The closed ordinal Ramsey number $R^{cl}(\omega^2,3) = \omega^6$

TL;DR

This paper determines the exact value of a topological variant of the ordinal Ramsey number for specific ordinals, advancing understanding in ordinal combinatorics.

Contribution

It provides the first precise calculation of the closed ordinal Ramsey number for , , a significant step in ordinal Ramsey theory.

Findings

Exact value of R^{cl}(\u03c9^2,3) = 

Advances the understanding of topological ordinal Ramsey numbers

Establishes a new benchmark for future ordinal Ramsey calculations

Abstract

Closed ordinal Ramsey numbers are a topological variant of the classical (ordinal) Ramsey numbers. We compute the exact value of the closed ordinal Ramsey number $R^{cl}(\omega^2,3) = \omega^6$.

Equations

$$\alpha ={\omega }^{{\gamma }_1}\cdot m_1+{\omega }^{{\gamma }_2}\cdot m_2+\cdots +{\omega }^{{\gamma }_l}\cdot m_l.$$

$$\{\{\omega \cdot k,\omega \cdot l+n\}:l>k>n>0\}\cup \{\{\omega \cdot k+k^{\prime },\omega \cdot l+l^{\prime }\}:k<l,l^{\prime }>k^{\prime }>0\}$$

$$E_1^n=\{(\alpha ,\beta )\in L_{n-1}\times L_n|\alpha {⊲}^{\ast }\beta \}$$

$$E_2^n=\{(\alpha ,\beta )\in L_n\times L_{n-2}|\alpha <\beta \}$$

$$E_3^n=\{(\alpha ,\beta )\in L_{n-3}\times L_n|\alpha <\beta ,\alpha /{<}^{\ast }\beta ,\max \{{\beta }_i\}<{\alpha }_{n-1}\}$$

$$E_4^n=\{(\alpha ,\beta )\in L_{n-4}\times L_n|\alpha <\beta ,\alpha /{<}^{\ast }\beta ,\max \{{\beta }_i\}>{\alpha }_{n-1}+{\alpha }_{n-2}\}$$

Full text

1. introduction

For ordinals $\beta$ and $\alpha$ write $\beta\to_{cl}(\alpha_{0},\alpha_{1})^{2}$ to mean that for every pair-colouring $\mathfrak{c}:[\beta]^{2}\to\{0,1\}$ there exist some $i\in\{0,1\}$ and $X\subseteq\beta$ of order type $\operatorname{\mathit{ord}}(X)=\alpha_{i}$ such that $X$ is closed in its supremum, and $[X]^{2}\subseteq\mathfrak{c}^{-1}(\{i\})$. Should such an ordinal exist, let $R^{cl}(\alpha_{0},\alpha_{1})$ denote the least such ordinal. Call $R^{cl}(\alpha_{0},\alpha_{1})$ the closed ordinal Ramsey number of $(\alpha_{0},\alpha_{1})^{2}$.

Caicedo and Hilton [CH17, Section 7] proved the upper bound $R^{cl}(\omega^{2},k)\leq\omega^{\omega}$, for every natural $k>0$. For $k=3$, the existing lower bound $R^{cl}(\omega^{2},3)\geq\omega^{3}$ is a consequence of [CH17, Proposition 3.1]. In this paper, we will calculate the exact value $R^{cl}(\omega^{2},3)=\omega^{6}$.

We achieve the bound $R^{cl}(\omega^{2},3)\leq\omega^{6}$ by a combinatorial analysis of any arbitrary “canonical” pair-colouring of $\omega^{6}$ in two colours. Canonical colourings were presented and discussed in [Mer19], where it was shown that, for our purposes, every pair-colouring can be assumed to be canonical.

The bound $R^{cl}(\omega^{2},3)\geq\omega^{6}$ is achieved by proving the more general result: for every natural $k$, $R^{cl}(\omega^{k+1},3)\geq\omega^{5k+1}$. This result is given by a single colouring $\mathfrak{c}:[\omega^{\omega}]^{2}\to\{0,1\}$ such that for each $k\in\omega$ and $\theta<\omega^{5k+1}$, the restriction $\mathfrak{c}\upharpoonright[\theta]^{2}$ demonstrates $R^{cl}(\omega^{k+1},3)>\theta$.

For a history of partition relations and Rado’s arrow notation see [HL10]. The ordinal partition calculus was introduced by Erdős and Rado in [ER56], and topological partition calculus was considered by Baumgartner in [Bau86]. Baumgartner’s work was continued in recent papers on topological (closed) ordinal partition relations by Hilton, Caicedo-Hilton, and Piña, see [Hil16],[CH17], and the sequence of works starting with [Pn15]. See also [OAW19] and the author’s [Mer19].

2. preliminaries

We use lowercase greek letters $\alpha,\beta,\gamma,\dots$ to denote ordinals. For $B$ a set of ordinals, write $\operatorname{\mathit{ord}}(B)$ for the order-type of $B$. Write $A\mathrel{\subseteq_{\rm cof}}B$, to mean that $A$ is a cofinal subset of $B$.

For any nonzero $\alpha$ there exist a unique $l\in\mathds{N}$, a sequence of ordinals $\gamma_{1}>\dots>\gamma_{l}$, and a sequence of nonzero natural numbers $m_{1},\dots,m_{l}$ such that

[TABLE]

Call this representation of $\alpha$ the Cantor normal form of $\alpha$. The Cantor-Bendixson rank (CB rank) of $\alpha$ is $\gamma_{l}$, and is denoted $\operatorname{CB}(\alpha)$. For the ordinal $\alpha=0$, we define $\operatorname{CB}(0)=0$.

We say that $\beta<^{*}\alpha$ whenever $\alpha=\beta+\omega^{\gamma}$ for some nonzero ordinal $\gamma$ with $\gamma>\operatorname{CB}(\beta)$. Equivalently, for some $\gamma>\operatorname{CB}(\beta)$, $\alpha$ is the least ordinal of $\operatorname{CB}$ rank $\gamma$ with $\beta\leq\alpha$. We write $\beta\lhd^{*}\alpha$ if $\alpha$ is the unique immediate successor of $\beta$ in $<^{*}$. Denote $\operatorname{T}(\alpha)=\{\alpha\}\cup\{\beta\ |\ \beta<^{*}\alpha\}$ and $\operatorname{T}^{=n}(\alpha)=\{\beta\in\operatorname{T}(\alpha)\ |\ \operatorname{CB}(\beta)=n\}$. If $\operatorname{CB}(\alpha)$ is a successor ordinal, denote $\operatorname{Fan}^{-}(\alpha)=\{\beta\ |\ \beta\lhd^{*}\alpha\}$.

It is useful to visualize $\omega^{k}+1$ under the $\lhd^{*}$ relation as an $\omega$-regular (bar the terminal nodes), rooted, directed tree of height $k+1$. The root is $\omega^{k}$, the unique point of CB rank $k$. The descendants of the root are $\{\omega^{k-1}\cdot i:i\in\omega\}$, all the points of CB rank $k-1$, and so on. The leaves are the points of CB rank [math]. In line with the standard order on ordinals, it is preferable to visualize the root as being on top and the leaves on the bottom. Then, the $n$-th level corresponds to the points of CB rank $n$. See Figure 1 for a visualization.

It is advisable that the reader takes a moment to locate in the figure the objects $\operatorname{T}(\alpha)$, $\operatorname{T}^{=n}(\alpha)$, $\operatorname{Fan}^{-}(\alpha)$ for some $\alpha\leq\omega^{3}$, $n\leq\operatorname{CB}(\alpha)$. Another suggestion is to find in the figure a few copies of $\omega^{2}$ – some closed in their supremum and some not.

When we “thin out” a set of ordinals $X$ of order type $\omega^{k}$, we mean that we take $Y\subseteq X$ such that $(Y,<^{*})$ is isomorphic to $(X,<^{*})$. Preserving the relation $<^{*}$ guarantees that if $X$ was closed in its supremum, then also $Y$ is, and furthermore $\operatorname{\mathit{ord}}(Y)=\operatorname{\mathit{ord}}(X)$. Unlike the actual order on the points, the order type $\omega^{k}$ can be read off of $<^{*}$.

Let $G=(V,E)$ be a graph. We only consider graphs where the edge relation is symmetric. We identify the graph $G$ with a colouring $\mathfrak{c}:[V]^{2}\to\{0,1\}$ by taking $\mathfrak{c}(v,u)=1$ if and only if $(v,u)\in E$. For $v\in V$, denote $N(v)=\{u\in V\ |\ (v,u)\in E\}$. For $U\subseteq V$, denote $N(U)=\bigcup_{v\in U}N(v)$. We say that a set of vertices $U\subseteq V$ is a clique if $[U]^{2}\subseteq E$ and that it is independent if $[U]^{2}\cap E=\emptyset$.

Definition 1**.**

Let $G=(\delta,E)$ be some graph on an ordinal $\delta$. Let $A,B\subseteq\delta$ be infinite, disjoint and without maxima.

• •

Write $A\mathrel{\perp}B$ to mean that for all $X$, if $X\mathrel{\subseteq_{\rm cof}}A$, then $B\setminus N(X)$ is finite.

• •

Write $A\mathrel{{}_{\omega}{\perp}}B$ if $A\mathrel{\perp}B$ and in addition $N(a)\cap B$ is finite for every $a\in A$.

Lemma 2** ([Mer19], Lemma 4.2).**

Let $G=(\delta,E)$ for $\delta$ countable. Let $A,B\subseteq\delta$ be such that $A\mathrel{{}_{\omega}{\perp}}B$. Then there is some $A_{0}\mathrel{\subseteq_{\rm cof}}A$ and some $B_{0}\subseteq B$, cofinite in $B$, such that $N(b)\cap A_{0}$ is cofinite in $A_{0}$, for all $b\in B_{0}$.

For the full definition of a canonical colouring, $\widetilde{\mathfrak{c}}$ and $\hat{\mathfrak{c}}$, refer to subsection 2.2 and section 3 of [Mer19]. In particular, see definitions 3.3, 3.9 and 3.10 there. For this paper, we specialize the definition and results we use to the special case of a colouring of $\omega^{k}$ in two colours, for a natural $k$.

Fix $\mathfrak{F}$ to be the filter of cofinite subsets of $\omega$. Define $\mathfrak{F}^{1}=\mathfrak{F}$ and $\mathfrak{F}^{n+1}$ on $\omega^{n+1}$ inductively by taking the product filter on $\omega^{n+1}\cong\omega\times\omega^{n}$. That is, $X\in\mathfrak{F}^{n+1}$ if and only if $\{\alpha\in\omega:\{\beta\in\omega^{n}:(\alpha,\beta)\in X\}\in\mathfrak{F}^{n}\}\in\mathfrak{F}^{1}$. For $X$, a set of ordinals with $\operatorname{\mathit{ord}}(X)=\omega^{k}$ witnessed by $\rho:X\to\omega^{k}$, we say that $Y\subseteq X$ is a $k$-large set in $X$ if $\rho[Y]\in\mathfrak{F}^{k}$. If $k$ is clear from context, we may omit it and simply say that $Y$ is large in $X$.

Say that a graph (colouring) on $\omega^{k}$ is canonical if

1. (i)

For every $\theta\leq\omega^{k}$, $\alpha<\omega^{k}$ and $l\leq k$, either $\operatorname{T}^{=l}(\theta)\cap N(\alpha)$ or $\operatorname{T}^{=l}(\theta)\setminus N(\alpha)$ is an $l$-large set in $\operatorname{T}^{=l}(\theta)$. 2. (ii)

Additionally, if $\alpha=\theta$, then the $l$-large set in $\operatorname{T}^{=l}(\theta)$ above is the entirety of $\operatorname{T}^{=l}(\theta)$. 3. (iii)

Finally, for $\theta=\omega^{k}$, whether the $l$-large set in $\operatorname{T}^{=l}(\theta)$ is contained in $N(\alpha)$ or disjoint from $N(\alpha)$ is determined only by $\operatorname{CB}(\alpha)$.

For $k>j>l$, denote by $\hat{\mathfrak{c}}(j,l)\in\{0,1\}$ the “colour” by which every $\alpha\in\operatorname{T}^{=j}(\omega^{k})$ is connected to every $\beta\in\operatorname{T}^{=l}(\alpha)$.

For $k>j,l$, denote by $\widetilde{\mathfrak{c}}(j,l)\in\{0,1\}$ the “colour” by which every $\alpha\in\operatorname{T}^{=j}(\omega^{k})$ is connected to an $l$-large set in $\operatorname{T}^{=l}(\omega^{k})$.

Example 3**.**

The following is the edge-set of a canonical graph on $\omega^{2}$:

[TABLE]

For this graph: $\hat{\mathfrak{c}}(1,0)=0$, $\widetilde{\mathfrak{c}}(1,1)=0$, $\widetilde{\mathfrak{c}}(0,0)=1$, $\widetilde{\mathfrak{c}}(1,0)=0$, $\widetilde{\mathfrak{c}}(0,1)=0$.

Remark 4**.**

In [Mer19], the filter $\mathfrak{F}^{n}$ was defined smaller, hence a canonical colouring there is more restrictive. In this paper, we will not need that extra strength.

The following theorem allows us, for our purposes, to assume that every arbitrary colouring we encounter is canonical.

Theorem 5** ([Mer19], Proposition 3.11).**

For every natural $k$ and colouring $\mathfrak{c}:[\omega^{k}]^{2}\to\{0,1\}$, there exists $X\subseteq\omega^{k}$, a subset of $\omega^{k}$ close in its supremum of order type $\omega^{k}$, such that the restriction of $\mathfrak{c}$ to $X$ is a canonical colouring.

Lemma 6** ([Mer19], Lemma 4.3).**

Fix some canonical triangle-free $G=(\delta,E)$, where $\delta=\omega^{k}$ for some $k$ natural, with corresponding colouring $\mathfrak{c}:[\delta]^{2}\to 2$. If there exists no independent $X\subseteq\delta$ closed in its supremum with $\operatorname{\mathit{ord}}(X)=\omega^{2}$, then the following statements hold

1. (1)

*For a fixed $l$, there is at most one $j<l$ such that $\hat{\mathfrak{c}}(j,l)=1$. * 2. (2)

*For a fixed $j$, there is at most one $l$ such that $\widetilde{\mathfrak{c}}(j,l)=1$. * 3. (3)

For a fixed $l$, there is at most one $j$ such that $\widetilde{\mathfrak{c}}(j,l)=1$.

Remark 7**.**

In the full definition of a canonical colouring of some arbitrary $\delta<\omega^{\omega}$, the functions $\hat{\mathfrak{c}}$, $\widetilde{\mathfrak{c}}$ take additional parameters besides the $\operatorname{CB}$-rank of the ordinals — indices of summands in the Cantor normal form of $\delta$. When colouring $\omega^{k}$, however, there is a unique summand in the Cantor normal form. With respect to the notation of the full definition, in this text $\hat{\mathfrak{c}}(j,l)$, $\widetilde{\mathfrak{c}}(j,l)$ are shorthand for $\hat{\mathfrak{c}}(1,j,l)$, $\widetilde{\mathfrak{c}}(1,j;1,l)$, respectively.

3. Upper bound

Lemma 8**.**

Let $G=(\delta,E)$ be a triangle free graph on some ordinal $\delta$. Let $A\subseteq\delta$ with $\operatorname{\mathit{ord}}(A)=\omega^{k}$ and let $B\subseteq\delta$ with $\operatorname{\mathit{ord}}(B)=\omega$ be such that $A\mathrel{{}_{\omega}{\perp}}B$. Then there exists some $b\in B$ such that $\operatorname{\mathit{ord}}(N(b)\cap A)=\omega^{k}$.

Proof.

For each $a\in A$, let $m_{a}=\min(N(a)\cap B)$.

Assume that for each $b\in B$ the set $\{a\in A\ |\ m_{a}>b\}$ is cofinal in $A$. Then whenever $Y\subseteq A$ is finite, there exists some arbitrarily large $a\in A$ such that $\{b\in B\ |\ \max(N(Y)\cap B)<b<m_{a}\}$ is not empty. Thus, extending $Y$ at each stage by such a sufficiently large element $a$, we can construct inductively a cofinal set $X\subseteq A$ such that $B\setminus N(X)$ is infinite in $B$. This contradicts $A\mathrel{{}_{\omega}{\perp}}B$.

Therefore, there must exist some $M\in B$ such that $\{a\in A\ |\ m_{a}>M\}$ is not cofinal in $A$. In particular, the set $\{a\in A\ |\ m_{a}\leq M\}$ contains a set of order type $\omega^{k}$. By additive indecomposability of $\omega^{k}$, there is some $b\leq M$ such that $\{a\in A\ |\ m_{a}=b\}$ is of order type $\omega^{k}$. So $\operatorname{\mathit{ord}}(N(b)\cap A)=\omega^{k}$. ∎

Fact 9** (Specker, [Spe57]).**

$\omega^{2}\to(\omega^{2},3)$, i.e., every triangle-free graph on $\omega^{2}$ contains an independent set of order type $\omega^{2}$.

Proposition 10**.**

$R^{cl}(\omega^{2},3)\leq\omega^{6}$**

Proof.

Let $G$ be a triangle-free canonical graph on $\omega^{6}$. Assume that

• $(*)$

There is no independent copy of $\omega^{2}$ closed in its supremum in $G$.

By Lemma 6

• •

There is at most one $t<5$ such that $\hat{\mathfrak{c}}(5,t)=1$;

• •

There is at most one $t<5$ such that $\widetilde{\mathfrak{c}}(5,t)=1$;

• •

There is at most one $t<5$ such that $\widetilde{\mathfrak{c}}(t,5)=1$.

Thus, there are $t_{1}<t_{2}<5$ such that $\hat{\mathfrak{c}}(5,t_{j})=\widetilde{\mathfrak{c}}(t_{j},5)=\widetilde{\mathfrak{c}}(5,t_{j})=0$ for $j\in\{1,2\}$. For each $i\in\omega$, denote $h_{i}=\omega^{5}\cdot i$ and $W^{t_{j}}_{i}=\operatorname{T}^{=t_{j}}(h_{i})$.

Since $\widetilde{\mathfrak{c}}(5,t_{j})=0$, for each $i\in\omega$ and $j\in\{1,2\}$, there are at most finitely many $k>i$ such that $W^{t_{j}}_{k}\setminus N(h_{i})$ is not large in $\operatorname{T}^{=t_{j}}(h_{k})$. Therefore, for every finite $I\subseteq\omega$, there are infinitely many $k\in\omega$ such that, for every $i\in I$, the set $W^{t_{j}}_{k}\setminus N(h_{i})$ is large in $\operatorname{T}^{=t_{j}}(h_{k})$. Iteratively choosing in this manner a subset of $\{h_{i}\ |\ i\in\omega\}$, we may assume that $W^{t_{j}}_{k}\setminus N(h_{i})$ is large in $\operatorname{T}^{=t_{j}}(h_{k})$, for all $i<k\in\omega$ and $j\in\{1,2\}$.

Now, as each $k\in\omega$ has finitely many $i<k$, and for each $i<k$ the set $W^{t_{j}}_{k}\setminus N(h_{i})$ is large in $\operatorname{T}^{=t_{j}}(h_{k})$, we may thin out $\operatorname{T}(h_{k})$ so that $W^{t_{j}}_{k}$ avoids $N(\{h_{i}\ |\ i<k\})$. Hence, we may assume there are no edges between $\{h_{i}\}$ and $W^{t_{j}}_{k}$, for all $i<k\in\omega$ and $j\in\{1,2\}$.

Claim. Fix $j\in\{1,2\}$. Then $G$ can be thinned out so that for every $k\in\omega$, it holds that $W^{t_{j}}_{k}\mathrel{\perp}\{h_{i}\ |\ i>k\}$.

proof of Claim.

Assume the contrary.

We will construct inductively $k_{n},X_{n},I_{n}$ such that at every stage:

1. (1)

$I_{n}\subseteq\omega$, $k_{n}\in I_{n}$, $X_{n}\mathrel{\subseteq_{\rm cof}}W^{t_{j}}_{k_{n}}$; 2. (2)

$I_{n+1}\subseteq\{i\in I_{n}\ |\ i>k_{n}\}$; 3. (3)

$N(X_{n})\cap\{h_{i}\ |\ i\in I_{n+1}\}=\emptyset$.

Set $I_{0}=\omega$. We describe the inductive step, given some $I_{n}\subseteq\omega$:

Thin out $G$ so that $\operatorname{Fan}^{-}(\omega^{6})=\{h_{i}\ |\ i\in I_{n}\}$. By assumption, there is some $k\in I_{n}$ such that $W^{t_{j}}_{k}\not\mathrel{\perp}\{h_{i}\ |\ i>k\}$. Fix $k_{n}$ to be such a $k$ and let $X_{n}\mathrel{\subseteq_{\rm cof}}W^{t_{j}}_{k_{n}}$ be of order type $\omega$ such that $\{h_{i}\ |\ i\in I_{n}\}\setminus N(X_{n})$ is infinite. Define $I_{n+1}=\{i\in I_{n}\ |\ i>k_{n},~{}h_{i}\notin N(X_{n})\}$.

Now, let $Y=\{h_{k_{n}}\ |\ n\in\omega\}$ and let $X=\bigcup_{n\in\omega}X_{n}$. Observe that $X\cup Y$ is a copy of $\omega^{2}$ closed in its supremum, with no edges crossing between $X$ and $Y$. By Ramsey’s theorem and triangle-freeness, $Y$ has an infinite independent subset. By thinning out $X\cup Y$, we may assume that already $Y$ is independent. By Fact 9, $X$ has an independent subset of order type $\omega^{2}$. Again by thinning out $X\cup Y$ we may assume $X$ is independent. Hence, by construction, $X\cup Y$ is independent in contradiction to $(*)$. ∎

Assume that $G$ was thinned out as guaranteed by the claim. Observe that since $\widetilde{\mathfrak{c}}(t_{j},5)=0$, in fact $W^{t_{j}}_{k}\mathrel{{}_{\omega}{\perp}}\{h_{i}\ |\ i>k\}$ for any $j\in\{1,2\}$ and $k\in\omega$.

Fix some $k$. Choose arbitrarily some $A_{2}\subseteq W^{t_{2}}_{k}$ and $A_{1}\subseteq W^{t_{1}}_{k}$ such that both $A_{1}$ and $A_{2}$ are cofinal in $\operatorname{T}(h_{k})$, and $A_{1}\cup A_{2}$ is a copy of $\omega^{2}$ closed in its supremum.

Since $A_{2}\mathrel{{}_{\omega}{\perp}}\{h_{i}\ |\ i>k\}$, by Lemma 2 we may thin out so that $N(h_{i})\cap A_{2}$ is cofinite in $A_{2}$ for every $i>k$. By Lemma 8, there exists some $i>k$ such that $N(h_{i})\cap A_{1}$ is of order type $\omega^{2}$. Then $N(h_{i})\cap(A_{1}\cup A_{2})$ contains an independent copy of $\omega^{2}$ closed in its supremum, which concludes the proof. ∎

4. Lower bound

For any ordinal $\alpha<\omega^{\omega}$, for each $i$, denote by $\alpha_{i}\in\omega$ the coefficient of $\omega^{i}$ in the Cantor normal form of $\alpha$. That is, $\alpha=\sum_{i\in\omega^{*}}\omega^{i}\cdot\alpha_{i}$. Denote $L_{n}=\operatorname{T}^{=n}(\omega^{\omega})$. For every natural $n$, consider the sets of edges:

[TABLE]

Let $G_{\omega}=(V_{\omega},E_{\omega})$ be the graph on $V_{\omega}=\omega^{\omega}$ with edges $E_{\omega}=\bigcup_{\begin{subarray}{c}n\in\omega\\ i\leq 4\end{subarray}}E^{n}_{i}\cup(E^{n}_{i})^{-1}$.

Lemma 11**.**

The graph $G_{\omega}$ is triangle-free.

Proof.

Assume to the contrary that $\{\alpha,\beta,\gamma\}$ is a triangle. Without loss of generality assume $\operatorname{CB}(\alpha)<\operatorname{CB}(\beta)<\operatorname{CB}(\gamma)=n$. Clearly $\operatorname{CB}(\alpha)\geq n-4$.

Consider the case $\operatorname{CB}(\beta)=n-1$, meaning $\beta\lhd^{*}\gamma$. We cannot have $\alpha<^{*}\gamma$, hence $\alpha\not<^{*}\beta$ and in particular $\operatorname{CB}(\alpha)\neq\operatorname{CB}(\beta)-1$. So $\operatorname{CB}(\alpha)<n-2$ and it must be that $\alpha<\gamma$. If $\beta<\alpha$, we will have $\beta<\alpha<\gamma$ and $\beta<^{*}\gamma$, implying $\alpha<^{*}\gamma$, which is false. So $\alpha<\beta$, resulting in $\operatorname{CB}(\alpha)\neq\operatorname{CB}(\beta)-2$. So $\operatorname{CB}(\alpha)=n-4$. Thus, $\max\{\beta_{i}\}<\alpha_{n-2}\leq\alpha_{n-1}+\alpha_{n-2}<\max\{\gamma_{i}\}$. But $\beta<^{*}\gamma$, therefore $\max\{\beta_{i}\}\geq\max\{\gamma_{i}\}-1$ in contradiction.

Consider the case $\operatorname{CB}(\beta)=n-2$. Then $\alpha<\gamma$ and $\gamma<\beta$. If $\alpha<^{*}\beta$, then we would have $\gamma<\alpha$, so $\alpha\not<^{*}\beta$ and $\operatorname{CB}(\alpha)\neq n-3$. Since $\beta\nless\alpha$, we have $\operatorname{CB}(\alpha)\neq n-4$. So $\operatorname{CB}(\alpha)$ cannot take any value, a contradiction.

We are left with $\operatorname{CB}(\beta)=n-3$ and $\alpha\lhd^{*}\beta$. But now $\alpha_{n-1}+\alpha_{n-2}<\max\{\gamma_{i}\}<\beta_{n-1}$, despite $\alpha_{n-1}=\beta_{n-1}$. We conclude that there are no triangles in $G_{\omega}$. ∎

Notation**.**

For $X\subseteq\omega^{\omega}$, denote $\operatorname{CB}(X)=\sup\{\operatorname{CB}(\alpha)\}_{\alpha\in X}$.

Lemma 12**.**

If $X\subseteq G_{\omega}$ is an independent copy of $\omega^{k+1}$, closed in its supremum, with $X$ not cofinal in $\omega^{\omega}$, then $\operatorname{CB}(X)\geq 5k$.

Proof.

We prove by induction on $k$. Let $X\subseteq G_{\omega}$ be an independent, closed in its supremum, copy of $\omega^{k+1}$ with $\sup X<\omega^{\omega}$. Let $\rho_{X}:\omega^{k+1}\to X$ be the bijection witnessing $\omega^{k+1}\cong X$. Due to $X$ being closed in its supremum, $\rho_{X}$ is continuous. Since $X$ is bounded, $\operatorname{CB}(X)$ is finite.

For each $i<k+1$, we may consider $\operatorname{CB}\circ\rho_{X}$ as a colouring of $\operatorname{T}^{=i}(\omega^{k+1})$ in $\operatorname{CB}(X)$ many colours. By iterating the pigeonhole principle, in $\operatorname{T}^{=i}(\omega^{k+1})$ there is a copy of $\omega^{k+1-i}$ on which $\operatorname{CB}\circ\rho_{X}$ is constant. By indecomposability of $\omega^{k+1-i}$, we may thin out $\omega^{k+1}$ (and $X$, accordingly) so that $\operatorname{CB}\circ\rho_{X}$ is constant on $\operatorname{T}^{=i}(\omega^{k+1})$. Iterating this thinning out $k+1$ times, we may assume $\operatorname{CB}\circ\rho_{X}$ is constant on $\operatorname{T}^{=i}(\omega^{k+1})$ for each $i<k+1$.

Denote $\alpha^{j}=\rho_{X}(\omega^{k}\cdot j)$, $n=\operatorname{CB}(\alpha^{1})$, $B_{j}=\rho_{X}[\operatorname{Fan}^{-}(\omega^{k}\cdot j)]$, and $m=\operatorname{CB}(\beta)$ for some $\beta\in B_{1}$. Observe that by continuity of $\rho_{X}$, whenever $A\subseteq\omega^{k+1}$ with $\sup A=a\in\omega^{k+1}$, the set $\rho_{X}[A]$ must intersect $\operatorname{T}(\rho_{X}(a))$.

Claim. $m\leq n-5$.

proof of claim.

By $\operatorname{Fan}^{-}(\alpha^{j})\subseteq N(\alpha^{j})$, it cannot be that $B_{j}$ intersects $\operatorname{Fan}^{-}(\alpha^{j})$. Therefore, $m=\operatorname{CB}(B_{j})\neq n-1$.

Similarly, since $\operatorname{T}^{=n-2}(\alpha^{j+1})\subseteq N(\alpha^{j})$, we have $m=\operatorname{CB}(B_{j+1})\neq n-2$.

For any $C\mathrel{\subseteq_{\rm cof}}\operatorname{T}^{=n-3}(\alpha^{j})$, the set $\{\gamma_{n-1}\ |\ \gamma\in C\}$ is unbounded in $\omega$. hence, there exists some $\gamma\in C$ with $\alpha^{j+1}\in N(\gamma)$. Thus, $B_{j}\cap\operatorname{T}^{=n-3}(\alpha^{j})$ cannot be a cofinal subset of $\operatorname{T}^{=n-3}(\alpha^{j})$ and so $m\neq n-3$.

Let $\gamma\in\operatorname{T}^{=n-4}(\alpha^{j})$. Since $X$ is not cofinal in $\omega^{\omega}$, there is some $r$ large enough such that $X\subseteq\operatorname{T}(\omega^{r})$. There are only finitely many elements $\alpha\in\operatorname{T}^{=n}(\omega^{r})$ with $\max\{\alpha_{i}\}\leq\gamma_{n-1}+\gamma_{n-2}$, so $N(\gamma)\cap\{\alpha^{j}\ |\ j\in\omega\}\neq\emptyset$. We conclude that $B_{j}$ is disjoint from $\operatorname{T}^{=n-4}(\alpha^{j})=\emptyset$ and so $m\neq n-4$. ∎

For each $j\in\omega$, the set $X\cap\operatorname{T}(\alpha^{j})\setminus\{\alpha^{j}\}$ contains an independent, closed in its supremum, copy of $\omega^{k}$, which is not cofinal in $\omega^{\omega}$. So by the induction hypothesis, $m\geq 5(k-1)$. Combined with the claim, this results in $n\geq 5k$. ∎

Corollary 13**.**

$R^{cl}(\omega^{k+1},3)\geq\omega^{5k+1}$**

Proof.

Consider the subgraph induced by $G_{\omega}$ on $\omega^{5k+1}=\operatorname{T}(\omega^{5k+1})\setminus\{\omega^{5k+1}\}$, call this subgraph $G$. Let $X$ be some independent, closed in its supremum copy of $\omega^{k+1}$ in $G$. Every final segment $X^{\prime}\subseteq X$ of $X$ contains an independent, closed in its supremum copy of $\omega^{k+1}$, so by the above lemma $\operatorname{CB}(X^{\prime})\geq 5k$. Then $X\mathrel{\subseteq_{\rm cof}}\omega^{5k+1}$ and in particular $X$ is not contained in the restriction of $G$ to any $\delta<\omega^{5k+1}$. Thus, for every $\delta<\omega^{5k+1}$, we have found a graph demonstrating $R^{cl}(\omega^{k+1},3)>\delta$. ∎

Theorem 14**.**

$R^{cl}(\omega^{2},3)=\omega^{6}$**

Proof.

Apply Corollary 13 to $k=1$ and combine with Proposition 10. ∎

5. Acknowledgments

The research was conducted at Ben-Gurion University of the Negev. The research was partially supported by ISF grant No. 181/16 and 1365/14.