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8/3/2019 Saharon Shelah- A Finite Partition Theorem with Double Exponential Bound
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A FINITE PARTITION THEOREMWITH DOUBLE EXPONENTIAL BOUND
DEDICATED TO PAUL ERDOS
Saharon Shelah
Institute of MathematicsThe Hebrew University
Jerusalem, Israel
Rutgers UniversityDepartment of Mathematics
New Brunswick, NJ USA
Abstract. We prove that double exponentiation is an upper bound to Ramsey the-orem for colouring of pairs when we want to predetermine the order of the differencesof successive members of the homogeneous set.
I thank Alice Leonhardt for excellent typing.
Research supported by the United States-Israel Binational Science FoundationPublication number 515; Latest Revision 02/95Done Erdos 80th birthday meeting in Keszthely, Hungary; July 1993
Typeset by AMS-TEX
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The following problem was raised by Jouko Vaananen for model theoretic rea-sons (having a natural example of the difference between two kinds of quantifiers,actually his question was a specific case), and propagated by Joel Spencer: Is there
for any n, c an m such that() for every colouring f of the pairs from {0, 1, . . . , m 1} by 2 (or even c)
colours, there is a monocromatic subset {a0, . . . , an1}, a0 < a1 < . . . suchthat the sequence ai+1 ai : i < n 1 is with no repetition and is withany pregiven order.
Noga Alon [Al] and independently Janos Pach proved that for every n, c thereis such an m as in (); Alon used van der Waerden numbers (see [Sh 329]) (soobtained weak bounds).Later Alon improved it to iterated exponential (Alan Stacey and also the authorhave later and independently obtained a similar improvement). We get a double
exponential bound. The proof continues [Sh 37]. Within the realm of doubleexponential in c, n we do not try to save.We thank Joel Spencer for telling us the problem, and Martin Goldstern for verycareful proof reading.
Notation. Let ,k,m,n,c,d belong to the set N of natural numbers (which includezero). A sequence is (0), . . . , (g1), also , are sequences. means that is a proper initial subsequence of. We consider sequences as graphs of functions(with domain of the form n = {0, . . . , n 1}, so e.g. means the largest initialsegment common to and . s is the sequence (0), . . . , (g 1), s (oflength g() + 1).
Let m :=
: g() = , and Rang() {0, . . . , m 1}
,
>m =km we write []m (or [] if and m are clear from the context) for theset { m : }.[A]n = {w A : |w| = n}.Intervals [a, b), (a, b), [a, b) are the usual intervals of integers. The proof is similarto [Sh 37] but sets are replaced by trees.
0 Definition. r = r(n, c) is the first number m such that:
()n,cm for every f : [{0, . . . , m 1}]2 {0, . . . , c 1} and linear order
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1 Definition. We say S is an (, m, m , u)-tree if:
(a) u {0, 1, . . . , 1} and m m
(b) S
(m
)(c) S is closed under initial segments(d) if S is -maximal then g() = (e) if S and g() u then for m numbers j < m we have j S
2 Claim. Suppose k,,m,p,m N satisfies
()k,,m,p,m m pmk+1,
then for every i() < , and u [0, 1), and |u| p and f : [(m)]2 {0, 1}there is T (m) closed under initial segments, (T, ) = (m, ) satisfying
for every 1, . . . , k T
(m) with 1 i(), . . . , k i() pairwisedistinct we can find a set S such that:
(a) S is an (, m, m , u \ i())-tree(b) if j [1, k) and S (m) then f({j, k}) = f({j , })(c) S (m) k i()
Remark.
(1) With minor change we can demand in
for any i() < .(2) We could use here f with range {0, . . . , c 1}, and in claim 3 get a longersequence : < n
such that f({1 , 2}) depends just on 1 2 , then usea partition theorem on such colouring.
Proof. For each >(m) choose randomly a set A [0, m), |A| = m, A =
{x1, . . . , xm} (pairwise distinct, chosen by order) (not all are relevant, some can be
fixed).
We define T = { : (m), and i < g() (i) Ai}.We have a natural isomorphism h from m onto T:
h() =
i
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Let Y = { (m):Prob[h(k) = ] = 0}, so |Y| = m|u|.
So h(1), . . . , h(k1) are determined. Now h(1), . . . , h(k1) and f induces anequivalence relation E on Y:
E iffk1j=1
f({h(j), }) = f({h(j),
}).
The number of classes is 2k1, let them be A1, . . . , A2k1 (they are pairwisedisjoint, some may be empty).
We call Aj large if there is S as required in clauses (a) and (c) of
such that()[ S (m) Aj ].
It is enough to show that the probability of h(k) belonging to a non-large equiva-lence class is < 1
(m
k ), hence it is enough to prove:
()2 Aj not large Prob(h(k) Aj) 1, without loss of generality Rang(f) = [0, d],let f : [m]2 {0, 1} be f({, }) = Min{f({, }, 1}. Let for j < n,
uj := [q + nd
j,q + nd
j + nd
). By downward induction on j [1, n] we try to defineTj such that:
(i) Tj is an (,m,m(q+nd+j),i
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DOUBLE EXPONENTIAL BOUND 7
there. (I.e. the tree in claim 2 is replaced by one isomorphic to it, levels outsidei |i2 j2 |.
Proof. (0) Check.(1) Let := i j and k := ||. We are looking for upper and lower bounds of thecardinality of the set (the order is lexicographic)
C := { (2m 1) : i < j}.
Clearly each C satisfies . Moreover, since j m, each element C
must satisfy (k) j(k) < 2m 1. Hence
C s
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5 Conclusion. If f : [0, (2m 1))[2] [0, c), = nc, m := 2(n+1)(c+1)n
, thenwe can find a0 < < an1 < m such that f [{a1, . . . , an1}]
2 constant andai+1 ai : i < n 1 in any pregiven order
Proof. As in fact 4, let i : i < (2m 1) enumerate (2m 1) in lexicographicorder, and let B := {i < (2m 1) : i
m} and for i B let i = i()2 : i < .Define a function f : [m]2 c by requiring f({i,
j}) = f({i, j}) for all i, j B.
Now the conclusion follows from lemma 3, Remark 3A(2) and Fact 4, particularlyclause (2).
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REFERENCES
[Al] Noga Alon. Notes.[Sh 329] Saharon Shelah. Primitive recursive bounds for van der Waerden num-
bers. Journal of the American Mathematical Society, 1:683697, 1988.
[GrRoSp] Ronald Graham, Bruce L. Rothschild, and Joel Spencer. Ramsey Theory.Willey Interscience Series in Discrete Mathematics. Willey, New York,1990. 2nd edition.
[Sh 37] Saharon Shelah. A two-cardinal theorem. Proceedings of the AmericanMathematical Society, 48:207213, 1975.