Computability and Logic

26.1. RAMSEY’S THEOREM: FINITARY AND INFINITARY 321
So let us set aside the difficult problem of finding the least m that is large enough,
and turn to proving that there is some m that is large enough. The proof of Theorem
26.1 that we are going to present will make a ‘detour through the infinite’. First we
prove the following infinitary analogue:
26.2 Theorem (Infinitary Ramsey’s theorem). Let r, s be positive integers. Then no
matter how the size-r subsets of the set X ={0, 1, 2, ...} are partitioned into s classes,
there will always be an infinite subset Y of X such that all size-r subsets of Y belong to the
same class.
Note that if the theorem holds as stated, then it clearly holds for any other enumerably
infinite set in place of {0, 1, 2, ...}. (If Zeus threw a party for an enumerable
infinity of guests, any two of whom either liked each other or disliked each other,
there would either be infinitely many guests, any two of whom liked each other, or
infinitely many, any two of whom disliked each other.) In fact Theorem 26.2 holds
for any infinite set X, because any such set has an enumerably infinite subset (though
it requires the axiom of choice to prove this, and we are not going to go into the
matter).
The proof of Theorem 26.2 will be given in this section, and the derivation of
Theorem 26.1 from it—which will involve an interesting application of the compactness
theorem—in the next. Before launching into the proof, let us introduce some
notation that will be useful for both proofs.
A partition of a set Z into s classes may be represented by a function f whose
arguments are the elements of Z and whose values are elements of {1,...,s}: the
ith class in the partition is just the set of those z in Z with f (z) = i. Let us write
f : Z → W to indicate that f is a function whose arguments are the elements of
Z and whose values are elements of W . Our interest is in the case where Z is the
collection of all the size-r subsets of some set X. Let us denote this collection [X] r .
Finally, let us write ω for the set of natural numbers. Then the infintary version of
Ramsey’s theorem may be restated as follows: If f :[ω] r →{1, ..., s}, then there is
an infinite subset Y of ω and a j with 1 ≤ j ≤ s such that f :[Y ] r →{j} (that is, f
takes the value j for any size-r subset of Y as argument).
Proof of Theorem 26.2: Our proof will proceed as follows. For any fixed s > 0,
we show by induction on r that for any r > 0 we can define an operation such that
if f :[ω] r →{1, ..., s}, then ( f ) is a pair ( j, Y ) with f :[Y ] r →{j}.
Basis step: r = 1. In this case the definition of ( f ) = ( j, Y ) is easy. For each of
the infinitely many size-1 sets {b}, f ({b}) is one of the finitely many positive integers
k ≤ s. We can thus define j as the least k ≤ s such that f ({b}) = k for infinitely many
b, and define Y as {b : f ({b}) = j}.
Induction step: We assume as induction hypothesis that has been suitably defined
for all g :[ω] r →{1, ..., s}. Suppose f :[ω] r+1 →{1, ..., s}. In order to define
( f ) = ( j, Y ), we define, for each natural number i, a natural number b i , infinite sets
Y i , Z i , W i , a function f i :[ω] r →{1,...,s}, and a positive integer j i ≤s. Let Y 0 = ω.
We now suppose Y i has been defined, and show how to define b i , Z i , f i , j i , W i , and
Y i+1 .