A blend of methods of recursion theory and topology: A П 0^1 tree of shadow points

Arch. Math. Logic 43, 991–1008 (2004) Digital Object Identifier (DOI): 10.1007/s00153-004-0244-0

Mathematical Logic

Iraj Kalantari · Larry Welch

A blend of methods of recursion theory and topology: A
01 tree of shadow points Received: 13 June 2003 / Revised version: 18 December 2003 / Published online: 28 May 2004 – © Springer-Verlag 2004 Abstract. This paper is a sequel to our [7]. In that paper we constructed a
01 tree of avoidable points. Here we construct a
01 tree of shadow points. This tree is a tree of sharp filters, where a sharp filter is a nested sequence of basic open sets converging to a point. In the construction we assign to each basic open set on the tree an address in 2<ω . One interesting fact is that while our
01 tree of sharp filters (a subtree of <ω ) is isomorphic to the tree of addresses (a subtree of 2<ω ), the tree of addresses is recursively enumerable but not recursive. To achieve this end we use a finite injury priority argument.

1. Introduction The present paper continues our work in [7]. Here, too, we assume that our space
X,  is resolvable and semi-recursively presentable. In our previous papers we also have required our spaces to be completely connected, that  be compactible, and that X contain at least two points. With these additional conditions, X is regular, second countable, and of second category. But these conditions can be relaxed somewhat here, as the construction we present will work in any resolvable, semi-recursively presentable space in which there is a uniform method by which a complete binary recursive tree of sharp filters can be formed inside any member of . For instance, 2<ω (Cantor space) is such a space, but it is not completely connected. This space has been much studied, and is the topological space whose recursive properties are best known. We will address the application of our theorem to Cantor space in a separate section after the proof of the main theorem. We showed in [7] that avoidable points are those nonrecursive points that can be excluded from the domain of a quantum recursive function (a recursive function that includes all recursive points in its domain); a shadow point is a nonrecursive point that is included in the domain of every quantum recursive function. The primary theorem of this paper (Theorem 3.1) is the building of a nonrecursively bounded
01 tree of shadow points. In capturing avoidable points on trees in [7], we blended the topological and recursion theoretic requirements without a Department of Mathematics, Western Illinois University, Macomb, IL 61455, USA. e-mail: {i-kalantari, l-welch}@wiu.edu Mathematics Subject Classification (2000): 03D45, 03D80, 03C57, 54A20 Key words or phrases: Recursion theory – Topology –
01 Trees – Recursive analysis – Recursive topology

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major conflict (see Section 9 of [7]), since we were able to keep the trees recursively bounded as per Jockusch & Soare [5]. However, as a recursively bounded
01 tree that contains no recursive points will necessarily contain avoidable points (see Theorem 9.1 of [7]), we have to employ a priority argument to prevent recursive boundedness and capture a recursive tree of shadow points. In building our tree, we make each node a basic open set, choosing those sets in such a way that the nodes on each infinite branch through the tree intersect in a single point. To help us determine the shape of the tree, we give each node an address which is a member of 2<ω , ordering the nodes according to the usual order of their addresses. To make the infinite branches nonrecursive, we will of course have to impose recursion-theoretic requirements on the tree. In order to be sure the branches converge to shadow points instead of avoidable points, we must impose topological requirements. In this way the proof of our theorem blends recursion theory and topology. Thus our construction produces two trees: a subtree of <ω and a tree of addresses in 2<ω ; these trees are order isomorphic but not recursively so. This occurs because the first tree is a nonrecursively bounded, recursive one, and the second tree is a recursively bounded, recursively enumerable, but nonrecursive one. In our [7] and in this paper, we study trees of sharp filters; it should be pointed out that a
01 tree of sharp filters corresponds to a strong
02 class. The connections between strong
02 and
01 classes are studied in Jockusch, Lewis & Remmel [4], while a deeper study of the same and connections between unbounded and recursively bounded
01 classes are studied in Cenzer & Remmel [2]. In particular, for our spaces, every strong
02 class can give rise to a
01 tree of sharp filters, and conversely, every
01 tree of sharp filters represents a strong
02 class. As suggested by Steve Simpson, a shadow point x can be defined as a point that is not recursive, yet for all
02 subsets S of X, if S includes all recursive points, then S contains x. In that context, our primary theorem implies, for example, that there exists a closed
02 set of such points. In Section 4 we examine avoidable and shadow points and some implications of our main theorem in the space 2ω . 2. Key definitions For convenience we repeat some key definitions and facts from [7]. Recall that a sharp filter in X is a sequence of basic open sets, closure of each contained in the previous one, converging to a single point. Definition 1. A sequence of partial recursive functions {ψn : n ∈ ω} is called an acceptable enumeration capturing all recursive sharp filters in X if for each n ∈ ω, ψn : ω → X : 1. (∀n, s, k)[ψns (k + 1) ↓⇒ ψns (0) ↓ ∧ · · · ∧ ψns (k) ↓], and 2. (∀n, k)[ψn (k + 1) ↓⇒ ψn (k + 1) ⊆ ψn (k)]. (Here s is a stage in the computation and ψns (k) denotes the outcome of ψn (k) at stage s.) It is clear that acceptable enumerations capturing all recursive sharp filters exist and for each recursive sharp filter A, there is n such that ψn = A.

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Definition 2. Let
X, X  be a resolvable space. Suppose there is a partial recursive function φ : ω → ω and x ∈ X, such that for any n 1. if ψn is a recursive sharp filter, then φ(n) ↓; and 2. if φ(n) ↓ and ψn (φ(n)) ↓ (which is true if ψn is a sharp filter), then x ∈ / ψn (φ(n)). We then say φ is an avoidance function for x, or x is avoidable via φ. If x is avoidable via some φ, we say x is avoidable. φ is an avoidance function if it is an avoidance function for some x. For an avoidance function φ, let Sφ = {x ∈ X : φ is an avoidance function for x}, and refer to Sφ as the spectrum of φ. Theorem 1. 1. Let φ : ω → ω be a recursive avoidance function. Then there is a recursive quantum function f such that dom(f ) = X − Sφ . 2. Let f be a recursive quantum function. Then exdom(f ) ⊆ Av(X), and thus dom(f ) ⊇ Shad(X). Definition 3. If a point x is nonrecursive and not avoidable, we say x is a shadow point. Similar to Rec(X) denoting the set of recursive points of X, we use Av(X) and Shad(X) to denote the set of recursively avoidable and shadow points of X respectively. Definition 4. Let  be the set of all finite binary strings on the set {0, 1}. We usually denote members of  by σ or τ . Length of σ is denoted by lh(σ ). If σ is a substring of τ , we write σ ⊆str τ . If σ is lexicographically before τ , we write σ ≤lex τ . A tree of sharp filters T is the range of a partial function :  →  satisfying the following conditions: 1. (∅) ↓; 2. for all σ, τ ∈ , if (τ ) ↓ and σ ⊆str τ , then (σ ) ↓ and (τ ) ⊆ (σ ); 3. if σ
str τ , τ
str σ , (σ ) ↓ and (τ ) ↓, then (σ ) ∩ (τ ) = ∅; and 4. if b : ω →  is a total function such that for every n, lh(b(n)) = n, (b(n)) ↓, and (b(n + 1)) ⊆ (b(n)) (i.e., if ◦ b is a branch through T), then ◦ b is a sharp filter. T is a complete tree of sharp filters if is a total function. For an infinite branch b = {βi : i ∈ ω} through T, which is a sharp filter and converges to a point, we denote that point by xb and say that the point lies on T. For notational convenience, for σ ∈  we denote (σ ) by θσ ; thus we have T = { (σ ) : σ ∈   } = {θσ : σ ∈   }, where   = dom( ) is a subset of  and a tree under the ordering ⊆str . Similarly, when we have a tree Ti we write it as Ti = {θi,σ : σ ∈ i }.

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For a tree T, let T = {xb : b is a branch in T}. In order to refer to the basic open sets at a fixed height on a tree T, as well as to their topological union, define the n-nodes and n-support of a tree T as follows: n-Nodes(T) = {θσ ∈ T : lh(σ ) = n}; and
n-Supp(T) = {θσ ∈ T : lh(σ ) = n}. It should be noted that 0-Supp(T) ⊇ 1-Supp(T) ⊇ 2-Supp(T) ⊇ · · · . Finally, define ω-Supp(T) =



n-Supp(T).

n

Notice that condition (3) make it clear that ω-Supp(T) = T . When we need to construct a tree T with desired topological properties, we simply proceed to describe by defining θσ for each σ in some   where   will become dom( ). We can think of the basic open set θσ as the node that lies on the tree T at the address σ . In this way, we blend recursion theoretic demands with topological ones. However, before we proceed, we introduce some recursion theoretic properties of trees and make some observations about these properties. Definition 5. T is recursive as a set if T = {θσ : σ ∈ dom( )} is a recursive subset of . Definition 6. T is recursive as a tree, or a
01 tree, if there is a recursive procedure to determine, for each n ∈ ω and each finite sequence
δ0 , · · · , δn  of members of , whether there is σ ∈ dom( ) with lh(σ ) = n, such that (∀τ ⊆str σ )(∀k ≤ n)[(lh(τ ) = k) ⇒ (θτ = δk )]. Theorem 2. Let T = {θσ : σ ∈ dom( )} be a tree of sharp filters that is recursive as a set, where is partial recursive. Then T is recursive as a tree. Definition 7. A tree T is recursively bounded if there is a recursive function f : ω → <ω such that (∀σ ∈ dom( ))(∀n ∈ ω)[(lh(σ ) = n) ⇒ (θσ ∈ f (n))]. This definition is equivalent to that of Jockusch & Soare [5] and [6]; Cenzer & Remmel [2] give another equivalent definition.

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3. Trees of shadow points: a priority argument In Theorem 4.5 of [7] we showed that Shad(X) is of second category, and hence dense in X. As a result, it is possible to build a tree of sharp filters all of whose branches converge to shadow points. But a pertinent question is how low such trees can lie in the analytic hierarchy. We answer that question here by growing a recursive tree. Our tree of sharp filters is a
01 tree. In contrast to the Theorem 9.1 in [7], this tree must necessarily be nonrecursively bounded. In order to produce such a tree, we use a finite injury priority argument. The argument is necessarily long but fruity, and depends on growing a tree of sharp filters while trying to satisfy conflicting requirements. Though most
01 trees are obtained by pruning a complete tree, ours is grown by a combination of pruning and grafting onto a tree that is not complete. Though this result captures a set of points in recursive topology or recursive analysis, and is not about degrees, we employ a priority argument because of the multiplicity of requirements to be met. This is a finite injury priority argument for the following reason. Each requirement is attended to at a specific level of our tree. If we have managed to satisfy a requirement Rm by attention to the nodes at level n, and then a higher priority argument requires pruning and grafting at a lower level, the pruned nodes that satisfy Rm then get removed from the tree, and the new grafted nodes may no longer satisfy Rm . In that sense Rm can be injured. The nodes on our tree, which are basic open sets, are matched to addresses to be followed strategically in forming the tree. Consequently, the
01 tree of sharp filters we finally obtain is isomorphic – nonrecursively – to the tree of addresses that we finally get. A node σ on the tree of addresses may be acted upon at a stage n > lh(σ ), and because of this, the tree of addresses that we end with is not
01 (it is recursively enumerable but not recursive). This will not damage our proof that the tree of sharp filters is a
01 tree, however. It is important throughout to keep in mind the distinction between the tree of addresses, which is a subtree of 2<ω , and the tree of sharp filters, which is a subtree of <ω . Theorem 3. Given any δ ∈ , there is a
01 tree T∞ of sharp filters having 2ℵ0 infinite branches, such that T∞ = {xb : b is a branch of T∞ } ⊆ δ ∩ Shad(X). Proof. In order to construct our tree T∞ , in addition to arranging for it to be a recursive tree of sharp filters, the other requirements are the usual contrasting ones in a priority argument. These requirements are: T∞ will have 2ℵ0 infinite branches (a positive requirement), T∞ will not have any recursive point as the limit of a branch (a negative requirement), and T∞ will not have any avoidable point as the limit of a branch (a negative requirement). Clearly, if T∞ satisfies the conditions and the requirements listed above, it is a tree of the desired kind for the theorem. Our technique is to start with a complete recursive tree of sharp filters T∗ , to obtain from it an initial subtree T−1 which is not complete, but of a particular form, and to successively modify T−1 to get recursive trees T0 , T1 , . . . . The trees T0 , T1 , . . . , which are subpartial orders of T∗ , are obtained, each from the preceding one, not just by pruning of branches and excision of points, but by a finite injury construction which involves both pruning and grafting new nodes

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taken from T∗ . It is important to note, though, that when a node is grafted onto a tree Tn , its height is reduced and many of the nodes that lie below it on T∗ are removed. This is necessary because otherwise T∞ would be recursively bounded. Typically, we will prune a branch on Tn−1 at some level below level n, grafting a new one on Tn . The tree we ultimately wish to obtain is the diagonal tree T∞ = {θn,σ : n < ω ∧ θn,σ ∈ Tn ∧ lh(σ ) ≤ n}. Let  ∗ be the set of all finite strings of 0’s and 1’s, and build a complete recursive tree of sharp filters T∗ = {θσ∗ : σ ∈  ∗ } with θ∅∗ ⊆ δ. Recall that {ψn : n ∈ ω} is an acceptable enumeration capturing all sharp filters. Then let f : T∗ → ω be a recursive function such that for each θσ∗ ∈ T∗ , the sharp filter which is the leftmost branch in T∗ above θσ∗ is ψf (θσ∗ ) . That is ψf (θσ∗ ) (0) = θσ∗
0 , and for each n ∈ ω, if ψf (θσ∗ ) (n) = θτ∗ , then put ψf (θσ∗ ) (n + 1) = θτ∗
0 . Thus for each θ ∗ ∈ T∗ , ψf (θ ∗ ) is a recursive sharp filter which is a branch of T∗ converging to a recursive point in θ ∗ ∩ T ∗ , where T ∗ = {xb : b is a branch of T∗ }. We will form the recursive tree T−1 ⊆ T∗ as described below, and will use it to form a recursive sequence of recursive trees {Tn : n ∈ ω} with each Tn ⊆ T∗ . For each Tn , n ∈ {−1} ∪ ω, there will be a set of strings n ⊆  ∗ such that Tn = {θn,σ : σ ∈ n }; furthermore, Tn will be a tree of sharp filters also formed as described below. Since Tn ⊆ T∗ , for each σ ∈ n there is some τ ∈  ∗ such that θn,σ = θτ∗ ; we will manage the construction so that lh(σ ) ≤ lh(τ ). We will also ensure that if θτ∗ ∈ Tn ∩ Tn+1 and θτ∗ = θn,σ , then σ ∈ n+1 and θn+1,σ = θn,σ = θτ∗ . From this we will be able to conclude that if θτ∗ ∈ Tm ∩ Tn and θτ∗ = θm,σ , then θn,σ = θm,σ = θτ∗ also. In other words, we ensure that each node θn,σ originates from T∗ and has an address σ on Tn that places it at a level of Tn not higher than the level at which it resides on T∗ . We also ensure that if a node θn,σ of Tn is retained on Tn+1 , its address does not change. For each n, s,we will define Tn,s = {θn,σ : (σ ∈ n ) ∧ (lh(σ ) ≤ s)}. We will then take T∞ = n Tn,n . Now suppose f : ω →  is a function infinitely many of whose finite initial segments lie in T∞ . As a result of the way Tn , n ∈ ω and T∞ are formed, each finite initial segment of f lies in cofinitely many of the trees Tn,n . But because of the grafting that occurs, it need not be the case that f is a branch through any of the trees Tn , n ∈ ω. This is somewhat different  from the construction of a recursively bounded
01 tree, where typically T∞ = n Tn . Nodes: Types and Attributes. Each tree Tn will contain three types of nodes θn,σ : ‘splitters’, ‘primary witnesses’ and ‘secondary witnesses’. 1. If lh(σ ) ≡ 0 (mod 3), then θn,σ is a splitter. 2. If lh(σ ) ≡ 1 (mod 3), then θn,σ is a primary witness. 3. If lh(σ ) ≡ 2 (mod 3), then θn,σ is a secondary witness.

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So what type of a node θn,σ is depends solely on lh(σ ). Note that, if θτ∗ ∈ Tm ∩ Tn , then, as mentioned above, for some string σ we will have θτ∗ = θm,σ = θn,σ ; so θn,σ will be the same type of node as θm,σ . Splitters are nodes meant to guarantee that T∞ will have 2ℵ0 infinite branches. All the nodes at levels 0, 3, 6, . . . are splitters. Any node at level 2, 5, . . . will have two child nodes, both splitters, and each splitter will be guaranteed to have an infinite branch above it in T∞ . (Of course this will cause uncountably many infinite branches to lie above it because of further splittings up the way.) Witnesses are nodes meant to force the branches of T∞ to converge to shadow points. There are two kinds of negative requirements to be met to accomplish this: that the points to which the branches of T∞ converge not be recursive, and that they not be avoidable. Each such requirement is assigned a splitter and a pair of witnesses on each infinite branch of each Tn . The splitter has two primary witnesses as children, and each primary witness has at most one child, which is a secondary witness. The primary and secondary witnesses work in tandem to help satisfy the requirement on their branch. The primary witnesses function as the finite injury switches that shunt action from left to right on the tree. Every primary witness will have exactly one of three attributes: it will be ‘dead’, ‘live’(or ‘alive’), or ‘dormant’. As long as a requirement does not receive attention on a branch, the left primary witness above its splitter is alive and grafting occurs above it, while the right primary witness is dormant and no grafting occurs onto it. When a requirement receives attention, its left primary witness becomes dead, all nodes above it are removed, and grafting onto it ceases forever, while the right primary witness becomes alive and grafting occurs above it. Dormant and dead primary witnesses are thus terminal nodes, while live ones are not. If a primary witness θn,σ is dead, then for each m ≥ n, we have σ ∈ m (so θm,σ exists and θm,σ = θn,σ ) and θm,σ is also dead. Furthermore, if θn,σ is dead, then for each m ≥ n and each τ str σ , τ ∈ / m (so θm,τ does not exist); we say the nodes above θn,σ are pruned. Note that if θn,σ is dead, so that the nodes above θn,σ are pruned, we will ensure that, for every m ≥ n, Tm will not include any basic open set η with η  θn,σ . If a primary witness θn,σ is dormant, then for every basic open set η  θn,σ we have η ∈ / Tn , so that if τ str σ , then θn,τ does not exist (and in fact θm,τ does not exist for any m ≤ n); we say the nodes above θn,σ are unborn. Also if θn,σ is dormant then for each m ≥ n, θm,σ (equal to θn,σ ) exists and is either dormant or alive, unless it has been pruned. Note that if θn,σ is dormant then at no stage m ≥ n is θm,σ ever dead. If a primary witness θn,σ is alive, then the secondary witness θn,σ
0 exists, as do the splitters θn,σ
0i , i ∈ {0, 1}. This fact helps guarantee that there will be 2ℵ0 infinite branches above θn,σ . Also if θn,σ is alive, then for each m ≥ n, θm,σ (which equals θn,σ ) exists and is either alive or dead, unless it is pruned due to the death of some node θm,τ with τ str σ . The secondary witnesses carry the bulk of the responsibility for ensuring that T∞ will be a recursive tree and that the points to which its branches converge will

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be shadow points rather than avoidable points. The idea is to pick each secondary witness from a high enough level on T∗ to help achieve these things. When a requirement receives attention, a dormant primary witness on Tn becomes alive on Tn+1 , and its child secondary witness must be created, at that stage T∗ is searched for a suitable candidate to put on Tn+1 . Modules. Here is a picture of a small part of the tree T−1 .

The boxed configuration, consisting of seven nodes, is a ‘module’. T−1 is built by stacking ‘basic’ modules iteratively. Whenever a new secondary witness is created on Tn+1 , it and all the nodes above it on Tn+1 are built by iteratively grafting modules lifted directly from T∗ , called ‘basic’ modules. Note that the module is ‘rooted’ at the secondary witness. When the secondary witness and its module are pasted into Tn+1 they are said to be ‘grafted onto’ the primary witness. Given θτ∗ ∈ T∗ , the module rooted at θτ∗ is the set of nodes {θτ∗ , θτ∗
i , θτ∗
ij : i, j ∈ {0, 1}}. Suppose θn,σ is a live primary witness on Tn . To say that the module rooted at θτ∗ lies above θn,σ means that the following three conditions hold: 1. θτ∗ ⊆ θn,σ ; 2. θn,σ
0 = θτ∗ , and for each i, j ∈ {0, 1}, θn,σ
0i = θτ∗
i and θn,σ
0ij = θτ∗
ij ; and 3. θn,σ
1 , is undefined. Thus if the module rooted at θτ∗ lies above the primary witness θn,σ , then θτ∗ = θn,σ
0 is a secondary witness and for each i, j ∈ {0, 1}, θτ∗
i = θn,σ
0i is a splitter and θτ∗
ij = θn,σ
0ij is a primary witness. To say that the module rooted at θτ∗ is grafted onto the primary witness θn,σ means that the following three conditions hold: 1. The module rooted at θτ∗ lies above θn,σ ; 2. θn,σ
0i0 is alive for each i ∈ {0, 1}; and 3. θn,σ
0i1 is dormant for each i ∈ {0, 1}.

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If θn,σ is a primary witness and θn,σ = θτ∗ , the basic module above θn,σ is the module rooted at θτ
0 . To say that the basic subtree rooted at θτ∗ is grafted onto θn,σ means that the following two conditions hold: 1. The module rooted at θτ∗ is grafted onto θn,σ ; and 2. for every live primary witness θn,υ with υ str σ , the basic module above θn,υ is grafted onto θn,υ . (We may imagine building the module rooted at θτ∗ , and then iteratively grafting a basic module onto the live primary witnesses to obtain the whole subtree.) Construction of the Initial Tree. We now define T−1 = {θ−1,σ : σ ∈ −1 }. T−1 is the tree whose nodes are defined recursively as follows: 1. 2. 3. 4.

θ−1,∅ = θ∅∗ . Thus θ−1,∅ ⊆ δ and θ−1,∅ is a splitter; θ−1,0 = θ0∗ is a live primary witness; θ−1,1 = θ1∗ is a dormant primary witness; and ∗ is grafted onto θ the basic subtree rooted at θ00 −1,0 .

Hence for each θσ∗ ∈ T∗ , if θσ∗ ∈ T−1 , then σ ∈ −1 and θ−1,σ = θσ∗ . In order to construct T∞ , we note that the process will ensure us that if there are infinitely many n with θσ∗ ∈ Tn then, letting N = min{n : θσ∗ ∈ Tn }, we will have θσ∗ ∈ Tn for all n ≥ N; and furthermore, if θσ∗ = θN,τ then for all n ≥ N and υ ⊆ τ we will haveυ ∈ n andθn,υ = θN,υ .  We let T∞ = n Tn,n = n {θn,σ : lh(σ ) ≤ n} and ∞ = n {σ ∈ n : lh(σ ) ≤ n}, and for each σ ∈ ∞ we let θ∞,σ = θn,σ where n is such that θn,σ exists and n ≥ lh(σ ). Then by considering the assertion in the previous paragraph we note that for each σ ∈ ∞ , θ∞,σ is well-defined, and T∞ = {θ∞,σ : σ ∈ ∞ }.

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Construction: The Requirements. The positive requirement is only one: P: T∞ has 2ℵ0 infinite branches. This will be fulfilled by the splitters. We will see that the mere existence of splitters at each level 3n, n ∈ ω, of T∞ will be enough to satisfy the positive requirement. There are two kinds of negative requirements: for k ∈ ω, let   R2k : If ψk is a sharp filter, then no branch of T∞ is equivalent to ψk . R2k+1 : If φk is an avoidance function, then no branch of T∞   lies in the spectrum of φk . Requirement Rm will be fulfilled by the primary witnesses θn,σ , lh(σ ) = 3m+1 and the secondary witnesses θn,σ , lh(σ ) = 3m + 2. The priority ranking will be: P, R0 , R1 , R2 , · · · . We say R2k requires attention through σ at stage n+1, n ≥ −1, if the following four conditions hold: 1. lh(σ ) = 3(2k) and σ ∈ n (so θn,σ is a splitter); 2. θn,σ
0 is a live primary witness and θn,σ
1 is a dormant primary witness; 3. ψkn (m) ↓ for some m ≥ 0 and for each m < m, ψkn (m + 1) ⊆ ψkn (m ) (see Definition 1); and 4. ψkn (m) ⊆ θn,σ
0 , where m is as in 3 above. We say R2k+1 requires attention through σ at stage n + 1, n ≥ −1, if the following three conditions hold: 1. lh(σ ) = 3(2k + 1) and σ ∈ n (so θn,σ is a splitter); 2. θn,σ
0 is a live primary witness and θn,σ
1 is a dormant primary witness; and 3. φkn (f (θn,σ
1 )) ↓. (Remember that f : T∗ → ω is a recursive function, and ψf (θn,σ
1 ) is the leftmost branch in T∗ above θn,σ
1 .) Suppose some Rm requires attention at stage n + 1, n ≥ −1, and let m be the least such m (i.e. highest priority). Also let σ be the lexicographically least string through which Rm requires attention at stage n + 1. Then Rm receives attention through σ at stage n + 1, and for each m > m , Rm is injured through σ at stage n + 1. Construction: The Trees. Note that a requirement Rm always requires attention through splitters. The construction below will ensure that when Rm receives attention through σ at stage n + 1 the primary witnesses θn,σ
0 and θn,σ
1 bear the brunt of the action, the subtree above θn,σ
0 is pruned, and a new subtree is grafted onto θn,σ
1 . At stage n we construct Tn . Stage -1: Let T−1 be as defined above. Stage n+1 (when n ≥ -1) : If no Rm requires attention at stage n + 1, let Tn+1 = Tn and let attributes of all primary witnesses be the same on Tn+1 as on Tn . If Rm receives attention through σ at stage n + 1 then define Tn+1 = {θn+1,σ : σ ∈ n+1 } as follows:

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1. For every τ ∈ n with τ str σ , let θn+1,τ = θn,τ , and if θn,τ is a primary witness let the attribute of θn+1,τ be the same as that of θn,τ . 2. Let θn+1,σ
0 = θn,σ
0 and declare this primary witness to be dead. / n+1 . 3. Prune the subtree above θn+1,σ
0 , so that if τ str σ
0, then τ ∈ 4. Let θn+1,σ
1 = θn,σ
1 and declare this primary witness to be alive. 5. If m = 2k, let the basic subtree rooted at ψf (θn,σ
1 ) (n + 1) be grafted onto θn+1,σ
1 . If m = 2k + 1, let the basic subtree rooted at ψf (θn,σ
1 ) (n ) be grafted onto θn+1,σ
1 , where n = max{n + 1, φk (f (θn,σ
1 ))}.

The intention of (5) is to ensure that Rm is satisfied and that if θn+1,σ
10 = θτ∗ then lh(τ ) ≥ n + 1 and τ ⊇str σ
10. (Though we may well have lh(σ ) < n + 1, this does not matter. It is lh(τ ) that is of importance to us and that we will use to establish recursiveness of Tn , and not lh(σ ).) Next, we will make a series of claims whose proofs establish the theorem. Claims 1 and 2 prove that the trees Tn behave as promised. Claims 3 through 7 prove that the injury to the requirements is finite, and that the nodes θn,σ thus eventually cease changing. Claims 8 through 12 prove that T∞ is a recursive tree of sharp filters. Claims 13 through 16 prove that all the requirements are satisfied. Claim 1. For each σ ∈  ∗ and each s, t ≥ −1, if σ ∈ s ∩ t , then θs,σ = θt,σ . Proof of Claim 1. Presume otherwise. Take σ, s, t with s < t, σ ∈ s ∩ t , and θs,σ = θt,σ . According to the construction, for every stage n with σ ∈ n ∩ n+1 we have θn,σ = θn+1,σ . So there must be some stage n such that s < n < t and σ ∈ / n . But again according to the construction, this can occur only if σ is pruned at stage n because there is some τ with τ ⊂str σ and θn,τ a dead primary witness. We may assume τ is the lexicographically least such string, so that no υ str τ is ever a dead primary witness at any stage n with s < n < t (and hence not at any stage n with s ≤ n ≤ t, because σ ∈ s ∩ t ). But then induction on the construction assures us that for all stages n ≥ n, θn ,τ exists and is dead, so for / n . Thus σ ∈ / t , contradicting the definition of σ .   every υ str τ , v ∈ Claim 2. Each tree Tn , n ≥ −1, satisfies the following conditions: 1. θn,∅ exists and is a splitter. 2. If θn,σ is a splitter then θn,σ
0 and θn,σ
1 exist and are primary witnesses. 3. If θn,σ is a splitter with lh(σ ) = 3m, and if Rm has not received attention through σ by the end of stage n, then θn,σ
0 is alive and θn,σ
1 is dormant. 4. If θn,σ is a splitter with lh(σ ) = 3m, and if Rm has received attention through σ at some stage n ≤ n, then θn,σ
0 is dead and θn,σ
1 is alive. 5. If θn,σ is a live primary witness, then θn,σ
0 exists and is a secondary witness, / n , so there is no node θn,τ on Tn . and for all τ ⊇str σ
1, τ ∈ 6. If θn,σ is a dormant primary witness, then each θn,τ , with τ str σ , is unborn; while if θn,σ is a dead primary witness, then each θn,τ , with τ str σ , is pruned. In / n . either case, if τ str σ , then τ ∈ 7. If θn,σ is a secondary witness, then θn,σ
0 and θn,σ
1 exist and are splitters. Proof of Claim 2. By induction based on the construction and the conditions satisfied by T−1 .  

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Claim 3. Suppose σ ∈ n and lh(σ ) = 3m. Also suppose that for each m ≤ m, Rm never receives attention through any τ str σ at any stage s ≥ n. Then for each τ ⊆str σ and each s ≥ n the following conditions hold: 1. τ ∈ s ; 2. θs,τ = θn,τ ; and 3. if θn,τ is a primary witness then θs,τ has the same attributes as θn,τ . Proof of Claim 3. This is an immediate consequence of Claims 1 and 2.

 

Claim 4. Suppose σ ∈ n and lh(σ ) = 3m. Also suppose that for each m ≤ m,  never receives attention through any τ ⊇ σ at any stage s ≥ n. Then for each Rm str s ≥ n and each τ str σ with lh(τ ) ≤ 3m + 4 the following conditions hold: 1. τ ∈ s iff τ ∈ n ; 2. if τ ∈ n , then θs,τ = θn,τ ; and 3. if lh(τ ) = 3m + 1, then θs,τ has the same attribute as θn,τ . Proof of Claim 4. This is an immediate consequence of Claims 1, 2, and 3.

 

 never receives attention at any Claim 5. Let m, n be such that for each m ≤ m, Rm stage s ≥ n. Then for each τ with lh(τ ) ≤ 3m + 4 and each s ≥ n the following conditions hold: 1. τ ∈ s iff τ ∈ n ; 2. if τ ∈ n then θs,τ = θn,τ ; and 3. if θn,τ is a primary witness, then θs,τ has the same attribute as θn,τ .

Proof of Claim 5. By induction on m, using Claim 4 and the fact that there are only finitely many binary strings with lh(τ ) ≤ 3m + 4. Claims 1 and 3 are also used in the induction.   Claim 6. Every negative requirement Rm receives attention at most finitely often, and every Rm is injured at most finitely often. Proof of Claim 6. Suppose that for each m < m, Rm receives attention at most finitely often, and let n be such that no Rm , m < m, ever receives attention at any stage s ≥ n. Then Rm is never injured at any stage s ≥ n. If Rm never receives attention at any stage s ≥ n, we are done. So suppose Rm receives attention through σ at some stage s ≥ n. Then lh(σ ) = 3m, and by Claim 2, θs,σ
0 and θs,σ
1 exist and are primary witnesses with θs,σ
0 dead and θs,σ
1 alive. Then by Claim 5 (2) applied to m − 1, for each t ≥ s we have σ
0 ∈ t and θt,σ
0 = θs,σ
0 , and by Claim 2 (4), θt,σ
0 is dead. So by the construction, Rm can never require attention through σ at any stage t > s. Since there are only finitely many binary strings σ with lh(σ ) = 3m, and since by the above argument Rm can receive attention through each such σ after stage n at most once, it follows that Rm receives attention at most finitely often.   Claim 7. For each σ ∈  there is N ∈ ω such that for each n ≥ N, the following conditions hold: 1. σ ∈ n iff σ ∈ N ; and

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2. if σ ∈ N , then θn,σ = θN,σ . Proof of Claim 7. For σ ∈  take m such that lh(σ ) ≤ 3m + 4 and N such that  never receives attention at any stage n ≥ N, as per Claim 6. for all m ⊆ m, Rm Then the conclusion follows from Claim 5.   Note that {Tn : n ≥ −1} is a uniformly recursive sequence of trees and that {n : n ≥ −1} is a uniformly recursive sequence of sets of strings. We now formally define ∞ and T∞ . Define ∞ = {σ : (∃n)[σ ∈ n ∧ lh(σ ) ≤ n]}, and for each σ ∈ ∞ define θ∞,σ = θn,σ , where n is such that σ ∈ n . Note that by Claim 1, θ∞,σ is well-defined. We let ∞ :  →  be a partial function defined as follows: θ∞,σ , if σ ∈ ∞ ; ∞ (σ ) = ↑, otherwise. We let T∞ = rng( ∞ ) = {θ∞,σ : σ ∈ ∞ }. As in the case of each Tn , we refer to the nodes of T∞ as splitters, primary witnesses, or secondary witnesses. Claim 8. T∞ is recursively enumerable, ∞ is a recursively enumerable set of strings and ∞ is a partial recursive function.  Proof of Claim 8. It is easily verified that T∞ = n Tn,n and that T∞ is recursively enumerable. The rest is also obvious from the definitions of ∞ and ∞ .   Claim 9. For every n ∈ ω, σ ∈ n , and τ ∈  ∗ , if θn,σ = θτ∗ , then σ ⊆str τ . Thus if σ ∈ ∞ , and θ∞,σ = θτ∗ , then σ ⊆str τ . Also, if σ, τ ∈ n and σ ⊇str τ , then θn,σ  θn,τ . Thus Tn is a subpartial order of T∗ . Proof of Claim 9. By induction on the construction.

 

Claim 10. T∞ is a recursively enumerable tree of sharp filters. Proof of Claim 10. That T∞ is recursively enumerable has been noted above. We shall show that T∞ satisfies the conditions of Definition 2.4, the definition of a tree of sharp filters. 1. 0,∅ exists by Claim 2 (1), and lh(∅) ≤ 0, so ∅ ∈ ∞ and hence ∞ (∅) ↓. 2. Let σ ∈  and suppose ∞ (σ
i) ↓ for some i ∈ {0, 1}. Then θ∞,σ
i exists, so for some n ≥ lh(σ
i) we have θn,σ
i defined. Because Tn is a tree, θn,σ is defined too, and because lh(σ ) ≤ n, we have σ ∈ ∞ and hence ∞ (σ ) ↓. 3. If σ, τ ∈  with σ
str τ and τ
str σ and if ∞ (σ ) ↓ and ∞ (τ ) ↓ then there are m, n with θ∞,σ = θm,σ and θ∞,τ = θn,τ . Now Tm , Tn ⊆ T∗ , so pick σ  , τ  ∈  ∗ such that θm,σ = θσ∗ and θn,τ = θτ∗ . Then by Claim 9, σ ⊆str σ  and τ ⊆str τ  . Hence σ 
str τ  and τ 
str σ  . By the properties of the tree of sharp filters T∗ , θ∞,σ ∩ θ∞,τ = θσ∗ ∩ θτ∗ = ∅; thus ∞ (σ ) ∩ ∞ (τ ) = ∅. 4. Let b : ω →  be a total function such that for every n, lh(b(n)) = n, ∞ (b(n)) ↓, and b(n + 1) ⊇str b(n). Then for each n there is mn such that

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θmn ,b(n+1) exists and n + 1 = lh(b(n + 1)) ≤ mn . So lh(b(n)) ≤ n ≤ mn and because Tmn is a tree, θmn ,b(n) exists. Because b(n + 1) ⊇str b(n), by Claim 9 we have θ∞,b(n+1) = θmn ,b(n+1)  θmn ,b(n) = θ∞,b(n) . For each n let σn ∈  ∗ be such that θ∞,b(n) = θσ∗n ∈ T∗ . Then θσ∗n+1  θσ∗n for each n, so ∞ ◦ b = {
n, ∞ (b(n) : n ∈ ω} is a subsequence of a branch in T∗ . Because each branch in T∗ is a sharp filter, and because any subsequence of a sharp filter is a sharp filter,   ∞ ◦ b is a sharp filter. Claim 11. T∞ is recursive as a set. Proof of Claim 11. To determine whether θ ∈ T∞ , first see if θ ∈ T∗ . If θ ∈ / T∗ , ∗ ∗ ∗ then θ ∈ / T∞ . But if θ ∈ T find τ ∈  such that θ = θτ , and suppose lh(τ ) = n. By Claim 9, θ  ∈ T∞ iff there is σ such that θ = θ∞,σ and lh(σ ) ≤ n, which occurs iff θ ∈ m≤n Tm,m . It is clear from the construction that {Tm : m ∈ ω} is a uniformlyrecursive set of trees, so we can determine, recursively in n and θ, whether θ ∈ m≤n Tm,m .   Claim 12. T∞ is recursive as a tree. Proof of Claim 12. This follows from Claim 8 (that is partial recursive), Claim 11 (that T is recursive as a set), and the fact that every tree which is recursive as set is recursive as a tree (see Theorem 7.4 of [7]).   Claim 13. Requirement P is satisfied, so T∞ has 2ℵ0 infinite branches. Proof of Claim 13. Let

T = {θ∞,σ : [lh(σ ) ≡ 0 (mod 3)] ∧ (∃N )(∀n ≥ N )[σ ∈ T has 2ℵ0 infinite branches. The Claim will of course follow n ]}. We will show

from this easily, since

T is a subtree of T∞ . T because ∅ ∈ n for all n. Now suppose θ∞,σ ∈

T. Let First note that θ∞,∅ ∈

lh(σ ) = 3m and pick N so large that for all n ≥ N, σ ∈ n , and for all m ≤ m, Rm never receives attention at any stage n ≥ N. Note that θN,σ is a splitter. By Claim 2 the primary witnesses θN,σ
0 and θN,σ
1 exist, and exactly one of them is alive. Call that live witness θN,σ
i . Since lh(σ
i) = 3m + 1, Claim 5 tells us that θn,σ
i is a live primary witness for all n ≥ N. Hence by Claim 2, for all n ≥ N , θn,σ
i0 is a secondary witness and θn,σ
i0j is a splitter for each j ∈ {0, 1}. It T. Since T∞ is follows that σ
i0j ∈ ∞ for each j ∈ {0, 1}, so each θ∞,σ
i0j ∈

a tree of sharp filters, θ∞,σ
i00 ∩ θ∞,σ
i01 = ∅. This proves that T splits at each   of its nodes and thus has 2ℵ0 infinite branches. Claim 14. Let b be an infinite branch through T∞ and let θ∞,τ be a primary witness lying on b. Then there is some N such that for all n ≥ N , θn,τ is alive. Proof of Claim 14. Presume otherwise. Then by induction on Claim 2, either θn,τ is a dormant for all n or there is N such that θn,τ is dead for all n ≥ N. In both cases T∞ can contain at most finitely many nodes above θn,τ , contradicting that b is an infinite branch.   Claim 15. For each k, requirement R2k is satisfied, so T∞ has no infinite branches equivalent to a recursive sharp filter.

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Proof of Claim 15. Presume otherwise. Let b be an infinite branch through T∞ , and suppose b ≡ ψk . Let τ be such that θ∞,τ = b(3(2k) + 1), so that θ∞,τ is a primary witness lying on b. By Claim 14, there is N such that for all n ≥ N , θn,τ is alive. Let σ and i be such that τ = σ
i, so θ∞,σ is a splitter. Also pick m1 , so that ψk (m1 ) ⊆ θ∞,τ .  Now pick N  ≥ N so large that ψkN (m1 ) ↓, N  ≥ 3(2k) + 1, and for all m ≤ 2k, Rm never receives attention at any stage n ≥ N  . If i = 0 then R2k requires attention through σ at stage N  , and since no Rm ,  m < 2k, receives attention at that stage, no such Rm requires attention. Thus R2k is the highest priority requirement requiring attention at stage N  . The construction then guarantees that R2k receives attention at stage N  , contradicting the definition of N  . On the other hand, if i = 1 then by Claim 2 (3 & 4), R2k has received attention through σ at some stage n < N  . That can occur only if there is m0 with ψk (m0 ) ⊆ θn−1,σ
0 . Now since N  ≥ 3(2k) + 1 and N  ≥ N we have θN  ,τ = θN  ,σ
1 exists and is alive, and θ∞,σ
1 = θN  ,σ
1 . Also since TN  is a tree, the splitter θN  ,σ exists and so by Claim 2, θN  ,σ
0 also exists. Hence θ∞,σ
0 = θN  ,σ
0 = θn,σ
0 by Claim 1. Thus ψk (m0 ) ⊆ θ∞,σ
0 and ψk (m1 ) ⊆ θ∞,τ = θ∞,σ
1 . But since T∞ is a tree of sharp filters, θ∞,σ
0 ∩ θ∞,σ
1 = ∅, contradicting that ψk is a sharp filter.   Claim 16. For each k, requirement R2k+1 is satisfied, so no infinite branch of T∞ lies in the spectrum of any avoidance function. Proof of Claim 16. Presume otherwise. Let b be an infinite branch through T∞ and let φk be an avoidance function such that xb ∈ Sφk . Let τ be such that θ∞,τ = b(3(2k + 1) + 1), so that θ∞,τ is a primary witness lying on b. By Claim 14, there is N such that for all n ≥ N, θn,τ is alive. Let σ and i be such that τ = σ
i, so θ∞,σ is a splitter.  Now pick N  > N so large that N  ≥ 3(2k + 1) + 1, φkN (f (θN  ,σ
1 )) ↓, and  for all m ≤ 2k + 1, Rm never receives attention at any stage n ≥ N  . If i = 0 then R2k+1 requires attention through σ at stage N  . Just as in Claim 15, this leads to a contradiction. But if i = 1 then by Claim 2 (3 & 4), R2k+1 has received attention through σ at some stage n ≤ N  . Then the construction guarantees that the basic subtree rooted at ψf (θN  −1,σ
1 ) (n ) is grafted onto θN  ,σ
1 , where n = max{N  , φk (f (θN  −1,σ
1 ))}. Since N  > N , we have θN  −1,σ
1 = θN,σ
1 = θ∞,σ
1 = θ∞,τ , so θ∞,σ
10 = θN  ,σ
10 ⊆ ψf (θN  −1,σ
1 ) (φk (f (θN  −1,σ
1 ))) = ψf (θ∞,τ ) (φk (f (θ∞,τ ))). Now θ∞,σ
10 is the unique secondary witness lying above σ
1 = τ whose associated string has length lh(τ ) + 1, so it follows that θ∞,σ
10 = b(3(2k1 ) + 2). But then b(3(2k1 ) + 2) ⊆ ψf (θ∞,τ ) (φk (f (θ∞,τ ))); so xb ∈ / Sφk , contradicting the definition of φk .  

Let T∞ = {xb : b is an infinite branch of T∞ }. Then Claims 13, 15, and 16 prove that T∞ is a set of 2ℵ0 points, none of which is recursive or avoidable. Thus all the points of T∞ are shadow points. Since θ∞,∅ ⊆ δ, it follows that   T∞ ⊆ δ ∩ Shad(X).

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4. The trees T∞ & ∞ , and the case of 2ω The tree T∞ and the set of addresses ∞ in the proof of the Theorem 3, thought of as a subtree of 2<ω , are of course order-isomorphic. Because we can recursively enumerate the nodes of T∞ and their addresses, ∞ is clearly a recursively enumerable tree. As a subtree of 2<ω , ∞ is recursively bounded. But ∞ can not be a
01 tree, because if it were, that would make T∞ a recursively bounded
01 tree too, so that its infinite branches would not all converge to shadow points. Thus ∞ is a recursively bounded, nonrecursive tree that is order-isomorphic to the nonrecursively bounded
01 tree T∞ . Now suppose we build a new tree T , along with its set of address   , as follows. For each node θσ of our original recursive tree T−1 , if θσ is on T∞ (possibly at a new address), put θσ on T at its original address σ , and also put onto T the entire initial segment of T−1 below θσ . Clearly the infinite branches of T converge to exactly the same points as do those of T∞ , which are of course shadow points. Also T and   are order isomorphic, and be can be recursively enumerated. Since   is a subtree of 2<ω , it is recursively bounded, and since T is a subtree of T−1 , it also is recursively bounded. If   were a
01 tree, then of course so would T be. But we can not have T to be a
01 tree, because that together with recursive boundedness would prevent all infinite branches from converging to shadow points. So both T and   are nonrecursive. Also T is topologically closed. In order to see the reformulation of the main result as explained in the introduction, suppose we play the same construction, not on a space of our kind, but on the topological space X = 2ω with subbasis  = 2<ω . Then T−1 is a recursive subtree of 2<ω , and T−1 = −1 (where −1 is the set of addresses of the nodes of T−1 , as per our the proof above). If we then form T∞ , ∞ , T , and   , we get T =   , and T is nonrecursive, recursively enumerable, and recursively bounded. Next, form the following definitions, thinking of 2ω as a tree. 1. A domain is a
02 class in 2ω that contains all recursive branches. 2. A branch of 2ω is avoidable if there is some domain that does not contain it. 3. A branch is a shadow branch if it is nonrecursive and contained in all domains. Then T =   is a closed, recursively bounded, nonrecursive, recursively enumerable subtree of 2<ω all of whose infinite branches are shadow branches in 2ω . This set of shadow points is thus a strong
02 class in 2ω . (See [2] for definition.) Let us take this a step farther. Given a branch in a subtree of 2ω , we can consider that branch to be a function from ω to 2<ω , just as we do for trees in topological spaces of our kind. For a branch b, let a subbranch of b be a subsequence of b or a finite initial segment of such a subsequence. Then T∞ is a nonrecursively bounded
01 tree all of whose branches are subbranches of T , and whose infinite branches are thus infinite subbranches of shadow branches of 2ω . But ∞ is a recursively bounded, nonrecursive, recursively enumerable subtree of  = 2<ω , so even though T∞ and ∞ are order-isomorphic, they are not recursively so. 5. Shadow points & condensation In this section we prove results parallel to the results of Section 10 of [7]. There we established findings regarding various complete recursive trees of pseudo-irratio-

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nals and boundary points of a basic open set (Theorems 8.2 and 8.3 of [7]), and used the findings of Section 9 on avoidable points to establish the ubiquity of avoidable points (see Theorems 10.5 and 10.6 of [7]). Here we prove similar ubiquity results for shadow points. Theorem 4. Let  = {δn : n ∈ ω}. For each α ∈ , there is a
01 complete tree of sharp filters T = {θσ : σ ∈ } such that for each σ ∈ , if lh(σ ) = n, then θσ ∩ ∂δn = ∅ and θσ ⊆ α. Thus T ⊆ Shad(X) ∩ α ∩ Irr(X). Proof. Construct a recursive tree of sharp filters T∗ as per Theorem 8.2 of [7] and continue the construction as in the proof of Theorem 3.   Theorem 5. Let α, δ ∈  be such that ∂α is connected and contains more than one point and ∂α ∩ δ = ∅. Then there is a
01 tree of sharp filters T = {θσ : σ ∈ } such that for each σ ∈ , θσ ∩ ∂α = ∅, θσ ⊆ δ, and T ⊆ Shad(X) ∩ ∂α ∩ δ. Proof. Construct a recursive tree of sharp filters T∗ as per Theorem 8.3 of [7] and continue the construction as in the proof of Theorem 3.   Theorem 6. Let α ∈ . Then every point of α is a point of condensation of Shad(X) ∩ Irr(X) ∩ α. Consequently ||Shad(X) ∩ Irr(X) ∩ α|| = 2ℵ0 . Proof. Let α ∈  and let x ∈ α. Form a tree T of shadow points in α ∩ Irr(X) as per Theorem 4. Since T ⊆ Shad(X) ∩ Irr(X) ∩ α and T contains 2ℵ0 points, this proves that x is a point of condensation of Shad(X) ∩ Irr(X) ∩ α and that thus ||Shad(X) ∩ Irr(X) ∩ α|| = 2ℵ0 .   Theorem 7. Let α ∈  be such that ∂α is connected and contains more than one point. Then every point of ∂α is a point of condensation of Shad(X) ∩ ∂α and hence of Shad(X) ∩ Rat(X). Consequently ||Shad(X) ∩ ∂α|| = 2ℵ0 . Proof. Fix such α. Let x ∈ ∂α and let δ ∈  be such that x ∈ δ. Form a tree T of shadow points on ∂α as per Theorem 5. Since T ⊆ Shad(X) ∩ ∂α and T contains 2ℵ0 points, this proves that x is a point of condensation of Shad(X) ∩ ∂α and therefore of Shad(X) ∩ Rat(X). Clearly T contains 2ℵ0 points, so ||Shad(X) ∩ ∂α|| = 2ℵ0 .   Acknowledgements. We are grateful to Rumen Dimitrov, John Chisholm, Linda Lawton, Joe Miller, and Steve Simpson for profitable conversations and encouragement since the distribution of the first manuscript. In particular we thank Doug Cenzer for his suggestions which have lead to clearer explanations, and to the referee for his helpful remarks.

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3. Goodstein, R.L.: Recursive Analysis. North-Holland, Amsterdam 1961 4. Jockusch, C.G., Lewis, A.A., Remmel, J.B.:
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