COMPUTABLE RAMSEY’S THEOREM FOR PAIRS NEEDS
INFINITELY MANY Π02 SETS
GREGORY IGUSA AND HENRY TOWSNER
Abstract. In Ramsey’s Theorem and Recursion Theory, Theorem 4.2,
Jockusch proved that for any computable k-coloring of pairs of integers,
there is an infinite Π02 homogeneous set. The proof used a countable
collection of Π02 sets as potential infinite homogeneous sets. In a remark
preceding the proof, Jockusch stated without proof that it can be shown
that there is no computable way to prove this result with a finite number
of Π02 sets.
We provide a proof of this claim, showing that there is no computable
way to take an index for an arbitrary computable coloring and produce
a finite number of indices of Π02 sets with the property that one of those
sets will be homogeneous for that coloring.
While proving this result, we introduce n-trains as objects with useful
combinatorial properties which can be used as approximations to infinite
Π02 sets.
1. Introduction
In [1], Jockusch initiated the study of the effective content of Ramsey’s
theorem, stated below. This study has proved to be enormously fruitful in
effective combinatorics, and also in reverse mathematics. In Theorem 4.2
of [1], Jockusch proved that for any computable k-coloring of pairs of integers,
there is an infinite Π02 homogeneous set. Before this proof, he made the
remark that even for 2-colorings of pairs of integers (basic recursive partitions,
in his language), it can be shown that there is no uniform computable way to
take an index for an arbitrary computable coloring, and to produce a finite
number of indices of Π02 sets with the property that one of those Π02 sets will
be an infinite homogeneous set for that coloring.
In 2013, Tahina Rakotoniaina inquired about the proof of this claim, and
Jockusch withdrew his claim, asking for a proof or refutation of the claim.
We present a proof here.
Our original draft gave a direct proof of Lemma 3.2; we are grateful to
the referee for suggesting that the proof might generalize to Theorem 3.3, a
combinatorial result that indicates a reason that n-trains (see Definition 3.1)
are useful as approximations to Π02 sets.
Date: October 28, 2016.
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GREGORY IGUSA AND HENRY TOWSNER
2. Definitions
Definition 2.1. A k-coloring of the n-element subsets of N is a function
c : [N]n → k, from the set of n-element subsets of N, to k.
We think of such a coloring as a rule that assigns a color to every n-element
subset of N, using up to k different colors.
Ramsey’s theorem is then the following theorem of combinatorics.
Theorem 2.2 (Ramsey’s Theorem). For any n, k ≥ 1, and any k-coloring
c : [N]n → k, there exists an infinite subset H ⊆ N such that c [H]n is a
constant function.
We call such an H an infinite homogeneous set for c.
In this paper, we will be primarily concerned with the case when n = k = 2.
In [1], Jockusch proves the following.
Theorem 2.3 (Jockusch, [1], Theorem 4.2). If c : [N]2 → k is a computable
k-coloring of pairs, then there exists an infinite Π02 homogeneous set for c.
We prove the following.
Theorem 2.4. There does not exist a partial computable f with the property
that for any e, if e is the code of a total computable 2-coloring c : [N]2 → 2
of pairs, then f (e) is the code for a finite set {a0 , a1 , . . . , ak } of indices for
Π02 sets with the property that at least one of those Π02 sets is an infinite
homogeneous subset for c.
Indeed, we prove slightly more: that there is no such f where f (e) is the
code for a c.e. set Wf (e) enumerating finitely many codes for Π02 sets.
3. Trains
Definition 3.1. An n-train is a sequence of (not necessarily distinct) sets
of natural numbers of size n, τ0 , τ1 , . . . , τm such that for every a ∈ τi+1 \ τi ,
a > τi .
For instance
{1, 2, 3}, {2, 3, 5}, {5, 7, 9}, {5, 9, 12}
is a 3-train.
If 0 ≤ k < n we write τ (k) for the (k + 1)-st element of τ in the usual
ordering of N.
Note that, given a sequence of sets S0 ⊇ S1 ⊇ · · · ⊇ Sk , of size ≥ n, if we
let τi be the first n elements of Si , then hτi i is an n-train. We will see in our
proof of Theorem 4.1 that an n-train can be used as an approximation to
the first n elements of an infinite Π02 set.
Our main tool for working with n-trains is the following combinatorial
result. We color elements using R, B, and if ι ∈ {R, B}, we write ι for the
opposite color: R = B and B = R.
COMPUTABLE RAMSEY’S THEOREM NEEDS INFINITELY MANY Π02 SETS
3
j is an (n + 1)-train.
Lemma 3.2. Suppose that for each j < n, τ0j , τ1j , . . . , τm
j
Let c : [N]2 → {R, B} be given. Then there is a coloring c∗ : τij → {R, B}
such that for each τij on which c [τij ]2 = ι homogeneously, there is an a ∈ τij
with c∗ (a) = ι.
S
We prove this by way of a more general result.
j is an (n + 1)Theorem 3.3. Suppose that for each j < n, τ0j , τ1j , . . . , τm
j
train. Let (V, E) be the hypergraph where V = i,j τij and E = {τij }. Then
S
(V, E) is bipartite, meaning that there is a coloring c∗ : τij → {R, B} such
that no τij is colored homogeneously by c∗ .
S
Note that this theorem provides a “colorblind” proof of Lemma 3.2: By
making sure that every τij has at least one element of each color, in particular
we ensure that if the pairs of τij had been colored homogeneously by c, then
it has at least one element that is colored the other way by c∗ . So it proves
Lemma 3.2 without needing to know the coloring c.
The proof of Theorem 3.3 is by a greedy algorithm. We think of each train
as a sequence of requirements ordered from back-to-front, and choose the
correct order in which to interleave the requirements from different trains
(the ordering ≺ defined below). When we consider a requirement τijkk , we
will need to color at most two of the elements to satisfy the requirement. If
any of the elements had been previously colored, then we only need to color
one element of τijkk to satisfy the requirement.
There are n + 1 possible elements we could color to satisfy the requirement,
some of which may already be colored by higher priority requirements.
However, our ordering ≺ will ensure that any fixed train can have colored
at most one element of τijkk . This ensures that there is at least one element
available for us to color in order to satisfy this requirement.
0
Proof. We define an ordering ≺ on the sets τij : we say τij ≺ τij0 if, taking
0
0
k < n + 1 largest such that τij (k) 6= τij0 (k), we have τij (k) > τij0 (k). (This is
the opposite of the reverse lexicographic order, which we choose not to refer
to as the reverse reverse lexicographic order.) Note that for a fixed j and
i > i0 , we have that τij (n) > τij0 (n), and so τij ≺ τij0 .
For each r, let jr , ir be such that τijrr is the r-th element in this ordering.
We define the coloring c∗ in stages, considering τijrr at the r-th stage. At
stage r, there are three cases.
(1) If c∗r is undefined on τijrr , then we assign the labels br and ar to τijrr (n)
and τijrr (n − 1), and we define cr+1 (br ) = B and cr+1 (ar ) = R.
(2) If c∗r is defined on some elements of τijrr , but c∗r τijrr is homogeneous,
with color ι, then let k be maximal such that c∗r (τijrr (k)) is undefined,
we assign the label ar to c∗r (τijrr (k)) and define cr+1 (an ) = ῑ
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GREGORY IGUSA AND HENRY TOWSNER
(3) If c∗r τijrr is not homogeneous, do nothing.
Assuming that in Case (2), there is always at least one element of τijrr that
is not yet colored by c∗r , this algorithm clearly proves the lemma.
The proof that there is always at least one such element will be by the
following claim.
Claim 1. Assume that x 6= y are elements of τijrr , colored at stages s ≤ t < r,
and that js = jt .
Then c∗r (x) 6= c∗r (y).
Proof. Let x, y, s, t be as in the statement of the claim.
Note first that if s = t then x and y must have been colored by Case (1)
at stage s, so we have that c∗r (x) 6= c∗r (y).
Now, assume that s < t. We show that x ∈ τijtt , and so at stage t, y must
have been colored by Case (2), and it must therefore have been colored the
opposite color from x.
Let z = τijrr (n), and note that, because x, y ∈ τijrr , we must have that
x, y ≤ z.
To see that x ∈ τijtt , note that s < t, and so, because js = jt , we must have
that is > it . Thus, every element of τijss that is not in τijtt must be larger
than every element of τijtt . However, the largest element of τijtt must be ≥ z,
because τijtt ≺ τijrr . In particular, because x ≤ z, x is not greater than the
largest element of τijtt , and so x cannot be in τijss \ τijtt . Thus, because x ∈ τijss ,
we must also have that x ∈ τijtt .
Therefore, x ∈ τijtt , and so at stage t, y must have been colored by Case
(2), and it must therefore have been colored the opposite color from x. We now show that if c∗r is homogeneous on dom(c∗r ) ∩ τijrr , then there must
be at least one element of τijrr that is not in dom(c∗r ).
To see this, assume c∗r is homogeneous on dom(c∗r ) ∩ τijrr , and let dom(c∗r ) ∩
τijrr = {x0 , . . . , xk−1 }. For each l < k, let sl be the stage when xl was colored.
By Claim 1, we must have that for l < l0 < k, jl 6= jl0 . There are only n
different values of j, and so we have that k ≤ n. But |τijrr | = n + 1 and so
there must be at least one element of τijrr that is not yet colored by c∗r .
Therefore, in Case (2), there is always at least one element
to color, and
S
so our algorithm is able to act at every stage. Let c∗∞ = r c∗r . (Note that
this is a finite union, because there are only finitely many τij .) Let c∗ be any
extension of c∗∞ that is defined on the union of the τij .
4. Construction
Theorem 4.1. Fix finitely many Π2 formulas ∀x∃yR0 (z, x, y), . . ., ∀x∃yRn−1 (z, x, y).
There is a computable 2-coloring of pairs c, computable uniformly from the
COMPUTABLE RAMSEY’S THEOREM NEEDS INFINITELY MANY Π02 SETS
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formulas R0 , . . . Rn−1 so that for each j < n, the set
Sj = {z | ∀x∃yRj (z, x, y)}
fails to be an infinite homogeneous set for c.
Proof. We describe how, for a given b, we define c(a, b) for all a < b. Fix the
value b and suppose we have defined c(a, a0 ) for all a < a0 < b.
For each j < n let Sj = {a : ∀x < i∃y < bRj (a, x, y)}, and let bj be the
largest number ≤ b such that Sj has at least n + 1 elements. We then define
an (n + 1)-train by taking the set τij for i ≤ bj to be the n + 1 smallest
numbers in Sj . Note that this is an (n + 1)-train, because it is the sequence
of the first n + 1 elements of a sequence of shrinking sets.
Let c∗ be given by Lemma 3.2, and extend c∗ to be defined on all a < b
by defining it arbitrarily where it is not already defined. Set c(a, b) = c∗ (a)
for all a < b.
Suppose that for some j < n, the set Sj is infinite. Let τ j be the n + 1
smallest elements of Sj . We claim that, for b sufficiently large, there is always
some i so that τij = τ j . For every a < τ j (n) such that a 6∈ τ j , there is some i
such that ∃x ≤ i∀y¬Rj (a, x, y), so certainly for every b, if i0 ≥ i and a ∈ τij ,
either a ∈ τ j or a > τ j . Let i be large enough to witness this bound for all
a < τ j (n).
For each a ∈ τ j and each x ≤ i, there is some y such that Rj (a, x, y). If b
is big enough to bound these finitely many values of y, it must be the case
that τij = τ j . Therefore for all sufficiently large b, τij = τ j .
Since Sj is infinite, let b be some element of Sj sufficiently large so that
j
τi = τ j . If c [τ j ]2 = ι then there is some a ∈ τ j with c(a, b) = c∗ (a) = ι.
Therefore Sj is not homogeneous.
We can now prove our main theorem:
Theorem 4.2. There is no partial computable f such that for any e, if e is
the code of a total computable 2-coloring c : [N]2 → 2 of pairs, then f (e) is
defined and the c.e. set Wf (e) is a finite set of indices for Π02 sets with the
property that at least one of those Π02 sets is an infinite homogeneous subset
for c.
Proof. Let f be a partial computable function such that for any e, if e is
the code of a total computable 2-coloring c : [N]2 → 2 of pairs, then f (e) is
defined.
We define a coloring c as follows. Via the recursion theorem, we obtain the
code for c, and begin evaluating f (e). If the computation for f (e) has not
halted after b steps, we define c(a, b) for a < b arbitrarily. If the computation
for f (e) has halted, then we begin enumerating Wf (e) . If Wf (e) is empty
after b steps, we continue to define c(a, b) for a < b arbitrarily. Each time
that Wf (e) enumerates a new number, we continue the construction of c as in
the proof of the previous theorem, assuming that Wf (e) will never enumerate
any additional numbers.
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GREGORY IGUSA AND HENRY TOWSNER
If Wf (e) is finite, then at some point this assumption will be true, and we
will be able to conclude that no ai is the code for an infinite homogeneous
Π02 subset for c. Note that c always produces a total computable 2-coloring,
whether or not f (e) is defined, so the recursion theorem must produce a
value e on which f (e) is defined.
References
[1] Jockusch, Carl, Ramsey’s theorem and recursion theory, J. Symbolic Logic 37 (1972),
268–280.