ n, if Wi has at least 2j+2 elements,then the first 2j+2 elements of W} belong to Wj m already. From the proof ofLemma 8.3 and the fact that V/< 2n + 2iq ^ m[W} m c Wi g] we conclude:Vy 2/i + 2V^ ^ m[a,(;) = aTO0)]. We define r := am(2/i') and ' s := am(2/z +1).Observe: V? ^ m[i*(r, #) A r(s, q)].

(id) Let /?eN. The set {xeN \x ^ p} is recursively enumerable and, being so, itoccurs in the sequence WQ,W^W2,.... Calculate neH such that Wn = {xeN\x^ p).As in (ic), we are striving for a negative conclusion. Strengthening our assumptionsand reasoning as in the proof of (ic), we find meN such that, for all j ^ n, if Wi hasat least 2y+2 elements, then the first 2y+2 elements of W^ belong to W} m already.Observe that p is the first element of the infinite set Wn and also the first element ofWnm. Therefore, we may calculate / < 2« + 2 such that p = OLm{i). Observe: Vg ^

mt a9 (0 = am(01- Either / is even, and Vg ^ m[R{p, q)], or / is odd and V r m[T(p, q)].We now define subsets A,B of N by: A := {pe N 13m\fq ^ m[R(p,q)]} and

5 := {/?e N 13m V^ ^ m[T(p, q)]}. Then (iia, b, c) follow straightforwardly from(ia, c, d). One easily sees that R, T are recursive, and A, B are £ ° .

8.6 We may draw some metamathematical conclusions from Theorem 8.5. Let EL bethe formal system for intuitionistic analysis which is explained in [14].

Let CT be Church's thesis in the following form:

Va 3e Vn 3z[T(e, n, z) A U(z) = a(«)].

(Tis the recursive subset of N3, introduced by Kleene, U is a total recursive functionfrom N to N, the so-called result-extracting function.)

Let CT0 be Church's thesis in the following schematic form:

For each arithmetically definable binary relation R on N:lf\/m3n[R(m,n)], then 3e\/mlz[T(e,m,z) A R(m, U(z))].

8.6.1 Observe that in EL-j-CT, Theorem 8.5 (i) plainly contradicts the IntuitionisticRamsey Theorem 6.5. (Observe that, in EL + CT, Theorem 8.5 (i) says that there existalmost full binary relations R, T on N such that R f] T=0.) Therefore:E L + C T + BIM is inconsistent. This is a well-known fact, first shown by Kleene, whoin [7] gave an example of a recursive subtree T of the binary tree {0,1}*, which hasarbitrarily long finite branches, but is such that we may calculate, for each recursiveinfinite branch, an initial part that does not belong to T.

Remark that, as EL + CT is consistent, it is impossible to prove the IntuitionisticRamsey Theorem 6.5 in EL.

8.6.2 Let A,B be subsets of ^ which have the properties mentioned in Theorem8.5 (ii). We claim that, in EL + CT0, the assumption 3a e Sf Vn[A(<x(n)) <-• ^(a(0))]leads to a contradiction.

Proof of this claim. Assume : a e y and Vn[A{<x{ri)) <-> A((x.(0))]. With a view to ournegative goal we distinguish two cases.

Case (i). A(oc(0)). Then Vn[A(a(n))], therefore Vm 3« > m[A(n)]. Applying CTo, wefind a total recursive function / from N to N such that Vm[f(m) > m A A(f(m))].Determine eeN such that We = {f(m)\meN}. Observe: #We^2e + 2 andV/2G W [,4(«)], in contradiction with Theorem 8.5(iia, b).


Case (ii). -i A(a(0)). Reasoning as in case (i), we find esN such that # We 2e + 2and Vn G We[-i A(n)], again in contradiction with Theorem 8.5 (iib). This proves theclaim.

It follows from [9] that the system EL + BI^ + CT0 is consistent. Therefore, we areunable to prove, in this system:

- - 3a e Se V/i[y4(a(0)) «->• A(ct(n))].

CT0 may be consistently added to much stronger systems for intuitionisticanalysis, for instance the system FIM of [8]. In no such system is the above-mentionedstatement provable.

8.6.3 Let us consider the following form of Markov's principle.

MP: For every arithmetically definable subset A of N:If Vn[A(n) V ->A(n)] and ->-, 3n[A(n)], then ln[A(ri)].

We claim that, in EL + CT0 + MP one may prove:

For every arithmetically definable subset A of CJ:If Vn[A{n) V ->A(n)], then

Proof of this claim. Assume: Vn[A(n) V ->A(n)]. According to Lemma 4.4:-i-i (A is weakly infinite v N\A is weakly infinite). As we have to prove a negativeconclusion, we distinguish two cases.

Case (i). A is weakly infinite, that is: Vp-« -• 3q[p < q A A(q)].Using MP, we conclude: Vp3q[p < q A A(q)]. Using CT0, we find a recursive

function/from N to M such that VpeN[p <f(p) A A(f(p))].Define a e ^ by: a(0) = / (0) and for all neN: <x(n+ 1) =/(a(/i)).

Then: Vn[i4(a(O))^v4(a(n))].Ca^e (ii). N\/4 is weakly infinite, is treated in a similar way.This proves the claim.

As EL + CT0 + MP is consistent, (cf. [10]), we conclude: We cannot prove in ELthat there exist decidable subsets A,BofN with the properties mentioned in Theorem8.5 (ii). Neither can we do so from any set of axioms to which CT0 + MP may beadded consistently, as, for instance, the system FIM.

8.7 We conclude this section with a remark on the intuitionistic continuity principle.The following strong version of the pigeonhole principle and, a fortiori, thecorresponding strong version of Ramsey's Theorem, are intuitionistically unprovable:

(•) V£G 2N3a G Sf Vw[/?(a(0)) = /?(a(w))].

This is clear from the example given in the introduction. It is possible to derive acontradiction from (*) by the following weak principle of continuity:

CP: For all subsets A of Jf x N:If V/?3n[A(fi,«)], then V/?3m 3n Vy[yw = 0m -> A(y,«)].

Here ym denotes the finite sequence <y(0),y(l), ...,y(m— 1)>.


Our claim is that EL + (*) + CP is inconsistent.

Sketch of the proof. Suppose (*) and CP.

Observe: V£e2*3«3aEy[a(0) = n A Vw[/?(a(m)) = 0(n)]]. Let 0e2N be such thatV/>[Q(/?) = 0]. Calculatep,neN such that

0) = n A Vm[y(a(m)) = y(/i)]].

Let N be the maximum of the numbers p, n +1 and consider /? = OiV* I , where l e 2*is such that V/?eN[!(/?) = 1].

This application of CP exemplifies a well-known technique in intuitionisticanalysis: weak counterexamples may be used, together with CP, to obtain proofs ofinconsistency.

9. An application

9.1 We will show how a recent result of Coquand [3] may be derived from theIntuitionistic Ramsey Theorem. We became aware of this result after the other sectionsof this paper had been completed. We first extend Definition 7.2.

9.2 Let R be a subset of the set S of finite strictly increasing sequences of naturalnumbers. We define

R is almost full :=V<xeSf3ae S[R(a. o a)].

(We drop the restriction that all sequences in R are of the same length.)If we consider R in the obvious way as a set of finite sets of natural numbers, then

R is almost full if and only if every infinite set of natural numbers has a subset thatbelongs to R.

An important example of an almost full subset of S is the following:

RPH :={aeS\a^ <> and length (a) > a(0)}.

The letters PH refer to Paris and Harrington [11].

9.3 Let R be a subset of S. Let ke N and let 8 be a function from Sk to {0,1}. Wedefine a subset R3 of R:

Rs := {a € R \ W> e Sk[b(k - 1) < length (a) -> S(a o b) = <5«a(0),..., a(k - 1)»]}.

If we consider 3 as a colouring of the A>element-subsets of N, then Rs is the set of^-monochromatic members of R.

9.4 THEOREM (Coquand).For every subset R of S, for every ke M,for every d:Sk->{0,\}.IfRis almost full,

then Rs is almost full.

Proof Let R be an almost full subset of S, and let 3:Sk-*{0,1}. LetLet QED be the proposition: 3aeS[Rs(<xoa)]. Let yeSf. Determine aeS such thatR(aoyoa). We may decide: either Rs(aoyoa) and therefore: QED, or3beSk[S(<xoyoaob) = 0]. Therefore: VyeSf 3aeSk[S(<xoyoa) = 0 V QED] and,similarly: Vye£f 3a€Sfc[<5(aoyoa) = 1 V Q.EZ>]. Apply Theorem 7.3 and concludeQED.


9.5 In order to derive a useful corollary from Theorem 9.4, we need a well-knownprinciple of intuitionistic analysis, the Fan Theorem. The version of this theorem thatwe use [8, *27.9, p. 76 or 14, FAN(T), p. 218] is a corollary of the principle ofinduction on monotone bars, BIlVI (cf. 2.4). Let # be intuitionistic Cantor space, thatis, the binary fan, the set of all functions from 11 to {0,1}.

9.5.1 FT (Fan Theorem).Let Q be a subset of the set of binary finite sequences such thatV a e # 3 « [ e « a ( 0 ) , . . . , a ( « - l ) » ] . Then 3WVae#3n ^ A^0(<a(O), . . . , a ( « -

9.5.2 COROLLARY.

Let R be an almost full subset of S, and let keN.

Proof. By enumerating Sk, we may identify ^ with the set of all functions fromSk to {0,1}. Now apply Theorem 9.4 and The Fan Theorem, Theorem 9.5.1.

Observe that Corollary 9.5.2 is an intuitionistic version of the compactnessargument, by which in [11] a sharpened version of the Finite Ramsey Theorem isproved. (In order to obtain the desired conclusion, specialize a, in the second line of9.5.2, to sequences of the form Xm-(m+p)).

9.6 Coquand observed that from Theorem 9.4, an important case of Theorem 7.3may be derived, as follows.

Let keN and suppose that R, Tare decidable almost full subsets of Sk.Let Q := {a e S | 36 e S[R(a o b)] A 36 e S[T(a o b)]}. Let 3: Sk -• {0,1} be the charac-

teristic function of T. Observe that Q is almost full, and that for all as S, if Q\a), then3beS[R(aob) A T(aob)]. As, by Theorem 9.4, Qs is almost full, R f] Tis almost full.

In the above argument, it suffices to assume that T is a decidable subset of S.In [3] another constructive version of Ramsey's Theorem, that avoids the

application of the intuitionistic principle of bar induction, is stated and proved.

10. Concluding remarks

10.1 As far as we know, the first person to study Ramsey's Theorem from anintuitionistic point of view was Mervyn Jansen who, in Nijmegen in 1974, wrote aMaster's Thesis on the subject, under the guidance of Johan J. de Iongh. He foundthe example mentioned in 5.3 and, formulating an admittedly unpleasant ad hoccondition, he proved, using the principle of bar induction, that every binary relationR on N which satisfies this condition has the property mentioned in the conclusionof 6.2, that is, it is impossible that both R and its complement are almost full.

The second author, who did not know about the earlier attempt by Jansen,obtained a stronger result, which he announced in the first 'Stelling', added to hisdissertation [1]. He proved for every decidable binary relation R on N that not bothR and (N x I^J)\i? are almost full. This result sparked off the research which led to thepresent paper. The first author studied the second author's proof and found Theorems6.5, 7.3 and 5.4. The negationless wording of these theorems owes something to aquestion posed by John Burgess. In a letter from 1983, he had asked for a constructiveargument establishing

< n A a(m) ^ <x(n) AjS(m) ^ £(n)].


(We challenge the reader to find the elementary proof of this special case ofTheorem 6.5).

In Section 4 of this paper one finds a paraphrase of the original argument of thesecond author: Lemma 4.6, Corollary 4.7 and a slight weakening of both Lemma 6.1and Corollary 6.2 are due to him.

Section 8 of this paper elaborates another observation of the second author, to theeffect that from consistency results about CT0, one may obtain unprovability resultsconcerning versions of Ramsey's Theorem.

The application of the Intuitionistic Ramsey Theorem in Section 9 is due to thefirst author, who also wrote the final version of the paper.

10.2 Further generalizations of the Intuitionistic Ramsey Theorem, such as ananalogue to the classical clopen Ramsey Theorem (cf. [4]), are possible. We hope totreat them in a sequel to the present paper.

References

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12. F. P. RAMSEY, 'On a problem of formal logic', Proc. London Math. Soc. 30 (1928) 264-286.13. E. SPECKER, 'Ramsey's Theorem does not hold in recursive set theory', Proceedings of the Summer

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15. W. VELDMAN, 'On the contraposition of two axioms of countable choice'. Brouwer CentenarySymposium, ed. A. S. Troelstra, D. van Dalen, Studies in Logic and the Foundations ofMathematics 110 (North Holland, Amsterdam, 1982) 513-523.

Mathematisch Instituut Faculteit der WijsbegeerteKatholieke Universiteit Rijksuniversiteit UtrechtToernooiveld P.O. Box 801266525 ED Nijmegen 3508 TC UtrechtThe Netherlands The Netherlandse-mail: [email protected] e-mail: [email protected]

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