Proof. First, suppose that there are only finitely many blocks that do not embed into a later block. After some finite initial segment, all blocks that occur embed into some later block. By non-uniformly fixing this initial segment, we can assume that every block embeds into another block. We show that for any tuple $\overline {c}$ , there is some tuple $\overline {a}$ which is d-free over $\overline {c}$ . Given $\overline {c}$ , let $\overline {a}$ be some f-block of size greater than one such that all its elements are greater than those of $\overline {c}$ . Since we assumed that f was not intrinsically computable, it cannot be the identity almost everywhere and so there must be infinitely many blocks of size greater than one. We claim that $\overline {a}$ is d-free over $\overline {c}$ .

Now, with $\overline {c}$ , $\overline {a}$ as above, suppose there is some quantifier-free formula $\varphi (\overline {x}, \overline {u}, \overline {v})$ and tuple $\overline {b}$ such that $\varphi (\overline {c}, \overline {a}, \overline {b})$ is true. We make some simplifying assumptions. First, we may assume that there is no repetition amongst the elements of all of the tuples. Second, by including in $\bar {c}$ any elements of $\bar {b}$ which are less than the elements of $\bar {a}$ , we may assume that all of the elements of $\bar {b}$ are greater than all of the elements of $\bar {a}$ . And third, we can assume that $\bar {b}$ consists of n distinct adjacent blocks $\overline {b} = \overline {b}_{1}\overline {b}_{2} \dots \overline {b}_{n}$ with $\overline {b}_1$ the first block after $\overline {a}$ .

Now define $\overline {a}' = \overline {a} + 1$ and $\overline {b}' = \overline {b} + 1$ . Then $\overline {c} \, \overline {a}' \overline {b}'$ has the same order type as $\bar {c},\bar {a},\bar {b}$ and so still satisfies $\varphi (\overline {x}, \overline {u}, \overline {v})$ . However the values of f on $\overline {a}$ are not the same as f on $\overline {a}'$ as $\overline {a}$ is a block and $\overline {a}'$ is not. Furthermore, suppose $\psi (\overline {x}, \overline {u}, \overline {v})$ is some existential formula satisfied by $\overline {c}$ , $\overline {a}'$ , $\overline {b}'$ . We can again replace this by a quantifier-free formula $\chi (\overline {x}, \overline {u}, \overline {v}, \overline {w})$ which is satisfied by $\overline {c}$ , $\overline {a}'$ , $\overline {b}', \overline {e}'$ for some tuple $\overline {e}'$ . Noting that there are no gaps between the elements of $\bar {a}',\bar {b}'$ , we may split $\bar {e}'$ into $\bar {e}_1'\bar {e}_2'$ where the elements of $\bar {e}_1'$ are less than the elements of $\bar {a}'$ and the elements of $\bar {e}_2'$ are greater than the elements of $\bar {a}'\bar {b}'$ .

Now let $\overline {a}"$ be the image of $\bar {a}$ in some block into which it embeds, and similarly let $\bar {b}_1",\ldots ,\bar {b}_n"$ be the images of $\bar {b}_1,\ldots ,\bar {b}_n$ in blocks into which they embed, choosing these images sufficiently large that $\bar {c} < \bar {e}_1' < \bar {a}" < \bar {b}_1" < \bar {b}_2" < \cdots < \bar {b}_n"$ . Finally, choose $\bar {e}_2"$ larger than all of these. Let $\bar {e}" = \bar {e}_1' \bar {e}_2"$ .

Our construction has ensured that $\overline {c} \, \overline {a}" \overline {b}" \overline {e}"$ has the same order type as $\overline {c} \, \overline {a}' \overline {b}' \overline {e}'$ and so $\overline {c} \, \overline {a}" \overline {b}"$ satisfies the formula $\psi (\overline {x}, \overline {u}, \overline {v})$ . Moreover, the values of f on $\overline {c} \, \overline {a}" \overline {b}"$ are the same as the values on $\overline {c} \, \overline {a} \, \overline {b}$ . Thus, $\overline {a}$ is d-free over $\overline {c}$ , as desired. By Theorem 3.2, the degree spectrum of f strictly contains the c.e. degrees on a cone.

In the other direction, suppose that there are infinitely many f-blocks which embed into no larger f-blocks. We work on the cone on which we can compute $\alpha _f$ and whether a given block embeds into a later block. We will show that, for any X on this cone, if $f^{\mathcal {A}}$ is the copy of f in an X-computable copy $(\mathcal {A}, <_{\mathcal {A}})$ of ${(\omega ,<)}$ then $f^{\mathcal {A}}$ is of X-c.e. degree. This implies that the degree spectrum of R on a cone is contained in the c.e. degrees. To simplify notation, we will suppress X, assuming that $\alpha _f$ is computable and that we can compute whether a given block embeds into any later block.

To show that $f^{\mathcal {A}}$ is of c.e. degree, we will show that for any k it computes the kth element of $\mathcal {A}$ . This implies that it computes the successor relation on $\mathcal {A}$ . Recall that the successor relation on $\mathcal {A}$ is always of c.e. degree and computes $f^{\mathcal {A}}$ , so this will imply that $f^{\mathcal {A}}$ is of c.e. degree.

Given k, look in ${(\omega ,<)}$ for a block, all of whose elements are greater than k, which does not embed into any greater block. Suppose that this block is the elements $(\ell ,\ldots ,\ell +m-1)$ of $\omega $ , so that there are $\ell $ elements less than this block. Let the block type be I. In $\mathcal {A}$ , look for a tuple of elements $a_0 < a_1 < \cdots < a_\ell < \cdots < a_{\ell +m-1}$ of $\mathcal {A}$ such that $f^{\mathcal {A}}$ on $(a_\ell ,\ldots ,a_{\ell +m-1})$ has block type I. (This means that f on $(\ell ,\ldots ,\ell +m-1)$ is the same as $f^{\mathcal {A}}$ on $(a_\ell ,\ldots ,a_{\ell +m-1})$ .) Once we find such elements, we know that $a_k$ is the kth element of $\mathcal {A}$ . This is because block type I does not embed into any later $f^{\mathcal {A}}$ -block (and noting that a block cannot embed into a sequence of blocks without embedding into one of them).

Proof. Suppose that we have a weak f-coding sequence with intervals $[a_1, b_1], [a_2, b_2], \ldots $ and maps $\varphi _i$ . Recall that all of the maps are non-decreasing.

We say that an element $x \in [a_1,b_1]$ eventually increases to infinity if, for each y, there is some composition of the $\varphi $ which, when applied to x, increase it above y. That is, if for every y there is some n such that $\varphi _n(\cdots (\varphi _1(x))\cdots )> y$ . If x does not eventually increase to infinity, then there must be some y such that for all sufficiently large n, $\varphi _n(\cdots (\varphi _1(x))\cdots ) = y$ . Note that $b_1$ eventually increases to infinity, as otherwise there would be n such that for all $m \geq n$ the map $\varphi _m$ would be the identity, contradicting (4). Also, we note that if x eventually increases to infinity, then so does every $y \geq x$ , and also every y in the same block as x.

Let $a_1' \in [a_1,b_1]$ be such that every element of $[a_1,a_1')$ does not eventually increase to infinity, and every element of $[a_1',b_1]$ does eventually increase to infinity. By the above remarks, $[a_1',b_1]$ is closed under blocks. First, we argue that $\varphi _1$ does not preserve f when restricted to $[a_1',b_1]$ . This is because it does preserve f when restricted to $[a_1,a_1')$ , as we now argue. There is n even such that for all $x \in [a_1,a_1'),$

$$\begin{align*}\varphi_n(\cdots(\varphi_1(x))\cdots) = \varphi_{n+1}(\varphi_n(\cdots(\varphi_1(x))\cdots)).\end{align*}$$

But $\varphi _{n+1} \circ \cdots \circ \varphi _2 = \varphi _{2 \mapsto n+1}$ preserves f since n is even. So $\varphi _1$ preserves f on $[a_1,a_1')$ , and so does not preserve f when restricted to $[a_1',b_1]$ .

Now since $a_1'$ eventually increases to infinity, there must be some odd n with $\varphi _n(\cdots (\varphi _1(a_1')))> b_1$ . (Since $b_1$ is the largest element of its block, this also implies that all elements of any block containing the image of an element $[a_1',b_1]$ under $\varphi _n \circ \cdots \circ \varphi _1$ is also greater than $b_1$ .) We now replace $[a_1,b_1]$ by $[a_1',b_1]$ and delete all of the blocks $[a_2,b_2],\ldots ,[a_n,b_n]$ so that the next block after $[a_1',b_1]$ is $[a_{n+1},b_{n+1}]$ . We still have a map $\varphi _1^* = \varphi _{1 \mapsto n+1}$ from $[a_1',b_1]$ to $[a_{n+1},b_{n+1}]$ . Now we replace $[a_{n+1},b_{n+1}]$ with the smallest interval $[b_{n+1}',a_{n+1}']$ which contains the image of $[a_1',b_1]$ under $\varphi _1^*$ and which is closed under blocks. By choice of n, we get that $a_{n+1}'> b_1$ .

Thus we have now obtained

This is a weak f-coding sequence with $a_{n+1}'> b_1$ . By continuing this process, now with the second interval $[a_{n+1}',b_{n+1}']$ , and so on, we eventually obtain an f-coding sequence.

Proof. Suppose there is an infinite f-coding sequence $[a_1, b_1], [a_2,b_2], [a_3,b_3], \ldots $ with functions $\varphi _i$ . We show that, on the cone above $\alpha _f$ and this coding sequence, the degree spectrum of f is all $\Delta _2^0$ degrees. As usual, we will suppress the base of the cone and assume that $\alpha _f$ and this coding sequence are computable. We will also assume, by fixing a finite initial segment, that every block embeds into a later block. Then we must show that the degree spectrum of f is all $\Delta ^0_2$ degrees. Since the degree spectrum of f is contained in the $\Delta ^0_2$ degrees, we must show that every $\Delta ^0_2$ is in the degree spectrum of f.

Fix a $\Delta _2^0$ set X. We will build a computable copy $\mathcal {A}$ of ${(\omega , <)}$ such that $f^{\mathcal {A}} \equiv _T X$ . During this construction, the elements we place into the linear order will fall into one of two categories: coding elements and padding elements. Padding elements are elements that are added to move coding elements, and play no other role in the coding. We will have to ensure that the approximation $f^{\mathcal {A}_s}$ to $f^{\mathcal {A}}$ stays the same on the padding elements; we can do this because every block embeds into a later block. For each $e \in \omega $ , there will be some collection of coding elements corresponding to e. We divide these into the initial coding elements which are the elements first chosen to code whether $e \in X$ . It is these elements on which we will check the values of $f^{\mathcal {A}}$ to determine whether e is in X. When we first start coding whether $e \in X$ , we will add initial coding elements to $\mathcal {A}$ corresponding (via the current guess at the isomorphism $\mathcal {A} \to {(\omega ,<)}$ ) to an interval $[a_i,b_i]$ with i even if $e \notin X$ or odd if $e \in X$ . If, at some later stage, we see that e has entered or exited X, we insert elements into $\mathcal {A}$ so that these initial coding elements now correspond to elements of some $[a_j,b_j]$ , $j> i$ , with the parity of j depending on whether $e \in X$ or $e \notin X$ ; moreover, the elements in this interval that they correspond to should be the images, under the $\varphi $ ’s, of the elements they previously corresponded to. It is also possible that some $e' < e$ will enter or exit X, causing us to add elements to $\mathcal {A}$ for the sake of $e'$ , and thus requiring us to take action for the coding elements for e as well. Because the embeddings $\varphi $ might not be surjective, as the construction progresses, the initial coding elements will “gather” other subsequent coding elements. The coding elements for e will be consecutive and we call them the coding segment for e.

The goal of the construction is to move the coding elements so that the values of $f^{\mathcal {A}}$ on them mirror whether the corresponding e is in X or not, while also ensuring that once padding elements have been added to the structure, the value of $f^{\mathcal {A}}$ on those elements does not change. Given $e' < e$ , the coding segment for e will be greater than the coding segment for $e'$ , so that we can move the segment for e without moving the segment for $e'$ .

Construction: For each e, starting from stage e where they are defined, we will have the initial coding elements $\bar {u}_e \in \mathcal {A}$ , the full coding segment $\bar {v}_e \in \mathcal {A}$ containing $\bar {u}_e$ , and a restraint $r_e = \max _{\mathcal {A}} \bar {v}_e \in \mathcal {A}$ which is the $\mathcal {A}$ -largest element of $\bar {v}$ . These are all tuples of $\mathcal {A}$ (not of $(\omega ,<)$ , though they do correspond to elements of $(\omega ,<)$ ), and we write each tuple in $\mathcal {A}$ -increasing order. The initial coding elements $\bar {u}_e$ will never change once defined, while the coding segment $\bar {v}_e$ may have more elements added, and the restraint $r_e$ may increase in the $\mathcal {A}$ order as, when $\bar {v}_e$ obtains more elements, $r_e = \max _{\mathcal {A}} \bar {v}_e$ may increase in the $\mathcal {A}$ -order. The previous sentence is referring to which elements of $\mathcal {A}$ these tuples consist of, rather than referring to which elements of $(\omega ,<)$ they correspond to as this may increase throughout the construction. So, for example, $\bar {u}_e$ is a fixed tuple of $\mathcal {A}$ , but may correspond to increasingly greater tuples from $(\omega ,<)$ as more and more elements are added to $\mathcal {A}$ . We will use $\bar {v}_e[s]$ and $r_e[s]$ to refer to these elements at stage s. We will always have that if $e' \leq e$ then $\bar {v}_e> r_{e'}$ in the $\mathcal {A}$ -ordering so that the coding segment for e is free of the restraint of $e'$ .

Recall that at each stage s, given a tuple $\bar {a} \in \mathcal {A}_s$ , such as the tuples $\bar {u}_e$ and $\bar {v}_e$ , $\pi _s(\bar {a})$ is the corresponding tuple in ${(\omega ,<)}$ . We also have $f^{\mathcal {A}_s}$ which is the function f on $\mathcal {A}_s$ .

From each stage to the next, we will maintain the following properties:

1. (1) At each stage $s \geq e$ , the coding elements $\bar {v}_e$ of $\mathcal {A}$ correspond to $\pi _s(\bar {v}_e)$ in ${(\omega ,<)}$ which is an interval $[a_i,b_i]$ , with i odd if $e \in X_s$ or i even if $e \notin X_s$ .

2. (2) If at stage $s \geq e$ the coding elements $\bar {v}_e[s]$ corresponded (via $\pi _s \colon \mathcal {A}_s \to {(\omega ,<)}$ ) to an interval $[a_i,b_i]$ , and at stage $s+1$ the coding elements $\bar {v}[s+1]$ correspond (via $\pi _{s+1} \colon \mathcal {A}_s \to {(\omega ,<)}$ ) to an interval $[a_j,b_j]$ , then the coding elements $\bar {v}_e[s]$ at stage s correspond at stage $s+1$ to the images, under the $\varphi $ , of the elements of ${(\omega ,<)}$ that they corresponded to at stage s. That is,

   $$\begin{align*}\pi_{s+1}(\bar{v}_e[s]) = \varphi_{i \mapsto j} (\pi_s(\bar{v}_e[s])).\end{align*}$$

3. (3) If, at stage s an element $a \in \mathcal {A}$ is a padding element, by which we mean that it is not a coding element (i.e., not part of any $\bar {v}_e[s]$ ), then at stage $s+1$ it is still not a coding element, and $f^{\mathcal {A}_s}(a) = f^{\mathcal {A}_{s+1}}(a)$ .

4. (4) From stage $s \geq e$ to stage $s+1$ , elements can only be added to $\mathcal {A}$ below $r_e$ if there is $e' \leq e$ which entered or left X at stage $s+1$ , i.e., with $e' \in X_{s} \triangle X_{s+1}$ . Thus, if $X_{s+1} \upharpoonright \upharpoonright _{e} = X_s \upharpoonright \upharpoonright _e$ ,Footnote ⁷ then the partial isomorphism $\pi _s \colon \mathcal {A}_s \to {(\omega ,<)}$ does not change at stage $s+1$ on elements below $r_e$ .

Stage 0: We begin with $\mathcal {A}_0$ being empty.

Stage s: At this stage in the construction our partial linear order $\mathcal {A}_s$ can be partitioned into a finite number of coding segments and padding blocks, each of which form an interval:

$$\begin{align*}\bar{p}_0 <_{\mathcal{A}} \bar{v}_0[s] <_{\mathcal{A}} \bar{p}_1 <_{\mathcal{A}} \bar{v}_1[s] <_{\mathcal{A}} \cdots .\end{align*}$$

We add new elements to $\mathcal {A}_s$ to code $X_{s+1}$ .

We do not do anything to $\bar {p}_0$ . We begin by handling $\bar {v}_0[s]$ . If $X_{s+1}(0) = X_s(0)$ , then we do nothing. But if $X_{s+1}(0) \neq X_s(0)$ , then we must act. Suppose that $0 \in X_{s}$ but $0 \notin X_{s+1}$ . Then $\bar {v}_0[s]$ corresponds to the elements of an interval $[a_i,b_i]$ for i even. We insert $a_{i+1} - a_i$ new padding elements above the elements of $\bar {p}_0$ and below the elements of $\bar {v}_0[s]$ . Also add new coding elements ( $b_{i+1} + a_i - a_{i+1} - b_i$ of them) to ensure that $\bar {v}_0[s]$ corresponds, in $\mathcal {A}_{s+1}$ , to $\varphi _i(\pi _s(\bar {v}_0[s])) \subseteq [a_{i+1},b_{i+1}]$ . These new elements will be subsequent coding elements; let $\bar {v}_0[s+1]$ consist of $\bar {v}_0[s]$ together with these new elements of $\mathcal {A}$ . If $0 \notin X_s$ but $0 \in X_{s+1}$ , then we act similarly but move $\bar {v}_0$ from an interval $[a_i,b_i]$ for i odd to $[a_{i+1},b_{i+1}]$ with $i+1$ even.

Now we have added several elements below $\bar {p}_1$ . We must ensure that the values of f on $\bar {p}_1$ at stage $s+1$ are the same as they were at stage s. For each block making up $\bar {p}_1$ , insert new padding elements below the least element of this block and possibly between the elements of this block to move them up to the image of the original f-block in some other f-block into which it embeds. (We use here the fact that each f-block embeds into infinitely many other blocks.) Further, we add enough padding elements on the end to complete these blocks. This ensures that for all padding elements present at stage s, the value of $f^{\mathcal {A}_{s+1}}$ on these elements at stage $s+1$ is the same as it was at stage s.

We continue this process, in turn, with the subsequence coding segments and padding blocks. In increasing order, we ensure that each of these segments from $\mathcal {A}_{s}$ still satisfy our requirements in $\mathcal {A}_{s+1}$ . However we must now take into account the fact that we may have already inserted elements.

• • Given a padding segment $\bar {p}_i$ , consider each set of elements which corresponded at stage s to a block. We must make sure that those same elements correspond in $\mathcal {A}_{s+1}$ to a block of the same type. For each block making up $\bar {p}_i$ , insert new padding elements below the least element of this block and possibly between the elements of this block to move them up to the image of the original f-block in some other f-block into which it embeds. Further, ensure we add enough padding elements on the end to complete this block. This ensures that for all padding elements present at stage s, the value of $f^{\mathcal {A}_s}$ on these elements is the same as at stage $s+1$ .

• • Given $\bar {v}_e[s]$ the coding segment corresponding to e, first check the value of $X_{s+1}(e)$ . Even if $X_{s+1}(e) = X_s(e)$ has not changed, we may have added new elements below $\bar {v}_e[s]$ so that it no longer corresponds to the same interval $[a_k,b_k]$ that it did at stage s. Let $\varphi _{k \mapsto \ell }$ be such that (i) $\ell $ is even if $e \in X_{s+1}$ and $\ell $ is odd if $e \notin X_{s+1}$ and (ii) $\ell $ is sufficiently large that by adding elements below and between the elements of $\bar {v}_e[s]$ that at stage $s+1$ the segment $\bar {v}_e[s]$ corresponds to the image, under $\varphi _{k \mapsto \ell }$ , of $[a_k,b_k] = \pi _s(\bar {v}_e[s])$ . (Note that (5) in the definition of a coding sequence—the difference between a weak coding sequence and a coding sequence—is key to obtain such an $[a_\ell ,b_\ell ]$ with $a_\ell $ large enough to accommodate the new elements that have been added to the linear order below $\bar {v}_e[s]$ .) Any new elements that were inserted and end up in the new interval are added to the collection of coding elements $\bar {v}_e[s+1]$ corresponding to e, and the other newly added elements are padding elements. Further, we ensure that enough new coding elements are after the final element to ensure we complete the entire interval, putting these elements in $\bar {v}_e[s+1]$ as well. Thus $\bar {v}_e[s+1]$ will now consist of $\pi _{s+1}^{-1}([a_\ell ,b_\ell ])$ .

• • Finally, we introduce the coding elements corresponding to s. After we have ensured all previously added elements still satisfy the requirements check the value of $X_{s+1}(s)$ (which we may assume is $0$ , i.e., $s \notin S_{x+1}$ ) and identify the next interval of the form $[a_i, b_i]$ such that i is even if $s \in X_{s+1}$ and i is odd if $s \notin X_{s+1}$ , and with $a_i$ greater than the length of the linear order at this stage. Insert enough new padding elements to the end of the linear order to extend it to have length $a_i$ , then add $b_i-a_i+1$ new coding elements corresponding to the interval $[a_i,b_i]$ . These are the initial coding elements $\bar {u}_s[s+1]$ for s, and make up the coding segment $\bar {v}_s[s+1]$ at this stage. This ensures we have satisfied requirement $R_s$ at stage s.

Verification: First, to see that $\mathcal {A}$ is really a computable copy of ${(\omega , <)}$ , observe that for any fixed $a \in \mathcal {A}$ , only finitely many elements are inserted below a. This follows from (4). Because a was added at stage s in the construction, then there can be at most $s-1$ coding blocks below it. An element is only inserted below a if some e corresponding to one of these coding blocks enters or leaves X, with a below the restraint. After these $s-1$ elements of the $\Delta _2^0$ set X stop changing values, elements will stop being inserted below a. Hence, since these are finitely many elements of a $\Delta _2^0$ set, this will occur in a finite number of stages.

To show that $f^{\mathcal {A}} \equiv _T X$ , we use the following claim which follows from (2).

Claim 3.6.1. Suppose that $s < t$ . If, at stage s, $\bar {v}_e[s]$ corresponds to $[a_m,b_m]$ and, at stage t, $\bar {v}_e[t]$ corresponds to $[a_n,b_n]$ , then the elements of $\bar {v}_e[s]$ corresponds at stage t to the images under $\varphi _{m \mapsto n}$ of the elements they corresponded to at stage s:

$$\begin{align*}[a_n,b_n] \supseteq \pi_{t}(\bar{v}_e[s]) = \varphi_{m \mapsto n} (\pi_s(\bar{v}_e[s])) = \varphi_{m \mapsto n} [a_m,b_m]. \end{align*}$$

In particular, if m and n have the same parity, then $f^{\mathcal {A}_t}(a) = f^{\mathcal {A}_s}(a)$ . If m and n have opposite parities, then there is $a \in \bar {u}_e$ such that $f^{\mathcal {A}_t}(a) \neq f^{\mathcal {A}_s}(a)$ .

We first use the lemma and (1) to show that $f^{\mathcal {A}} \ge _T X$ . Given some element e run the above construction, which is computable, until stage $e+1$ when the initial coding elements $\bar {u}_e$ corresponding to e are introduced. Now, compute the true value of $f^{\mathcal {A}}$ on these elements. If $f^{\mathcal {A}}$ is the same on $\bar {u}_e$ as it was at stage $e+1$ , then $X(e) = X_e(e+1)$ . Otherwise, if $f^{\mathcal {A}}$ is different on $\bar {u}_e$ than it was at stage $e+1$ , then $X(e) = 1- X_e(e+1)$ .

Finally, we show that $X \ge _T f^{\mathcal {A}}$ . This uses the lemma as well as properties (1) and (3). Given some element $a \in \mathcal {A}$ run the above construction until a is added to the linear order. Using (3) if a is added as a padding element then the construction ensures that $f^{\mathcal {A}}(a)$ does not change and so we take the value at this stage. If a is a coding element corresponding to e then we use (1) and the lemma. We will see that $f^{\mathcal {A}}(a)$ takes on one of two values depending on $X(e)$ . Say that a first appears in $\mathcal {A}$ at a stage s in a coding block $\bar {v}_e[s]$ , corresponding at that stage to $[a_i,b_i]$ and with $X_s(e) = k$ . If $X(e) = X_s(e) = k$ , then $f^{\mathcal {A}}(a)$ is the same as it was at stage s. Otherwise, suppose that $X(e) \neq X_s(e) = k$ . Then, at some stage $t> s$ , we find that $X(e) = X_t(e) = 1-k$ . At this stage, a is an element of $\bar {v}_e[t]$ which corresponds to $[a_j,b_j]$ with j of a different parity from i. Then $f^{\mathcal {A}}(a)$ is the same as it was at stage t.

Proof. We work on the cone above $\alpha _f$ (which we suppress) and show that if no infinite f-coding sequences exist then we can produce $\Delta ^0_2$ set C such that there is no computable copy $\mathcal {L}$ of ${(\omega ,<)}$ with $f^{\mathcal {L}} \equiv _T C$ . To do this, we meet the following requirements:

$$\begin{align*}R_{e, i, j}: \text{ If } \mathcal{L}_e \cong {(\omega,<)}, \text{ then either } \Phi_i^{f^{\mathcal{L}_e}}\neq C \text{ or } \Psi_j^C \neq f^{\mathcal{L}_e},\end{align*}$$

where $\mathcal {L}_e$ is a computable listing of the (possibly partial) linear orders. The construction is a finite injury priority construction. Given a set X, we make use of the notation $X[0,\ldots ,k]$ for the sequence $\langle X(0),\ldots ,X(k) \rangle $ .

The strategy for $R_{e, i, j}$ is as follows. First, if the eth program fails to code a linear order the requirement is automatically satisfied and so we will assume that at all stages s, $\mathcal {L}_{e,s}$ is a linear order. When we say that a requirement $\mathcal {R}$ places a restraint on C, this means that it is restraining that part of C from changing for the sake of a lower-priority requirement. $\mathcal {R}$ itself may cause C to change on the restraint.

1. (1) To initialize this requirement choose some x that has not yet been restrained, restrain it, and assign it to this requirement. (This requirement will not cause C to change on any value other than x, and so this is the only function of the restraints placed by the higher-priority requirements.) We begin with $C(x) = 0$ . We say the strategy is in Phase 0.

2. (2) At stage s, if the requirement is in Phase 0, say this requirement requires attention if there are computations

   $$\begin{align*}\Phi_{i, s}^{f^{\mathcal{L}_{e, s}}}(x) = 0 = C_s(x) \quad \text{with use } u_0\end{align*}$$

   and

   $$\begin{align*}\Psi_{j, s}^{C_s}[0, \ldots, m_0] = f^{\mathcal{L}_{e, s}}[0, \ldots, m_0] \quad \text{with use } v_0, \end{align*}$$

   where $m_0 \ge u_0$ is such that all f-blocks that intersect $[0, \ldots , u_0]$ in $\mathcal {A}_{e,s}$ , and all elements less than them in $\mathcal {L}_{e,s}$ , are completely contained in $[0, \ldots , m_0]$ .

   When this requirement acts, restrain $[0, \ldots , v_0]$ in C and define $C_{s+1}(x) =1$ . Finally, move the requirement to Phase 1.

   

3. (3) At stage s, if the requirement is in Phase $n+1$ , say this requirement requires attention if there are computations

   $$\begin{align*}\Phi_{i, s}^{f^{\mathcal{L}_{e, s}}}(x) = C_s(x) \quad \text{with use } u_{n+1}\end{align*}$$

   and

   $$\begin{align*}\Psi_{j, s}^{C_s}[0, \ldots, m_{n+1}] = f^{\mathcal{L}_{e, s}}[0, \ldots, m_{n+1}] \quad \text{with use } v_{n+1}, \end{align*}$$

   where $m_{n+1} \ge \max \{m_n, u_{n+1}\}$ such that all f-blocks that intersect the set of elements ${[0, \ldots , \max \{m_n, u_{n+1}\}]}$ , and all elements less than them in $\mathcal {L}_{e,s}$ , are completely contained in $[0, \ldots , m_{n+1}]$ .

   When this requirement acts, restrain $[0, \ldots , v_{n+1}]$ in C and define $C_{s+1}(x) = 1-C_s(x)$ . Finally, move the requirement to Phase $n+2$ .

Construction of C: At stage s of the construction, consider the first s requirements, in order of decreasing priority. If any requirement requires attention then the one with the highest priority acts according to its strategy, injuring and resetting all lower priority strategies. If no requirement acts, then initialize the lowest priority requirement that has not yet been initialized.

Verification: It is not obvious that the construction is a finite injury construction, as even if a requirement $R_{e,i,j}$ is not injured, it appears on the surface that it might go through infinitely many phases. In this case, our approximation for C would also not come to a limit. We will argue by induction on the requirements that each requirement acts only finitely many times and is eventually satisfied. This argument will use the fact that there is no infinite coding sequence.

Consider a requirement $R_{e, i, j}$ and assume that after some stage it is no longer injured. If ${\mathcal {L}}_e$ is partial (or not a linear order), then $R_{e, i, j}$ is automatically satisfied and will no longer act after some stage. So we may assume that $\mathcal {L}_e$ is a linear order. If there is some n such that the strategy enters Phase n but never enters Phase $n+1$ , then we are also done since $R_{e,i,j}$ will have acted only finitely many times and will also be satisfied. (Otherwise, if $\Phi _i^{f^{\mathcal {L}_e}} = C$ and $\Phi _j^C = f^{\mathcal {L}_e}$ , then we would eventually enter Phase $n+1$ .) So it is enough to show that the strategy enters finitely many phases as this will also show that it requires attention finitely many times. To do this we show that if, after a strategy is no longer injured, it enters Phase n for arbitrarily large n then we can produce an infinite f-coding sequence; in fact we produce a weak f-coding sequence and then use Lemma 3.5 to obtain an f-coding sequence.

Let $s_n$ be the stage at which the strategy enters Phase n, if it exists, and recall that $v_n$ is the restraint placed at Phase n. We begin by arguing that the restraints are maintained, and that C (up to the restraint) cycles back and forth between two possible configurations depending on whether the phase is odd or even.

It will be helpful to consider, at each stage s, the guess $\pi _s \colon \mathcal {L}_{e,s} \to {(\omega ,<)}$ at the isomorphism $\mathcal {L}_{e} \to {(\omega ,<)}$ . Note that (by definition), $f^{\mathcal {L}_{e,s}}$ is the image of f under $\pi _s$ .

Finally, to produce the weak f-coding sequence we proceed as follows:

• • Let $[a_0, b_0]$ be the smallest initial segment of ${(\omega ,<)}$ which contains all of the $\pi _{s_0}$ -images of the elements $0,\ldots ,u_0$ in $\mathcal {L}_{e,s_0}$ and all of the blocks which they intersect. Note that all of $[a_0,b_0]$ is contained within the $\pi _{s_0}$ -images of $[0,\ldots ,m_0]$ .

• • Given $[a_i, b_i]$ , let $[a_{i+1}, b_{i+1}]$ be the minimal interval in ${(\omega ,<)}$ that contains all of the $\pi _{s_{i+1}}$ -images of the elements $0,\ldots ,\max \{m_{i},u_{i+1}\}$ in $\mathcal {L}_{e,s_{i+1}}$ and all of the blocks which they intersect. Note that all of $[a_i,b_i]$ is contained within the $\pi _{s_{i+1}}$ -images of $[0,\ldots ,m_{i+1}]$ .

• • Define the maps $\varphi _i: [a_i, b_i] \to [a_{i+1}, b_{i+1}]$ by $\varphi _i = \pi _{i+1} \circ \pi _{i}^{-1}$ . (The way to think about these maps is that, when $[a_i,b_i]$ was defined at stage $s_{i}$ , these elements of ${{(\omega ,<)}}$ corresponded to certain elements of $\mathcal {L}_e$ . However by stage $s_{i+1}$ , we have seen more elements enter $\mathcal {L}_e$ , and now those elements of $\mathcal {L}_e$ are instead in correspondence with other elements of ${(\omega ,<)}$ , namely some elements of $[a_{i+1},b_{i+1}]$ . The map $f_i$ keeps track of how this correspondence has changed. The element of $\mathcal {L}_e$ that corresponded to $n \in [a_i,b_i]$ now corresponds to $f_i(n) \in [a_{i+1},b_{i+1}]$ .)

This sequence satisfies most of the requirements by construction. We just need to ensure that the $\varphi _i$ are not f-preserving but the $\varphi _{i+1} \circ \varphi _i$ are. But this is exactly the content of Lemma 3.6.1. Hence, assuming we enter Phase n for arbitrarily large n we have produced an infinite weak coding sequence. From an infinite weak coding sequence, using Lemma 3.5 we can produce an infinite coding sequence. Thus, by our assumption about the absence of such a sequence it follows that after a strategy is no longer injured it only enters finitely many phases.

Proof. Let $L_k$ be the block corresponding to the loop of length k, i.e., the block isomorphic to $[1, \ldots , k] \to [1, \ldots , k]$ via $x \mapsto x+1$ for $x<k$ and $k \mapsto 1$ . Consider the block function f where the odd blocks are given by the sequence $L_1, L_2, L_3, L_4, L_5, \ldots $ and the even blocks are given by the sequence $L_1, L_1, L_2, L_1, L_2, L_3, \ldots $ . An initial segment looks like:

$$\begin{align*}L_1 + L_1 + L_2 + L_1 + L_3 + L_2 + L_4 + L_1 + L_5 + L_2 + L_6 + L_3 + L_7 + \cdots \end{align*}$$

This function is constructed so that it satisfies the following:

• • all blocks that occur in the function occur infinitely often,

• • no two block types embed in another different block type,

• • no two different block types that occur in the function have the same size, and

• • no two blocks types are adjacent (in the same order) more than once.

Since each block occurs infinitely often the degree spectrum of f on a cone must contain a non-c.e. degree. To show that the degree spectrum of f on a cone does not contain all $\Delta _2^0$ degrees we show that there is no infinite f-coding sequence. Indeed we will show that any f-coding sequence has length at most five. By Theorems 3.3 and 3.6 this proves the theorem. (For the moreover statement, one must observe that for this particular f no cone is required in the proofs of these theorems.)

Consider any (possibly finite) f-coding sequence $[a_1, b_1], [a_2, b_2], \ldots $ and corresponding maps $\varphi _i$ . We make the following definitions:

1. (1) Say that $l_1 < \cdots < l_p$ form a link in an interval $[a_i,b_i]$ if they form a block of length p in $[a_i,b_i]$ . Say that this interval witnesses this link.

2. (2) Say that a link $l_1 < \cdots < l_p$ in $[a_i,b_i]$ is vulnerable in $[a_j,b_j]$ , $j> i$ , if the images $\varphi _{i \mapsto j}(l_1) < \cdots < \varphi _{i \mapsto j}(l_p)$ are no longer contained in a single block and now intersect two or more different blocks.

3. (3) Say that a link $l_1 < \cdots < l_p$ in $[a_i,b_i]$ is broken in $[a_j,b_j]$ , $j> i$ , if some element is inserted between the images $\varphi _{i \mapsto j}(l_1)$ and $\varphi _{i \mapsto j}(l_p)$ , i.e., the $l_r$ are no longer adjacent after applying $\varphi _{i \mapsto j}$ .

We first claim that either there is a link witnessed in $[a_1,b_1]$ that becomes vulnerable in $[a_2,b_2]$ , or there is a link witnessed in $[a_2,b_2]$ that becomes vulnerable in $[a_3,b_3]$ . The map $\varphi _1$ from $[a_1,b_1]$ to $[a_2,b_2]$ is not f-preserving, and so there is a link $l_1 < \cdots < l_p$ witnessed in $[a_1,b_1]$ such that $\varphi _1$ is not f-preserving on these elements. This link forms a block of size p in $[a_1,b_1]$ , and f has only one block type of each size, so either the images of $l_1 < \cdots < l_p$ in $[a_2,b_2]$ lie in two different blocks or are contained in some larger block. In the former case, the link $l_1 < \cdots < l_p$ has become vulnerable in $[a_2,b_2]$ . In the second case, let $k_1 < \cdots < k_q$ be this larger block in $[a_2,b_2]$ containing the images of $l_1,\ldots ,l_p$ . Then $k_1 < \cdots < k_q$ is a link witnessed in $[a_2,b_2]$ which becomes vulnerable in $[a_3,b_3]$ . This is because $\varphi _{1 \mapsto 3}$ from $[a_1,b_1]$ to $[a_3,b_3]$ is f-preserving and so the images of $l_1 < \cdots < l_p$ form a block in $[a_3,b_3]$ ; this block is contained in, but not all of, the image of $k_1 < \cdots < k_q$ in $[a_3,b_3]$ .

Now suppose that we have a link $l_1,\ldots ,l_p$ which is witnessed in $[a_i,b_i]$ and which becomes vulnerable in $[a_{i+1},b_{i+1}]$ . We claim that it must break in either $[a_{i+2},b_{i+2}]$ or $[a_{i+3},b_{i+3}]$ . Indeed, in $[a_{i+1},b_{i+1}]$ the images of the link $l_1,\ldots ,l_p$ lie in at least two different blocks, and so the same is true in $[a_{i+3},b_{i+3}]$ . Moreover, since the map $\varphi _{i+1 \mapsto i+3}\colon [a_{i+1},b_{i+1}] \to [a_{i+3},b_{i+3}]$ is f-preserving, in $[a_{i+3},b_{i+3}]$ the images of $l_1,\ldots ,l_p$ lie in blocks of the same block types. However, since $\varphi _{i \mapsto i+2} \colon [a_{i},b_{i}] \to [a_{i+2},b_{i+2}]$ is f-preserving, the images of $l_1,\ldots ,l_p$ form a block in $[a_{i+2},b_{i+2}]$ . Thus—as each $\varphi _j$ is order-preserving— $\varphi _{i+1 \mapsto i+3}\colon [a_{i+1},b_{i+1}] \to [a_{i+3},b_{i+3}]$ cannot be the identity. Since each pair of f-blocks only appears adjacent to each other once in f, some element must be inserted between the images in $[a_{i+3},b_{i+3}]$ of $l_1$ and $l_p$ ; thus the link must be broken in $[a_{i+3},b_{i+3}]$ if not earlier.

Now we argue that after a link is broken, the coding sequence must terminate. Suppose that a link $l_1,\ldots ,l_p$ is witnessed in $[a_i,b_i]$ but broken in $[a_j,b_j]$ . Then i and j must be of opposite parities, as if they were of the same parity then the map $\varphi _{i \mapsto j} \colon [a_i,b_i] \to [a_j,b_j]$ would be f-preserving, and the images of $l_1,\ldots ,l_p$ would form a block in $[a_j,b_j]$ , and hence would be a sequence of successive elements. Thus, as i and j have opposite parities, the map $\varphi _{i \mapsto j+1}$ is f-preserving, so the images of $l_1,\ldots ,l_p$ in $[a_{j+1},b_{j+1}]$ must form a block of size p as $l_1,\ldots ,l_p$ did in $[a_i,b_i]$ . However, in $[a_j,b_j]$ and hence in $[a_{j+1},b_{j+1}]$ the images of $l_1,\ldots ,l_p$ are no longer successive elements, preventing them from forming such a block. Thus the coding sequence must terminate with $[a_j,b_j]$ .

Thus we have a link in $[a_1,b_1]$ or $[a_2,b_2]$ which becomes vulnerable in $[a_2,b_2]$ or $[a_3,b_3]$ , and hence broken by $[a_5,b_5]$ or before. There can be no more elements of the coding sequence.