Proof Let $C\in \boldsymbol {a}$ be c.e., and fix a 1-1 enumeration $\{C_s\}_{s\in \mathbb {N}}$ of C. We build a perfect $\Pi ^0_1$ class P such that $\boldsymbol {W}[P]=\{\boldsymbol {a}\}$ using a stage by stage construction and satisfying requirements.

Proof Let $V_{e,i}$ be a uniform sequence of c.e. sets such that $S=\{e:\exists i|V_{e,i}|=\infty \}$ . Fix an $e_0\in S$ , and let $T_{j_0}$ be the computable tree that gives $W[T_{j_0}]=G(\{e_0\})$ in Theorem 2.1. If $\sigma _0\subset \sigma _1\subset \cdots \subset \sigma _n$ is a sequence of finite strings, and f is a total computable function, we define the tree $Tree(f,\sigma _0,\ldots ,\sigma _n)$ to be the tree which is the result of copying $T_{f(n)}$ above ${\sigma _n}* 0,{\sigma _n}* 1$ and copying $T_{f(m)}$ above ${\sigma _m}* 1$ for each ${\sigma _m}* 0\subseteq \sigma _n$ . Formally take the downwards closure of the set of strings $\{{\sigma _n}* 0* T_{f(n)}\}\cup \{{\sigma _n}* 1* T_{f(n)}\}\cup \{{\sigma _m}* 1* T_{f(m)}:m<n$ and ${\sigma _m}* 0\subseteq \sigma _n\}$ . We can extend this to an infinite $\sigma $ -sequence $Tree(f,\sigma _0,\sigma _1,\ldots )$ in a natural way, by copying $T_{f(m)}$ above ${\sigma _m}* 1$ for each m such that ${\sigma _m}* 0\subseteq \sigma _{m+1}$ . The tree $Tree(f,\sigma _1,\ldots )$ might not be computable unless the sequence of $\sigma _m$ is computable.

In the following we fix $\langle f, e,i,\rangle $ , where $e,i\in \mathbb {N}$ and f is a total computable function. We describe the construction of a $\Pi ^0_1$ class $P_{f,e,i}$ (the relationship $\langle e,i,f\rangle \mapsto P_{f,e,i}$ is an effective one). For ease of notation we temporarily drop $f,e,i$ . We build P by defining a computable tree U, in stages. At stage s we define U up till level s. To make U computable we only enumerate strings of length s at stage s. We also define a sequence of finite strings $\sigma _{0,s}\subset \sigma _{1,s}\subset \cdots $ . The intention is that if $|V|=\infty $ then $\cup _n\lim _s\sigma _{n,s}$ will be a path in $P=[U]$ which is of the same Turing degree as $W_e$ , with all other paths in P having degrees of paths occurring in the $T_{f(n)}.$

If $|V|< \infty $ , we still make sure that P is perfect, now containing only paths whose degrees are realized in the $T_{f(n)}$ . We do this by aiming to make $P=[Tree(f,\lim \sigma _0,\lim \sigma _1,\ldots )]$ , by making $U=Tree(f,\lim \sigma _0,\lim \sigma _1,\ldots )$ except on the dead ends introduced by the coding of $W_e$ .

The plan: The idea is to break down the membership question $e\in S$ into infinitely many $\Pi ^0_2$ questions (about the cardinality of $V_{e,i}$ ). We approximate the truth of $e\in S$ locally by examining if $|V_{e,i}|=\infty $ , and building a class $P_{e,i}$ . We use $P_{e,i}$ to denote $P_{f,e,i}$ when f is simply the function $f(x)=j_0$ . At the end we combine the different $\Pi ^0_1$ classes $P_{e,i}$ together. Locally if $|V_{e,i}|=\infty $ , then we must make some $\alpha \in P_{e,i}$ such that $\alpha \equiv _T W_e$ . The standard way of doing this is to make $\alpha $ the complement of the retraceable $\Pi ^0_1$ set associated with $W_e$ ; we refer the reader to Cenzer, Downey, Jockusch, and Shore [Reference Cenzer, Downey, Joskusch and Shore2], where they showed that $\{\boldsymbol {w},\boldsymbol {0}\}$ is realizable for any $\boldsymbol {w}$ . The problem is that this gives a rank $1$ class in which some $\alpha \equiv _T W_e$ sits, and contains computable (in fact, isolated) paths. Inside $P_{e,i}$ we have to be careful not to introduce c.e. degrees which are not in $G(S)$ . The only reason why the isolated paths show up is because we have to allow a different path to be extendible in the tree; this is to allow for the later coding of some $n\in W_e$ . Observe that for the purpose of making $\alpha \equiv _T W_e$ , it would not matter if we keep many paths (associated with $n\in W_e$ ) extendible on the tree instead of only two. Hence if $\alpha _s$ is the current approximation of an initial segment of $\alpha $ , we would copy the tree $T_{j_0}$ above every node distinct from $\alpha _s$ which we currently want to keep extendible. If later on some m enters $W_e$ and we need to switch $\alpha _{s+1}$ to a different string $\sigma $ , then we simply take the leftmost extension of $\sigma $ of length s. On the other hand if $\sigma $ is such a string which is never used, then we will copy the entire tree $T_{j_0}$ above $\sigma $ . Doing so puts $deg(W_{e_0})$ into $W[P_{e,i}]$ —no other c.e. degree is introduced. Of course we only extend $\alpha _s$ if new numbers enter $V_{e,i}$ . If $|V_{e,i}|<\infty $ then we will copy $T_{j_0}$ above every terminal node.

Following the plan above gives effectively the class $P_{e,i}$ for each pair $e,i$ , which introduces no new c.e. degree and contains $deg(W_e)$ iff $|V_{e,i}|=\infty $ . Finally we want to glue the separate $P_{e,i}$ ’s into a single class P. A slight technical issue arises if we take the simple amalgamation $P=\{0^n 1* P_{n}:n\in \mathbb {N}\}$ ; even though none of the $P_n$ introduces new c.e. degrees, however, the support $X=\emptyset $ will be in P. To get around this problem we nest the construction of each $P_x$ within the outer construction of $P_{e_0}$ . This will create X in P with Turing degree $X\equiv _T W_{e_0}$ , which supports the trees associated with $P_0,P_1,\ldots $ .

Construction of U: At stage $1$ declare $\sigma _0=\langle \rangle $ and $\sigma _1,\sigma _2,\ldots $ undefined, and enumerate $\langle \rangle ,0,1$ into U. For a $\sigma \in U$ , when we say we copy the next level of $T_{j}$ above $\sigma $ at stage s we mean the following. For each $\sigma *\tau \in U$ such that $|\sigma *\tau |=s-1$ , we enumerate $\sigma *\tau * 0$ into U if $\tau * 0$ is in $T_{j}$ ; similarly for $\sigma *\tau * 1$ . Clearly if $\sigma \in U_{|\sigma |}$ and we issue this command for $\sigma $ at every stage after $|\sigma |$ , then $T_{j}$ will be copied successfully above $\sigma $ ; i.e., $U\cap \{\tau :\tau \supseteq \sigma \}=\sigma * T_{j}$ . A string is said to require copying at stage s, if it is either of the form ${\sigma _m}* 1$ for some convergent ${\sigma _m}* 0\subseteq \sigma _{m+1}$ , or it is ${\sigma _m}* 0$ or ${\sigma _m}* 1$ for the maximal convergent $\sigma _m$ . When we say we copy $T_f$ above such a $\sigma $ we mean that we copy the next level of $T_{f(m)}$ for the corresponding m. The actions at stage $s+1$ involve the following. Look for the largest n such that $\sigma _n\downarrow $ . See if there is some least $2m<n$ such that $m\in W_{e,s+1}-W_{e,s}$ :

1. (i) m exists. Let $\tau $ be the leftmost string of length s extending ${\sigma _{2m}}* 1$ , such that $\tau \in U$ . Move $\sigma _{2m+1}=\tau $ and put both $\tau * 0,\tau * 1$ in U. Declare $\sigma _{2m+2},\sigma _{m+3},\ldots $ undefined.

2. (ii) m does not exists. If $s+1\in V$ we let $\tau $ be the leftmost string of length s extending ${\sigma _n}* r$ which is in U. Here $r=0$ if n is odd, and $r=W_{e,s+1}(\frac {1}{2}n)$ if n is even. Define $\sigma _{n+1}=\tau $ and enumerate both $\tau * 0,\tau * 1$ into U, otherwise if $s+1\not \in V$ do nothing.

Copy $T_{f}$ above every string which needs copying, and go to the next stage. This ends the construction. It is easy to see that U is a computable tree.

We are ready to show that there is a perfect $\Pi ^0_1$ class P such that $W[P]=G(S)$ . Let $f(x)=j_0$ for every x; we want to combine all the classes $P_{f,e,i}$ . Let g be a computable function defined by the following: $g(2x)=j_0$ and for the odd inputs we select an index such that $[T_{g(2\langle e,i\rangle +1)}]=P_{f,e,i}$ . From Lemmas 3.2 and 3.3, and the fact that $T_{j_0}$ is perfect, we may conclude that each $T_{g(x)}$ is perfect. Since $e_0\in S$ pick $i_0$ such that $|V_{e_0,i_0}|=\infty $ , and let $P=P_{g,e_0,i_0}$ . It is not hard to verify that P itself is also perfect, and $W[P]=G(S)$ .

Proof Suppose S contains an index for $\emptyset $ . We want to produce P of rank $2$ . We run the argument in the proof of Theorem 3.1, except now we let $T_{j_0}=\{0^m:m\in \mathbb {N}\}$ , and set $f(x)=j_0$ for all x. Each of the $P_{f,e,i}$ has rank at most $1$ —if $V_{e,i}$ is finite then so is $P_{f,e,i}$ and if $V_{e,i}$ is infinite then $P_{f,e,i}$ has rank $1$ . We let $P=\{0^{\langle e, i \rangle } 1* P_{f,e,i}:\langle e, i \rangle \in \mathbb {N}\}$ , which is of rank at most $2$ , and $W[P]=G(S)$ .

Proof For any c.e. set $L>_T\emptyset $ , one can effectively obtain a c.e. set D and reduction $\Psi $ such that $D=\Psi ^L$ and $D>\emptyset $ and $D\not \geq _T L$ . If we iterate this process starting with some noncomputable low $_2$ c.e. set L, we get computable increasing functions g and $\ell $ such that $L=W_{g(0)}>_T W_{g(1)}>_T W_{g(2)}>_T\cdots $ , and $W_{g(n)}=\Phi _{\ell (n)}^L$ for all n. Let $T_e$ be the eth primitive recursive tree. Since the $W_{g(e)}$ are uniformly computable from L via $\ell $ , it follows that there are L-computable relations $R_1$ and $R_2$ such that

$$\begin{align*}R_1(e,i,x,s)&\Leftrightarrow \Phi_i^{W_{g(e)}}\upharpoonright{x}[s]\downarrow \text{and is on } T_e,\\R_2(e,i,j,x,m,s,\sigma)&\Leftrightarrow \Phi_i^{W_{g(e)}}\upharpoonright{m}[s]\downarrow=\sigma\textrm{ and }\Phi_j^\sigma\upharpoonright{x}[s]\downarrow=W_{g(e)}\upharpoonright{x}.\end{align*}$$

We define $n\not \in S$ iff whenever $n=g(e)$ , then there are some $i,j$ for which:

1. (i) $\Phi _i^{W_{g(e)}}$ is total and is in $[T_e]$ , and

2. (ii) $\Phi _j(\Phi _i^{W_{g(e)}})$ is total and equals $W_{g(e)}$ .

The quantifier for e is bounded since g is increasing. Clause (i) can be expressed as $\forall x\exists sR_1(e,i,x,s)$ . Since this is $\Pi ^0_2(L)$ , and L is low $_2$ it is also $\Sigma ^0_3$ . Similarly clause (ii) can be expressed as $\forall x\exists m\exists s\exists \sigma R_2(e,i,j,x,m,s,\sigma )$ which is also $\Sigma ^0_3$ . Hence S is $\Pi ^0_3$ , and it is easy to verify that there can be no $\Pi ^0_1$ class P such that $W[P]=G(S)$ . To see that S can be made to contain $\emptyset $ , note that $S\cup \tilde {S}$ is also good for any $\Pi ^0_3$ set $\tilde {S}$ where $G(\tilde {S})\cap G(W_{g(n)})=\emptyset $ for every n.

Proof For simplicity of notation, we will pretend that $\mathcal {C}=\omega $ and also write $\Phi _{j(e)}=\Gamma _e$ for the domination function for e.

In Theorem 3.1, we realized each $\Sigma _3^0$ set of e’s. Here we will be dealing with a $\Sigma _4^0$ set S. Thus, $e\in S$ iff $\exists p \forall m \exists s \forall t R(e,p,m,s,t)$ . For a fixed p, should the $\Pi _3^0$ condition $\forall m \exists s \forall t R(e,p,m,s,t)$ hold, we need to code $W_e$ .

Again, the idea will be to devote a unique part of the $\Pi _1^0$ class, to the pair $\langle e,p\rangle .$ These indices indicate that we are concerned with $W_e$ (with domination function $g_e^{W_e}),$ and p is the $\Sigma _4^0$ witness.

We will be making the class to be of rank 2. Concentrating on this fixed pair, $\langle e,p\rangle $ we are thinking of the part of the class being a rank 1 subclass extending some fixed $\sigma *1=\nu $ , say. The idea is again to more-or-less follow the coding of Cenzer, Downey, Jockusch, and Shore [Reference Cenzer, Downey, Joskusch and Shore2], where they showed that $\{\mathbf {w},\mathbf {0}\}$ is realizable for any $\mathbf {w}$ . In this construction, it is convenient to allow $\langle m,0\rangle $ to correspond to the (initial) coding location for “ $m\in W_{e,s}$ ”. That is, initially we will think of $\nu *0^{\langle m, j\rangle }$ as the coding location $c(\langle m,j\rangle ,s)$ ; and in the construction, $c(\langle m,0\rangle ,s)$ for “ $m\in W_e$ ”. Should m enter $W_{e,s}$ , and conditions are opportune, we would code by changing this to $\nu *0^{\langle m, 0\rangle -1}*1^{s-\langle m,0\rangle }*0^i$ and re-assign the coding location for $\langle m',j'\rangle $ to be $c(\langle m', j'\rangle , s)$ accordingly for $\langle m,0\rangle <\langle m',j'\rangle $ . (Of course all other paths in this subclass are isolated ones of the form $\rho *1^\omega $ , as in Theorem 3.1.) Again this makes a rank 1 subclass extending $\nu $ .

As we see below, there are other reasons we might move a coding location. In the construction, we would only allow $c(n,s)$ to move if $W_{e,s}\upharpoonright \gamma (n,s)$ changes. This is to keep the degree below that of $W_e$ . Without loss of generality we will assume that for $i\le s$ , $\gamma (i,s), \Gamma ^{W_e,s}(i)\downarrow [s]<s.$

We only want to code $W_e$ if $\Sigma _4^0$ is correct. We do not want to code $W-e$ should the $\Pi _4^0$ outcome be correct. Thus, each time $\Sigma _3^0$ of the triple corresponding to $e,p,m$ in $\neg R(e,p,m,s,t)$ looks correct, what we would like to do is to move the coding location for $c(\langle m+i,0\rangle s)$ for $i\ge 0$ spontaneously to $\nu *0^{\langle m,0\rangle -1}*1^{s-\langle m,0\rangle }*0^i$ . We call this action kicking. Then, if $e\not \in S$ , so that the $\Pi _4^0$ outcome is correct and has witness m for this p, eventually all the paths in in the class extending $\nu $ will be computable, as they will be driven to infinity. If $e\in S$ , and p is the correct $\Sigma _4^0$ witness, then since each triple $e,p,m$ has a $\Sigma _2^0$ witness, for each m the above process will fire only finitely often, and hence only move the coding location for $m\in W_e$ a finite number of times.

Now, of course there is a small problem with all of this, as it upsets the computation of the coding locations from $W_e$ . In the $\Sigma _3^0$ case in Theorem 3.1, locations $c(m,s)$ only moved when $m'\le m$ entered $W_{e,s'} $ for some $s'\le s$ .

The solution is to use the highness of $W_e$ . At the same time as the above, we will be monitoring $\Gamma _e^{W_{e,s}}(k)$ for $k\le s$ , and building a sequence of potential computable functions $f_k$ . The reader should note that if the $\Pi _4^0$ outcome is true, it is enough the force almost all coding markers to infinity. That is, even if $\langle e,p,m\rangle $ is the witness of the failure of the $\Sigma _4^0$ condition, it is enough that some $c(n',s)\to \infty $ for some $n'\ge \langle m,0\rangle $ .

So suppose that the we have a fixed $e,p,m$ , and m seems to be a witness $\forall s\exists t\neg R(e,p,m,s,t)$ because this fires. At this stage we would define $f_{\langle e,p,m\rangle }(\langle m,0\rangle ) $ Footnote ² to be large and fresh, and continue to do this for $f_{\langle e,p,m\rangle }(\langle m,k\rangle )$ (i.e., for more and more k each time it continues to fire) until we see some $W_e$ change on $\gamma (\langle m,k\rangle )$ to make $g_e^{W_e}(\langle m,k\rangle )>f_{\langle e,p,m\rangle }(\langle m,k\rangle )$ , which it must as $\Gamma ^{W_e}$ is dominant, and $f_{\langle e,p,m\rangle }$ will be total if it fires infinitely often. Note that in the construction, for each k where we get a $W_e$ change on $\gamma (\langle m,k\rangle )$ at some $t\ge s$ , we would kick the coding markers $c(z,t)$ for $z\ge \langle m,k\rangle $ . Additionally, we would kick all coding markers $c(\langle m', q\rangle ,s)$ for $m'>m,k $ to also extend $c(\langle m, s\rangle t+1)$ . Here s is the stage where we first saw a $\Gamma ^{W_e}$ change for some $\langle m,k\rangle $ , and hence this is last defined value of $f_{\langle e,p,m\rangle }$ . Notice that now the original ordering of the coding markers is no longer valid. If we need to kick for m because the $\Pi _4^0$ outcome seems correct, then we will move all coding markers for $m'>m,k$ , and they will now be above $c(\langle m,s+1\rangle , t+1)$ . If the process begins anew, and k was the place where $\Gamma ^{W_e}$ is dominant above $\langle m, k\rangle $ , then if it fires again, all coding markers corresponding to $m'>m,k$ will again be moved as we’d get a change of $W_e$ on $\gamma (\langle m,s+1\rangle ,t+1).$

The key point is that if $\langle e,p,m\rangle $ fires infinitely often, than for almost all $m'$ coding markers corresponding to $m'$ will be driven to infinity, as required. The remaining details work as in the $\Sigma _3^0$ case.

Proof (iii) $\Rightarrow $ (ii): Immediate.

(iv) $\Rightarrow $ (v): It is easy to see that (iv) is equivalent to the condition “there is a computable function $\hat {g}$ such that for every n, $n\in G(S)\Leftrightarrow \Phi ^{W_n}_{\hat {g}(n,m)}$ is total for some m”. For any inputs $X,n,a,b$ and c, let $R^X(n,a,b,c)$ hold iff $\Phi ^{W_n}_{\hat {g}(n,a)}(b)[c]\downarrow $ with use u and $W_n[s]\upharpoonright u\subset X$ . Then $R^X$ is clearly total and X-computable for all X, and has the required property.

(v) $\Rightarrow $ (iv): Let $\Phi ^X_{\hat {g}(n,a)}(b)\downarrow $ iff $R^X(n,a,b,c)$ holds for some c.

(ii) $\Rightarrow $ (v): The predicate within square brackets in the proof of Proposition 1.3 is computable in $W_e$ . As above, this is obviously witnessed by a truth-table functional by comparing $W_e[s]$ with the oracle.

(i) $\Rightarrow $ (iii): This is Theorem 3.1.

(iv) $\Rightarrow $ (i): Fix $\hat {g}$ such that for every e, $e\in G(S)\Leftrightarrow \Phi ^{W_e}_{\hat {g}(e,q)}$ is total for some q. If $S=\emptyset $ then there is nothing to do, so fix a c.e. set D with an index in S.

For each pair $e,q$ such that $e,q\in \omega $ , we build (effectively) a c.e. set $V_{e,q}$ . At the end we will take R to be the set of all indices for the sets $V_{e,q}$ (ranging over all pairs $e,q$ ) and show that R represents $G(S)$ . R is of course computable.

Fix a pair $e,q$ . We describe how to build $V=V_{e,q}$ . We also define markers $m_0[s]<m_1[s]<\cdots $ for V. At each stage s define $n_i[s]=\varphi ^{W_e}_{\hat {g}(e,q)}(i)[s]$ , if this converges. We assume that $n_{i+1}[s]\downarrow \Rightarrow n_i[s]\downarrow $ and is less than $n_{i+1}[s]$ . We shall ensure that $m_k[s]\downarrow $ if and only if $n_k[s]\downarrow $ for all $k<s$ . At stage $0$ set all our parameters undefined and set $V=\emptyset $ . Now assume we are at stage $s+1$ . Let $z\in W_e[s+1]-W_e[s]$ be the least such, if it exists. If $z<n_k[s]$ , pick the least such k with $n_k[s]\downarrow $ and enumerate $m_k[s]$ into V. Note that $n_0[s+1],\ldots ,n_{k-1}[s+1]$ are all still defined since there has been no $W_e$ change below them, and therefore $m_0,\ldots , m_{k-1}$ are all still defined. For every number x larger than $m_{k-1}$ that has been assigned to code $D(i)$ for some i, we set $D(i)=V(x)$ . See if $n_k[s+1]\downarrow $ , and if so set $m_k$ to be a large fresh number, and assign a new number less than $m_k$ to code the next $D(i)$ . Otherwise set $m_k\uparrow $ .

If z does not exist, or if z is larger than any $n_k[s]$ seen at the previous stage s, we do the following. For the least k such that $n_k[s]\uparrow $ , and for every number x larger than $m_{k-1}$ that has been assigned to code $D(i)$ for some i, we set $D(i)=V(x)$ . Assign a new number to code the next $D(i)$ . Check and see if $n_{k}[s+1]\downarrow $ . If so, set $m_k$ to be a large fresh number, otherwise do nothing else. This ends the definition of V.

Now we state some properties about V. There are two possibilities:

1. (I) $\Phi _{\hat {g}(e,q)}^{W_e}$ is total. Then $\lim _s n_k[s]$ and $\lim _s m_k[s]$ exist for every k. In this case it is easy to check that $W_e\equiv _T V$ . To compute $W_e\leq _T V$ , fix x and wait for a stage large enough such that $m_x[s]\downarrow $ and V is correct up to $m_x[s]$ . Then $x\in W_e$ iff $x\in W_e[s]$ . To compute $V\leq _T W_e$ we wait for a stage large enough such that $n_k[s]\downarrow $ and $W_e$ is correct up to $n_k[s]$ . This means that $m_k[s]\downarrow $ and the construction will never again change V below $m_k[s]$ .

2. (II) $\Phi _{\hat {g}(e,q)}^{W_e}$ is not total. Then there must be some least k such that either $\lim _s n_k[s]=\lim _s m_k[s]=\infty $ , or $n_k[s], m_k[s]\uparrow $ for almost every s. Let r be the final value of $m_{k-1}$ . (If $k=0$ set $r=0$ ). In the first case there are infinitely many stages where $m_k$ is enumerated into V, and at every such stage we will assign for every i, some number larger than r to code $D(i)$ . We will also be able to ensure that $V(x)=D(i)$ for every $x>r$ assigned to code some $D(i)$ . In the second case $m_{k}[s]\uparrow $ for almost all s, and the construction will attend to the coding of D above r at almost every stage. In either case, we see that every bit of V larger than r is either assigned to code some $D(i)$ , or is never again modified after $m_k$ is lifted past it. Thus, $V\equiv _T D$ .

Now we verify that $G(R)=G(S)$ . Suppose that $V_{e,q}$ is included in R. Then either (I) or (II) holds for the construction of $V_{e,q}$ . In case (II) $V_{e,q}\equiv _T D$ and so $V_{e,q}$ is represented in $G(S)$ . In case (I) $\Phi _{\hat {g}(e,q)}^{W_e}$ is total and $V_{e,q}\equiv _T W_e$ . This also means that $V_{e,q}$ is represented in $G(S)$ . Hence $R\subseteq G(S)$ .