2.1.1 The requirements.

Let us consider now the goals of the construction. As discussed, the balls that are never removed from the machine will be precisely the elements of $M^\complement $ . Each such ball will eventually settle as a “resident” of some node $\beta $ on the tree. For a node $\alpha $ on the tree, we will let $Y {(\alpha) }$ denote the collection of balls which are permanent residents of nodes $\beta \succcurlyeq \alpha $ . The rules about ball movement will ensure that $Y {(\alpha) }$ is a d.c.e. set. To enter $Y {(\alpha) }$ , a ball x will need to first pass through $\alpha $ , i.e., be a resident of $\alpha $ at some stage. It may later move to extensions of $\alpha $ . It is possible that it will be removed from $Y {(\alpha) }$ by being pulled by some node, that lies to the left of $\alpha $ on the tree. In any case, balls can be removed from $Y {(\alpha) }$ by enumerating them into M. We will ensure, however, that $Y {(\alpha) }\cup M$ is in fact a c.e. set, though not uniformly in $\alpha $ . This is because either only finitely many balls outside M will ever enter $Y {(\alpha) }$ , or finitely many balls outside M ever leave $Y {(\alpha) }$ , or $Y {(\alpha) }= \emptyset $ .

We will ensure the following:

1. (a) For each node $\beta $ on the true path, $Y {(\beta)} =^* {M}^\complement $ . That is, all but finitely many elements of ${M}^\complement $ will have passed through $\beta $ at some stage and settled at $\beta $ or some extension of $\beta $ , i.e., below $\beta $ (recall that our trees grow downwards).

2. (b) For each e, if $\beta $ is the node on the true path of length $e+1$ , then $Y {(\beta)} $ is either almost contained in, or almost disjoint from, $W_e$ .

3. (c) We also need to ensure that M is coinfinite.

For (a), we will ensure that only finitely many balls have permanent residence to the left of $\beta $ , to the right of $\beta $ , or above $\beta $ .

For (b), we will ensure that either all balls that ever pass through $\beta $ are already seen to be in $W_e$ when they do, or that we know that only finitely many elements of $W_e$ will ever be available to pass through $\beta $ . In the first case, the node $\beta $ will ensure that all but finitely many balls outside $W_e$ will be enumerated into M (we will say that they are eliminated from the machine).

For (c), we will ensure that infinitely many nodes $\beta $ on the true path hold on to balls and ensure that they are not eliminated, i.e., keep them out of M. Similar action will ensure that balls indeed only move finitely much, i.e., they do eventually settle at some node.

2.2.1 The tree.

We let the tree of strategies be the collection of all sequences of the symbols ${\texttt {fin}}_n$ and $\infty _n$ . These symbols are called outcomes. We write $\alpha \preccurlyeq \beta $ to indicate that $\alpha $ is a prefix of $\beta $ .

In addition to extension, we will use the Kleene–Brouwer ordering on the tree. To define this we fix an ordering of the outcomes, say

$$\begin{align*}\infty_0 < {\texttt{fin}}_0 < \infty_1 < {\texttt{fin}}_1 < \dots. \end{align*}$$

We say that $\alpha $ lies to the left of $\beta $ , and write $\alpha <_L\beta $ , if $\alpha $ and $\beta $ are incomparable (neither is a prefix of the other), and $\alpha (k)< \beta (k)$ for the least k such that ${\alpha (k)\ne \beta (k)}$ . We write $\alpha \leqslant \beta $ if either $\alpha <_L\beta $ or $\beta \preccurlyeq \alpha $ . As this is the only linear ordering of strategies that we will use, we use the notation $\alpha \leqslant \beta $ rather than $\alpha \leqslant _{\text {KB}} \beta $ .

We let ${\lambda }$ denote the root of the tree (the empty sequence).

2.2.2 Notation for the pinball machine.

As discussed, at each stage, finitely many balls x will reside at some nodes of the machine. We let

$$\begin{align*}S{(\alpha)}_{s} \end{align*}$$

denote the collection of x that reside at the node $\alpha $ at the beginning of stage s. We let $Y {(\alpha)}_{s}$ be the collection of all balls that at the beginning of stage s reside at $\alpha $ or below $\alpha $ , i.e., at some extension of $\alpha $ :

$$\begin{align*}Y{(\alpha)}_{s} = \bigcup_{\beta\succcurlyeq \alpha} S{(\beta)}_{s}. \end{align*}$$

So $Y {{(\lambda)}}_{s}$ is the collection of balls that are on the machine at the beginning of stage s.

2.2.3 True stage, and delayed, enumerations.

Since the given set A is $\text {low}_2$ , it is not high. C.e. sets that are not high have “true stage” enumerations with respect to any given computable growth rate: for any computable function f, there is a computable enumeration $(A_r)$ Footnote ⁵ of A such that for infinitely many stages r, the first $f(r)$ many elements of the stage r complement ${A}^\complement _r$ are “correct”: they are elements of the complement ${A}^\complement $ (see [Reference Soare17, Lemma XI.1.6]).Footnote ⁶

We fix an enumeration $(\tilde {A}_r)$ of A, that is true with respect to the function $f(r) = r^2$ . During the construction, though, it will be useful to delay this enumeration. We will define, during the construction, an enumeration $(A_s)$ of A, such that for all s there is some r such that $A_s = \tilde A_r$ . We will write $n(s)=r$ .

Let s be a stage, and suppose that $n(s)=r$ . We will let the construction run: balls will move around, or be enumerated into M, and it will be useful to keep track of these actions as happening during several stages: s, $s+1$ , $s+2$ , and so on. While we are doing this, we keep our enumeration of A fixed: that is, $\tilde A_r = A_s = A_{s+1} = A_{s+2} = \dots $ . After finitely many stages, we will reach a stage $t\geqslant s$ at which we decide that there is nothing we want to do. We then declare t to be a “new balls” stage, and set $A_{t+1} = \tilde {A}_{r+1}$ ,Footnote ⁷ i.e., set $n(t+1) = r+1$ .

Thus, the function $s\mapsto n(s)$ will be weakly increasing and onto; we will have $n(s+1)\ne n(s)$ exactly when s is a new balls stage, in which case $n(s+1) = n(s)+1$ . What is important is that the enumeration $(\tilde A)_r$ is given to us, whereas the enumeration $(A_s)$ is defined by us during the construction: based on how the construction develops, we decide whether to declare a “new balls” stage, i.e., when to increase n by 1 and thus enumerate more numbers into A. We start with $n(0)=0$ . Observe that for any stage s, $n(s)$ is the number of new balls stages $t<s$ . We will ensure that there are infinitely many new balls stages, so that indeed $\bigcup _s A_s = A$ .

Definition 2.2. We let

$$\begin{align*}Q_s = \{\text{the smallest } n(s)^2 \text{ many elements of }{A}^\complement_s\}. \end{align*}$$

A stage s is A-true if $Q_s \subseteq {A}^\complement $ .

Thus, if $s<t$ are successive new balls stages, then $Q_{s+1} = Q_{s+2} = \cdots = Q_t$ , and $s+1$ is A-true if and only if all the stages $s+1,s+2,\dots , t$ are A-true. A stage s is A-true if and only if the stage $r = n(s)$ is true for the enumeration $(\tilde A_r)$ with respect to $r\mapsto r^2$ . By assumption, there are infinitely many such r. As we will ensure that $s\mapsto n(s)$ is onto, there will be infinitely many A-true stages s.

We will ensure that at every stage s, $Y {{(\lambda)}}_{s}\smallsetminus M_s = Q_s$ . So at a new balls stage s, we will place all the elements of $Q_{s+1}$ that have not yet been put on the machine, at the root. As stages go by, some of these elements may be enumerated into M.

The reason for using true stages for the function $f(r) = r^2$ is the following. If s is a new balls stage then

$$\begin{align*}|Q_{s+1}| = |Q_s| + 2n(s) +1> |Q_s| + n(s). \end{align*}$$

Thus, at a new balls stage s we will be adding more than $n(s)$ -many new balls to the root of the machine; and we will use the fact that $n(s)\to \infty $ . As time goes by, we will be getting balls at the root in increasing number, and infinitely often, all of these balls will be A-correct.

2.2.4 $\Delta _3$ guessing.

By assumption, A is $\text {low}_2$ , i.e., $\Pi _2(A)\subseteq \Delta _3$ . Thus, any Boolean combination of $\Pi _2(A)$ statements is equivalent to a $\Sigma _3$ statement. Since there is a universal $\Pi _2(A)$ set, this is effective: given $\psi $ , a Boolean combination of $\Pi _2(A)$ statements, we can effectively find a $\Sigma _3$ statement $\chi $ such that $\psi $ holds if and only if $\chi $ holds.

In turn, from $\chi $ we can effectively produce a uniformly computable collection of nondecreasing sequences $\bar \ell (\psi ,n) = {(\ell ({\psi ,n})_{s})}$ for $n\in \mathbb N$ such that:

• • $\psi $ holds if and only if for some n, $\bar \ell (\psi ,n)$ is unbounded.

To see this, write $\chi $ as $\exists n\forall x\exists y\,\theta (n,x,y)$ . Then we let $\ell (\psi ,n)_{s}$ be the greatest $x<s$ such that for all $x'\leqslant x$ , there is some $y<s$ such that $\theta (n,x',y)$ holds.

Shortly, for each node $\alpha $ on the tree of strategies, we will formulate an “ $\alpha $ question” $\psi (\alpha )$ , which will be a $\Pi _2(A)$ statement. In other words, the collection of nodes $\alpha $ such that $\psi (\alpha )$ holds is $\Pi _2(A)$ . Using the transformation above for the statements $\psi (\alpha )$ and $\lnot \psi (\alpha )$ (both are Boolean combinations of $\Pi _2(A)$ statements), we obtain families of sequences $\bar \ell (\psi (\alpha ),n)$ and $\bar \ell (\lnot \psi (\alpha ),n)$ . By the recursion theorem, we know a computable index for the construction, and so can have access to these families of sequences during the construction.Footnote ⁸

2.2.5 The $\alpha $ question.

Definition 2.3. Let $\alpha $ be a node. The statement $\psi (\alpha )$ is:

For every $k\in \mathbb N$ there is an A-true stage s for which

$$\begin{align*}| Y{(\alpha)}_{s}\cap W_{|\alpha|,s}|\geqslant k. \end{align*}$$

As just discussed, during the construction we have access to the sequences $\bar \ell (\psi (\alpha ),n)$ and $\bar \ell (\lnot \psi (\alpha ),n)$ . Now for each node $\beta $ we define a (single) sequence $\bar \ell (\beta )$ by induction on the length $|\beta |$ :

• • $\ell ({\lambda })_{s}= s$ ;

• • $\ell (\beta \hat {\,\,}\infty _n)_{s} = \min \{ \ell (\beta )_{s}, \ell (\psi (\beta ),n)_{s} \}$ ;

• • $\ell (\beta \hat {\,\,}{\texttt {fin}}_n)_{s} = \min \{ \ell (\beta )_{s}, \ell (\lnot \psi (\beta ),n)_{s} \}$ .

Thus, via the sequences $\bar \ell (\gamma )$ , the children $\gamma $ of $\beta $ together try to answer the $\beta $ question.

Also, by definition, $\bar \ell ({\lambda })$ is unbounded. We can therefore define the following.

In other words, ${\lambda }$ is on the true path, and if $\alpha $ is on the true path, then the child of $\alpha $ on the true path is the leftmost child $\beta $ for which $\bar \ell (\beta )$ is unbounded. This is the leftmost child that guesses a correct witness for the $\Sigma _3$ version of $\psi (\alpha )$ or its negation. Lemma 2.4(c) implies that the true path is infinite.

2.2.6 Pulling and eliminating.

We will now discuss the heart of the construction: how to decide where balls move, and which are enumerated into M. The following terminology will be useful:

• • a fin -node is a nonzero node $\beta $ whose last entry is ${\texttt {fin}}_n$ for some n (that is, $\beta = (\beta ^-)\hat {\,\,} {\texttt {fin}}_n$ for some n); and similarly,

• • an $\infty $ -node is a nonzero node whose last entry is $\infty _n$ for some n.

Here note that for a nonzero node $\alpha $ , we let $\alpha ^-$ denote $\alpha $ ’s parent, the result of removing the last entry of $\alpha $ .

The next definition will govern the movement and enumeration of balls. We will explain the details after we give the definition. The rough idea, though, is the following. If $\beta $ is an $\infty $ -node of length $e+1$ , then we will ensure that $Y {(\beta)} \subseteq W_e$ by only allowing elements that have already entered $W_e$ to enter $Y {(\beta)} $ . On the other hand, if $\beta $ is a ${\texttt {fin}}$ -node, then we get $Y {(\beta)} \cap W_e =^* \emptyset $ “for free”, by the fact that the $\beta ^-$ question $\psi (\beta ^-)$ fails. As discussed, in the first case, an $\infty $ -node $\beta $ will want to enumerate into M any balls it sees that are outside $W_{e,s}$ . If $\beta $ is correct about its guess, then this will not cause $\mathbb N\subseteq ^* M$ , because $\psi (\beta ^-)$ holds: sufficiently many balls will not be enumerated into M. The complexity of the definitions comes from the need to deal with nodes to the left of the true path, that have wrong opinions.

For any node $\beta $ , let

$$\begin{align*}Y{(<_{L}\!\beta)}_{s} = \bigcup \,\,\{Y{(\gamma)}_{s}\,:\, \gamma<_L\beta\}, \end{align*}$$

and let

$$\begin{align*}Y{(\leqslant\! \beta)}_{s} = Y{(<_{L}\!\beta)}_{s}\,\cup \,Y{(\beta)}_{s} = \bigcup \,\,\{Y{(\gamma)}_{s}\,:\, {\gamma}\leqslant {\beta}\}. \end{align*}$$

Here recall that $\gamma \leqslant \beta $ denotes the Kleene–Brouwer ordering, not the lexicographic ordering.

Let us explain. (i) and (v) mean that x resides at $\beta $ ’s parent $\beta ^-$ , or at or below one of $\beta $ ’s `of $\beta $ . These are the regions from which $\beta $ is allowed to pull. Thus, balls on the machine will only move downward in the Kleene–Brouwer ordering: either to the left, or drop from a parent to a child.

(ii) will help ensure that each ball outside M eventually stops moving: it cannot keep being pulled by longer and longer nodes. (vi) will help show that M is coinfinite. A node $\beta ^-$ on the true path will guard one ball (the smallest that it can see), and ensure that it is not enumerated into M.

The fact that an $\infty $ -node $\beta $ is only allowed to pull balls in $W_e$ is in conflict with the requirement that $Y {(\beta)} =^* {M}^\complement $ . It is simple to reconcile the conflict: if at a stage s, x is in a region from which $\beta $ can pull, and $x\notin W_{e,s}$ , then $\beta $ can simply enumerate x into M and remove it from the machine. The danger, of course, is that numbers go into $W_e$ slowly, and we may be too hasty in enumerating them into M, risking $M=^*\mathbb N$ . To avoid this, we use the sequences $\bar \ell (\gamma )$ , which will ensure that each node $\gamma $ to the left of the true path eventually stops enumerating into M numbers from ${A}^\complement $ . If $\beta $ is an $\infty $ -node on the true path, we need to ensure that $Y {(\beta)} $ is infinite, i.e., that it sees enough numbers in $W_e$ sufficiently early. We will show that this follows from $\psi (\beta ^-)$ being true.

Thus, (iii) and (vii) ensure that nodes to the left of the true path eventually stop acting: they stop eliminating or pulling balls $x\in {A}^\complement $ . (Balls $x\in A$ are “phantom” balls; the right way to think about them is as if they were never there, and so pulling them does not count as acting.) This uses the fact that if $\gamma $ lies to the left of the true path, then $\bar \ell (\gamma )$ is bounded. Of course, these restrictions also restrain the action of a node $\beta $ on the true path. Here, we note the difference between (iii) and (vii), i.e., the difference between the conditions $x<\ell (\beta )_{s}$ and $|Y {(\beta)}_{s}|<\ell (\beta )_{s}$ . The issue is of timing.

The main step of the verification will be to ensure that if $\beta $ lies on the true path, then the number of balls in $Y {(\beta)}$ during A-true stages is unbounded (this is Lemma 2.14 below). Suppose that this is known for $\beta ^-$ (either by induction, or directly from $\psi (\beta ^-)$ , if $\beta $ is an $\infty $ -node). We need to ensure that if the opportunity arises, in a true stage, then $\beta $ will act by pulling balls. It could be that at such stages, $\ell (\beta )_{s}$ is too small, compared to the size of the balls that need pulling. This is why we do not require that $x< \ell (\beta )_{s}$ when deciding whether to pull x or not; we need to consider the number of balls we currently have, i.e., $|Y {(\beta)}_{s}|$ .

On the other hand, there is no reason for haste in eliminating balls; we just need to ensure that in the limit, all but finitely many “deviant” balls do end up in M (these are the balls outside $W_e$ , when $\beta $ ensures that ${M}^\complement \subseteq ^* W_e$ ). So it is legitimate for us to require the stronger condition $x< \ell (\beta )_{s}$ when considering elimination. And in fact, it is important that we do so. The reason is a little delicate. Suppose that $\alpha $ , the node on the true path, is a ${\texttt {fin}}$ -node, but that $\gamma $ is an $\infty $ -child of $\alpha ^-$ that lies to the left of $\alpha $ . Since $\gamma $ is allowed to only pull elements of $W_e$ (again $e = |\alpha ^-|$ ), and $W_e$ may be small (indeed, may be empty), it may be the case that $Y {(\gamma)}$ never reaches the size $\lim _s \ell (\gamma )_{s}$ . Thus, $\gamma $ is always “hungry” for balls. If we allowed $\gamma $ to eliminate balls whenever $|Y {(\gamma)}_{s}| < \ell (\gamma )_{s}$ , it may never stop doing so. For pulling balls, the weaker condition $|Y {(\gamma)}_{s}| < \ell (\gamma )_{s}$ is sufficient, since if $\gamma $ keeps pulling balls $x\notin {A}^\complement $ , we will argue that eventually the size of $Y {(\gamma)}$ reaches $\lim _s \ell (\gamma )_{s}$ , and $\gamma $ will stop acting.

We will need the following for the construction.

2.4.1 Finite ball movement.

As mentioned above, we say that x is a permanent resident of a node $\beta $ if $x\in S{(\beta)}_{s}$ for all but finitely many stages s. We let $S{(\beta)}$ be the set of permanent residents of $\beta $ .

Lemma 2.9 implies the following.

Thus, as $n(s)\to \infty $ , every $x\notin M$ is eventually put on the machine and is never removed from the machine. This shows the following.

We let $Y {(\alpha) } = \lim _s Y {(\alpha)}_{s} = \bigcup _{\beta \succcurlyeq \alpha }S{(\beta)}$ ; so $Y {{(\lambda)}} = {M}^\complement $ .

2.4.2 True path.

Recall that the true path (Definition 2.5) is the collection of nodes $\alpha $ such that $\bar \ell (\alpha )$ is unbounded, but for all $\gamma <_L\alpha $ , $\bar \ell (\gamma )$ is bounded. We observed that the true path is an infinite path of the tree.

Proof. This is proved by induction on the length of $\alpha $ ; it is vacuously true for $\alpha = {\lambda }$ . Suppose that this is known for $\alpha ^-$ . Let

$$\begin{align*}l = \max \left\{\lim_{s\to \infty} \ell(\beta)_{s} \,:\, \beta<_L\alpha\text{ is a child of}\ \alpha^-\right\}. \end{align*}$$

By Lemma 2.4(b),

$$\begin{align*}l = \max \Big\{\lim_{s\to \infty} \ell(\gamma)_{s} \,:\, \gamma<_L\alpha\ \&\ \gamma \succ \alpha^-\Big\}. \end{align*}$$

Then by (vii) of Definition 2.6, no number $x\geqslant l$ is ever eliminated by any $\gamma <_L\alpha $ extending $\alpha ^-$ (indeed, whether it is an element of ${A}^\complement $ or not). By induction (from left to right) on the children $\beta $ of $\alpha ^-$ which lie to the left of $\alpha $ , we show that $\beta $ pulls only finitely many $x\in {A}^\complement $ . Suppose that this is true for all $\gamma <_L\beta $ . Then after some stage, every ball $x\notin A$ which is pulled by $\beta $ , is not later pulled by a node to the left of $\beta $ , and is not eliminated by any descendant of $\beta $ , and so is an element of $Y {(\beta)}$ . So if $\beta $ pulls infinitely many $x\in {A}^\complement $ , then $Y(\beta)$ is infinite. But then, $|Y {(\beta)}_{s}|\geqslant l$ for all but finitely many stages s. By (iii) of Definition 2.6, at such stages, $\beta $ pulls no balls, a contradiction.

As a result:

The main lemma is the following.

Proof. We prove this by induction on the length of $\alpha $ . First we consider $\alpha = {\lambda }$ . If s is a new balls stage and $s+1$ is A-true, then, as discussed above, at least $n(s)$ many new balls from $Q_{s+1}$ are placed in $S {{(\lambda)}}_{s+1}$ ; and $n(s)\to \infty $ .

Suppose that $\alpha \ne {\lambda }$ lies on the true path, and that the lemma holds for $\alpha ^-$ . Let $s_0$ be a stage after which no $\gamma <_L\alpha $ pulls or eliminates any ball $x\in {A}^\complement $ ; in particular, no $\gamma <_L\alpha $ pulls any ball during an A-true stage $s>s_0$ . Further, if $s>s_0$ is A-true then $Y {(<_{L}\!\alpha)}_{s} = Y {(<_{L}\!\alpha)}$ , as all balls on the machine at stage s are from ${A}^\complement $ . Let $m = |Y {(<_{L}\!\alpha)}|$ , which by Lemma 2.13 is finite.

Let $k\in \mathbb N$ . There are two cases.

First, suppose that $\alpha $ is a ${\texttt {fin}}$ -node. By induction, there is an A-true stage $s>s_0$ for which

$$\begin{align*}|Y{(\alpha^-)}_{s}| \geqslant k + m + 1. \end{align*}$$

Choose such s sufficiently large so that $\ell (\alpha )_{s}>k$ . If $|Y {(\alpha)}_{s}|\geqslant k$ then we are done. Otherwise, $|Y {(\alpha)}_{s}|<\ell (\alpha )_{s}$ . Every $x\in Y {(\alpha ^-)}_{s}\smallsetminus Y {(\leqslant \!\alpha)}_{s}$ larger than $|\alpha |$ is pullable by $\alpha $ at stage s. Now $Y {(\alpha ^-)}_{s}$ contains at most one $x\leqslant |\alpha |$ , namely, $x= |\alpha |$ (as every $x\in Y {(\alpha ^-)}_{s}$ has size $> |\alpha ^-|$ )Footnote ¹⁰ ; and as discussed, $|Y {(<_{L}\!\alpha)}_{s}|=m$ . Hence, there remain at least $k-|Y {(\alpha)}_{s}|$ many balls in $Y {(\alpha ^-)}_{s}$ that are pullable by $\alpha $ . Since $s>s_0$ , and s is A-true, no node to the left of $\alpha $ pulls such balls at stage s. Hence, at stage s, all balls pullable by $\alpha $ will actually be pulled by $\alpha $ , and moved to $Y {(\alpha)}_{s+1}$ . Also, no balls already in $Y {(\alpha)}_{s}$ will be pulled to the left. Since balls are pulled at stage s, no balls are eliminated (anywhere on the machine) at stage s—we only eliminate balls if no balls are pullable. So $Y {(\alpha)}_{s}\subseteq Y {(\alpha)}_{s+1}$ , and overall, we see that $|Y {(\alpha)}_{s+1}|\geqslant k$ . Since s is not a new balls stage, $s+1$ is also A-true, and so is as required.

If $\alpha $ is an $\infty $ -node, then the argument is the same, except that we need an A-true stage $s>s_0$ such that $\ell (\alpha )_{s}>k$ and such that

$$\begin{align*}|Y{(\alpha^-)}_{s}\cap W_{|\alpha^-|,s}| \geqslant k + m + 1, \end{align*}$$

since $\alpha $ is allowed to pull only balls from $W_{|\alpha ^-|,s}$ . However, since $\bar \ell (\alpha )$ is unbounded, $\psi (\alpha ^-)$ holds, which gives us exactly what we need. Note that in this case, we do not need to use the inductive hypothesis on $\alpha ^-$ , nor is it sufficient for our purposes.Footnote ¹¹

2.4.3 M is not everything.

As a result:

2.4.4 Maximality.

The proof of Theorem 2.1 is concluded with the following lemma.

Proof. Let $e\in \mathbb N$ ; let $\alpha $ be the node on the true path of length $e+1$ .

First, suppose that $\alpha $ is an $\infty $ -node. Every ball pulled by $\alpha $ is already an element of $W_e$ , so $Y {(\alpha)}\subseteq W_e$ . By Lemma 2.18, ${M}^\complement \subseteq ^* W_e$ .

Next, suppose that $\alpha $ is a ${\texttt {fin}}$ -node. Since $\bar \ell (\alpha )$ is unbounded, we know that $\psi (\alpha ^-)$ fails: there is some k such that for every A-true stage s,

$$\begin{align*}|Y{(\alpha^-)}_{s}\cap W_{e,s} | \leqslant k. \end{align*}$$

Then $|Y {(\alpha ^-)}\cap W_e| \leqslant k$ . For otherwise, there would be a set $D\subseteq Y {(\alpha ^-)}\cap W_e$ of size $k+1$ . But then, for all but finitely many stages s, $D\subseteq Y {(\alpha ^-)}_{s}\cap W_{e,s}$ , so $|Y {(\alpha ^-)}_{s}\cap W_{e,s} | \geqslant k+1$ for some A-true stage s, which is not the case. It follows that ${M}^\complement \cap W_e =^*\emptyset $ .

3.0.1 Boolean algebras and binary trees.

Recall that we can generate Boolean algebras from trees. If $T\subseteq 2^{<\omega }$ is a tree, then $B(T)$ is the quotient of the free Boolean algebra with generators $\sigma \in T$ , modulo the relations:

• • every $\tau \in T$ is the join of its children;

• • if $\sigma ,\tau \in T$ are incomparable then $\sigma \wedge \tau = 0_{B(T)}$ .

Equivalently, we can think of $B(T)$ as the collection of finite unions of the sets $[\sigma ]\cap [T]$ , for all $\sigma \in T$ , ordered by set inclusion; in this version, $\tau $ is identified with $[\tau ]\cap [T]$ .

Note that this implies that if $\tau $ is a leaf (or more generally, if $\tau $ is not extendible to an infinite path on T), then $\tau = 0_{B(T)}$ ; and that ${\lambda }=1_{B(T)}$ . Every countable Boolean algebra can be presented in such a way (up to isomorphism) by considering a tree of nonzero finite Boolean combinations of any sequence of generators.Footnote ¹³ For our purposes in this section, we rely on the fact that the atomless Boolean algebra is $B(2^{<\omega })$ , the one generated by the full binary tree.Footnote ¹⁴ Thus, given a coinfinite $\text {low}_2$ c.e. set A, we will enumerate a superset $H\supseteq A$ , and construct a tree of sets $\langle {Z(\rho )\,:\, \rho \in 2^{<\omega }}\rangle $ satisfying:

1. (i) $Z({\lambda })=^* {H}^\complement $ ;

2. (ii) for each $\rho $ , $Z(\rho )$ is infinite;

3. (iii) for each $\rho $ , $H\cup Z(\rho )$ is c.e.;

4. (iv) for all $\rho \in 2^{<\omega }$ , $Z(\rho \hat {\,\,} 0)$ and $Z(\rho \hat {\,\,} 1)$ are disjoint, and $Z(\rho ) =^* Z(\rho \hat {\,\,} 0)\cup Z(\rho \hat {\,\,} 1)$ ;

5. (v) for every c.e. set $W\supseteq H$ there is a finite set $D\subset 2^{<\omega }$ such that $W\smallsetminus H=^*\bigcup \left \{ Z(\rho ) \,:\, \rho \in D \right \}$ .

This suffices: to see that $\mathcal {L}^*(H)$ is a Boolean algebra, it suffices to show that it is complemented (see [Reference Soare17, Section X.2]). Indeed, if $W\supseteq H$ is c.e., by (v), let $D\subseteq 2^{<\omega }$ be a finite set such that $W\smallsetminus H =^* \bigcup \left \{ Z(\rho ) \,:\, \rho \in D \right \}$ . By (iv), we may assume that all elements of D have the same length, say k. Let $E = \{0,1\}^k\smallsetminus D$ ; then $H\cup \bigcup \left \{ Z(\rho ) \,:\, \rho \in E \right \}$ is a complement of W in $\mathcal {L}^*(H)$ ; by (iii), it is c.e. (ii) and (iv) also ensure that $\mathcal {L}^*(H)$ has no atoms.

As we are using the $\Delta _3$ machinery, the sets $Z(\rho )\cup H$ will not be uniformly c.e.; rather, they will be $=^*$ to sets whose indices we can read off the true path of the construction.

3.1.1 The requirements.

In the current construction, we will have two kinds of requirements. The analogues of the maximality requirements from the previous construction are the requirements for meeting (v) above. For each $\rho \in 2^{<\omega }$ of length e, we will ensure that either $Z(\rho )\subseteq ^* W_e$ or $Z(\rho )\cap W_e=^*\emptyset $ . Thus, ${D = \left \{ \rho \in \{0,1\}^e \,:\, Z(\rho )\subseteq ^* W_e \right \}}$ will show that (v) holds for $W_e$ . The way to meet these requirements will be very similar to the previous construction, except that the $\alpha $ question will be more complicated; for each $\rho \in \{0,1\}^e$ , we will need to guess whether we will see enough balls to make $Z(\rho )\subseteq W_e$ while ensuring that $Z(\rho )$ is infinite. The mechanism of pulling and eliminating will be used to achieve that, but we will see that the definitions of “pullable” and “eliminable” need to be more complicated.

The other requirements are new: they ensure that (iv) above holds. Namely, for each $\rho $ , we need to ensure that $Z(\rho \hat {\,\,} 0)$ and $Z(\rho \hat {\,\,} 1)$ form a splitting of $Z(\rho )$ (up to finite differences), into two infinite sets. In the construction, we will have nodes devoted to such a requirement. The task for such a node is to take balls with label $\rho $ , and decide whether to extend their label to either $\rho \hat {\,\,} 0$ or $\rho \hat {\,\,} 1$ . The difficulty, of course, is that during the construction, we do not know whether a ball x is in A or not. It would be very bad if all but finitely many balls x that we direct to $\rho \hat {\,\,} 0$ , for example, will end up in A, as that would make $Z(\rho \hat {\,\,} 0)$ finite. Unlike the other kind of requirement, $\Delta _3$ guessing is not sufficient for this task: we know that there will be infinitely many $x\notin A$ with label $\rho $ . We need to somehow obtain individual balls that we have a good reason to guess are not in A.

Our main contribution is precisely this: a new method for certifying that groups of balls are not in A. As we will explain later, in fact, to perform this certifying and splitting, it will be notationally convenient to have not just one level of nodes devoted for each requirement, but two; nodes and their children together will perform these tasks.

So we will have two kinds of nodes.

1. (i) Decision nodes, of length $3e$ , whose task, for each $\rho \in \{0,1\}^e$ , to decide $W_e$ on $Y {(\alpha ,\rho)}$ . We call a node of length $3e$ an e-decision node.

2. (ii) Splitting nodes and their children, of lengths $3e+1$ and $3e+2$ . The children of a decision node $\alpha $ of length $3e+1$ will pull balls with labels $\rho $ of length e, and decide to extend their label to either $\rho \hat {\,\,} 0$ or $\rho \hat {\,\,} 1$ . We call a node of length $3e+1$ a parent e-splitting node, and a node of length $3e+2$ a child e-splitting node.

3.1.2 True balls rather than true stages.

As above, we will make use of a true-stage enumeration $(\tilde A_r)$ of A with respect to the function $f(r)=r^2$ . We will use the same mechanism as above to slow down this enumeration to an enumeration $(A_s)$ defined during the construction, again letting $n(s)=r$ when $A_s = \tilde A_r$ . As in the previous construction, if we move or enumerate balls, or perform any other action, at a stage s, then will set $A_{s+1}=A_s$ (by setting $n(s+1)=n(s)$ ). When no other action is taken at stage s, we will declare s to be a new balls stage, and set $n(s+1)=n(s)+1$ .

As in the very first part of the proof of Lemma 2.14, this ensures that infinitely often, the root of the tree receives large collections of balls which are all outside A.

However, unlike the previous construction, there will be an element of delay in ball movement. In order to ensure that both $Y {(\alpha ,\rho \hat {\,\,} 0)}$ and $Y {(\alpha ,\rho \hat {\,\,} 1)}$ receive many balls outside A, a child of a splitting node $\alpha $ will hold on to many balls in $Y {(\alpha ,\rho)}$ until it receives confirmation from $\emptyset '$ that these balls are indeed outside A. (Infinitely often, this confirmation will be incorrect, but infinitely often it will be correct, and this will be sufficient for our purposes; we will see that what is important, is that each individual attempt at certification makes only finitely many mistakes.) The stages at which such confirmation is received do not need to line up with the A-true stages. So we will not be able to directly prove Lemma 2.14 for the current construction. Rather, instead of looking for “completely true” stages, at which every ball on the machine is correct, we will simply ensure that we get more and more true balls.

Definition 3.1. A number $x\in {A}^\complement _s$ is A-true at stage s if

$$\begin{align*}A\restriction{\,{x+1}} = A_s\restriction{\,{x+1}}. \end{align*}$$

That is, not only x will not enter A in the future, but no number $y\leqslant x$ currently outside A will enter A in the future. If we get enough of these, we do not really care that larger balls are not A-true at the same stage.

3.1.3 The questions and the tree.

A decision node $\alpha $ needs to decide, for each $\rho $ of length e, whether to make $Z(\rho )$ almost contained in $W_e$ or almost disjoint from $W_e$ . This means that the $\alpha $ question is more complicated. Let $\alpha $ be an e-decision node. For each set $D\subseteq \{0,1\}^e$ , the statement $\psi (\alpha ,D)$ says:

D is the set of $\rho \in \{0,1\}^{e}$ for which for every k there is some s such that

$$\begin{align*}|Y{(\alpha,\rho)}_{s}\cap W_{e,s} \cap C_s | \geqslant k. \end{align*}$$

The outcomes of $\alpha $ are

$$\begin{align*}D_n \end{align*}$$

for each $D\subseteq \{0,1\}^{e}$ and each $n\in \mathbb N$ , ordered in order-type $\omega $ in some way. Note that $\psi (\alpha ,D)$ will hold for precisely one $D\subseteq \{0,1\}^e$ , and we will ensure that $Z(\rho )\subseteq ^* W_e$ if and only if $\rho \in D$ (and otherwise, $Z(\rho )\cap W_e =^*\emptyset $ ).

Since each $\psi (\alpha ,D)$ is a finite Boolean combination of $\Pi _2(A)$ statements, it is equivalent to a $\Sigma _3$ statement, so as in the previous construction, we obtain uniformly computable, nondecreasing sequences $\bar \ell (\psi (\alpha ,D),n)$ , such that $\psi (\alpha ,D)$ holds if and only if for some n, $\bar \ell (\psi (\alpha ,D),n)$ is unbounded. As above, we use these to define sequences $\bar \ell (\alpha )$ for nodes $\alpha $ ; these will only be used for nodes $\alpha $ that are the children of decision nodes. So we recursively define:

• • If $\alpha = (\alpha ^-)\hat {\,\,} D_n$ is the child of a decision node $\alpha ^-$ , then we let $\ell (\alpha )_{s}$ be the minimum of $\ell (\psi (\alpha ^-,D),n)_{s}$ , and $\ell (\beta )_{s}$ for any that is also the child of a decision node.

It is more difficult to explain now why we need both parent and children splitting nodes. If $\alpha $ is a parent e-splitting node, then its children are $\alpha \hat {\,\,} n$ for all $n\in \mathbb N$ , ordered naturally. If $\beta $ is such a child, then $\beta $ has a unique child $\beta ^+$ on the tree (which will in turn be an $e+1$ -decision node). The outcomes n of $\alpha $ do not quite represent guesses about the behaviour of $\alpha $ ; we will discuss them later.

3.1.4 Updating previous definitions.

We update notation from the previous construction. For a node $\alpha $ and $\rho \in 2^{<\omega }$ , and a stage s, we let:

• • $Y {(<_{L}\!\alpha ,\rho)}_{s}$ be the collection of balls that at the beginning of stage s, reside at some node $\beta $ that lies to the left of $\alpha $ , and have label extending $\rho $ . That is, $Y {(<_{L}\!\alpha ,\rho)}_{s} = \bigcup _{\beta <_L\alpha } Y {(\beta ,\rho)}_{s} = Y {(<_{L}\!\alpha)}_{s}\cap Y {({\lambda },\rho)}_{s}$ ;

• • $Y {(\leqslant \!\alpha ,\rho)}_{s} = Y {(<_{L}\!\alpha ,\rho)}_{s}\cup Y {(\alpha ,\rho)}_{s}$ .

We need to comment on the new part of this definition, namely, the second part of (iii). The issue is that, as mentioned above, we will not be able to directly show that $Y {(\alpha)}_{s}$ is “full” during A-true stages (in the sense of Lemma 2.14), only that it gets “filled” by A-true balls (see Proposition 3.13(b) below). Consider a node $\alpha $ on the true path (a child of a decision node). At a stage s, the parent $\alpha ^-$ may have many A-true balls. At the same stage, $\alpha $ already has larger balls and so does not feel “hungry” ( $|Y {(\alpha ,\rho)}_{s}| \geqslant \ell (\alpha )_{s}$ ). But because s is not an A-true stage, the balls that $\alpha $ has may be false. The problem is solved by allowing $\alpha $ to pull balls smaller than ones it already has, even if it does not feel “hungry”. This modification is sufficiently tame so that nodes on the left of the true path will again only pull finitely many balls from ${A}^\complement $ .Footnote ¹⁵

3.1.5 Certification.

Since A is $\text {low}_2$ and c.e., some $\emptyset '$ -computable function dominates all A-computable functions (this follows from Martin’s characterisation [Reference Martin10] of the high degrees as those that compute functions that dominate all computable functions).

Let $\alpha $ be a parent e-splitting node, and fix $\rho \in \{0,1\}^e$ . The very rough idea of splitting is that $\alpha $ will define blocks of balls, and hold on to these balls until $\varphi $ gives confirmation to release these balls to nodes below. The $k{}^{\text {th}}$ block will contain at least $2k$ many balls. For now, let us forget about $\alpha $ ’s children, and imagine that released balls are passed to the next splitting node (if $\alpha $ did not have multiple children then there would be a unique $(e+1)$ -splitting node extending $\alpha $ ). When the $k{}^{\text {th}}$ block of balls is certified by $\varphi $ , then k of the balls in that block will receive the label $\rho \hat {\,\,} 0$ , and the rest, the label $\rho \hat {\,\,} 1$ .

For this purpose, we will define an A-computable function $f^{\alpha ,\rho }$ . An input $k\in \mathbb N$ for $f^{\alpha ,\rho }$ indicates an attempt to capture and test the $k{}^{\text {th}}$ block of balls. Suppose that at some stage r, for all $m<k$ , the $m{}^{\text {th}}$ -block of balls is currently defined, and that $\alpha $ holds $2k$ many balls with label $\rho $ (that are not associated with any existing block). We would then declare these balls to constitute the $k{}^{\text {th}}$ block, and define $f^{\alpha ,\rho }_{r+1}(k)> \varphi _{r}(k)$ . The $A_{r+1}$ -use $u = u^{\alpha ,\rho }_{r+1}(k)$ of this computation will bound at least $2k$ -many elements of the block.

We then wait for a stage $s>r$ at which one of two things happens. Either ${A_s\restriction {\,{u}}\ne A_{r+1}\restriction {\,{u}}}$ , in which case $f^{\alpha ,\rho }_s(k)\operatorname {\mathrm {\uparrow }}$ . This A-change likely involves some elements of the block entering A. We will then wait for $\alpha $ to obtain new balls, that will allow us to redefine a new $k{}^{\text {th}}$ block.

Otherwise, we hope to see $\varphi _s(k)> f^{\alpha ,\rho }_s(k) = f^{\alpha ,\rho }_{r+1}(k)$ . This means that ${\varphi _s(k)> \varphi _r(k)}$ . We will regard this as evidence that elements from the block will not later enter A, and say that the block is certified. We then process the block as described above, splitting it evenly between $\rho \hat {\,\,} 0$ and $\rho \hat {\,\,} 1$ , and passing the balls in the block to the next node below.

Of course, it is possible that neither of these events happens: the computation $f^{\alpha ,\rho }_{r+1}(k)$ is never undefined, but $\varphi (k)$ never exceeds the value of that computation. Therefore, while we wait, we will try to define the $(k+1){}^{\text {th}}$ -block, and so on.

The argument that this works will be by contradiction. We want to argue that infinitely many blocks will be released, and infinitely many of those will be A-correct, i.e., only contain balls from ${A}^\complement $ . Assuming this is not the case, we will want to show that $f^{\alpha ,\rho }$ is a total function. If this is shown, then we obtain the desired contradiction from the fact that $\varphi $ dominates $f^{\alpha ,\rho }$ , showing that for almost all k, the correct $k{}^{\text {th}}$ block will in fact be certified and released. To show that $f^{\alpha ,\rho }$ is total (under the assumption for contradiction), we use the fact that $\varphi _s(k)$ changes only finitely many times. This implies that the $k{}^{\text {th}}$ block will be released only finitely many times, and so eventually, an A-correct $k{}^{\text {th}}$ block will be defined, showing that $f^{\alpha ,\rho }(k)\operatorname {\mathrm {\downarrow }}$ .

Thus, the use of the certification process is in the delay it imposes on the release of balls. If we didn’t wait for certification, then it is possible that we would keep redefining and releasing some $k{}^{\text {th}}$ block, none of whose versions is A-correct. It would then be possible that of the blocks that we release, all but finitely many of the balls that we target to $\rho \hat {\,\,} 0$ , say, end up in A, resulting in $Z(\rho \hat {\,\,} 1)=^*Z(\rho )$ , so the splitting requirement is not met.

We now come to the need for multiple children. The reason is delicate. We want to argue that an A-correct definition of some $f^{\alpha ,\rho }(k)$ will eventually be made. Imagine the following sequence of events:

1. (1) $f^{\alpha ,\rho }(k)$ is defined at some stage.

2. (2) While waiting for certification, we also define $f^{\alpha ,\rho }(k+1)$ (after all, we do not know where $\varphi $ starts dominating).

3. (3) The $(k+1){}^{\text {th}}$ block is certified and released, but not the $k{}^{\text {th}}$ one.

4. (4) Some x in the $k{}^{\text {th}}$ block enters A, making both $f^{\alpha ,\rho }(k)$ and $f^{\alpha ,\rho }(k+1)$ undefined.

Now, all we know, by induction, is that an analogue of Lemma 2.14 holds for $\alpha $ : for all m, there is some stage s at which $C_s\cap Y {(\alpha ,\rho)}_{s}$ has size at least m. If s is such a stage after these events unfolded, then it is possible that the bulk of $C_s\cap Y {(\alpha ,\rho)}_{s}$ is actually lying well below $\alpha $ —it is part of the $(k+1)$ -block that was released at step (3) above. Since these balls are not currently residing at $\alpha $ , we cannot use them to define a new $k{}^{\text {th}}$ -block. Unless…we pull them back up to $\alpha $ from wherever they are currently residing.

Now, in all similar constructions, pulling balls from below is a source of innumerable problems. Just as a simple example, it becomes much more difficult to argue that balls outside H will eventually reach a permanent residence, since pulling balls back up is increasing in the Kleene–Brouwer ordering. More seriously, it will be difficult to show that the sets $H\cup Y(\gamma ,\rho )$ , for $\gamma $ on the true path, are c.e. So we will not do this. A work-around is to allow $\alpha $ to have infinitely many children $\alpha \hat {\,\,} n$ . While waiting for certification, each block will “reside” at one of the children. Indeed the block waiting at a child $\beta $ will simply be $S {(\beta ,\rho)}_{s}$ .

While the $k{}^{\text {th}}$ block is waiting at $\alpha \hat {\,\,} n$ , all blocks defined later will reside at children of $\alpha $ that lie to the right of $\alpha \hat {\,\,} n$ , namely, children $\alpha \hat {\,\,} m$ for various $m>n$ . When the $k{}^{\text {th}}$ block is dissolved (at step (4) above), it is now fine for $\alpha \hat {\,\,} n$ to pull all balls from the right, even those that were released to lower nodes, and use them to constitute a new $k{}^{\text {th}}$ block.

One can think of $\alpha \hat {\,\,} n$ as representing the guess that $\varphi $ starts majorising $f^{\alpha ,\rho }$ from input n. This is not exactly right but the intuition is not far off.

Another point to make is that for distinct $\rho $ , the function $\varphi $ may start dominating the various functions $f^{\alpha ,\rho }$ at different locations. The argument above will show that for each $\rho \in \{0,1\}^e$ , there is some child $\beta $ that releases $\rho $ -blocks infinitely often. However, to obtain a child of $\alpha $ on the true path, we need the same $\beta $ to work for all $\rho $ . Thus, in the definition of certification below, we will let a child $\beta $ hold on to its blocks, until all of them (one for each $\rho $ ) are certified, and only then will it release them all.

We can now give the details. Let $\alpha $ be a parent e-splitting node; let $\rho \in \{0,1\}^e$ . As discussed, we will define a function $f^{\alpha ,\rho }$ , with intended oracle A. So at various stages s, for various k, we may have $f^{\alpha ,\rho }_s(k)\operatorname {\mathrm {\downarrow }}$ or not, and if it is defined, then we will declare a use $u = u^{\alpha ,\rho }_s(k)$ for this computation. The usual rule applies: if $f^{\alpha ,\rho }_s(k)\operatorname {\mathrm {\downarrow }}$ , with use u, and $A_s\restriction {\,{u}} =A_{s+1}\restriction {\,{u}}$ , then $f^{\alpha ,\rho }_{s+1}(k)\operatorname {\mathrm {\downarrow }} = f^{\alpha ,\rho }_s(k)$ with the same use. If, on the other hand, $A_s\restriction {\,{u}} \ne A_{s+1}\restriction {\,{u}}$ , then we declare that $f^{\alpha ,\rho }_{s+1}(k)\operatorname {\mathrm {\uparrow }}$ . Note that this will only happen if s is a new balls stage, and during such stages, we do not define new computations $f^{\alpha ,\rho }_{s+1}(k)$ .

We will define $f^{\alpha ,\rho }_s(k)$ only for $k\geqslant 1$ , since there is no point in dealing with a block of size 0.Footnote ¹⁶

We will ensure that if $k\geqslant 1$ and $f^{\alpha ,\rho }_s(k+1)\operatorname {\mathrm {\downarrow }}$ then $f^{\alpha ,\rho }_s(k)\operatorname {\mathrm {\downarrow }}$ as well, that is, the domain of $f^{\alpha ,\rho }_s$ is a (finite) initial segment of $\mathbb N\smallsetminus \{0\}$ (see Lemma 3.15 below).

When we define a computation $f^{\alpha ,\rho }_r(k)$ (at some stage $r-1$ ), then we will declare that the $k{}^{\text {th}}$ block is currently waiting at some child $\beta $ of $\alpha $ ; we will record this by setting $k^{\rho }_s(\beta ) = k$ (note that $\alpha $ is determined by $\beta $ ). This notation implies that at most one $\rho $ -block is waiting at $\beta $ at a given stage.

If $k^{\rho }_s(\beta )\operatorname {\mathrm {\downarrow }}=k$ then $f^{\alpha ,\rho }_s(k)\operatorname {\mathrm {\downarrow }}$ . Thus, if $k^{\rho }_s(\beta )\operatorname {\mathrm {\downarrow }}=k$ but $f^{\alpha ,\rho }_{s+1}(k)\operatorname {\mathrm {\uparrow }}$ then we declare that $k^{\rho }_{s+1}(\beta )\operatorname {\mathrm {\uparrow }}$ . That is, if the $k{}^{\text {th}}$ block is waiting at $\beta $ at (the beginning of) stage s, but is dissolved when passing from stage s to stage $s+1$ , then no block is waiting at $\beta $ at stage $s+1$ .

If $\beta $ releases balls at stage s, then we will also set $k^{\alpha ,\rho }_{s+1}(\beta )\operatorname {\mathrm {\uparrow }}$ , even though $f^{\alpha ,\rho }_{s+1}(k)\operatorname {\mathrm {\downarrow }}$ ; at stage $s+1$ , the block does not wait at $\beta $ anymore.

We will ensure that if $k^{\rho }_s(\beta )\operatorname {\mathrm {\downarrow }}$ , then for every child $\gamma $ of $\alpha $ that lies to the left of $\beta $ , we have $k^{\rho }_s(\gamma )\operatorname {\mathrm {\downarrow }}$ as well (and $k^{\rho }_s(\gamma )<k^{\rho }_s(\beta )$ ). Again, see Lemma 3.15.

3.3.1 Defining the true path.

We define the true path inductively. The root ${\lambda }$ is declared to lie on the true path. Let $\alpha $ be a node on the true path.

• • If $\alpha $ is a decision node, then the leftmost child $\beta $ of $\alpha $ with $\bar \ell (\beta )$ unbounded is on the true path.

• • If $\alpha $ is a parent splitting node, then the leftmost child $\beta $ of $\alpha $ that releases balls infinitely often, lies on the true path.

• • If $\alpha $ is a child splitting node, then its unique child is also on the true path.

As above, if $\alpha $ is a decision node, then as one of the statements $\psi (\alpha ,D)$ is true, one of $\alpha $ ’s children will lie on the true path. On the other hand, we will need to work to show that if $\alpha $ is a parent splitting node on the true path, then one of its children lies on the true path. For now, we do not know that the true path is infinite.

In fact, to show that a parent splitting node $\alpha $ which lies on the true path has a child on the true path, we will need to already know that the analogues of Lemmas 2.12 and 2.14 hold for $\alpha $ . Thus, we will need to prove both by simultaneous induction.

Recall (Definition 3.2) that we let

$$\begin{align*}C_s = \left\{x\in Q_s \,:\, A_s\restriction{\,{x+1}} = A\restriction{\,{x+1}}\right\}, \end{align*}$$

the collection of balls $x\in Q_s$ which are A-true at stage s.

The proof of Proposition 3.13 will take some work. As discussed, we prove it by induction on the length of $\alpha $ . For now, we note:

3.3.2 Case I: The root

We start the inductive verification of Proposition 3.13 by verifying that it holds for the root. (a) is vacuous in this case; (b) is as in Lemma 2.14. We note that this is the only part of the proof in which we use the fact that we are using the true-stage enumeration of A.

3.3.3 Case II: Decision nodes

Suppose that $\alpha $ is an e-decision node that lies on the true path, and suppose that Proposition 3.13 holds for $\alpha $ . Let $\beta $ be $\alpha $ ’s child that lies on the true path; we show that Proposition 3.13 holds for $\beta $ as well. Let $D\subseteq \{0,1\}^e$ such that $\beta = \alpha \hat {\,\,} D_n$ for some n. That is, $\rho \in D$ if and only if for all k there is some s such that $|Y {(\alpha ,\rho)}_{s}\cap W_{e,s}\cap C_s|\geqslant k$ .