Proof. The reduction from $\mathsf {TT}^1_k$ - $\mathsf {Ext}$ to $\mathsf {WF}$ is straightforward. Given an instance $\langle f,\sigma \rangle $ of the former, checking whether there exists $H \cong 2^{<\omega }$ monochromatic for f and containing $\sigma $ as an initial segment is a uniformly $\Sigma ^1_1$ property in f and $\sigma $ , and thus can be decided using $\mathsf {WF}$ .

In the other direction, let $T \subseteq \omega ^{<\omega }$ be a tree. We define a uniformly T-computable coloring $f : 2^{<\omega } \to 2$ with $f(\langle \rangle ) = 0$ and such that T is well-founded if and only if there is no $H \cong 2^{<\omega }$ monochromatic for f with color $0$ . Thus, T is well-founded if and only if the $\mathsf {TT}^1_2$ - $\mathsf {Ext}$ -solution to $\langle f,\{ \langle \rangle \} \rangle $ is $0$ . Define $S = T * \omega ^{<\omega } = \{ \langle \alpha ,\beta \rangle : \alpha \in T, \beta \in \omega ^{<\omega },|\alpha |=|\beta | \}$ , viewed as a subtree of $\omega ^{<\omega }$ by thinking of $\langle \alpha ,\beta \rangle $ as a prefix of $\langle \alpha ',\beta ' \rangle $ if and only if $\alpha \preceq \alpha '$ and $\beta \preceq \beta '$ . Observe that $[T] \neq \emptyset $ if and only if $|[S]| = 2^{\aleph _0}$ . Let R be the collection of all strings of the form $0^{\gamma (0)}10^{\gamma (1)}1\cdots 0^{\gamma (|\gamma |-1)}1 0^s$ for some $\gamma \in S$ and some $s \in \omega $ . Then R is a T-computable subtree of $2^{<\omega }$ , and we have $\langle \rangle \in R$ . Although $[R]$ is non-empty (since it contains, at the very least, the constantly $0$ function) we have that $|[R]| = 2^{\aleph _0}$ if and only if $|[S]| = 2^{\aleph _0}$ . We define f by setting $f(\sigma ) = 0$ if $\sigma \in R$ and $f(\sigma ) = 1$ if $\sigma \notin R$ . Thus $f(\langle \rangle ) = 0$ , and there exists $H \cong 2^{<\omega }$ monochromatic for f and containing $\langle \rangle $ if and only if there exists a $H \cong 2^{<\omega }$ monochromatic for f with color $0$ , which in turn holds in if and only if $|[R]| = 2^{\aleph _0}$ , which, by our previous observations, holds if and only if $[T] \neq \emptyset $ . This completes the proof.

Proof. As already noted, $\mathsf {V}_{0} \leq _{\mathrm {W}} \mathsf {V}_{1}$ . It thus remains to show that $\mathsf {V}_{1} \leq _{\mathrm {W}} \mathsf {TC}_{\mathbb {N}} \leq _{\mathrm {W}} \mathsf {V}_{0}$ . To see that $\mathsf {V}_{1} \leq _{\mathrm {W}} \mathsf {TC}_{\mathbb {N}}$ , fix any instance $f : 2^{<\omega } \to 2$ of $\mathsf {V}_{1}$ . Consider the $\Pi ^{0,f}_1$ set $A = \{ \langle i,\sigma \rangle : (\forall \tau \succeq \sigma )[f(\tau ) = i] \}$ , regarded as an instance of $\mathsf {TC}_{\mathbb {N}}$ uniformly computable from f. Clearly, $A \neq \emptyset $ if and only if $\{ \tau : f(\tau ) = 0 \}$ and $\{ \tau : f(\tau ) = 1 \}$ are not both dense. Thus, any $\mathsf {TC}_{\mathbb {N}}$ -solution to A is a $\mathsf {V}_{1}$ -solution to f.

To conclude the proof, we show that $\mathsf {TC}_{\mathbb {N}} \leq _{\mathrm {W}} \mathsf {V}_{0}$ . Fix an instance A of $\mathsf {TC}_{\mathbb {N}}$ , specified by a co-enumeration $e : \omega \to \omega $ . From e we can uniformly compute the function $\ell : \omega \to \omega $ such that $\ell (s) = (\mu x)[x+1 \notin e \operatorname {\mathrm {\upharpoonright }} s]$ . Clearly, $A \neq \emptyset $ if and only if $\lim _s \ell (s)$ exists, in which case $\lim _s \ell (s) \in A$ . We now define a uniformly e-computable coloring $f : 2^{<\omega } \to 2$ , as follows. Let $f(\langle \rangle ) = 0$ , and for each $s \in \omega $ , each string $\sigma $ of length s, and each $i < 2$ let

$$\begin{align*}f(\sigma i) = \begin{cases} i & \text{if } \ell(s) \neq \ell(s+1),\\ f(\sigma) & \text{otherwise}. \end{cases} \end{align*}$$

Now, let $\langle i,\sigma \rangle $ be any $\mathsf {V}_{0}$ -solution to f. We claim that $\ell (|\sigma |)$ is a $\mathsf {TC}_{\mathbb {N}}$ -solution to A. Indeed, suppose $A \neq \emptyset $ and fix the least s such that $\ell (s) = \ell (t)$ for all $t \geq s$ . If $|\sigma | < s$ , then it follows from the definition of f that neither $\{ \tau : f(\tau ) = 0 \}$ nor $\{ \tau : f(\tau ) = 1 \}$ is dense above $\sigma $ , even though we assumed $\{ \tau : f(\tau ) = i \}$ is. Thus, $|\sigma | \geq s$ , which proves the claim.

Proof. As already mentioned, $\mathsf {V}_{2} \leq _{\mathrm {W}} \mathsf {V}_{3} \leq _{\mathrm {W}} \mathsf {V}_{4}$ , so we have only to show that $\mathsf {V}_{4} \leq _{\mathrm {W}} \mathsf {sTC}_{\mathbb {N}} \leq _{\mathrm {W}} \mathsf {V}_{2}$ . For the first reduction, fix $f : 2^{<\omega } \to 2$ and define $A = \{ \sigma : (\forall \tau \succeq \sigma )[f(\tau ) = 1] \}$ . We can regard A as an instance of $\mathsf {sTC}_{\mathbb {N}}$ , uniformly computable from f. Then $A \neq \emptyset $ if and only if $\{ \tau : f(\tau ) = 0 \}$ is not dense. Given $-1$ as an $\mathsf {sTC}_{\mathbb {N}}$ -solution to A, we can then take $\langle 0, \langle \rangle \rangle $ as a $\mathsf {V}_{4}$ -solution to f. And given some $\mathsf {sTC}_{\mathbb {N}}$ -solution to A different from $-1$ , coding a string $\sigma \in 2^{<\omega }$ , we can take $\langle 1,\sigma \rangle $ as a $\mathsf {V}_{4}$ -solution to f.

Now, to show that $\mathsf {sTC}_{\mathbb {N}} \leq _{\mathrm {W}} \mathsf {V}_{2}$ , fix an instance A of $\mathsf {sTC}_{\mathbb {N}}$ , specified by a co-enumeration $e : \omega \to \omega $ . Let $\ell $ be the function defined in the proof of Proposition 3.6. Define $f : 2^{<\omega } \to 2$ as follows: for every s, let $f(0^s) = 0$ ; for every s and every $\sigma \succeq 0^s1$ , let

$$\begin{align*}f(\sigma) = \begin{cases} 0 & \text{if } (\exists t)[s \leq t < |\sigma| \wedge \ell(t) \neq \ell(t+1)],\\ 1 & \text{otherwise}. \end{cases} \end{align*}$$

Note that $A \neq \emptyset $ if and only if $\lim _s \ell (s)$ exists (and belongs to A), if and only if there is a $\sigma $ such that $f(\tau ) = 1$ for all $\tau \succeq \sigma $ . So suppose $\langle i,\sigma \rangle $ is a $\mathsf {V}_{2}$ -solution to f. If $i = 0$ , then $\{ \tau : f(\tau ) = 0 \}$ is dense, hence by the previous observation we must have $A = \emptyset $ . If $i = 1$ , then we claim that $\ell (|\sigma |) \in A$ . Suppose not, so that for some $t \geq |\sigma |$ we have $\ell (t) \neq \ell (t+1)$ . Fix $\sigma ' \succeq \sigma 1$ with $|\sigma '|> t$ . Then $\sigma ' \succeq 0^s1$ for some $s \leq t$ , hence $s \leq t < |\sigma '|$ . By definition of f, we have that $f(\tau ) = 0$ for all $\tau \succeq \sigma '$ . But since $\langle 1,\sigma \rangle $ is a $\mathsf {V}_{2}$ -solution to f, we also have that $\{ \tau : f(\tau ) = 1 \}$ is dense above $\sigma $ , a contradiction. Thus, from any $\mathsf {V}_{2}$ -solution to f we can uniformly computably determine whether or not A is empty, and if it not, determine an element of A, which completes the proof.

Proof of Theorem 3.4.

We have only to justify that $\mathsf {V}_{0} \nleq _{\mathrm {W}} \mathsf {TT}^1_2$ and that $\mathsf {V}_{2} \nleq _{\mathrm {W}} \mathsf {V}_{1}$ . The former is a direct consequence of Corollary 5.9, to be proved later, which establishes the stronger fact that $\mathsf {TT}^1_2$ is not Weihrauch equivalent to any first-order problem. The latter follows from Propositions 3.6 and 3.7, and the fact that $\mathsf {sTC}_{\mathbb {N}} \nleq _{\mathrm {W}} \mathsf {TC}_{\mathbb {N}}$ . To justify this last fact, consider for example the problem $\mathsf {isFinite}$ , whose instances are characteristic functions of subsets of $\omega $ having, as their unique solution, either $1$ or $0$ depending as this subset is finite or infinite. Clearly, $\mathsf {isFinite} \leq _{\mathrm {W}} \mathsf {sTC}_{\mathbb {N}}$ since the set of upper bounds of a given subset of $\omega $ is $\mathbf {\Pi }^0_1$ . But $\mathsf {isFinite} \nleq _{\mathrm {W}} \mathsf {TC}_{\mathbb {N}}$ by a result of Neumann and Pauly [Reference Neumann and Pauly21, Proposition 24].

Proof. The proof is quite similar to the proof that $\mathsf {RT}^1_k \nleq _{\mathrm {W}} \mathsf {RT}^1_j$ . We spell out the details. Suppose $\mathsf {RT}^1_k \leq _{\mathrm {W}} \mathsf {TT}^1_j$ via $\Phi $ and $\Psi $ . We define a sequence $\alpha _0 \preceq \cdots \preceq \alpha _j$ of elements of $k^{<\omega }$ , regarded as initial segments of an instance of $\mathsf {RT}^1_k$ , along with a sequence of strings $\sigma _0 \preceq \cdots \preceq \sigma _j$ of $2^{<\omega }$ .

Let $\alpha _0 = \langle \rangle $ and $\sigma _0 = \langle \rangle $ . Now fix $c < j$ and suppose $\alpha _c$ and $\sigma _c$ have been defined. Search for numbers $m,n \in \omega $ and a finite set S satisfying the following:

• • $S \cong 2^{< n}$ , and the root of S extends $\sigma _c$ ;

• • $\Phi (\alpha _c c^m)(\sigma ) \downarrow $ for all $\sigma \in S$ and has the same value;

• • $\Psi (\alpha _c c^m,S)(x) \downarrow = 1$ for some $x \geq |\alpha _c|$ .

The search must succeed, because $g = \alpha _c c^\omega $ is an instance of $\mathsf {RT}^1_k$ , so $\Phi (g)$ is in turn an instance f of $\mathsf {TT}^1_j$ . Moreover, for any $H \cong 2^{<\omega }$ monochromatic for f with root extending $\sigma _c$ we must have that $\Psi (g,H)$ is an infinite monochromatic set for g. In this case, any sufficiently large initial segment of g can serve as $\alpha _c c^m$ , and any sufficiently large initial segment of H can serve as S.

Having found m and n, set $\alpha ^{\prime }_{c} = \alpha _c c^m$ . Without loss of generality, we may assume $m+|\alpha _c|> x$ for the x witnessing the third property above, so $\alpha ^{\prime }_{c}(x)$ is defined and equal to c. Now, call $\alpha \in k^{<\omega }$ c-good if $\alpha \succeq \alpha _c^{\prime }$ and $\alpha (x)> c$ for all $x \geq |\alpha _c^{\prime }|$ . Since $c < j < k$ , c-good strings exist. Similarly, call $g : \omega \to k$ c-good if $g \succ \alpha _c^{\prime }$ and $g(x)> c$ for all $x \geq |\alpha _c^{\prime }|$ . If g is c-good, then $\Phi (g)(\sigma ) \downarrow $ for all $\sigma \in S$ and has the same value, and $\Psi (g,S)(x) \downarrow = 1$ for some x with $g(x) = c$ . But g, being c-good, has no infinite monochromatic set of color c, so S cannot be extended to any $H \cong 2^{<\omega }$ monochromatic for $\Phi (g)$ . In particular, it cannot be that for every leaf $\lambda $ of S the set $\{ \sigma : \Phi (g)(\sigma ) = \Phi (g)(\lambda ) \}$ is dense above $\lambda $ . It follows that there exists a c-good $\alpha $ , a leaf $\lambda $ of S, and a string $\sigma \succeq \lambda $ such that for all c-good $\beta \succeq \alpha $ and all $\tau \succeq \sigma $ , if $\Phi (\beta )(\tau ) \downarrow $ then $\Phi (\beta )(\tau ) \neq \Phi (\beta )(\lambda )$ . Fix some such $\alpha $ and $\sigma $ , and let $\alpha _{c+1} = \alpha $ and $\sigma _{c+1} = \sigma $ .

Now, fix any $g : \omega \to k$ extending $\alpha _j$ which is c-good for all $c < j$ . Let $f = \Phi (g)$ . By assumption, f is a j-coloring. But by induction, $|\{ f(\tau ) : \tau \succeq \sigma _c \}| \leq j-c$ for all $c \leq j$ . In particular, this means $|\{ f(\tau ) : \tau \succeq \sigma _j \}| = 0$ , which is impossible.

Proof. The proof is by induction on n. For $n=1$ , the result is trivial, so assume $n> 1$ . Let $S = \cup _{i < n} S_i$ . Fix $\sigma \in S$ such that $\text {rk}_S(\sigma )$ is maximal. By reindexing the $S_i$ sets if necessary, we can assume that $\sigma \in S_{n-1}$ . Let

$$\begin{align*}S' = S \setminus \{ \tau \in S : \tau \text{ and } \sigma \text{ are comparable} \} \text{ and } S_i^{\prime} = S_i \cap S' \text{ for } i < n-1. \end{align*}$$

Because $\text {rk}_S(\sigma )$ is maximal, there are no strings $\tau \in S$ such that $\sigma \prec \tau $ . Since the sets $S_i$ consist of incomparable strings, it follows that $|S_i \setminus S_i^{\prime }| \leq 1$ , and hence $|S_i^{\prime }| \geq n-1$ , for each $i < n-1$ . Therefore, we can apply the inductive hypothesis to choose a set of incomparable strings $\sigma _i \in S^{\prime }_i$ for $i < n-1$ . The set $\{ \sigma _0, \ldots , \sigma _{n-2}, \sigma \}$ is the desired set of incomparable strings.

Proof. It suffices to prove the result for $k = 2$ . Seeking a contradiction, suppose ${}^1 \mathsf {TT}^1_2 \leq _{\mathrm {W}} \mathsf {RT}^1_j$ via $\Phi $ and $\Psi $ . Let $\Gamma $ be a functional such that, given $f : 2^{<\omega } \to 2$ and $S \cong 2^{<\omega }$ , $\Gamma (f,S)(0)$ outputs $\langle \sigma _0,\ldots ,\sigma _{j-1} \rangle $ for the least j many pairwise incomparable strings $\sigma _0,\ldots ,\sigma _{j-1} \in S$ . In particular, for every $f : 2^{<\omega } \to 2$ , $\langle f,\mathrm {Id},\Gamma \rangle $ is an instance of ${}^1 \mathsf {TT}^1_2$ . Hence by assumption, $\Phi (f,\mathrm {Id},\Gamma )$ is an instance g of $\mathsf {RT}^1_j$ . Moreover, note that for each $c < j$ , either $g(x) \neq c$ for all sufficiently large x, or there is a finite set E monochromatic for g with color c such that $\Psi (f,\mathrm {Id},\Gamma ,E)(0) \downarrow = \langle \sigma _0,\ldots ,\sigma _{j-1} \rangle $ for some pairwise incomparable strings $\sigma _0,\ldots ,\sigma _{j-1}$ .

Consider Cohen forcing, with conditions $\alpha \in 2^{<\omega }$ regarded as initial segments of a $2$ -coloring of $2^{<\omega }$ . Let $\check {f}$ be a name in the forcing language for a generic such $2$ -coloring, and let $\check {g}$ be a name for $\Phi (\check {f},\mathrm {Id},\Gamma )$ . From the above discussion, it follows that we can find a condition $\alpha $ along with a number $n_0$ , a non-empty set $C \subseteq j$ and, for each $c \in C$ , a finite set $E_c$ , such that $\alpha $ forces each of the following:

• • $E_c$ is monochromatic for $\check {g}$ with color c;

• • $\Psi (\check {f},\mathrm {Id},\Gamma ,E_c)(0) \downarrow $ ;

• • $\check {g}(x) \in C$ for all $x> n_0$ .

Notice that the first two clauses are $\Sigma ^0_1$ , while the third is $\Pi ^0_1$ . Therefore, if $f : 2^{<\omega } \to 2$ is any coloring extending $\alpha $ , whether generic or not, each of the above clauses holds if $\check {f}$ is replaced by f and $\check {g}$ by $g = \Phi (f,\mathrm {Id},\Gamma )$ .

By definition of $\Gamma $ , for each $c \in C$ we have $\Psi (\alpha ,\mathrm {Id},\Gamma ,E_c)(0) = \langle \sigma _{c,0},\ldots ,\sigma _{c,j-1} \rangle $ for some pairwise incomparable strings $\sigma _{c,0},\ldots ,\sigma _{c,j-1} \in 2^{<\omega }$ . By standard use conventions, we may assume $\sigma _{c,0},\ldots ,\sigma _{c,j-1} < |\alpha |$ for all c. Apply Lemma 4.3 to choose $\sigma _c \in \{ \sigma _{c,0},\ldots ,\sigma _{c,j-1} \}$ for each $c \in C$ such that $\{ \sigma _c : c \in C \}$ is a set of incomparable strings.

Finally, we are ready to define f. For all $\sigma < |\alpha |$ , let $f(\sigma ) = \alpha (\sigma )$ . As noted above, this defines $f(\sigma _c) = \alpha (\sigma _c)$ for all $c \in C$ . If $\sigma \geq |\alpha |$ and $\sigma \nsucceq \sigma _c$ for all $c \in C$ , let $f(\sigma ) = 0$ . If $\sigma \geq |\alpha |$ and $\sigma \succeq \sigma _c$ for some $c \in C$ , then $\sigma \nsucceq \sigma _d$ for all $d \neq c$ in C since $\sigma _d$ and $\sigma _c$ are incomparable. In this case, we let $f(\sigma ) = 1 - f(\sigma _c)$ . This completes the definition. To complete the proof, let $g = \Phi (f,\mathrm {Id},\Gamma )$ . Since f extends $\alpha $ it follows by our earlier remark that each $E_c$ is monochromatic for g with color c, and every infinite monochromatic set for g has color some $c \in C$ . It follows that for some $c \in C$ , $E_c$ is extendible to an $\mathsf {RT}^1_j$ -solution to g, and therefore $\langle \sigma _{c,0},\ldots ,\sigma _{c,j-1} \rangle $ is a ${}^1 \mathsf {TT}^1_2$ -solution to $\langle f,\mathrm {Id},\Gamma \rangle $ . This means that $\langle \sigma _{c,0},\ldots ,\sigma _{c,j-1} \rangle = \Gamma (f,H)$ for some $H \cong 2^{<\omega }$ monochromatic for f. But for every $\sigma \succeq \sigma _c$ with $|\sigma | \geq \alpha $ , and in particular, for every such $\sigma \in H$ , we have that $f(\sigma ) \neq f(\sigma _c)$ . Thus, H cannot be monochromatic for f, a contradiction.

Proof. It suffices to prove the result for $k = 3$ . Suppose to the contrary that ${}^1 \mathsf {TT}^1_3 \leq _{\mathrm {W}} \mathsf {RT}^1_+$ , say via $\Phi $ and $\Psi $ . We build a $3$ -coloring f of $2^{<\omega }$ and a Turing functional $\Gamma $ so that $\langle f,\mathrm {Id},\Gamma \rangle $ is an instance of ${}^1 \mathsf {TT}^1_3$ witnessing a contradiction. The way we define $\Gamma $ is as follows. Let $\Delta _0,\Delta _1,\ldots $ be the standard enumeration of all oracle Turing functionals (so named in order to avoid notationally clashing with the fixed functional $\Phi $ ). We define a certain computable function h and then choose a fixed point $e_0$ for h using the recursion theorem, so that $\Delta _{h(e_0)}(A) = \Delta _{e_0}(A)$ for all oracles A. We then let $\Gamma = \Delta _{e_0}$ .

For the definition of h, we decompose each oracle A as $\langle f,S \rangle $ , with our interest being in the situation where f is a $3$ -coloring of $2^{<\omega }$ and S is a monochromatic set for f. Of course, for an arbitrary A, either f or S or both may fail to have these properties, but we do not care about these cases. We obtain h using the s-m-n theorem so as to satisfy the following properties, for all $e \in \omega $ and all oracles $A = \langle f,S \rangle $ . First, find the least $\sigma \in S$ . If $f(\sigma ) = 0$ , then let $\Delta _{h(e)}(A)(0) \downarrow = \sigma $ . If $f(\sigma )> 0$ , then wait for $\Phi (f,\mathrm {Id},\Delta _e)(0)$ to converge and output a number $j \geq 1$ . (The idea is that if $\langle f,\mathrm {Id},\Delta _e \rangle $ is an instance of ${}^1 \mathsf {TT}^1_3$ , then $\Phi (f,\mathrm {Id},\Delta _e)$ is an instance $\langle j,g \rangle $ of $\mathsf {RT}^1_+$ , so $\Phi (f,\mathrm {Id},\Delta _e)(0) = j$ .) Then, find the first j many incomparable strings $\sigma _0,\ldots ,\sigma _{j-1} \in S$ and let $\Delta _{h(e)}(A)(0) \downarrow = \langle \sigma _0,\ldots ,\sigma _{j-1} \rangle $ .

Let $\Gamma $ be as described above, and consider the trivial coloring $f_0 : 2^{<\omega } \to 3$ such that $f_0(\sigma ) = 0$ for all $\sigma \in 2^{<\omega }$ . If $H \cong 2^{<\omega }$ is monochromatic for $f_0$ then it is clear from the definition that $\Gamma (f_0,H)(0) \downarrow $ . (Namely, $\Gamma (f_0,H)(0)$ outputs the least string in H.) Thus, $\langle f_0,\mathrm {Id},\Gamma \rangle $ is an instance of ${}^1 \mathsf {TT}^1_3$ . By assumption, $\Phi (f_0,\mathrm {Id},\Gamma )(0) \downarrow = j$ for some $j \geq 1$ . Fix a finite initial segment $\alpha _0 \prec f_0$ such that $\Phi (\alpha _0,\mathrm {Id},\Gamma )(0) \downarrow = j$ . Say $f : 2^{<\omega } \to 3$ is good if f extends $\alpha _0$ and $f(\sigma ) \neq 0$ for all $\sigma \geq |\alpha _0|$ . Similarly, say $\alpha \in 3^{<\omega }$ , regarded as a $3$ -coloring of a finite subset of $2^{<\omega }$ , is good if $\alpha _0 \preceq \alpha $ and $\alpha (\sigma ) \neq 0$ for all $\sigma \geq |\alpha _0|$ .

Now, consider any good coloring $f : 2^{<\omega } \to 3$ . Then $\langle f,\mathrm {Id},\Gamma \rangle $ is an instance of ${}^1 \mathsf {TT}^1_3$ . Indeed, we have that $\Phi (f,\mathrm {Id},\Gamma )(0) \downarrow = j$ since f extends $\alpha _0$ . Moreover, any $H \cong 2^{<\omega }$ monochromatic for f must have color $1$ or $2$ . Thus, by definition of $\Gamma $ , we have that $\Gamma (f,H)(0) \downarrow = \langle \sigma _0,\ldots ,\sigma _{j-1} \rangle $ for the least j many incomparable strings $\sigma _0,\ldots ,\sigma _{j-1} \in H$ . And $\Phi (f,\mathrm {Id},\Gamma ) = \langle j,g \rangle $ for some instance g of $\mathsf {RT}^1_j$ . So we can now proceed precisely as in the proof of Theorem 4.4, only working with good conditions in $3^{<\omega }$ , and building a good $3$ -coloring f.

Proof. Again, the proof is based on that of Theorem 4.4, but with some small differences. As there, we can take $k = 2$ , and then argue by contradiction. So, asume that $\mathsf {TT}^1_2 \leq _{\mathrm {W}} \mathsf {RT}^1_{\mathbb {N}}$ , via $\Phi $ and $\Psi $ . Consider Cohen forcing in $2^{<\omega }$ , let $\check {f}$ be as in Theorem 4.4, and let $\check {g}$ be a name in the forcing language for $\Phi (\check {f})$ . Since a generic here is an instance of $\mathsf {TT}^1_2$ , which is in turn mapped by $\Phi $ to an instance of $\mathsf {RT}^1_{\mathbb {N}}$ , it follows that we can fix a $j \geq 1$ and a condition $\alpha _0$ forcing that $\check {g}$ is a j-coloring. This can be expressed as a $\Pi ^0_1$ property, hence it will be preserved by any $2$ -coloring of $2^{<\omega }$ that extends $\alpha _0$ , generic or not. We can now proceed basically just like in Theorem 4.4, but working below $\alpha _0$ . More precisely, we find a condition $\alpha \geq \alpha _0$ along with a number $n_0$ , a non-empty set $C \subseteq j$ and, for each $c \in C$ , a finite set $E_c$ , such that $\alpha $ forces each of the following:

• • $E_c$ is monochromatic for $\check {g}$ with color c;

• • $\Psi (\check {f},E_c)(\sigma ) \downarrow = 1$ for j many pairwise incomparable strings $\sigma $ ;

• • $\check {g}(x) \in C$ for all $x> n_0$ .

The rest of the construction of a $2$ -coloring $f \succeq \alpha $ of $2^{<\omega }$ is now just as before.

Proof. The result is trivial for $k = 1$ , so we may assume $k \geq 2$ . Fix an instance $f : 2^{<\omega } \to k$ of $\mathsf {TT}^1_k$ . For each $n \in \omega $ and each $c < k-1$ we define a string $\rho _{c,n} \in 2^{<\omega }$ by induction on c. Let $\rho _{-1,n} = \langle \rangle $ for definiteness, and assume that $\rho _{c-1,n}$ has been defined for some $c < k-1$ . Then, let $\rho _{c,n}$ be the least extension of $\rho _{c-1,n}$ (in some fixed enumeration of $2^{<\omega }$ ) such that $f(\sigma ) \neq c$ for all $\sigma \succeq \rho _{c,n}$ with $|\sigma | < n$ . Notice that $\rho _{c,n}$ is always defined, and that it is uniformly computable from c and n. If $\lim _n \rho _{c,n}$ exists, we denote the limit by $\rho _c$ .

We next define a coloring $g : \omega \to k$ . Given $n \in \omega $ , choose the largest $c < k$ such that $\rho _{d,n} = \rho _{d,m}$ for all $d < c$ and all $m \geq n$ . (There is such a c because, in particular, $\rho _{-1,n} = \rho _{-1,m}$ for all $m \geq n$ .) Then, let $g(n) = c$ . Observe that g is uniformly computable from $f'$ , and so it is an instance of $\mathsf {D}^2_k$ .

Fix the largest $c < k$ such that $\rho _d$ is defined for all $d < c$ . (Again, there is such a c because $\rho _{-1}$ is defined.) Then by definition of g, for all sufficiently large n we have that $g(n) = c$ . Moreover, the set $\{ \sigma : f(\sigma ) = c \}$ must be dense above $\rho _{c-1}$ . Suppose not. Then we may fix the least $\rho \succeq \rho _{c-1}$ such that $f(\sigma ) \neq c$ for all $\sigma \succeq \rho $ . Since $\rho _{-1,n} \preceq \cdots \preceq \rho _{c-1,n}$ for all n, we have $\rho _{-1} \preceq \cdots \preceq \rho _{c-1}$ . Hence, by construction, $f(\sigma )> c$ for all $\sigma \succeq \rho $ . It follows that $c < k-1$ and that $\rho _{c,n} = \rho $ for all sufficiently large n. But then $\rho _c$ is defined, a contradiction.

To complete the proof, let M be any infinite monochromatic set for g, and let $n = \min M$ . By the argument above, the color of M under g must be c, so also $g(n) = c$ . By definition of g, this means that $\rho _{c-1,n} = \rho _{c-1,m}$ for all $m \geq n$ , hence $\rho _{c-1,n} = \rho _{c-1}$ . Since $\{ \sigma : f(\sigma ) = c \}$ is dense above $\rho _{c-1}$ , we can uniformly compute from $f \oplus M$ an $H \cong 2^{<\omega }$ monochromatic for f with color c and root extending $\rho _{c-1}$ .

Proof. We use a $\emptyset '$ oracle to enumerate $W_{f,C,\sigma }$ , letting $W_{f,C,\sigma }[s]$ denote the set of elements enumerated by stage s. In parallel, we define $\tau $ by finitely many finite extensions of $\sigma $ . At stage $0$ , we set $\tau _0 = \sigma $ . At stage $s> 0$ , we assume inductively that we have already defined $\tau _{s-1}$ . We now check if there exists $\tau ' < s$ extending $\tau _{s-1}$ such that for some $c \in C \setminus W_{f,C,\sigma }[s]$ we have that $f(\rho ) \neq c$ for all $\rho \succeq \tau '$ . If so, we let $\tau _s$ be the least such $\tau '$ , and we enumerate all corresponding c into $W_{f,C,\sigma }$ . Otherwise, we let $\tau _s = \tau _{s-1}$ . This completes the construction. We let $\tau = \lim _s \tau _s$ . It is then easy to see that $W_{f,C,\sigma }$ has the desired properties, as witnessed by $\tau $ .

Proof. Let $W_{f,\operatorname {ran}(f),\langle \rangle }$ be as given by Lemma 5.2, as witnessed by $\tau $ . Define $C = \operatorname {ran}(f) \setminus W_{f,\operatorname {ran}(f),\langle \rangle }$ . As noted above, C is non-empty. By property (2), we can computably construct a C-rake R for f of height $\omega $ with root $\tau $ . By property (3), R is good.

Proof. Say $|C| = k$ , so that $\mathrm {ht}(R) = mk$ for some $m \geq 1$ . We define a map $g : R \to C$ by reverse induction on rank. For each element of R of maximum rank, which is to say, for each leaf $\lambda $ , we let $g(\lambda ) = c_\lambda $ . Next, fix $\sigma \in R$ that is not a leaf and suppose we have defined g on all elements of R of larger rank. If $\sigma $ has a unique successor $\tau $ in R, let $g(\sigma ) = g(\tau )$ . Otherwise, since R is a C-rake, $\sigma $ has $k+1$ many successors $\tau _0,\ldots ,\tau _{k}$ in R. We can thus computably choose a $c \in C$ such that $g(\tau _i) = g(\tau _j) = c$ for some distinct $i,j \leq k$ . Let $g(\sigma ) = c$ . This completes the definition. Notice that for all $\sigma \preceq \tau $ in R, if $\mathrm {bl}_R(\sigma ) = \mathrm {bl}_R(\tau )$ then $g(\sigma ) = g(\tau )$ .

Now fix the color $c \in C$ that g assigns to the root of R. We define $h : 2^{< m} \to R$ by induction on finite strings. Let $h(\langle \rangle )$ be the shortest $\sigma \in R$ with $f(\sigma ) = c$ . Since R is a C-rake, we have that $\mathrm {bl}_R(h(\langle \rangle )) = 0$ , and so also $g(h(\langle \rangle )) = c$ , by construction of g. Suppose next that we have defined $h(\alpha ) \in R$ for some string $\alpha $ with $|\alpha | < m-1$ . Assume inductively that $\mathrm {bl}_R(h(\alpha )) = |\alpha |$ and that $f(h(\alpha )) = g(h(\alpha )) = c$ . Thus $h(\alpha )$ is not a leaf of R, nor in the same block as a leaf, since $\mathrm {bl}_R(\lambda ) = m-1$ for all leaves $\lambda $ of R. By construction of g, $h(\alpha )$ has two extensions $\tau ,\tau ' \in R$ such that $\mathrm {bl}_R(\tau ) = \mathrm {bl}_R(\tau ') = \mathrm {bl}_R(h(\alpha )) + 1$ and $g(\tau ) = g(\tau ') = c$ , and we may assume that $\tau ,\tau '$ are of minimal length with these properties (i.e., they are the shortest elements of the next block after that of $h(\alpha )$ ). Let $h(\alpha 0)$ and $h(\alpha 1)$ be the shortest incomparable $\sigma \succeq \tau $ and $\sigma ' \succeq \tau '$ in R with $f(\sigma ) = f(\sigma ') = c$ . As above, we have for each $i < 2$ that $\mathrm {bl}_R(h(\alpha i)) = \mathrm {bl}_R(h(\alpha )) + 1 =|\alpha | + 1$ and $f(h(\alpha i)) = g(h(\alpha i)) = c$ , so the inductive assumptions are maintained.

Let $S = \operatorname {ran}(h)$ . Then $S \cong 2^{< m}$ as witnessed by h, and S is monochromatic for f with color c. It follows from the construction that $S \trianglelefteq R$ . Consider any leaf $\sigma $ of S, and let $\lambda $ be the unique leaf of R extending $\sigma $ . Since $\mathrm {bl}_R(\sigma ) = \mathrm {bl}_R(\lambda )$ , we have that $c = f(\sigma ) = g(\sigma ) = g(\lambda ) = c_{\lambda }$ . Moreover, since $c = c_{\lambda } \in C \setminus W_{C,f,\lambda }$ , it follows by Lemma 5.2 (2) that $\{ \rho \in 2^{<\omega } : f(\rho ) = c \}$ is dense above some extension of $\lambda $ .

For each leaf $\lambda $ of R extending a leaf of S there therefore exists an $H_\lambda \cong 2^{<\omega }$ monochromatic for f with color c and with root $\rho _\lambda \succeq \lambda $ . Define

$$\begin{align*}H = S \cup \left( \bigcup_{\substack{\lambda \text{ a leaf of } R\\ \text{ extending a}\\ \text{leaf of } S}} H_\lambda \setminus \{ \rho_{\lambda} \} \right). \end{align*}$$

Then $H \cong 2^{<\omega }$ , H is monochromatic for f with color c, and $H^{\mathrm {rk} < \mathrm {ht}(R)/|C|} = H^{\mathrm {rk} < m} = S$ satisfies the conclusion of part (1).

Finally, note that k, m, and the function g are all uniformly computable from canonical indices for C, R, and the set $\{ c_\lambda : \lambda \text { a leaf of } R \}$ . Hence, so is the color assigned by g to the root of R, which is c, and from here we can uniformly compute the function $h : 2^{< m} \to R$ . Since $H^{\mathrm {rk} < \mathrm {ht}(R)/|C|} = \operatorname {ran}(h)$ , and h is strictly increasing under $\preceq $ , this completes the proof.

Proof. Apply Lemma 5.3 to obtain a non-empty finite set $C \subseteq \omega $ and a computable good C-rake $R_0$ for f of height $\omega $ . Let $\mathcal {P}$ be the class of all $H \trianglelefteq R_0$ such that H has height $\omega $ and is monochromatic for f. Notice that $\mathcal {P}$ is a $\Pi ^0_1$ class. The only part of the definition for which this may not be immediately apparent is the stipulation that each $H \in \mathcal {P}$ has height $\omega $ , and so in particular, that each element of H has a pair of incomparable successors in H. But since we want $H \trianglelefteq R_0$ , the latter condition can be expressed as

$$\begin{align*}(\forall \sigma \in H)(\exists \tau,\tau' \in H)[\sigma \preceq \tau,\tau' \wedge \tau \mid \tau' \wedge \mathrm{bl}_{R_0}(\tau) = \mathrm{bl}_{R_0}(\tau') = \mathrm{bl}_{R_0}(\sigma)+ 1]. \end{align*}$$

And since the set of all elements in $R_0$ in any given block is finite, the above is $\Pi ^0_1$ .

Since $R_0$ has height $\omega $ , it follows that for each $c \in C$ there exists $H \in \mathcal {P}$ monochromatic for f with color c, so in particular $\mathcal {P} \neq \emptyset $ . Now, let $\mathcal {Q}$ be the class of all $H \in \mathcal {P}$ such that $\neg \varphi (H^{\mathrm {rk} < n})$ holds for all n. This is again a $\Pi ^0_1$ class, and by hypothesis, $\mathcal {Q} = \emptyset $ . By compactness of Cantor space, we can fix an $m \in \omega $ such that for every $H \in \mathcal {P}$ there exists $n \leq m$ for which $\varphi (H^{\mathrm {rk} < n})$ holds. Let $R = R_0^{\mathrm {rk} < m|C|}$ , a good C-rake of finite height. If $H \cong 2^{<\omega }$ is any set monochromatic for f and satisfying $H^{\mathrm {rk} < m} \trianglelefteq R$ , then there must be a $G \in \mathcal {P}$ so that $H^{\mathrm {rk} < m} = G^{\mathrm {rk} < m}$ . Thus, $\varphi (H^{\mathrm {rk} < n})$ holds for some $n \leq m = \mathrm {ht}(R)/|C|$ , as was to be shown. By our use conventions (Section 2), this implies that $\varphi (H^{\mathrm {rk} < \mathrm {ht}(R)/|C|})$ holds as well.

Proof. Consider an arbitrary instance $\langle A,\Delta ,\Gamma \rangle $ of ${}^1\mathsf {TT}^1_{\mathbb {N}}$ . We have that $\Delta (A)$ is a coloring f of $2^{<\omega }$ , and if $H \cong 2^{<\omega }$ is any monochromatic set for f then $\Gamma (A, H)(0) \downarrow $ , and the output of this computation is a solution to $\langle A,\Delta ,\Gamma \rangle $ . We define an instance d of $\mathsf {RT}^1_{\mathbb {N}}$ , uniformly computable from $\langle A,\Delta ,\Gamma \rangle $ , such that if M is any infinite homogeneous set for d then $\langle A,\Delta ,\Gamma \rangle \oplus M$ uniformly computes $\Gamma (A, H)(0)$ for some such H.

To begin, we claim that there exists a non-empty finite set $C \subseteq \omega $ , a good C-rake R for f of finite height, and an $s \in \omega $ with the following properties. For each leaf $\lambda $ of R, and each choice of $c_\lambda \in C \setminus W_{f,C,\lambda }[s]$ , we apply Lemma 5.5 (2), relativized to A. This specifies an A-computable procedure for determining a $c \in \{ c_\lambda : \lambda \text { a leaf of } R \}$ and a finite set $S \trianglelefteq R$ of height $\mathrm {ht}(R)/|C|$ that is monochromatic for f with color c and such that $c_\lambda = c$ for every leaf $\lambda $ of R extending a leaf of S. The properties of C, R, and s are that this procedure halts, and that for the resulting S we have that $\Gamma (A,S)(0) \downarrow $ .

Note that the procedure from the lemma need not halt for arbitrary s, since we may have $W_{f,C,\lambda }[s] \subset W_{f,C,\lambda }$ for some $\lambda $ and therefore it could be that some $c_\lambda $ actually belongs to $W_{f,C,\lambda }$ . (Thus, the hypotheses of the lemma would not be met.) Likewise, nothing about the lemma guarantees that if the procedure halts and outputs c and S then necessarily $\Gamma (A,S)(0) \downarrow $ .

To prove the claim, note that if we take $\varphi (X)$ to be the $\Sigma ^0_1(A)$ formula stating that $\Gamma (A,X)(0) \downarrow $ , then f and $\varphi $ are precisely as in the hypothesis of Lemma 5.6, relativized to A. Thus, there exists a non-empty finite set $C \subseteq \omega $ and a good C-rake R for f of finite height such that for for every $H \cong 2^{<\omega }$ monochromatic for f with $H^{\mathrm {rk} < \mathrm {ht}(R)/|C|} \trianglelefteq R$ we have $\Gamma (A,H^{\mathrm {rk} < \mathrm {ht}(R)/|C|}) \downarrow $ .

At the same time, by Lemma 5.5, for every choice of $c_\lambda \in C \setminus W_{f,C,\lambda }$ for $\lambda $ a leaf of R, there is a $c \in \{ c_\lambda : \lambda \text { a leaf of } R \}$ and an $H \cong 2^{<\omega }$ monochromatic for f with color c such that $H^{\mathrm {rk} < \mathrm {ht}(R)/|C|} \trianglelefteq R$ and $c_\lambda = c$ for every leaf $\lambda $ of R extending a leaf of $H \cong 2^{<\omega }$ . Thus, in particular for this H, we have that $\Gamma (A,H^{\mathrm {rk} < \mathrm {ht}(R)/|C|}) \downarrow $ . But c and $H^{\mathrm {rk} < \mathrm {ht}(R)/|C|}$ are the number and finite set produced by the computable procedure in part (2) of that lemma. Hence, if s is large enough so that $W_{f,C,\lambda }[s] = W_{f,C,\lambda }$ for all leaves $\lambda $ of R, then C, R, and s are as desired. The claim is proved.

The definition of C, R, and s in the statement of the claim is uniformly $\Delta ^0_2$ in $\langle A,\Delta ,\Gamma \rangle $ . Hence, uniformly computably in $\langle A,\Delta ,\Gamma \rangle $ , we can approximate the least such C, R, and s by some $\langle C_t: t \in \omega \rangle $ , $\langle R_t: t \in \omega \rangle $ , and $\langle s_t: t \in \omega \rangle $ . That is, for each t, $C_t$ and $R_t$ are canonical indices of finite sets, $s_t$ is a number, and we have that $\lim _t C_t = C$ , $\lim _t R_t = R$ , and $\lim _t s_t = s$ . Of course, $R_t$ need not be good, or even a $C_t$ -rake, other than in the limit. Let $n_t$ denote the number of leaves of $R_t$ .

By Lemma 5.2 (1), relativized to A, we have that for each $t \in \omega $ and $i < n_t$ the set $W_{f,C_t,\lambda _i}$ is $\Sigma ^0_2(A)$ uniformly in f, $C_t$ and $\lambda _i$ . This means $W_{f,C_t,\lambda _i}$ can be A-computably approximated by a sequence of finite sets $\langle V_{t,i,u} : u \in \omega \rangle $ , uniformly in $C_t$ and $\lambda _i$ . More precisely, $V_{t,i,u} \subseteq C_t$ for all u, and for each $c \in C_t$ we have that $c \in V_{t,i,u}$ for cofinitely many u if and only if $c \in W_{f,C_t,\lambda _i}$ . By removing the least element whose membership in $V_{t,i,u}$ has changed most recently (with respect to u), or the least element if none such exists, we may further assume that $C_t \setminus V_{t,i,u} \neq \emptyset $ for all u. (If t happens to be large enough that $C_t = C$ and $R_t = R$ , then $C_t \setminus W_{f,C_t,\lambda _i} \neq \emptyset $ since R is a good C-rake for f. Then since $V_{t,i,u} \supseteq W_{f,C_t,\lambda _i}$ for all sufficiently large u, any such element removed will indeed belong to $C_t \setminus V_{t,i,u}$ for any such u.)

We now construct a coloring $d : \omega \to \omega $ . (We will argue below that this is actually an instance of $\mathsf {RT}^1_{\mathbb {N}}$ .) Fix t; we define $d(t)$ . First, compute $C_t$ , $R_t$ , and $s_t$ , and let $\lambda _0,\ldots ,\lambda _{n_t-1}$ be the leaves of $R_t$ , enumerated in lexicographic order. Then, for each $i < n_t$ , let $c_i$ be the least element of $C_t \setminus V_{t,i,t}$ , which is non-empty as discussed above. Finally, we define $d(t) = \langle C_t,R_t,c_0,\ldots ,c_{n_t-1} \rangle $ . Clearly, d is uniformly computable from $\langle A,\Delta ,\Gamma \rangle $ .

Now fix $t_0$ so that $C_t = C$ , $R_t = R$ , and $s_t = s$ for all $t \geq t_0$ . Let $n = n_{t_0}$ , the number of leaves of R, and enumerate the leaves in lexicographic order as $\lambda _0,\ldots ,\lambda _{n-1}$ . For each $t \geq t_0$ and $i < n$ , the approximations $\langle V_{t,i,u} : u \in \omega \rangle $ to $W_{f,C_t,\lambda _i} = W_{f,C,\lambda _i}$ now only depend on i (and not t). Thus, we may fix $t_1 \geq t_0$ so that additionally, for every $i < n$ and every $t \geq t_1$ , we have $V_{t,i,t} \supseteq W_{f,C,\lambda _i}$ . By construction of d, we have that if $t \geq t_1$ then $d(t) = \langle C,R,c_0,\ldots ,c_{n-1} \rangle $ for some $c_0,\ldots ,c_{n-1}$ with $c_i \in C \setminus V_{t,i,t}$ for each $i < n$ . The value of each $c_i$ may depend on t, but since $V_{t,i,t} \supseteq W_{f,C,\lambda _i}$ we have that $c_i \in C \setminus W_{f,C,\lambda _i}$ . Hence, the range of d is bounded, so d is an instance of $\mathsf {RT}^1_{\mathbb {N}}$ , as desired.

Now consider any infinite homogeneous set $M \subseteq \omega $ for d. As noted above, the color of M under d must be $\langle C,R,c_0,\ldots ,c_{n-1} \rangle $ , where $c_i \in C \setminus W_{f,C,\lambda _i}$ for each $i < n$ . By the choice of C and R, we can apply the A-computable procedure of Lemma 5.5 to find a $c \in \{ c_i : i < n \}$ and a finite set $S \trianglelefteq R$ of height $\mathrm {ht}(R)/|C|$ that is monochromatic for f with color c and such that $c_i = c$ for every $\lambda _i$ that extends a leaf of S, and $\Gamma (A,S)(0) \downarrow $ .