Fact 2.5. Let f and g be problems.

1. 1. If $f \leq _{\mathrm {sW}} g$ , then $f' \leq _{\mathrm {sW}} g'$ . If $f \not \leq ^\ast _{\mathrm {W}} g$ , then $f' \not \leq ^\ast _{\mathrm {W}} g'$ .

2. 2. If $f \leq _{\mathrm {W}} g$ , then $\widehat {f} \leq _{\mathrm {W}} \widehat {g}$ .

3. 3. $\widehat {f \times g} \equiv _{\mathrm {sW}} \widehat {f} \times \widehat {g}$ .

Proof. The first part of (1) was proved in [Reference Brattka, Gherardi and Marcone4, Proposition 5.6]. The second part of (1) follows from [Reference Brattka, Hölzl and Kuyper6, Theorem 11] and the fact that every continuous function is computable relative to some $p \in \mathbb {N}^{\mathbb {N}}$ . (2) and (3) were proved in [Reference Brattka and Gherardi2, Propositions 4.2 and 4.5].

Proof. Let $f_0$ and $f_1$ be problems. To show that $f_0 \sqcap f_1 \leq ^{\mathrm {t}}_{\mathrm {W}} f_i$ , pre-process with $(p_0,p_1) \mapsto p_i$ , and post-process with $((j,q),(p_0,p_1)) \mapsto q$ . Suppose g is a problem such that $g \leq ^{\mathrm {t}}_{\mathrm {W}} f_0,f_1$ . For each i, fix $\Phi _i$ and $\Psi _i$ which witness $g \leq ^{\mathrm {t}}_{\mathrm {W}} f_i$ . Then $g \leq ^{\mathrm {t}}_{\mathrm {W}} f_0 \sqcap f_1$ : Pre-process with $p \mapsto (\Phi _0(p),\Phi _1(p))$ and post-process with

$$\begin{align*}((i,q),p) \mapsto \begin{cases} \Psi_i(q,p) & \text{if }i=0,1 \\ 0^{\mathbb{N}} & \text{otherwise}. \end{cases} \end{align*}$$

To show that $f_i \leq ^{\mathrm {t}}_{\mathrm {W}} f_0 \sqcup f_1$ , pre-process with $p \mapsto (i,p)$ and post-process with $((j,q),p) \mapsto q$ . Suppose g is a problem such that $f_0,f_1 \leq ^{\mathrm {t}}_{\mathrm {W}} g$ . For each i, fix $\Phi _i$ and $\Psi _i$ which witness $f_i \leq ^{\mathrm {t}}_{\mathrm {W}} g$ . Then $f_0 \sqcup f_1 \leq ^{\mathrm {t}}_{\mathrm {W}} g$ : Pre-process with

$$\begin{align*}(i,p) \mapsto \begin{cases} \Phi_i(p) & \text{if }i=0,1 \\ 0^{\mathbb{N}} & \text{otherwise} \end{cases} \end{align*}$$

and post-process with

$$\begin{align*}(q,(i,p)) \mapsto \begin{cases} \Psi_i(q,p) & \text{if }i=0,1 \\ 0^{\mathbb{N}} & \text{otherwise}. \end{cases} \end{align*}$$

Proof. (1): In all these desired reductions, we will pre-process with the identity. Fix $k\in \mathbb {N}_{\geq 2}\cup \{<\!\infty ,\mathbb {N},\infty \}$ . To show $\mathsf {RT}^{1}_{k}\mathsf {e}\leq ^{\mathrm {t}}_{\mathrm {W}}\mathsf {RT}^{1}_{k}\mathsf {pe}$ , post-process with the projection. For $\mathsf {RT}^{1}_{k}\mathsf {c}\leq ^{\mathrm {t}}_{\mathrm {W}}\mathsf {RT}^{1}_{k}\mathsf {e}$ , the post-processing takes the given colouring c and an enumeration f and returns $c(f(0))$ . If f enumerates an $\mathsf {RT}^{1}_{k}\mathsf {e}$ -solution to c, then $c(f(0))$ is an $\mathsf {RT}^{1}_{k}\mathsf {c}$ -solution to c. Finally, to see that $\mathsf {RT}^{1}_{k}\chi \leq ^{\mathrm {t}}_{\mathrm {W}}\mathsf {RT}^{1}_{k}\mathsf {c}$ , the post-processing takes the given colouring c and a number j and returns the characteristic function $\chi $ defined by $\chi (x) = 1$ if $c(x) = j$ , otherwise $\chi (x) = 0$ . If j is an $\mathsf {RT}^{1}_{k}\mathsf {c}$ -solution to c, then $\chi $ is an $\mathsf {RT}^{1}_{k}\chi $ -solution to c.

(2): In all these desired reductions, we will post-process with the projection. To show $\mathsf {RT}^{n}_{k}\chi \leq ^{\mathrm {t}}_{\mathrm {W}} \mathsf {RT}^{n}_{k+1}\chi $ , pre-process with the identity. To show $\mathsf {RT}^{n}_{k+1}\chi \leq ^{\mathrm {t}}_{\mathrm {W}} \mathsf {RT}^{n}_{<\infty }\chi $ , the pre-processing function produces the given colouring and the bound $k+1$ . To show $\mathsf {RT}^{n}_{<\infty }\chi \leq ^{\mathrm {t}}_{\mathrm {W}} \mathsf {RT}^{n}_{\mathbb {N}}\chi $ , pre-process by forgetting the given bound.

(3): First, we show that $\mathsf {RT}^{1}_{\mathbb {N}}\mathsf {e} \leq ^{\mathrm {t}}_{\mathrm {W}} \mathsf {RT}^{1}_{\mathbb {N}}\mathsf {c}$ and $\mathsf {RT}^{1}_{\infty }\mathsf {e} \leq ^{\mathrm {t}}_{\mathrm {W}} \mathsf {RT}^{1}_{\infty }\mathsf {c}$ . Given $c: \mathbb {N} \to \mathbb {N}$ , define $d: \mathbb {N} \to \mathbb {N}$ by taking $d(x)$ to be the least number ( $\leq x$ ) which has the same c-colour as x. If c has bounded range, so does d. Also, if c is an $\mathsf {RT}^{1}_{\infty }\mathsf {e}$ -instance (i.e., some colour appears infinitely often in c), then d is an $\mathsf {RT}^{1}_{\infty }\mathsf {c}$ -instance. The post-processing takes a colouring c and a colour $s_0$ and produces the following enumeration f:

$$\begin{align*}f(x) = \begin{cases} \max\{s \leq x: c(s) = c(s_0)\} & \text{if }x \geq s_0 \\ s_0 &\text{otherwise}. \end{cases} \end{align*}$$

If $s_0$ appears infinitely often in d, then infinitely many $s_i$ will be enumerated.

(4): First notice that for any given infinite c-homogeneous set H, the principal enumeration of H can compute the characteristic function of H, an enumeration of H (via the identity function) and the colour of H (by computing the first n elements of H). Therefore, by using such a scheme for the post-processing, and the identity map for the pre-processing, one may obtain that $\mathsf {RT}^{n}_{k}\mathsf {c},\mathsf {RT}^{n}_{k}\chi ,\mathsf {RT}^{n}_{k}\mathsf {e}\leq ^{\mathrm {t}}_{\mathrm {W}}\mathsf {RT}^{n}_{k}\mathsf {pe}$ .

(5): To show that $\mathsf {RT}^{n}_{\infty }\mathsf {c} \leq ^{\mathrm {t}}_{\mathrm {W}} \mathsf {RT}^{2}_{\infty }\mathsf {e}$ , let $\{p_{i}\}_{i\in \mathbb {N}}$ denote the primes in increasing order and $\lceil \cdot \rceil :[\mathbb {N}]^{<\mathbb {N}}\to \mathbb {N}$ be the function such that $\lceil A\rceil =\prod _{i\in A}p_{i}$ . This is evidently a primitive recursive embedding of $[\mathbb {N}]^{<\mathbb {N}}$ into $\mathbb {N}\setminus \{0\}$ . Given $c: [\mathbb {N}]^n \to \mathbb {N}$ , define $d(x,y) = 0$ if all of the following hold: (1) $x=\lceil A\rceil $ , $y=\lceil B\rceil $ for some $A,B$ finite subsets of $\mathbb {N}$ ; (2) $A \subseteq B$ or $B \subseteq A$ ; (3) A and B both have size at least n; and (4) A and B are c-homogeneous. Otherwise, define $d(x,y) = \lceil \{x,y\} \rceil $ .

Assuming that c is an $\mathsf {RT}^{n}_{\infty }\mathsf {c}$ -instance, we first show that there exists a d-homogeneous set. Let J be an infinite c-homogeneous set. Then for all sufficiently large m, $\{\lceil J_\ell \rceil : \ell \geq m\}$ , where $J_\ell =\{x\in J: x\leq \ell \}$ is infinite d-homogeneous (with colour $0$ ).

For the post-processing, given an enumeration f and a colouring c, first compute $f(0)$ . If $f(0) = \lceil A \rceil $ for some A of size at least n, then apply c to the first n elements of A and return the resulting colour. Otherwise, return $0$ . If f enumerates an infinite d-homogeneous set H, then H must have d-colour $0$ and $\bigcup _{\lceil A\rceil \in H}A$ is an infinite c-homogeneous set J. The above procedure then produces the c-colour of J.

(6): Given $c: [\mathbb {N}]^n \to \mathbb {N}$ , the pre-processing function computes $d: [\mathbb {N}]^{n+1} \to 2$ by $d(A) = 0$ if $c\restriction [A]^n$ is constant, otherwise $d(A) = 1$ . If c has bounded range, then all infinite d-homogeneous sets have colour $0$ because of Ramsey’s theorem. The post-processing can be the projection because every d-homogeneous set of colour $0$ is clearly also c-homogeneous.

(7): The pre-processing function takes a colouring $c: [\mathbb {N}]^n \to \mathbb {N}$ and computes the colouring $d: [\mathbb {N}]^{n+1} \to \mathbb {N}$ defined by $d(A) = c(A\setminus \{\max A\})$ . If the pre-processing function is given some bound b, it outputs said bound as well. Notice that if b bounds the range of c, then b bounds the range of d as well (as they are equal). Also, it is easy to see that an infinite subset of $\mathbb {N}$ is c-homogeneous of colour j if and only if it is d-homogeneous of colour j. Therefore we can post-process with the projection.

Proof. Suppose for a contradiction that $\Phi $ and $\Psi $ witness $\mathsf {C}_{2} \leq ^{\mathrm {t}}_{\mathrm {W}} \mathsf {RT}^{n}_{\infty }\chi $ . Let g denote the empty enumeration, i.e., the $\mathsf {C}_{2}$ -instance with solution set $\{0,1\}$ . Since $\Psi $ is total computable, we may fix some number s such that $\Psi (0^s,g \restriction s)\downarrow = j$ . We extend $g\restriction s$ to a $\mathsf {C}_{2}$ -instance f by enumerating $0$ (if $j \neq 1$ ) or $1$ (if $j = 1$ ). Then j is not a $\mathsf {C}_{2}$ -solution for f, yet $0^s$ may be extended to an $\mathsf {RT}^{n}_{\infty }\chi $ -solution to $\Phi (f)$ , yielding a contradiction.

Proof. By the previous proposition, it suffices to show that $\mathsf {C}_{k} \leq ^{\mathrm {t}}_{\mathrm {W}} \mathsf {RT}^{1}_{k}\mathsf {c}$ . The pre-processing function sends a given enumeration f to the following colouring c: $c(x)$ is the least number which has not been enumerated by f by stage x. If f is an instance of $\mathsf {C}_{k}$ , then c is an instance of $\mathsf {RT}^{1}_{k}\mathsf {c}$ . We post-process by taking the projection.

Proof. Suppose towards a contradiction that $\Phi $ and $\Psi $ witness $\mathsf {RT}^{1}_{k+1}\leq ^\ast _{\mathrm {W}} g$ . We shall construct an initial segment $c_k: \subseteq \mathbb {N} \to k+1$ of an $\mathsf {RT}^{1}_{k+1}$ -instance by iterating through $j < k$ , as follows. Let $c_0$ be the empty colouring. Having constructed $c_j$ , we extend it to $c_{j+1}$ as follows: If there is some finite extension $c_{j+1}$ of $c_j$ such that $\Psi (B(\Phi (c_{j+1}))_j,c_{j+1})(x)\downarrow = 1$ for some x, then extend $c_j$ to the least such $c_{j+1}$ . Otherwise, define $c_{j+1} = c_j$ .

For each $j < k$ , let $x_j$ be the least number such that $\Psi (B(\Phi (c_k))_j,c_k)(x_j)\downarrow = 1$ , if one exists. The set $\{c_k(x_j): j < k\}$ is a proper subset of $k+1$ , so we may fix some $i < k+1$ which differs from every $c_k(x_j)$ . Define $c: \mathbb {N} \to k+1$ by extending $c_k$ with $i^{\mathbb {N}}$ . Then $\Phi (c) \in \mathrm {dom}(g)$ . Fix some $j < k$ such that $B(\Phi (c))_j \in Y$ is a g-solution for $\Phi (c)$ . Either $x_j$ is undefined and for all x, it is not the case that $\Psi (B(\Phi (c))_j,c)(x)\downarrow = 1$ , or $x_j$ is defined and $\Psi (B(\Phi (c))_j,c)(x_j) = 1$ . In both cases, $\Psi (B(\Phi (c))_j,c)$ is not an $\mathsf {RT}^{1}_{k+1}$ -solution for c.

To prove the second statement in the proposition, consider $B(\Phi (0^{\mathbb {N}})) \in \overline {Y}^\ast $ . Fix some $s \in \mathbb {N}$ such that for all $\mathsf {RT}^{1}_{\mathbb {N}}$ -instances p extending $0^s$ , $B(\Phi (p))$ has the same length k as $B(\Phi (0^{\mathbb {N}}))$ . Let $X_0$ be the set of all $x \in \mathrm {dom}(g)$ such that $B(x)$ has length k. Since $\mathsf {RT}^{1}_{\mathbb {N}}$ is a fractal, we have $\mathsf {RT}^{1}_{\mathbb {N}} \leq ^\ast _{\mathrm {W}} g\restriction X_0$ . However, $\mathsf {RT}^{1}_{k+1} \not \leq ^\ast _{\mathrm {W}} g\restriction X_0$ by applying the first statement in the proposition. This yields the desired contradiction.

Proof. For $\mathsf {RT}^{n}_{k}\mathsf {c}$ , define B to be the constant function $01\dots (k-1)$ . For $\mathsf {RT}^{n}_{<\infty }\mathsf {c}$ , if B is given an $\mathsf {RT}^{n}_{<\infty }\mathsf {c}$ -instance with bound b on the colours used, it returns the sequence $01\dots (b-1)$ of length b.

Proof. Suppose towards a contradiction that $\Phi $ and $\Psi $ witness $\mathsf {RT}^{1}_{2}\mathsf {e} \leq ^{\mathrm {t}}_{\mathrm {W}} g$ . Consider the finite sequence $B(\Phi (0^{\mathbb {N}}))$ . Since $\Phi $ is computable, we may fix some $s_0$ such that for all c extending $0^{s_0}$ , the sequence $B(\Phi (c))$ has the same length k. Since $\Psi $ is total, for each $j < k$ , $\Psi (B(\Phi (0^{\mathbb {N}}))_j,0^{\mathbb {N}})$ enumerates some number. Since $\Psi $ is computable, we may fix some $s> s_0$ such that for each $j < k$ , $\Psi (B(\Phi (0^s))_j,0^s)$ enumerates some number below s. Then $0^s1^{\mathbb {N}}$ is an $\mathsf {RT}^{1}_{2}\mathsf {e}$ -instance, $\Phi (0^s1^{\mathbb {N}})$ is a g-instance, and for every $j < k$ , $\Psi (B(\Phi (0^s1^{\mathbb {N}}))_j,0^s1^{\mathbb {N}})$ enumerates a number of colour $0$ and is therefore not an $\mathsf {RT}^{1}_{2}\mathsf {e}$ -solution for $0^s1^{\mathbb {N}}$ . This contradicts the fact that some $B(\Phi (0^s1^{\mathbb {N}}))_j$ is a g-solution for $\Phi (0^s1^{\mathbb {N}})$ .

Proof. (1): Given a total computable function $\Psi $ , we shall construct $c: \mathbb {N} \to 2$ such that for every $n \in \mathbb {N}$ , there is some s such that $\Psi (n^s,c)$ does not extend to an $\mathsf {RT}^{1}_{2}\mathsf {pe}$ -solution for c. Begin by fixing $t_0$ such that $\Psi (0^{t_0},0^{t_0})$ enumerates some number (less than $t_0$ ). Now $\Psi (0^{\mathbb {N}},0^{t_0}1^{\mathbb {N}})$ either enumerates some number greater than $t_0$ , or enumerates some numbers out of order. Fix some $s_0$ such that $\Psi (0^{s_0},0^{t_0}1^{s_0})$ witnesses either of these. Define $\sigma _1 = 0^{t_0}1^{s_0}$ .

In general, suppose that at stage n, we have constructed some $\sigma _n$ . Fix $t_n$ such that $\Psi (n^{t_n},\sigma _n 0^{t_n})$ enumerates some number x (less than $|\sigma _n 0^{t_n}|$ ). Let $i < 2$ denote the colour $(\sigma _n 0^{t_n})(x)$ . Then fix some $s_n$ such that $\Psi (n^{s_n},\sigma _n 0^{t_n} (1-i)^{s_n})$ enumerates some number greater than $|\sigma _n 0^{t_n}|$ , or enumerates some numbers out of order. Define $\sigma _{n+1} = \sigma _n 0^{t_n} (1-i)^{s_n}$ .

Finally, define $c = \bigcup _n \sigma _n$ . For any non-empty set $X \subseteq \mathbb {N}$ , there is some $n \in X$ and some enumeration g of X which extends $n^{s_n}$ . Since $\Psi (n^{s_n},c)$ cannot extend to an $\mathsf {RT}^{1}_{2}\mathsf {pe}$ -solution for c, $\Psi (g,c)$ is not an $\mathsf {RT}^{1}_{2}\mathsf {pe}$ -solution for c.

(2): Given a total computable function $\Psi $ , we shall construct an instance $f = (f_i)_i$ of $\widehat {\mathsf {C}_{2}}$ such that for every $n \in \mathbb {N}$ , there is some s such that $\Psi (n^s,f)$ does not extend to a $\widehat {\mathsf {C}_{2}}$ -solution for f.

At the beginning of stage n, we have fully defined $f_i$ for $i < n$ . There may also be finitely many $f_i$ which are partially defined as a finite initial segment of the empty enumeration. At stage n, let $g = (g_i)_i$ denote the $\widehat {\mathsf {C}_{2}}$ -instance defined by $g_i = f_i$ if $f_i$ is fully defined, otherwise $g_i$ is the empty enumeration.

Since $\Psi $ is total, $\Psi (n^{\mathbb {N}},g)(n)$ is defined. We define $f_n$ by enumerating $0$ (if $\Psi (n^{\mathbb {N}},g)(n) \neq 1$ ) or $1$ (if $\Psi (n^{\mathbb {N}},g)(n) = 1$ ). We also extend finitely many $f_i$ (without enumerating any number) to ensure that $\Psi (n^{\mathbb {N}},f)(n) = \Psi (n^{\mathbb {N}},g)(n)$ .

The remainder of the verification proceeds as above.

Proof. The second statement follows from the first, by applying Theorem 3.1(5).

Proof. It is known that $\mathsf {RT}^{n}_{k+1} \not \leq _{\mathrm {W}} \mathsf {RT}^{n}_{k}$ and $\mathsf {RT}^{n}_{\mathbb {N}} \not \leq _{\mathrm {W}} \mathsf {RT}^{n}_{<\infty }$ ([Reference Brattka and Rakotoniaina7, Theorem 4.22 and Corollary 4.23] and [Reference Hirschfeldt and Jockusch10, Theorem 3.3]). The desired results follow from the facts that $\mathsf {RT}^{n}_{k}\chi \equiv _{\mathrm {W}} \mathsf {RT}^{n}_{k}$ , $\mathsf {RT}^{n}_{<\infty }\mathsf {pe} \equiv _{\mathrm {W}} \mathsf {RT}^{n}_{<\infty }$ , and $\mathsf {RT}^{n}_{\mathbb {N}}\chi \equiv _{\mathrm {W}} \mathsf {RT}^{n}_{\mathbb {N}}$ .

Proof. Given a sequence $(c_i)_i$ (not necessarily convergent), we begin by replacing each $c_i$ with $d_i: [\mathbb {N}]^n \to k$ , defined by $d_i(A) = c_i(A)$ if $c_i(A) < k$ , otherwise $d_i(A) = 0$ . The result of pre-processing is the colouring $d: [\mathbb {N}]^{n+1} \to k$ , defined by $d(A) = d_{\max A}(A\setminus \{\max A\})$ .

We claim that if $(c_i)_i$ converges to a colouring $c: [\mathbb {N}]^n \to k$ , then every infinite d-homogeneous set H is also c-homogeneous, with the same colour. This would mean that if we post-process by taking the projection, we obtain the claimed reductions. Suppose $A \in [H]^n$ . Then for all $i> \max A$ in H, $d(A \cup \{i\}) = d_i(A)$ has the same colour, say $j < k$ . Choose such i which is sufficiently large so that $c_i(A) = c(A) < k$ . Then $d_i(A) = c_i(A) = c(A)$ , so $c(A) = j$ .

Proof. Suppose for a contradiction that $\Phi $ and $\Psi $ witness $\mathsf {LPO}' \leq ^{\mathrm {t}}_{\mathrm {W}}\mathsf {RT}^{n}_{\mathbb {N}}\mathsf {e}$ . We shall construct an $\mathsf {LPO}'$ -instance $(f_s)_s$ in stages. Begin by defining $f_0 = 1^{\mathbb {N}}$ . At stage s, suppose we have determined $f_t$ for $t \leq s$ . Below, we will abuse notation and use $(f_t)_{t \leq s}$ to denote an oracle for $(f_t)_{t \in \mathbb {N}}$ with the partial information $(f_t)_{t \leq s}$ . That is, if a computation queries any $f_t$ for $t> s$ , then it diverges.

Case 1. If there is some $m \in \mathbb {N}$ such that $\Psi (m^s,(f_t)_{t \leq s})\downarrow = 1$ and it is not the case that $\Psi (m^{s-1},(f_t)_{t \leq s-1})\downarrow = 1$ , then we define $f_{s+1}$ as follows: $f_{s+1}(x) = 0$ if $x \leq s$ and $f_{s+1}(x) = 1$ otherwise.

Case 2. If no such m exists, we define $f_{s+1} = f_s$ .

This completes the construction of $(f_s)_s$ . We claim that if Case 1 occurs infinitely often, then $f = \lim _s f_s$ equals $0^{\mathbb {N}}$ ; otherwise, f contains some $1$ . Suppose first that Case 1 occurs infinitely often. Given $x \in \mathbb {N}$ , fix some stage $s> x$ at which Case 1 occurs. We define $f_{s+1}(x) = 0$ . For all $t> s+1$ , $f_t(x) = 0$ as well, so $f(x) = 0$ .

On the other hand, suppose Case 1 occurs only finitely often. For all $x \in \mathbb {N}$ which is greater than all stages at which Case 1 occurs, we have $f_s(x) = 1$ for all $s \in \mathbb {N}$ .

It remains to argue that the failure of $\Phi $ and $\Psi $ is witnessed in either situation. If Case 1 occurs infinitely often, there must be infinitely many m for which $\Psi (m^{\mathbb {N}},(f_s)_s)=1$ . By Ramsey’s theorem applied to the restriction of $\Phi (f)$ to the set of such m, some such m must be contained in some infinite $\Phi (f)$ -homogeneous set. Fix such an m and fix some t such that $\Psi (m^t,(f_s)_s)\downarrow = 1$ . Then, $m^{t}$ may be extended to an $\mathsf {RT}^{n}_{\mathbb {N}}\mathsf {e}$ -solution for $\Phi (f)$ , but $\mathsf {LPO}(f) = 0$ .

If Case 1 only occurs finitely often, then let us fix an arbitrary infinite $\Phi (f)$ -homogeneous set H. Since $\Psi $ is total, $\Psi (m^{\mathbb {N}},(f_s)_s)$ is defined for each $m \in H$ . Fix some $m \in H$ such that $\Psi (m^{\mathbb {N}},(f_s)_s) \neq 1$ . Fix some t such that $\Psi (m^t,(f_s)_s)\downarrow = \Psi (m^{\mathbb {N}},(f_s)_s)$ . Then, $m^t$ may be extended to an $\mathsf {RT}^{n}_{\mathbb {N}}\mathsf {e}$ -solution for $\Phi (f)$ , but $\mathsf {LPO}(f) = 1$ .

Proof. Given $(f_s)_s$ , the pre-processing function defines a colouring $c: \mathbb {N} \to \mathbb {N}$ as follows. Given $x \in \mathbb {N}$ , if $\min \{y < x: f_x(y) = 1\}$ exists and equals $\min \{y < x-1: f_{x-1}(y) = 1\}$ , then define $c(x) = y+1$ . Otherwise, define $c(x) = 0$ .

Suppose $(f_s)_s$ is an $\mathsf {LPO}'$ -instance. Let $f = \lim _s f_s$ . If $\mathsf {LPO}(f) = 1$ , let y be the least number such that $f(y) = 1$ . Then $c(s) = y+1$ for all sufficiently large s. On the other hand, suppose $f = 0^{\mathbb {N}}$ . For each $y \in \mathbb {N}$ , $c(s) \neq y+1$ for all sufficiently large s. Furthermore, there are infinitely many s such that $c(s) = 0$ : either $\{\min \{y<x: f_x(y) = 1\}: x \in \mathbb {N}\}$ is infinite, in which case $c(x) = 0$ whenever the minimum changes value, or $\{\min \{y<x: f_x(y) = 1\}: x \in \mathbb {N}\}$ is finite, in which case $c(s) = 0$ for all sufficiently large s. Therefore c is an $\mathsf {RT}^{1}_{\infty }\mathsf {e}$ -instance.

The post-processing function works as follows. Given a colour j, we return $0$ if $j = 0$ and return $1$ otherwise. Our analysis in the previous paragraph implies that this works correctly.

Proof. It is known that $\mathsf {RT}^{1}_{2}\mathsf {c}$ is not continuously Weihrauch reducible to $\mathsf {C}_{\mathbb {N}}$ . Since we did not find a source (specifically for the topological separation), we give a proof. Given continuous functions $\Phi $ and $\Psi $ which witness the reduction, we shall construct a colouring $c: \mathbb {N} \to 2$ as follows.

Case 1. For all n and all finite partial colourings $\sigma : \subseteq \mathbb {N} \to 2$ , there is some $\tau \supseteq \sigma $ such that $\Phi (\tau )$ enumerates n. Then we construct c in stages, as follows. At stage n, suppose we have constructed a finite partial colouring $\sigma _n$ . Define $\sigma _{n+1}$ to be the least extension of $\sigma _n$ such that $\Phi (\sigma _{n+1})$ enumerates n. Define $c = \bigcup _n \sigma _n$ . Then $\Phi (c)$ is not a $\mathsf {C}_{\mathbb {N}}$ -instance because it enumerates every $n \in \mathbb {N}$ .

Case 2. Otherwise, fix some n and some $\sigma : \subseteq \mathbb {N} \to 2$ such that for all $\tau \supseteq \sigma $ , $\Phi (\tau )$ does not enumerate n. In particular, n is a $\mathsf {C}_{\mathbb {N}}$ -solution for $\Phi (\sigma ^{\frown } 0^{\mathbb {N}})$ , so $\Psi (n,\sigma ^{\frown } 0^{\mathbb {N}})$ produces the unique $\mathsf {RT}^{1}_{2}\mathsf {c}$ -solution $0$ for $\sigma ^{\frown } 0^{\mathbb {N}}$ . Fix some sufficiently large s such that $\Psi (n,\sigma ^{\frown } 0^s)$ produces $0$ . Then $\Psi (n,\sigma ^{\frown } 0^s 1^{\mathbb {N}})$ produces $0$ but $0$ is not an $\mathsf {RT}^{1}_{2}\mathsf {c}$ -solution for $\sigma ^{\frown } 0^s 1^{\mathbb {N}}$ , yielding a contradiction.

Since $\mathsf {RT}^{1}_{2}\mathsf {c}$ is not continuously Weihrauch reducible to $\mathsf {C}_{\mathbb {N}}$ , we conclude by [Reference Brattka, Hölzl and Kuyper6, Theorem 11] that $(\mathsf {RT}^{1}_{2}\mathsf {c})' \not \leq _{\mathrm {W}} \mathsf {C}_{\mathbb {N}}' \equiv _{\mathrm {W}} \mathsf {RT}^{1}_{\infty }$ .

Proof. Let g be a problem with codomain $(\mathcal {P}(\mathbb {N})\setminus \{\emptyset \},\delta _e)$ . Suppose that $\Phi $ and $\Psi $ witness $(\mathsf {RT}^{1}_{2}\chi )'\leq ^{\mathrm {t}}_{\mathrm {W}} g$ . We employ a finite injury priority argument, constructing for each s, $f_{s}:\mathbb {N}\to \{0,1\}$ such that $f(x)=\lim _{s}f_{s}(x)$ exists for all x. In the informal description, we shall think of f as a $\Delta _{2}^{0}$ function instead of specifying the definition of $f_{s}$ for each s. Fix some ordering of all non-empty finite sets, $D_{0},D_{1},\dots $ , such that if $D_{i}\subseteq D_{j}$ , then $i\leq j$ .

Intuitively, the ‘advantage’ that an enumeration has over the characteristic function is that an enumeration may delay revealing (small) elements for arbitrarily long. Since $\Psi $ is total, it must continue to reveal more bits of the characteristic function even with constant enumerations as its oracle. However, if the constant enumeration does not name a valid g-solution, $\Psi $ is not obliged to produce a valid solution to $f=\lim _{s}f_{s}$ . The tension here would be to make the enumeration ‘sufficiently slow’ whilst producing a potential solution to g in the limit.

The idea is that we will diagonalise against any possible g-solution X (a finite or infinite set) as follows. Initially let f be the all zero colouring. Let the sequence of finite sets $D_{i_{1}}\subsetneq D_{i_{2}}\subsetneq \dots \subseteq X$ be such that $X=\bigcup _{j}D_{i_j}$ , $|D_{i_{j}}|=j$ and $\max D_{i_{j+1}}>\max D_{i_{j}}$ for each j. We shall denote $\max D_{k}=d_{k}$ for each $k\in \mathbb {N}$ . The goal now is to construct some enumeration $\delta $ of X so that $\Psi (\delta ,(f_{s})_{s})$ is not the characteristic function of an f-homogeneous set. First consider the function $\Psi (d_{i_{1}}^{\mathbb {N}},(f_{s})_{s})$ . If there is some $t\in \mathbb {N}$ and $x,y\in \mathbb {N}$ such that $x<y$ and $\Psi (d_{i_{1}}^{t},(f_{s})_{s})(x)=\Psi (d_{i_{1}}^{t},(f_{s})_{s})(y)=1$ , then we define $f(y)=1-f(x)$ , ensuring that $\{x,y\}$ cannot be extended to a homogeneous set. Therefore, we have successfully diagonalised against X.

However, as observed earlier, $\Psi (d_{i_{1}}^{\mathbb {N}},(f_{s})_{s})$ is not at all obliged to produce any elements as $D_{i_{1}}$ may not be a g-solution. So suppose $\Psi (d_{i_{1}}^{\mathbb {N}},(f_{s})_{s})(x)=1$ for at most one $x\in \mathbb {N}$ . While we must give ‘new information’ in the enumeration (to ensure it is a valid solution), we can exploit the fact that we can force $\Psi $ to commit on arbitrarily long initial segments of $\mathbb {N}$ .

Then once we have seen that $|\{x<z\mid \Psi (d_{i_{1}}^{\mathbb {N}},(f_{s})_{s})(x)=1\}|<2$ for some ‘sufficiently large’ z (chosen by the construction), we wish to begin adding $d_{i_{2}}$ to the enumeration. That is, we begin to consider $D_{i_2}$ . Now we carefully choose the oracle that we will consider. The idea is for us to repeat the first bit, $d_{i_1}$ , long enough to force $\Psi $ to commit on the string that we have seen so far before adding the next bit. That is, we will monitor $\Psi (d_{i_{1}}^{t}\!^{\frown } d_{i_{2}}^{\mathbb {N}},(f_{s})_{s})$ , where t is such that $\Psi (d_{i_1}^{t},(f_{s})_{s})(x)\downarrow $ for sufficiently many x. If the characteristic function now produces at least two elements, say $x<y$ , we may then define $f(y)=1-f(x)$ as before, ensuring that $\Psi $ never produces an f-homogeneous set on enumerations beginning with ${d_{i_{1}}^{t}}^{\frown } d_{i_{2}}^{t'}$ for some sufficiently large $t'$ . If the second bit does not produce such an $x,y$ , then we will repeat this strategy to add the next bit (of the potential solution to g).

To ensure no injury to higher priority requirements, each finite set $D_{i}$ of size n is given some ‘restraint’ $r(i)$ that it must respect. Additionally this restraint defines the length of the string we wait for $\Psi $ to commit to on the enumeration of the first $n-1$ elements. That is, we are looking for $\Psi $ to produce two elements $x,y$ such that $x<y$ and $y\geq r(i)$ . Then the element y cannot have been ‘used’ by any higher priority requirement, and we may safely define $f(y)=1-f(x)$ without injuring any higher priority requirements. That is, the colour of $x<r(i)$ can only be changed by $D_{i'}$ with $i'<i$ . The restraints will be maintained stage-wise and have the property that for all $i,s\in \mathbb {N}$ , $r(i)[s]\leq r(i)[s+1]$ and $r(i)[s]<r(i+1)[s]$ . When we act for $D_i$ , each $r(i')$ will be kicked to some ‘large’ value for each $i'>i$ as is standard in a movable marker construction.

In the construction, we ensure that any solution $X\subseteq \mathbb {N}$ , possibly finite, has an enumeration $\delta $ such that $\Psi (\delta ,f)$ does not output a valid solution by doing the strategy as described for each finite set. The idea is that to deal with $D_i$ , we consider the (unique) sequence of sets that build up the set one by one in order of its principal enumeration. We then use the strategy as described for this sequence of sets. We now detail the formal construction.

Construction: Fix some ordering of all finite non-empty sets, $D_{0},D_{1},\dots $ , such that if $D_{i}\subseteq D_{j}$ , then $i\leq j$ . For each $D_i$ we will consider the following unique sequence of sets: $D_{i_1}\subsetneq D_{i_2}\subsetneq \cdots \subsetneq D_{i_n}=D_i$ , where $n=|D_i|$ and $d_{i_1}<d_{i_2}<\cdots <d_{i_n}$ (recall that $d_{i_j}=\max D_{i_j}$ ). During the construction, each $D_{i}$ will be either active, inactive, or waiting. These states correspond to when the strategy for $D_{i}$ is currently in progress, or has already acted, or not yet begun, respectively. For $D_{i}$ which are currently active, we will also use $\delta _{i}[s]\in \mathbb {N}^{\mathbb {N}}$ to denote the current enumeration of $D_{i}$ being used as an oracle in the computation $D_{i}$ is monitoring for its strategy.

1. At stage 0: Declare all $D_i$ to be in the waiting state and leave each $\delta _i[0]$ undefined. Define $r(j)[0]=j$ for all $j\in \mathbb {N}$ . Let f be the all zero colouring.

2. At stage s + 1: Let $i\leq s+1$ be the least such that one of the following holds:

   1. (i) $D_{i}$ is currently active and there exists $x,y$ such that $x>y\geq r(i-1)[s]$ , and $\Psi (\delta _{i}[s],(f_{t})_{t\leq s})(x)=\Psi (\delta _{i}[s],(f_{t})_{t\leq s}) (y)=1$ . For such an i, we act for $D_{i}$ , declaring $f(y)=1-f(x)$ . Also define for each e, $r(e)[s+1]$ to be $r(e)[s]$ if $e<i$ , and the least number $>\max \{r(e)[s],r(e-1)[s+1]\}$ otherwise. Declare $D_{i}$ to now be inactive, and for each $j>i$ , declare $D_{j}$ to be in the waiting state and $\delta _{j}[s+1]$ to be undefined. Finally, for $e\leq i$ , define $\delta _{e}[s+1]=\delta _{e}[s]$ and proceed to the next stage.

   2. (ii) $D_{i}$ is in the waiting state and for each $j<n$ , where $n=|D_{i}|$ , $D_{i_{j}}$ is currently active. Then do the following for the least $i<s$ such that one of the following applies (do nothing and go to the next stage if no such i exists):

      • • If $|D_i|>1$ , then let l be the least such that for each $x<r(i)[s]$ , $\Psi (\delta _{i_{n-1}}[s]\restriction l,(f_{t})_{t\leq s})(x)\downarrow $ . Such an l must exist, since $\Psi $ is assumed to be total. Then define $\delta _{i}[s+1]=(\delta _{i_{n-1}}[s]\restriction l)^{\frown } d_{i}^{\mathbb {N}}$ . Finally, define for each e, $r(e)[s+1]=r(e)[s]$ , and if $e\neq i$ , $\delta _{e}[s+1]=\delta _{e}[s]$ , whenever defined, and proceed to the next stage.

      • • If $|D_i|=1$ , then declare $D_i$ active.

Verification: Now we verify that the construction works; each $D_{i}$ acts only finitely many times; $f:\mathbb {N}\to \{0,1\}$ is $\Delta _{2}^{0}$ ; and for any set X that is a solution to the instance $\Phi ((f_{s})_{s})$ , there is some enumeration $\delta $ of X such that $\Psi (\delta ,(f_{s})_{s})$ is not a solution to f.

From the construction, once $D_{i}$ acts, it is declared inactive and never again acts unless it gets initialised and placed in the waiting state by some higher priority requirement. Evidently, this ensures that each $D_{i}$ acts at most finitely often and thus, for each i, $\lim _{s}r(i)[s]<\infty $ . Additionally, after stage $s^{*}$ , where $r(i)[s^*]=\lim _{s}r(i)[s]$ , $f(x)$ is never changed again for all $x<r(i)[s^{*}]$ . Therefore, f is $\Delta _{2}^{0}$ and $(f_{s})_{s}$ is a valid instance to $(\mathsf {RT}^{1}_{2}\chi )'$ .

Now suppose that $X\subseteq \mathbb {N}$ is a solution to the g-instance $\Phi ((f_{s})_{s})$ . Note that X can be infinite or finite. Let $D_{i_1}\subsetneq D_{i_{2}}\subsetneq \dots \subseteq X$ be a sequence of finite sets such that for each n, $|D_{i_{n}}|=n$ , $d_{i_{n}}<d_{i_{n+1}}$ , and $X=\bigcup _{j}D_{i_j}$ . For each j, let $t_{j}$ be the least stage such that $r(i_{j})[t_{j}]=\lim _{s}r(i_{j})[s]$ .

From the construction, for stages $s>t_{j}$ , no $D_{k}$ for any $k\leq i_{j}$ acts at stage s. That is to say, if $D_{i_{j}}$ was declared inactive at stage $t_{j}$ , then it must have acted during stage $t_{j}$ . Since $D_{i_j}$ acted, there exists $x,y$ such that $x>y\geq r(i_j-1)[t_{j}-1]$ and $\Psi (\delta _{i_j}[t_{j}-1],f)(x)=\Psi (\delta _{i_j}[t_j-1],f)(y)=1$ and $f(y)[t_{j}]=1-f(x)$ . Then for all $e\geq i_j$ and $s\geq t_{j}$ , $r(e)[s]>x,y$ . Therefore no $D_e$ with $e>i_j$ can further change $f(x)$ or $f(y)$ . Similarly, by the definition of $t_j$ , none of $D_{k}$ for any $k<i_j$ acts after stage $t_j$ . This implies that $f(y)=1-f(x)$ and therefore $\Psi (\delta _{i_j}[t_{j}],(f_{s})_{s})$ does not output a valid solution. In particular, for any set $X\supseteq D_{i_{j}}$ , $\Psi ((\delta _{i_j}[t_j]\restriction l)^{\frown } \delta ^{X},(f_{s})_{s})$ , where $\delta ^X$ is the principal enumeration of X and l is chosen such that $\Psi (\delta _{i_j}[t_j]\restriction l,(f_{s})_{s})(z)\downarrow $ for each $z< r(i_j)[t_j]$ cannot be a valid solution to f.

On the other hand, if $D_{i_{j}}$ does not act during stage $t_j$ , its state never again changes for all stages $s\geq t_j$ . That is, either $D_{i_{j}}$ remains forever in the waiting state, implying that some $D_{i_{k}}$ , where $k<j$ is inactive (since $D_{i_{1}}$ cannot be in the waiting state for cofinitely many stages), or $D_{i_{j}}$ itself remains in the active state for cofinitely many stages. In the former case, $D_{i_j}$ is necessarily a superset of some $D_{i_k}$ which remains inactive and thus we are done. Thus, we may assume that for all j, $D_{i_j}$ remains in the active state for all stages $s\geq t_j$ . For each j, since we never act for any higher priority set after stage $t_j$ , or for $D_{i_j}$ itself, then for all j, $|\{x\mid \Psi (\delta _{i_j}[t_j],(f_{s})_{s})(x)=1\}\cap \{y\mid y\geq \lim _{s}r(i_{j}-1)[s]\}|=0$ . Therefore $\Psi (\lim _{j}\delta _{i_j}[t_j],(f_{s})_{s})$ produces only a finite set which cannot be a valid solution to $(f_{s})_{s}$ but $\lim _{j}\delta _{i_j}[t_j]$ is an enumeration of X.

Proof. Suppose $\Psi $ is a continuous map such that if $(c_s)_s$ is an $(\mathsf {RT}^{1}_{2})'$ -instance, then $\Psi ((c_s)_s) = (p_n)_n$ , where each $p_n \in \overline {\mathbb {N}^{\mathbb {N}}}$ . We shall construct an $(\mathsf {RT}^{1}_{2})'$ -instance $(c_s)_s$ with limit colouring c such that all requirements

$$\begin{align*}R_n: p_n \text{ is not infinite } c\text{-homogeneous} \end{align*}$$

are satisfied.

Construction: Arrange these requirements in order of priority. At stage s of our construction, we shall define $c_s: \mathbb {N} \to 2$ . During the construction, we will refer to $p_n$ , with the understanding that we only have partial information about $p_n$ given by applying $\Psi $ to the colourings $c_s$ defined thus far. For each n, the strategy for $R_n$ may act to protect certain numbers which have been enumerated into $p_n$ . We will use $P_n[s]$ to denote the set of numbers protected by $R_n$ at stage s.

Begin by defining all $P_n[0]$ to be empty.

1. Stage s: Define $c_s$ as follows: $c_s(x)=0$ unless there are some $y<x$ and $n<s$ such that $x,y\in P_n[s]$ and $x,y\notin P_m[s]$ for any $m<n$ . If such y (and n) exist, then pick the least such y and define $c_s(x)=1-c_s(y)$ .

   For $n \leq s$ , we define $P_n[s+1]$ as follows. If x is enumerated into $p_0$ and $|P_0[s]|\leq 1$ then define $P_0[s+1]=P_0\cup \{x\}$ . For each $0<n\leq s$ , if some x is enumerated into $p_n$ then take the least such x and do as follows:

   • • If there are some $y<z$ such that $y,z\in P_n[s]$ and $y,z\notin P_m[s]$ for any $m<n$ , then define $P_n[s+1]=P_n[s]$ .

   • • Otherwise, define $P_n[s+1]=P_n[s]\cup \{x\}$ .

   This completes stage s of the construction.

Verification: We first show that for each requirement $R_{n}$ , the limit $P_n = \lim _s P_n[s]$ exists and is finite. It is clear from construction that $P_0$ exists and $|P_0| \leq 2$ . Suppose for all $m < n$ , $P_m = \lim _s P_m[s]$ exists and is finite. We claim that $P_n = \lim _s P_n[s]$ exists and $\left |P_{n}\cup \bigcup _{m<n}P_{m}\right |\leq \left |\bigcup _{m<n}P_{m}\right |+2$ . Indeed, after every $P_m$ for $m < n$ has stabilised, at most two numbers outside $\bigcup _{m<n} P_m$ can be added to $P_n$ .

Next, we prove by induction on x that for each x, $c(x) = \lim _s c_s(x)$ exists. We have $c_s(0) = 0$ for all s. As for $x> 0$ , if there are some $y < x$ and n such that $x,y \in P_n\setminus \bigcup _{m<n} P_m$ , then for the least such y, we have $\lim _s c_s(x) = 1-c(y)$ .

Finally, we show that each $R_{n}$ is satisfied. Suppose $p_n$ is infinite, otherwise we are done. Then there must be exactly two elements in $P_{n}\setminus \bigcup _{m<n}P_{m}$ , say $y < x$ . As shown above, we must have $c(x) = 1-c(y)$ and thus $p_n$ is not c-homogeneous.

Proof. $\mathsf {RT}^{n}_{\infty }\mathsf {c}$ and $\mathsf {RT}^{1}_{\infty }$ are both computably true, hence continuously countably guessable.

Proof. By [Reference Soldà and Valenti15, Theorem 7.20], it is known that $\mathsf {C}_{\mathbb {N}}' \not \leq _{\mathrm {W}} \mathsf {RT}^{2}_{2}$ . The desired result follows from the equivalences $\mathsf {RT}^{1}_{\infty }\chi \equiv _{\mathrm {W}} \mathsf {RT}^{1}_{\infty } \equiv _{\mathrm {W}} \mathsf {C}_{\mathbb {N}}'$ and $\mathsf {RT}^{2}_{2}\mathsf {pe} \equiv _{\mathrm {W}} \mathsf {RT}^{2}_{2}$ .

Proof. We first show that $\mathsf {RT}^{1}_{\infty }\mathsf {e}$ is equivalent to $\mathsf {RT}^{1}_{\infty }\mathsf {e}^{*}$ , which we define as $\mathsf {RT}^{1}_{\infty }\mathsf {e}$ with the restriction that instances have a unique colour which occurs infinitely often. Evidently, $\mathsf {RT}^{1}_{\infty }\mathsf {e}^{*}\leq ^{\mathrm {t}}_{\mathrm {W}}\mathsf {RT}^{1}_{\infty }\mathsf {e}$ . Let $f:\mathbb {N}\to \mathbb {N}$ , an instance of $\mathsf {RT}^{1}_{\infty }\mathsf {e}$ be given. We construct $f^{*}:\mathbb {N}\to \mathbb {N}$ , an instance of $\mathsf {RT}^{1}_{\infty }\mathsf {e}^{*}$ as follows.

For each $i,s\in \mathbb {N}$ , we maintain that at each stage s, $x_{i,s}<x_{i+1,s}$ , and $x_{i,s}\leq x_{i,s+1}$ (if defined). The intention here is that there is exactly one $x_{i}=\lim _{s}x_{i,s}$ such that $x_{i}<\infty $ , and $x_{i}$ is part of an infinite f-homogeneous set. Initially set $x_{i,0}\uparrow $ for all $i\in \mathbb {N}$ . At each stage s, let $x_{i,s}$ be the least x in $(\max _{j<i}x_{j,s}, s]$ such that $f(x)=i$ and for each $y \in [x,s]$ , $f(y)\geq i$ , if such x exists, and let $x_{i,s}\uparrow $ otherwise.

For each $y\in \mathbb {N}$ , define $f^{*}(y)=x_{i,y}$ if $f(y)=i$ and $x_{i,y}\downarrow $ . Otherwise, define $f^{*}(y)=y$ . Now we verify that $f^{*}$ is a valid instance of $\mathsf {RT}^{1}_{\infty }\mathsf {e}^{*}$ . First, observe that if there are infinitely many y such that $f(y)=i$ , then for each $j>i,\,\lim _{s}x_{j,s}=\infty $ . Second, if there are only finitely many y for which $f(y)=j$ , then either $x_{j,s}\uparrow $ for cofinitely many s, or $\lim _{s}x_{j,s}$ exists and is bounded by $\max \{y\mid f(y)=j\}$ .

Let i be the least i such that there are infinitely many y for which $f(y)=i$ . Clearly, $\lim _{s}x_{i,s}=y$ for the least y such that $f(y)=i$ and $f(z)\geq i$ for each $z\geq y$ . It is thus evident that y is the unique $f^{*}$ -colour repeated infinitely often.

Given an $\mathsf {RT}^{1}_{\infty }\mathsf {e}^{*}$ -solution $S^{*}:\mathbb {N}\to \mathbb {N}$ for $f^{*}$ , we define an $\mathsf {RT}^{1}_{\infty }\mathsf {e}$ -solution $S:\mathbb {N}\to \mathbb {N}$ for f as follows. Let $S(0)=S^{*}(0)$ . For each $x\in \mathbb {N}$ , if $f(x)=f(S^{*}(0))$ , then define $S(x)=x$ , otherwise define $S(x)=S^{*}(0)$ . Clearly, the range of S is an infinite f-homogeneous set of colour $f(S^{*}(0))$ , and S is computable uniformly in $S^{*}$ and f.

Now we show that $\mathsf {RT}^{1}_{\infty }\mathsf {e}^{*}\leq ^{\mathrm {t}}_{\mathrm {W}}\mathsf {RT}^{3}_{2}\mathsf {pe}$ . Given $f: \mathbb {N} \to \mathbb {N}$ , we compute the following colouring $g:[\mathbb {N}]^{3}\to 2$ :

$$ \begin{align*}g(x,y,z)=\begin{cases} 0&\text{if }|\{f(x')\mid x'\leq x\}\cap\{f(y')\mid y\leq y'\leq z\}|=1,\\ 1&\text{otherwise.} \end{cases} \end{align*} $$

Assuming f is an $\mathsf {RT}^{1}_{\infty }\mathsf {e}^{*}$ -instance, we shall prove that all infinite g-homogeneous sets have colour $0$ . Suppose $\{x_{0}<x_{1}<x_{2}<\dots \}$ is g-homogeneous. Pick the least i such that there is some $x\leq x_{i}$ for which $f(x)$ is the unique f-colour, n, repeated infinitely often. Also pick $j>i$ such that for all $x\geq x_{j}$ , either $f(x)=n$ , or $f(x)\neq f(x')$ for any $x'\leq x_{i}$ . Finally, pick $k>j$ such that there is some x, where $x_{j}\leq x\leq x_{k}$ and $f(x)=n$ . Since n is the unique f-colour that repeats infinitely often, such $i,j$ , and k must exist. By definition of g, we have $g(x_{i},x_{j},x_{k})=0$ as desired.

For the post-processing, suppose we are given some $S: \mathbb {N} \to \mathbb {N}$ . We shall compute some $H: \mathbb {N} \to \mathbb {N}$ as follows. First compute and compare the sets $\{f(x')\mid x' \leq S(0)\}$ and $\{f(y') \mid S(1) \leq y' \leq S(2)\}$ . If they do not have exactly one colour in common, then S cannot be an $\mathsf {RT}^{3}_{2}\mathsf {pe}$ -solution for g, so we simply return $H = \mathrm {id}$ . Otherwise, we may identify the unique colour n in common and the least $x \leq S(0)$ such that $f(x) = n$ . Define $H(0)=x$ . For each $i>0$ , define $H(i)=i$ if $f(i)=n$ , and x otherwise. In this case, it is clear that the range of H is f-homogeneous of colour n.

Suppose S is an $\mathsf {RT}^{3}_{2}\mathsf {pe}$ -solution for g. We shall show that the range of H is infinite, in other words, that n appears infinitely often in f. It suffices to verify that for any $0<i<j$ ,

$$ \begin{align*}\{f(x')\mid x'\leq S(0)\}\cap\{f(y')\mid S(i)\leq y'\leq S(j)\}=\{n\}. \end{align*} $$

Suppose for a contradiction that there exists $i,j$ such that the above fails. In particular, $\{f(x')\mid x'\leq S(0)\}\cap \{f(y')\mid S(i)\leq y'\leq S(j)\}$ contains some $n'\neq n$ . However, this implies that $\{f(x')\mid x'\leq S(0)\}\cap \{f(y')\mid S(1)\leq y'\leq S(j)\}$ contains both n and $n'$ , contrary to our assumption on S.

Proof. Suppose that $\Phi $ and $\Psi $ witness $\big(\mathsf {RT}^{1}_{\mathbb {N}}\big)'\leq _{\mathrm {W}}\mathsf {RT}^{2}_{\mathbb {N}}$ . In the discussion that follows, we directly construct a $\Delta _{2}^{0}$ instance f of $\mathsf {RT}^{1}_{\mathbb {N}}$ , which may be turned into a computable instance $(f_{s})_{s\in \mathbb {N}}$ of $\big (\mathsf {RT}^{1}_{\mathbb {N}}\big)'$ in the obvious way. Let $g=\Phi ((f_{s})_{s\in \mathbb {N}})$ . Throughout this discussion, we will also think of $\Psi $ as a principal enumeration of an f-homogeneous set given as oracle a g-homogeneous set. To facilitate the discussion, we make the following definitions. Fix a finite set $D=\{x_{0}<x_{1}<\dots <x_{n-1}\}$ and a string $\sigma \in \mathbb {N}^{<\mathbb {N}}$ of length n.

• • Define $H(D,\sigma )=\{y>\max D\mid g(x_{i},y)=\sigma (i)\text { for each }i\}$ .

• • The pair $(D,\sigma )$ is consistent if for each $i,j$ , where $i<j<n$ , $g(x_{i},x_{j})=\sigma (i)$ .

• • Finally, for each $k\in \mathbb {N}$ , define $P_{k}(D,\sigma )=\{x_{i}\in D\mid \sigma (i)=k\}$ .

It is easy to see that if $(D,\sigma )$ is consistent, then $P_{k}(D,\sigma )$ is a g-homogeneous set (possibly empty) of colour k for each $k\in \mathbb {N}$ .

We first restrict ourselves to the case, where g uses only two colours, say $0,1$ . Following from the definitions, if $H(D,\sigma )$ is infinite, then for any $y\in H(D,\sigma )$ , at least one of $H(D\cup \{y\},\sigma ^{\frown }0)$ or $H(D\cup \{y\},\sigma ^{\frown }1)$ is infinite. Moreover, if $(D,\sigma )$ is also consistent, then one of $P_{0}(D,\sigma )$ or $P_{1}(D,\sigma )$ must be extendable to an infinite g-homogeneous set. In particular, by ensuring that f colours only finitely many bits with the colours of $\Psi (P_{0}(D,\sigma ),(f_{s})_{s\in \mathbb {N}})(0)$ or $\Psi (P_{1}(D,\sigma ),(f_{s})_{s\in \mathbb {N}})(0)$ if they have converged, ensures that $\Phi $ and $\Psi $ fail to witness $(\mathsf {RT}^{1}_{\mathbb {N}})'\leq _{\mathrm {W}}\mathsf {RT}^{2}_{\mathbb {N}}$ . Evidently, it may be that neither $\Psi (P_{0}(D,\sigma ),(f_{s})_{s\in \mathbb {N}})(0)$ nor $\Psi (P_{1}(D,\sigma ),(f_{s})_{s\in \mathbb {N}})(0)$ has converged as we have yet to provide ‘sufficient information’ in the oracle. We ‘add’ information by defining a new finite set $D'=D\cup \{\mu y\in H(D,\sigma )\}$ . Let $\sigma '=\sigma ^{\frown } i$ such that $H(D',\sigma ^{\frown } i)$ is infinite. Note that provided that $H(D,\sigma )$ is infinite, at least one of $H(D',\sigma ^{\frown } 0)$ or $H(D',\sigma ^{\frown } 1)$ is infinite. Observe also that if $(D,\sigma )$ is consistent, then so are $(D',\sigma ^{\frown }i)$ for both $i=0,1$ . In this way, given a consistent $(D,\sigma )$ so that $H(D,\sigma )$ is infinite, we can obtain a sequence of consistent $(D_m,\sigma _m)$ for which $H(D_{m},\sigma _{m})$ is infinite for each $m\in \mathbb {N}$ . Then at least one of $\bigcup _{m}P_{0}(D_{m},\sigma _{m})$ or $\bigcup _{m}P_{1}(D_{m},\sigma _{m})$ must be an infinite g-homogeneous set. That is, if $\Psi (P_{0}(D_{m},\sigma _{m}),((f_{s})_{s\in \mathbb {N}})(0),\Psi (P_{1}(D_{m},\sigma _{m}),(f_{s})_{s\in \mathbb {N}})(0)$ never converges, $\Phi $ and $\Psi $ must also fail to witness the reduction $(\mathsf {RT}^{1}_{\mathbb {N}})'\leq _{\mathrm {W}}\mathsf {RT}^{2}_{\mathbb {N}}$ .

We extend the ideas above to the general case as follows. During stage s of the construction, we say that a pair $(D_{i},\sigma )$ fires if the following conditions hold:

• • $(D_{i},\sigma )$ is consistent.

• • $|H(D_{i},\sigma )\restriction s|>|H(D_{i},\sigma )\restriction (s-1)|$ .

• • For any $k<|D_{i}|$ , $|\{n\in rng(g[s])\mid \Psi (P_{n}(D_{i},\sigma ),f)(0)\downarrow \}|>|\{n\in rng(g[s])\mid \Psi (P_{n}(D_{i}\restriction k,\sigma \restriction k),f)(0)\downarrow \}|$ .

Then we wish to define $f(s)$ to be the least colour which avoids the current colour of each of $\Psi (P_{n}(D_{j},\tau ),(f_{t})_{t<s})(0)\downarrow $ for ‘higher priority’ $(D_{j},\tau )$ . If we are not careful, we may end up in a situation, where f defined by us has no infinite homogeneous set. In particular, g may continually use new colours n, and pick new elements via $\Psi (P_{n}(D,\sigma ),(f_{t})_{t\in \mathbb {N}})(0)$ , causing us to restrain infinitely many colours from f. Whilst g is also not a valid instance of $\mathsf {RT}^{2}_{\mathbb {N}}$ in this case, $\Phi $ and $\Psi $ do not necessarily fail since f as constructed by us is also not a valid instance of $(\mathsf {RT}^{1}_{\mathbb {N}})'$ . To circumvent this, we will define $f(x)=0$ for each $x<s$ at stage s whenever we witness that $|rng(g[s])|>|rng(g[s-1])|$ . This ensures that if $|rng(g)|=\infty $ , then f is the all zero colouring and thus a valid instance of $(\mathsf {RT}^{1}_{\mathbb {N}})'$ . Finally, provided that $|rng(g)|<\infty $ , there must be some ‘highest priority’ $(D_{i},\sigma )$ which fires infinitely often during the construction. For such a pair, we shall argue that for at least one $n\in rng(g)$ , $P_{n}(D_{i},\sigma )$ is extendable to an infinite g-homogeneous set and that f colours only finitely many bits with one of the colour of $\Psi (P_{n}(D_{i},\sigma ),(f_{s})_{s\in \mathbb {N}})(0)$ whenever defined. We now detail the construction.

Construction: Let $\{D_{i}\}_{i\in \mathbb {N}}$ be a list of the finite sets with the property that if $D_{i}\subseteq D_{j}$ , then $i\leq j$ , and arrange the pairs $(D_{i},\sigma )$ , where $\sigma \in \mathbb {N}^{|D_{i}|}$ in order of priority as follows. $(D_{i},\sigma )<(D_{j},\tau )$ if $i<j$ , or if $i=j$ and $\sigma <_{lex}\tau $ . (Note that if $i=j$ , then $|\sigma |=|\tau |=|D_{i}|$ .) Define $C(\emptyset ,\epsilon )[s]=0$ for all s.

1. Stage s: For a consistent pair $(D_{i},\sigma )$ , do the following that applies.

   • • If for each $(D_{j},\tau )\leq (D_{i},\sigma )$ , where $|D_{j}|=|\tau |<s$ , and each $n\in rng(g[s])$ , $\Psi (P_{n}(D_{j},\tau ),(f_{t})_{t<s})(0)\uparrow $ . Define $C(D_{i},\sigma )[s]=0$ .

   • • Otherwise, define $C(D_{i},\sigma )[s]$ to be the least y larger than the current colours of $\Psi (P_{n}(D_{j},\tau ),(f_{t})_{t<s})(0)$ for any $(D_{j},\tau )\leq (D_{i},\sigma )$ and $n\in rng(g[s])$ , whenever defined.

   If $|rng(g[s])|>|rng(g[s-1])|$ , then for each $x\leq s$ , define $f_{s}(x)=0$ . Otherwise, there must be some $(D_{i},\sigma )$ , where $0<|D_{i}|=|\sigma |<s$ which fires. Let the pair $(D_{i},\sigma )$ be the highest priority pair which fires and also such that for any consistent pair $(D_{j},\tau )$ , where $|D_{j}|=|\tau |<s$ , $D_{j}\supsetneq D_{i}$ , and $\tau \supsetneq _{pref}\sigma $ , $(D_{j},\tau )$ does not fire at stage s. Then let y be the least such that y is larger than $C(D_{j},\tau )[s]$ (wherever defined) for each $D_{j},\tau $ such that $D_{j}\supseteq D_{i}$ , $\tau \supseteq _{pref}\sigma $ , and $|D_{j}|=|\tau |<s$ . Define $f_{s}(s)=y$ and for each $x<s$ , if $f_{s-1}(x)>y$ , define $f_{s}(x)=y$ . Otherwise, let $f_{s}(x)=f_{s-1}(x)$ .

Verification: Now we verify that the construction works. It is easy to see from the construction that $f=\lim _{s}f_{s}$ exists, as we have that $f_{s}(x)\geq f_{s+1}(x)$ for any $x,s\in \mathbb {N}$ . Then we have that $\lim _{s}C[s]$ also exists. Now, if $|rng(g)|=\infty $ , then we obtain that f is the all zero colouring, and thus, $\Phi $ and $\Psi $ must fail to witness $(\mathsf {RT}^{1}_{\mathbb {N}})'\leq _{\mathrm {W}}\mathsf {RT}^{2}_{\mathbb {N}}$ . We may thus assume that $|rng(g)|=N<\infty $ . Now let $(D_{i},\sigma )$ be the highest priority pair satisfying the following: