Equations and Infinite Colorings Exposition by Stephen Fenner and William Gasarch

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1 1 Introduction Equations Infinite Colorings Exposition by Stephen Fenner William Gasarch Do you think the following is TRUE or FALSE? For any ℵ 0 -coloring of the reals, COL : R N there exist distinct e 1, e 2, e 3, e 4 such that COL(e 1 ) = COL(e 2 ) = COL(e 3 ) = COL(e 4 ), It turns out that this question is equivalent to the negation of CH. Komjáth (3) claims that Erdős proved this result. The prove we give is due to Davies (1). Definition 1.1 The Continuum Hypothesis (CH) is the statement that there is no order of infinity between that of N R. It is known to be independent of Zermelo-Frankel Set Theory with Choice (ZFC). Definition 1.2 ω 1 is the first uncountable ordinal. ω 2 is the second uncountable ordinal. Fact If CH is true, then there is a bijection between R ω 1. This has the counter-intuitive consequence: there is a way to list the reals: x 0, x 1, x 2,..., x α,... as α ω 1 such that, for all α ω 1, the set {x β β < α} is countable. 2. If CH is false, then there is an injection from ω 2 to R. This has the consequence that there is a list of distinct reals: where α ω 1 β [ω 1, ω 2 ). 2 CH FALSE x 0, x 1, x 2,..., x α,..., x ω1, x ω1 +1,..., x β,... Definition 2.1 Let X R. Then CL(X) is the smallest set Y X of reals such that a, b, c Y a + b c Y. Lemma If X is countable then CL(X) is countable. 1

2 2. If X 1 X 2 then CL(X 1 ) CL(X 2 ). Proof: 1) Assume X is countable. CL(X) can be defined with an ω-induction (that is, an induction just through ω). C 0 = X C n+1 = C n {a + b c a, b, c C n } One can easily show that CL(X) = i=0 C i that this set is countable. 2) This is an easy exercise. Theorem 2.3 Assume CH is true. There exists an ℵ 0 -coloring of R such that there are no distinct e 1, e 2, e 3, e 4 such that COL(e 1 ) = COL(e 2 ) = COL(e 3 ) = COL(e 4 ), Proof: Since we are assuming CH is true, we have, by Fact 1.3.1, there is a bijection between R ω 1. If α ω 1 then x α is the real associated to it. We can picture the reals as being listed out via x 0, x 1, x 2, x 3,..., x α,... where α < ω 1. Note that every number has only countably many numbers less than it in this ordering. For α < ω 1 let X α = {x β β < α}. Note the following: 1. For all α, X α is countable. 2. X 0 X 1 X 2 X 3 X α 3. α<ω 1 X α = R. We define another increasing sequence of sets Y α by letting Note the following: Y α = CL(X α ). 1. For all α, Y α is countable. This is from Lemma Y 0 Y 1 Y 2 Y 3 Y α. This is from Lemma α<ω 1 Y α = R. 2

3 We now define our last sequence of sets: For all α < ω 1, Z α = Y α Note the following: 1. Each Z α is finite or countable. 2. The Z α form a partition of R. β<α Y β. We will now define an ℵ 0 -coloring of R. For each Z α, which is countable, assign colors from ω to Z α s elements in some way so that no two elements of Z α have the same color. Assume, by way of contradiction, that there are distinct e 1, e 2, e 3, e 4 such that COL(e 1 ) = COL(e 2 ) = COL(e 3 ) = COL(e 4 ) Let α 1, α 2, α 3, α 4 be such that e i Z αi. Since all of the elements in any Z α are colored differently, all of the α i s are different. We will assume α 1 < α 2 < α 3 < α 4. The other cases are similar. Note that e 4 = e 1 + e 2 e 3. e 1, e 2, e 3 Z α1 Z α2 Z α3 Y α1 Y α2 Y α3 = Y α3. Since Y α3 = CL(X α3 ) e 1, e 2, e 3 Y α3, we have e 4 Y α3. Hence e 4 / Z α4. This is a contradiction. What was it about the equation e 1 + e 2 = e 3 + e 4 that made the proof of Theorem 2.3 work? Absolutely nothing: Theorem 2.4 Let n 2. Let a 1,..., a n R be nonzero. Assume CH is true. There exists an ℵ 0 -coloring of R such that there are no distinct e 1,..., e n such that COL(e 1 ) = = COL(e n ), Proof sketch: n a i e i = 0. i=1 Since this prove is similar to the last one we just sketch it. Definition 2.5 Let X R. CL(X) is the smallest superset of X such that the following holds: For all m {1,..., n} for all e 1,..., e m 1, e m+1,..., e n, e 1,..., e m 1, e m+1,..., e n CL(X) (1/a m ) a i e i CL(X). i {1,...,n} {m} 3

4 Let X α, Y α, Z α be defined as in Theorem 2.3 using this new defintion of CL. Let COL be defined as in Theorem 2.3. Assume, by way of contradiction, that there are distinct e 1,..., e n such that COL(e 1 ) = = COL(e n ) n a i e i = 0. i=1 Let α 1,..., α n be such that e i Z αi. Since all of the elements in any Z α are colored differently, all of the α i s are different. We will assume α 1 < α 2 < < α n. The other cases are similar. Note that n 1 e n = (1/a n ) a i e i CL(X) i=1 e 1,..., e n 1 Z α1 Z αn 1 Y αn 1. Since Y αn 1 = CL(X αn 1 ) e 1,..., e n 1 Y αn 1, we have e n Y αn 1. Hence e n / Z αn. This is a contradiction. Note 2.6 For most linear equations, CH is not needed to get a counterexample. 3 CH TRUE Theorem 3.1 Assume CH is false. Let COL be an ℵ 0 -coloring of R. There exist distinct e 1, e 2, e 3, e 4 such that COL(e 1 ) = COL(e 2 ) = COL(e 3 ) = COL(e 4 ), Proof: By Fact 1.3 there is an injection of ω 2 into R. If α ω 2, then x α is the real associated to it. Let COL be an ℵ 0 -coloring of R. We show that there exist distinct e 1, e 2, e 3, e 4 of the same color such that We define a map F from ω 2 to ω 1 ω 1 ω 1 ω. 1. Let β ω Define a map from ω 1 to ω by α COL(x α + x β ). 3. Let α 1, α 2, α 3 ω 1 be distinct elements of ω 1, i ω, such that α 1, α 2, α 3 all map to i. Such α 1, α 2, α 3, i clearly exist since ℵ 0 + ℵ 0 = ℵ 0 < ℵ 1. (There are ℵ 1 many elements that map to the same element of ω, but we do not need that.) 4

5 4. Map β to (α 1, α 2, α 3, i). Since F maps a set of cardinality ℵ 2 to a set of cardinality ℵ 1, there exists some element that is mapped to twice by F (actually there is an element that is mapped to ℵ 2 times, but we do not need this). Let α 1, α 2, α 3, β, β, i be such that β β F (β) = F (β ) = (α 1, α 2, α 3, i). Choose distinct α, α {α 1, α 2, α 3 } such that x α x α / {x β x β, x β x β }. We can do this because there are at least three possible values for x α x α. Since F (β) = (α 1, α 2, α 3, i), we have Since F (β ) = (α 1, α 2, α 3, i), we have Let Then COL(x α + x β ) = COL(x α + x β ) = i. COL(x α + x β ) = COL(x α + x β ) = i. e 1 = x α + x β e 2 = x α + x β e 3 = x α + x β e 4 = x α + x β. COL(e 1 ) = COL(e 2 ) = COL(e 3 ) = COL(e 4 ) Since x α x α x β x β, we have {e 1, e 2 } {e 3, e 4 } =. Moreover, the equation e 1 = e 2 is equivalent to x α x α = x β x β, which is ruled out by our choice of α, α, so e 1 e 2. Similarly, e 3 e 4. Thus e 1, e 2, e 3, e 4 are all distinct. Remark. All the results above hold practically verbatim with R replaced by R k, for any fixed integer k 1. In this more geometrical context, e 1, e 2, e 3, e 4 are vectors in k- dimensional Euclidean space, the equation e 1 + e 2 = e 3 + e 4 says that e 1, e 2, e 3, e 4 are the vertices of a parallelogram (whose area may be zero). 5

6 4 More is Known! To state the generalization of this theorem we need a definition. Definition 4.1 An equation E(e 1,..., e n ) (e.g., e 1 +e 2 = e 3 +e 4 ) is regular if the following holds: for all colorings COL : R N there exists e = (e 1,..., e n ) such that COL(e 1 ) = = COL(e n ), e 1,..., e n are all distinct. E(e 1,..., e n ), If we combine Theorems we obtain the following. Theorem 4.2 e 1 + e 2 = e 3 + e 4 is regular iff 2 ℵ 0 > ℵ 1. Jacob Fox (2) has generalized this to prove the following. Theorem 4.3 Let s N. The equation is regular iff 2 ℵ 0 > ℵ s. e 1 + se 2 = e e s+3 (1) Fox s result also holds in higher dimensional Euclidean space, where it relates to the vertices of (s + 1)-dimensional parallelepipeds. Subtracting (s + 1)e 2 from both sides of (1) rearranging, we get e 1 e 2 = (e 3 e 2 ) + + (e s+3 e 2 ), which says that e 1 e 2 are opposite corners of some (s + 1)-dimensional parallelepiped P where e 3,..., e s+3 are the corners of P adjacent to e 2. Of course, there are other vertices of P besides these, Fox s proof actually shows that if 2 ℵ 0 > ℵ s then all the 2 s+1 vertices of some such P must have the same color. 5 Acknowledgments We would like to thank Jacob Fox for references for writing the paper that pointed us to this material. References [1] R. O. Davies. Partioning the plane into denumerably many sets without repeated differences. Proceedings of the Cambridge Philosophical Society, 72: , [2] J. Fox. An infinite color analogue of Rado s theorem. ~publications.html. [3] P. Komjáth. Partitions of vector spaces. Periodica Mathematica Hungarica, 28: ,

20 Ordinals. Definition A set α is an ordinal iff: (i) α is transitive; and. (ii) α is linearly ordered by. Example 20.2.

20 Ordinals. Definition A set α is an ordinal iff: (i) α is transitive; and. (ii) α is linearly ordered by. Example 20.2. 20 Definition 20.1. A set α is an ordinal iff: (i) α is transitive; and (ii) α is linearly ordered by. Example 20.2. (a) Each natural number n is an ordinal. (b) ω is an ordinal. (a) ω {ω} is an ordinal.