Computability and Logic

322 RAMSEY’S THEOREM
Let b i be the least member of Y i .
Let Z i = Y i −{b i }. Since Y i is infinite, so is Z i . Let the members of Z i in increasing
order be a i0 , a i1 ,....
For any size-r set x of natural numbers, where x ={k 1 ,...,k r }, with k 1 < ···<
k r , let f i (x) = f ({b i , a ik1 , ..., a ikr }). Since b i is not one of the a ik and f is defined
on all size-(r + 1) sets of natural numbers, f i is well defined.
By the induction hypothesis, for some positive integer j i ≤ s and some infinite set
W i ,( f i ) = ( j i , W i ) and for every size-r subset x of W i , we have f i (x) = j i .We
have thus defined j i and W i , and we define Y i+1 ={a ik : k ∈ W i }.
Since W i is infinite, Y i+1 is infinite. Y i+1 ⊆ Z i ⊆ Y i , and thus if i 1 ≤ i 2 , then
Y i2 ⊆ Y i1 . And since b i is less than every member of Y i+1 ,wehaveb i < b i+1 , which
is the least member of Y i+1 . Thus if i 1 < i 2 then b i1 < b i2 .
For each positive integer k ≤ s, let E k ={i: j i = k}. As in the basis step, some E k
is infinite, and we let j be the least k such that E k is infinite, and let Y ={b i : i ∈ E j }.
This completes the definition of .
Since b i1 < b i2 if i 1 < i 2 , Y is infinite. In order to complete the proof, we must
show that if y is a size-(r + 1) subset of Y , then f (y) = j. So suppose that y =
{b i , b i1 ,...,b ir }, with i < i 1 < ···< i r and i, i 1 ,...,i r all in E j . Since the Y i are
nested, all of b i1 ,...,b ir are in Y i . For each m, 1 ≤ m ≤ r, let k m be the unique
member of W i such that b im = a ikm . And let x ={k 1 ,...,k r }. Then x is a subset
of W i , and since i 1 < ···< i r , we have b i1 < ···< b ir , a ik1 < ···< a ikr , and
k 1 < ···< k r , and thus x is a size-r subset of W i . But ( f i ) = ( j i , W i ) and thus
f i (x) = j i . Since i is in E j , j i = j. Thus
as required.
f (y) = f ({ b i , b i1 ,...,b ir
})
= f
({
bi , a ik1 ,...,a ikr
})
= fi (x) = j
Before moving on to the next section and the proof of Theorem 26.3, let us point
out that the following strengthening of Theorem 26.2 is simply false: Let s be a
positive integer. Then no matter how the finite sets of natural numbers are partitioned
into s classes, there will always be an infinite set Y of natural numbers such that all
positive integers r, all size-r subsets of Y belong to the same one of the s classes.
Indeed, this fails for s = 2. Let f (x) = 1 if the finite set x contains the number that
is the number of members in x; and f (x) = 2 otherwise. Then there is no infinite set
Y such that for every r, either f (y) = 1 for all size-r subsets y of Y or f (y) = 2 for
all such y. For if r is a positive integer that belongs to Y and b 1 ,...,b r are r other
members of Y , then f ({r, b 2 ,...,b r }) = 1, while f ({b 1 ,...,b r }) = 2.
26.2 König’s Lemma
In order to derive the original, finitary version of Ramsey’s theorem from the infinitary
version, we will establish a principle known as König’s lemma, concerning objects
called trees. For present purposes, a tree consists of: (i) a nonempty set T of elements,
called the nodes of the tree; (ii) a partition of T into finitely or infinitely many sets
T = T 0 ∪ T 1 ∪ T 2 ∪···