Homogeneous Dual Ramsey Theorem

TL;DR

This paper proves a positive answer to a question about homogeneous partitions and colorings, extending the finite Dual Ramsey Theorem and establishing the Ramsey property for certain finite measure algebras.

Contribution

It introduces a homogeneous version of the finite Dual Ramsey Theorem, resolving an open problem and linking it to measure algebra Ramsey properties.

Findings

Confirmed the existence of homogeneous partitions satisfying coloring conditions

Extended the finite Dual Ramsey Theorem to a homogeneous setting

Proved the Ramsey property for finite measure algebras with dyadic rational measures

Abstract

For positive integers $k < n$ such that $k$ divides $n$, let $(n)^k_{\hom}$ be the set of homogeneous $k$-partitions of $\{1, \dots, n\}$, that is, the set of partitions of $\{1, \dots, n\}$ into $k$ classes of the same cardinality. In the article "Ramsey properties of infinite measure algebras and topological dynamics of the group of measure preserving automorphisms: some results and an open problem" by Kechris, Sokic, and Todorcevic, the following question was asked: Is it true that given positive integers $k < m$ and $N$ such that $k$ divides $m$, there exists a number $n>m$ such that $m$ divides $n$, satisfying that for every coloring $(n)^k_{\hom}=C_1\cup\dots\cup C_N$ we can choose $u\in (n)^m_{\hom}$ such that $\{t\in (n)^k_{\hom}: t\mbox{ is coarser than } u\}\subseteq C_i$ for some $i$? In this note we give a positive answer to that question. This result turns out to be a homogeneous version of the finite Dual Ramsey Theorem of Graham-Rothschild. As explained by Kechris, Sokic, and Todorcevic in their article, our result also proves that the class $\mathcal{OMBA}_{\mathbb Q_2}$ of naturally ordered finite measure algebras with measure taking values in the dyadic rationals has the Ramsey property.

Equations

$$(n{)}_{\mathrm{hom}}^k=\text{mbox}thesetofallhomogeneousk\text{mbox}-partitionsof\{1,\dots ,n\};$$

$$i<j⟹\min (A_i)<\min (A_j).$$

$$A\le B⟺A\text{mbox}iscoarserthanB.$$

$$i<j⟹\min (X_i)<\min (X_j),\text{mbox}for1\le i,j\le k.$$

$$(n{)}^k=\text{mbox}thesetofallthek\text{mbox}-partitionsofn,$$

$$r(0,A)=\varnothing \&r(i,A)=\{A_j\cap \{1,\dots ,\min (A_i)\}:1\le j\le i\};i>0.$$

$$\pi (A)(i)=\{\begin{array}{ll}A(i)& \text{if 1≤A(i)≤k}\\ 1& \text{otherwise}.\end{array}$$

$$s(i,X{)}_j=X_j\cap \{1,\dots ,\min (X_k)\}$$

$$s(i,X{)}_j=s(i-1,X{)}_j\cup \{\min (X_j\setminus s(i-1,X{)}_j)\}.$$

$$(<\omega {)}^k\subset {\mathcal{S}}_k,\text{but }{\mathcal{S}}_k\setminus (<\omega {)}^k\ne \varnothing .$$

$${\mathcal{S}}_k^{\#}=\{b\in {\mathcal{S}}_k:\cup b\text{mbox}isaninitialsegmentof\mathbb{N}\}.$$

$$a=s_k(b)$$

$$T(a,k,n)=\{b\in {\mathcal{S}}_k^{\#}:a\ne b,a=s_k(b),\#b=\#a+n\}.$$

$$\phi :\{1,\dots ,n\}\to \{\#a+1,\dots ,\#a+n\}.$$

$$b_t(j)=\{\begin{array}{ll}a(j)& \text{if 1 ≤j≤\#a}\\ t({\phi }^{-1}(j))& \text{if \#a<j≤\#a+n}\end{array}$$

$$t\mapsto b_t$$

$$t(j)=b(\phi (j)),1\le j\le n.$$

$$i_0=\min \{i>m:m\text{mbox}divides|\cup s(i,X)|-\#a\}.$$

$$n(X)=|\cup s(i_0,X)|-\#a$$

$$b_X(j)=\{\begin{array}{ll}a(j)& \text{if 1 ≤j≤\#a}\\ s(i_0,X)(x_j)& \text{if \#a<j≤\#a+n(X)}\end{array}$$

$$X\in C_j\text{mbox}ifandonlyift(b_X)\in C_j^{n(X)}.$$

$$l=\min \{i>m:m\text{mbox}divides|\cup s(i,Y)|-\#a_0\}$$

$$b_Y(j)=\{\begin{array}{ll}{a}_0(j)& \text{if 1 ≤j≤\#a0}\\ s(l,Y)(y_j)& \text{if \#a0<j≤\#a0+n}\end{array}$$

Full text

Is it true that given positive integers $k<m$ and $N$ such that $k$ divides $m$, there exists a number $n>m$ such that $m$ divides $n$, satisfying that for every coloring $(n)^{k}_{\hom}=C_{1}\cup\dots\cup C_{N}$ we can choose $u\in(n)^{m}_{\hom}$ such that $\{t\in(n)^{k}_{\hom}:t\mbox{ is coarser than }u\}\subseteq C_{i}$ for some $i$?

In this note we give a positive answer to that question in Theorem 1 below. This result turns out to be a homogeneous version of the finite Dual Ramsey Theorem of Graham-Rothschild [3]. As explained in [4], our result also proves that the class $\mathcal{OMBA}_{\mathbb{Q}_{2}}$ of naturally ordered finite measure algebras with measure taking values in the dyadic rationals has the Ramsey property.

1. Homogeneous Dual Ramsey Theorem

1.1. Homogeneous partitions.

Let $k<n$ be positive integers. As customary, write $k|n$ if $k$ divides $n$, and in that case let

[TABLE]

that is, the set of all the partitions of $\{1,\dots,n\}$ into $k$ classes of the same cardinality. If $t$ and $u$ are partitions of $\{1,\dots,n\}$, not necessarily homogeneous, then we say that $t$ is coarser than $u$ if every class of $u$ is a subset of some class of $t$. Denote by $(u)^{k}_{\hom}$ the set of all homogeneous $k$-partitions of $\{1,\dots,n\}$ that are coarser than $u$.

Remark 1**.**

In [4], the notation $EQ_{k}^{\hom}$ is used to denote $(n)^{k}_{\hom}$.

The following is our main result.

Theorem 1** (Homogeneous Dual Ramsey Theorem).**

Let integers $k<m$ and $N$ be given, with $k|m$. There exists a positive integer $n>m$ such that $m|n$, satisfying that for every coloring $(n)^{k}_{\hom}=C_{1}\cup\dots\cup C_{N}$ there exists $u\in(n)^{m}_{\hom}$ such that $(u)^{k}_{\hom}\subseteq C_{i}$, for some $i\in\{1,\dots,N\}$.

We will prove Theorem 1 in Section 2 below. In our proof, we will make use of the infinite Dual Ramsey Theorem due to Carlson and Simpson [1], which we will state in the next subsection.

1.2. Dual Ramsey Theorem.

We will essentially go back to the notation introduced in [1]. Let $\mathbb{N}=\{1,2,3,\dots\}$ be the set of positive integers and let $(\omega)^{\omega}$ be the set of all the infinite partitions $A=\{A_{i}:i\in\mathbb{N}\}$ of $\mathbb{N}$ into infinite classes such that

[TABLE]

Given $A,B\in(\omega)^{\omega}$, we say that $A$ is coarser than $B$ if every class in $B$ is a subset of some class in $A$. Pre-order $(\omega)^{\omega}$ as follows:

[TABLE]

Likewise, given a positive integer $k$, let $(\omega)^{k}$ be the set of all the partitions $X=\{X_{i}:1\leq i\leq k\}$ of $\mathbb{N}$ into $k$ infinite classes satisfying

[TABLE]

Given $X,Y\in(\omega)^{k}$ and $A\in(\omega)^{\omega}$, define “$X$ is coarser than $Y$” (resp. “$X$ is coarser than $A$”) in the same way as above; and write $X\leq Y$ (resp. $X\leq A$). For $A\in(\omega)^{\omega}$, let $(A)^{k}=\{X\in(\omega)^{k}:X\leq A\}$.

We will regard the sets $(\omega)^{\omega}$ and $(\omega)^{k}$ as topological spaces with the usual topologies as defined in [1].

Theorem 2** (Infinite Dual Ramsey Theorem; Carlson-Simpson [1]).**

Let $k$ be a positive integer. For every finite Borel-measurable coloring of $(\omega)^{k}$ there exists $A\in(\omega)^{\omega}$ such that $(A)^{k}$ is monochromatic.

For positive integers $k$ and $n$, let

[TABLE]

i.e., partitions of $\{1,\dots,n\}$ into $k$ classes. For every such partition $u$, denote by $(u)^{k}$ the set of all $k$-partitions of $\{1,\dots,n\}$ that are coarser than $u$. Also, let $(<\omega)^{k}=\bigcup_{n\in\mathbb{N}}(n)^{k}$.

Theorem 1 is a homogeneous version of the following well-known result due to Graham and Rothschild, which can be obtained as a corollary of Theorem 2:

Theorem 3** (Dual Ramsey Theorem; Graham-Rothschild [3]).**

Let integers $k<m$ and $N$ be given. There exists a number $n>m$ satisfying that for every coloring $(n)^{k}=C_{1}\cup\dots\cup C_{N}$ there exists $u\in(n)^{m}$ such that $(u)^{k}\subseteq C_{i}$ for some $i\in\{1,\dots,N\}$.

2. Proof of Theorem 1

Let $A=\{A_{i}:i\in\mathbb{N}\}\in(\omega)^{\omega}$ be given. We will borrow some notation from [5]. Define

[TABLE]

Note that $r(i,A)$ is a partition of $\{1,\dots,\min(A_{i})\}$. We think of it as the $i$-th approximation of $A$. If a finite partition $b$ is the $i$-th approximation of some $B\leq A$, we write $b\in(<\omega)^{i}\upharpoonright A$.

It will be useful to understand elements $A=\{A_{i}:i\in\mathbb{N}\}\in(\omega)^{\omega}$ as surjective functions$A:\mathbb{N}\rightarrow\mathbb{N}$ where $A_{i}=A^{-1}(\{i\})$. Likewise, elements $X=\{X_{i}:1\leq i\leq k\}\in(\omega)^{k}$ will be as well understood as surjective functions $X:\mathbb{N}\rightarrow\mathbb{\{}1,\dots k\}$ where $X_{i}=X^{-1}(\{i\})$, for $1\leq i\leq k$. In the same spirit, we will regard partitions $u$ of a finite set $F\neq\emptyset$ into $k$ classes as surjections $u:F\rightarrow\{1,\dots,k\}$. We will shift between understanding partitions as surjective functions and sets of disjoint classes throughout the rest of this article. Now, keep that in mind and fix a positive integer $k$ for a while. We will borrow some notation and ideas from [2]. Given $A\in(\omega)^{\omega}$, let $\pi:(\omega)^{\omega}\rightarrow(\omega)^{k}$ be defined as follows:

[TABLE]

Notice that $\pi$ is a surjection. We understand $\pi$ as a projection function from $(\omega)^{\omega}$ onto $(\omega)^{k}$. Following [2], define an approximation function $s$ with domain $\{0\}\cup\mathbb{N}\times(\omega)^{k}$ as follows. For $i\in\{0\}\cup\mathbb{N}$ and $X=\{X_{1},\dots,X_{k}\}\in(\omega)^{k}$, the output $s(i,X)$ is a $k$-partition of some finite subset of $\mathbb{N}$ whose $j$-th class, $s(i,X)_{j}$ ($1\leq j\leq k$), will be defined by cases as follows:

For $i\leq k$, let

[TABLE]

So, for $i\leq k$, $s(i,X)_{j}$ is the unique initial segment of $X_{j}$ included as a subset in $\{1,\dots,\min(X_{k})\}$. Note that for any $A\in(\omega)^{\omega}$ such that $\pi(A)=X$, if $i\leq k$ we have $s(i,X)=r(i,A)$. In particular, $s(0,X)=\emptyset$ and for $0<i\leq k$, $s(i,X)$ is a partition of $\{1,\dots,\min(X_{i})\}$.

For $i>k$, let

[TABLE]

i.e., for $i>k$, we obtain the class $s(i,X)_{j}$ by adding to $s(i-1,X)_{j}$ the minimum element of $X_{j}$ that is not an element of $s(i-1,X)_{j}$. (Therefore, in this case, $s(i,X)_{j}$ is as well an initial segment of $X_{j}$ like in the case $i\leq k$). This completes the definition of the approximation function $s$. We think of $s(i,X)$ as the $i$-th approximation of $X$. Actually, each $X\in(\omega)^{k}$ can be identified with the sequence $(s(i,X))_{i\in\mathbb{N}}$ of its approximations.

Let $\mathcal{S}_{k}$ denote the range of $s$. We have

[TABLE]

Now, note that for every $a\in(<\omega)^{k}$ the union of $a$ is an initial segment of $\mathbb{N}$. That is not necessarily true for all elements of $\mathcal{S}_{k}$. Given $b\in\mathcal{S}_{k}$, if the union of $b$ is an initial segment of $\mathbb{N}$, then denote by $\#b$ the unique $l\in\mathbb{N}$ such that $b$ is a partition of $\{1,\dots,l\}$. Denote by $\cup b$ the union of $b$. Let

[TABLE]

If $a\in(<\omega)^{k}$ and $b\in\mathcal{S}_{k}$, write

[TABLE]

if there exist $X\in(\omega)^{k}$ and an integer $i>k$ such that $s(k,X)=a$ and $s(i,X)=b$. Now, let an integer $n$ with $k|n$ and $a\in(<\omega)^{k}$ be given. There exists a bijective correspondence between $(n)^{k}_{\hom}$ and the set

[TABLE]

To see that such a correspondence exists, fix a bijective function

[TABLE]

Now, given $t\in(n)^{k}_{\hom}$, define $b_{t}\in T(a,k,n)$ by

[TABLE]

Then the correspondence

[TABLE]

is bijective. Obviously, the inverse of a partition $b\in T(a,k,n)$ under this correspondence is the partition $t\in(n)^{k}_{\hom}$ defined by

[TABLE]

From now on, for all positive integers $k<n$ with $k|n$ and every $a\in(<\omega)^{k}$ we will fix one such bijective correspondence $t\mapsto b_{t}$ between $(n)^{k}_{\hom}$ and $T(a,k,n)$ throughout the rest of this section.

Now we are ready to prove our main result.

Proof of Theorem 1.

Fix positive integers $k<m$ and $N$, with $k|m$, and suppose that the conclusion in the statement fails. For each positive integer $n>m$ such that $m|n$, choose a coloring $(n)^{k}_{\hom}=C^{n}_{1}\cup\dots\cup C^{n}_{N}$ admitting no monochromatic set of the form $(u)^{k}_{\hom}$ with $u\in(n)^{m}_{\hom}$.

Given $X\in(\omega)^{k}$, let $a=s(k,X)$. Let

[TABLE]

Here, for $i>m$, $|\cup s(i,X)|$ denotes the cardinality of the union of the partition $s(i,X)$. Define the positive integer

[TABLE]

Note that $k|n(X)$. List the elements of  $\cup s(i_{0},X)$ in their natural increasing order as$x_{1}<\dots<x_{\#a+n(X)}$, and let $b_{X}\in T(a,k,n(X))$ be defined by

[TABLE]

Denote by $t(b_{X})$ the unique element $t\in(n(X))^{k}_{\hom}$ such that $b_{t}=b_{X}$. Define a coloring $(\omega)^{k}=C_{1}\cup\dots\cup C_{N}$ as follows:

[TABLE]

Note that each $C_{j}$ is Borel. So by Theorem 2 there exists $A\in(\omega)^{\omega}$ such that $(A)^{k}\subseteq C_{j_{0}}$ for some $j_{0}\in\{1,\dots,N\}$. Fix $a_{0}\in(<\omega)^{m}\upharpoonright A$ and let $Y\in(A)^{m}$ be such that $s(m,Y)=a_{0}$. Let

[TABLE]

and set $n=|\cup s(l,Y)|-\#a_{0}$. List the elements of  $\cup\,s(l,Y)$ in their natural increasing order as $y_{1}<\dots<y_{\#a_{0}+n}$. Define $b_{Y}\in T(a_{0},m,n)$ by

[TABLE]

Now denote by $u$ the unique element $t\in(n)^{m}_{\hom}$ such that $b_{t}=b_{Y}$.

Finally, for $t\in(u)^{k}_{\hom}$, define $X\in(A)^{k}$ with $X\leq Y$ by merging classes in $Y$ according to how the corresponding classes in $u$ were merged to build $t$. Let $a=s(k,X)$ and note that $n(X)=n$. If $b_{t}$ is the unique element of $T(a,k,n)$ corresponding to $t$, then $b_{X}=b_{t}$ and therefore $t=t(b_{X})$. Thus, $t\in C^{n}_{j_{0}}$, by the choice of $A$. Since $t$ was arbitrary, we get $(u)^{k}_{\hom}\subseteq C^{n}_{j_{0}}$. A contradiction. This completes the proof. ∎