1 Introduction

A subset S of Euclidean space R n is said to be co-computably enumerable (co-c.e.) if its complement R n ∖ S can be effectively covered by open balls. Equivalently, a set S ⊆R n is co-c.e. if there exists a computable function f :R n →R such that S = f −1({0}).

It is known that there exists a computable function f : R →R which has zero-points, but none of them is computable [14]. Hence there exists a nonempty co-c.e. subset S of R which does not contain a computable point.

On the other hand, under certain assumptions we can conclude that a co-c.e. set contains a computable point. For example, if S ⊆R is a co-c.e. set with an isolated point, that point must be computable. It entails: if S ⊆R is a nonempty co-c.e. set such that R ∖ S has finitely many connected components, then S contains a computable point.

What happens if we replace R by R n, i.e. if S ⊆R n is a co-c.e. set such that R n ∖ S has finitely many connected components? Then, it is possible that R n ∖ S is connected, and, in this case, we cannot conclude that S contains a computable point. To see this, let us take a nonempty co-c.e. set S ⊆R which has no computable point. Then S ×{0}⊆R 2 is easily seen to be a co-c.e. set, it has no computable point and its complement R 2 ∖ (S ×{0}) is connected.

However, if S is a co-c.e. subset of R n such that its complement R n ∖ S is disconnected and it has finitely many components, then S contains a computable point [5]. In fact, the assumption that the complement of S has finitely many components can be replaced by the more general assumption that the components of R n ∖ S can be in a certain sense effectively distinguished [5]. Moreover, this holds not just in Euclidean space, but also if the ambient space is any computable metric space (X, d, α) which has the effective covering property and in which closed balls are compact and connected [5].

In this paper we generalize this result by removing the assumptions of the effective covering property and compact closed balls. Moreover, we replace the condition that (X, d) has connected closed balls by the weaker condition that (X, d, α) is effectively locally connected.

Together with the general question of: under which conditions the implication

$$ S\text{ co-c.e. }\Longrightarrow S\text{ contains a computable point} $$
(1)

holds, we have another general question: under which conditions the implication

$$ S\text{ co-c.e. }\Longrightarrow S\text{ computable} $$
(2)

holds? That S is computable means that S is co-c.e. and computably enumerable (c.e.), and the latter means that we can effectively enumerate all rational open balls in (X, d, α) which intersect S.

In Euclidean space this notion can be characterized in the following way: a nonempty closed subset of R n is computable if and only if it can be effectively approximated by a finite set of points with rational coordinates with arbitrary precision on an arbitrary bounded region of R n.

If the metric space (X, d) is complete, then each nonempty computable set in (X, d, α) contains computable points, moreover they are dense in it.

Topology plays an important role regarding conditions under which (1) and (2) hold. Some topological properties of S ensure that (1) or (2) hold, for example if S is a cell or topological sphere [4, 6], a graph of a certain function [1], a chainable or circularly chainable continuum [5, 9], a (compact) manifold with boundary [8, 10] or a 1-manifold [3]. Also, in view of (1), it should be mentioned that there exists a contractible co-c.e. set in R 2 which does not contain any computable point [12].

For example, if S ⊆R n is a co-c.e. set which is a topological m-sphere, i.e. homeomorphic to the unit m-sphere S m = {x ∈R m+1∣∥x∥ = 1}, then S is computable [4, 6, 8].

Let us consider the special case when n = m + 1, i.e. when S is a topological (n − 1)-sphere in R n. By the generalized Jordan curve theorem, R n ∖ S has two components and each point of S is in the boundary of each of these components.

The question is: if S is any subset of R n, such that R n ∖ S has two components and each point of S is in the boundary of each of these components, does then the implication (2) hold?

The answer is affirmative. By [5] we have the following result: if (X, d, α) is a computable metric space, which has the effective covering property and compact and connected closed balls, and if S ⊆ X is such that the components of X ∖ S can be effectively distinguished and each point of S lies in the boundary of at least two components of X ∖ S, then (2) holds.

In this paper we show that this result also holds if we remove all assumptions on (X, d, α) and only assume that (X, d, α) is effectively locally connected.

Here is how the paper is organized. In Section 2 we provide some basic notions which are necessary for the rest of the paper. In Section 3 we examine the notion of an effectively locally connected computable metric space and in Section 4 we prove the main results of the paper.

2 Basic Notions and Techniques

We use the standard notions of computable real numbers and real number functions (see e.g. Section 2 in [11]).

If (X, d) is a metric space, x ∈ X and r > 0, we denote by B(x, r) the open ball of radius r centered at x.

A computable metric space is a tuple (X, d, α), where (X, d) is a metric space and α : N → X is a sequence whose range is dense in (X, d) and such that the function N 2 →R,

$$(i,j)\mapsto d(\alpha_{i} ,\alpha_{j} ) $$

is computable (see [1, 2, 11, 15]).

Let (X, d, α) be a computable metric space. A rational ball in the computable metric space (X, d, α) is a ball of the form B(α n , r), where n ∈N and r ∈Q, r > 0.

We now fix some effective enumeration (I i ) of all rational open balls in (X, d, α). More precisely, let q : N →Q be some fixed computable function whose range is Q ∩〈0, ∞〉 and let τ 1, τ 2 : N →N be some fixed computable functions such that {(τ 1(i), τ 2(i))∣i ∈N} = N 2. We define the sequence (λ i ) in X and the sequence (ρ i ) in Q by

$$\lambda_{i} =\alpha_{\tau_{1} (i)},\quad \rho_{i} =q_{\tau_{2} (i)}. $$

Note that for each n ∈N and each positive rational number r the pair (α n , r) is equal to (λ i , ρ i ) for some i ∈N.

For i ∈N we define

$$I_{i}=B(\lambda_{i} ,\rho_{i} ). $$

Clearly, {I i ∣i ∈N} is the family of all rational open balls in (X, d, α).

Let σ :N 2 →N and η : N →N be some fixed computable functions with the following property: {(σ(j,0),…, σ(j, η(j)))∣j ∈N} is the set of all nonempty finite sequences in N, i.e. {(a 0,…, a n )∣n ∈N, a 0,…, a n ∈N}. We are going to use the following notation: (j) i instead of σ(j, i) and \(\overline {j}\) instead of η(j). Hence

$$\{((j)_{0} ,{\dots} ,(j)_{\overline{j}})\mid j\in \mathbf{N}\} $$

is the set of all finite sequences in N.

For j ∈N the set \(\{(j)_{i} \mid 0\leq i\leq \overline {j}\}\) will be denoted by [j].

Let (X, d, α) be a computable metric space. Let i, j ∈N. We say that I i is formally contained in I j and we write I i ⊆ F I j if

$$d(\lambda_{i} ,\lambda_{j} )+\rho_{i} <\rho_{j} . $$

If I i ⊆ F I j , then I i ⊆ I j .

The following proposition easily follows from Proposition 2.1 in [11] (see also [5]).

Proposition 1

Let (X, d, α)be a computable metric space. Then the following sets are computably enumerable:

  1. (i)

    {(i, j) ∈N 2∣I i ⊆ F I j };

  2. (ii)

    {(i, j) ∈N 2∣I i ∩ I j ≠∅};

  3. (iii)

    {(i, j) ∈N 2∣α i ∈ I j }.

Let (X, d, α) be a computable metric space. We say that x ∈ X is a computable point in (X, d, α) if there exists a computable function f : N →N such that

$$d(x,\alpha_{f(k)})<2^{-k}$$

for each k ∈N. Similarly, a sequence (x i ) in X is said to be computable in (X, d, α) if there exists a computable function F :N 2 →N such that d(x i , α F(i, k)) < 2−k for all i, k ∈N.

For example, if α : N →Q n is a computable surjection (computable in the sense that the component functions of α are computable) and d is the Euclidean metric on R n, then (R n, d, α) is a computable metric space (see e.g. Proposition 2.1 in [11]). If (x i ) is a sequence in R n, then (x i ) is computable in (R n, d, α) if and only if the component sequences of (x i ) are computable (as functions N →R). In particular, a point x ∈R n, x = (x 1,…, x n ), is computable in (R n, d, α) if and only if x 1,…, x n are computable numbers.

A closed subset S of (X, d) is said to be computably enumerable in (X, d, α) if

$$\{i\in \mathbf{N}\mid S \cap I_{i} \neq\emptyset \}$$

is a c.e. subset of N.

If S is a nonempty c.e. set in (X, d, α) and the metric space (X, d) is complete, then S contains computable points, moreover there exists a computable sequence (x i ) in (X, d, α) such that S = Cl({x i ∣i ∈N}) [2, 7]. Here Cl(A), for A ⊆ X, denotes the closure of A in (X, d).

Let U ⊆ X. We say that U is a computably enumerable open set in (X, d, α) if there exists a c.e. subset A of N such that

$$U=\underset{i\in A}{\bigcup} I_{i}. $$

A set S ⊆ X is said to be co-computably enumerable (closed) in (X, d, α) if X ∖ S is a c.e. open set in (X, d, α).

We say that S is a computable (closed) set in (X, d, α) if S is both computably enumerable and co-computably enumerable [2, 15].

Let (X, d) be a metric space, A, B ⊆ X and ε > 0. We write A≈ ε B if for each x ∈ A there exists y ∈ B such that d(x, y) < ε and for each y ∈ B there exists x ∈ A such that d(x, y) < ε.

If D is a dense set in (X, d) and K a nonempty compact set in (X, d), then for each ε > 0 there exists a finite subset A of D such that K≈ ε A.

Suppose (X, d, α) is a computable metric space. For i ∈N we define

$${\Lambda}_{i} =\{\alpha_{j} \mid j\in [i]\}. $$

Note that {Λ i ∣i ∈N} is the family of all nonempty finite subsets of {α i ∣i ∈N}.

Let K be a compact set in (X, d). Note that, if K≠∅, for each k ∈N there exists i ∈N such that \(K\approx _{2^{-k}}{\Lambda }_{i} \). We say that K is a computable compact set in (X, d, α) if K = ∅ or there exists a computable function f : N →N such that

$$K\approx_{2^{-k}}{\Lambda}_{f(k)}$$

for each k ∈N.

Each computable compact set is a computable (closed) set [8]. Regarding the converse direction, a computable set which is compact need not be a computable compact set. However, if (X, d, α) has the effective covering property (for the definition see [2]) and compact closed balls, the following equivalence holds for a compact set S in (X, d) (Proposition 3.6 in [3]):

$$ S\text{ computable compact set }\Leftrightarrow S\text{ computable closed set}. $$
(3)

In particular, (3) holds if the ambient space is Euclidean space (if we consider Euclidean space as a computable metric space in the way previously described).

3 Effective Local Connectedness

A topological space \((X,\mathcal {T})\) is locally connected if for each x ∈ X and each open neighborhood U of x there exists an open connected neighborhood V of x such that V ⊆ U. In other words, \((X,\mathcal {T})\) is locally connected if there exists a basis \(\mathcal {B}\) for the topology \(\mathcal {T}\) such that each element of \(\mathcal {B}\) is connected.

In order to define that a computable metric space (X, d, α) is effectively locally connected we could concentrate on the rational open balls I i , i ∈N, and say that (X, d, α) is effectively locally connected if there exists a certain basis for the induced topology whose elements are connected rational open balls. However, as noted in [1], this definition would not be satisfactory since it is possible that in a locally connected metric space there exists no basis for the induced topology which consists of connected open balls.

So, we will not restrict ourself to connected open balls and therefore we need a certain notion that a sequence of open (connected) sets (U i ) is computable in (X, d, α). Using that notion, we will define the notion of an effectively locally connected computable metric space (see [1]).

Suppose (A i ) and (B i ) are sequences of sets. We say that (A i )effectively refines (B i ) and write (A i ) ≼ (B i ) if there exists a c.e. set \(\mathcal {C}\subseteq \mathbf {N}^{2}\) such that

$$ B_{i} =\underset{(j,i)\in \mathcal{C}}\bigcup A_{j} $$
(4)

for each i ∈N. Equivalently, (4) can be expressed by the following two statements:

  1. ∘

    if j, i ∈N are such that \((j,i)\in \mathcal {C}\), then A j ⊆ B i ;

  2. ∘

    if i ∈N and x ∈ B i , then there exists j ∈N such that x ∈ A j and \((j,i)\in \mathcal {C}\).

We also write \((A_{i} )\preceq _{\mathcal {C}}(B_{i})\).

Two sequences of sets (A i ) and (B i ) are said to be computably equivalent if (A i ) ≼ (B i ) and (B i ) ≼ (A i ).

We say that a computable metric space (X, d, α) is effectively locally connected if there exists a sequence (D i ) of open connected subsets of (X, d) which is computably equivalent to the sequence (I i ).

Clearly, if (D i ) is such a sequence of sets, then {D i ∣i ∈N} is a basis for the topology induced by the metric d.

If (X, d, α) is effectively locally connected, then it is easy to see that (X, d, α) is also effectively locally connected in the sense of [1]. The converse of this statement is also true (see Remark 1 below), hence these two notions coincide.

Example 1

Let (X, d, α) be a computable metric space. If each open ball in (X, d) is connected, then (X, d, α) is effectively locally connected (each sequence of sets is clearly computably equivalent to itself).

Moreover, if (X, d) is such that all open balls in (X, d) with sufficiently small radius are connected, then (X, d, α) is also effectively locally connected. More precisely, suppose that there exists r > 0 such that B(α n , s) is connected for all n ∈N and s ∈Q, 0 < s < r. We claim that (X, d, α) is effectively locally connected. First, we may assume that r is rational. Choose i 0 ∈N such that \(\rho _{i_{0} }<r\). Let f : N →N be defined by f(i) = i if ρ i < r, and f(i) = i 0 otherwise. For i ∈N let D i = I f(i). The function f is computable and so the set

$$\mathcal{C}=\{(j,i)\in \mathbf{N}^{2}\mid j=f(i)\}$$

is also computable. We have \((I_{i} )\preceq _{\mathcal {C}}(D_{i} )\).

On the other hand, let

$$\mathcal{F}=\{(j,i)\in \mathbf{N}^{2}\mid I_{f(j)}\subseteq_{F}I_{i} \}. $$

By Proposition 1 the set \(\mathcal {F}\) is c.e. If \((j,i)\in \mathcal {F}\), then clearly D j ⊆ I i .

Suppose i ∈N and x ∈ I i . Then d(x, λ i ) < ρ i and we may choose a positive rational number s such that s < r and d(x, λ i ) + 2s < ρ i and n ∈N such that d(α n , x) < s. Let l ∈N be such that (α n , s) = (λ l , ρ l ). We have

$$d(\lambda_{i} ,\lambda_{l} )+\rho_{l} \leq d(\lambda_{i} ,x)+d(x,\lambda_{l} )+s<d(\lambda_{i} ,x)+2s<\rho_{i}, $$

hence I l ⊆ F I i . Clearly x ∈ I l . We have f(l) = l and D l = I f(l) = I l . So x ∈ D l and \((l,i)\in \mathcal {F}\). We conclude that \((D_{i} )\preceq _{\mathcal {F}}(I_{i} )\).

Thus (D i ) and (I i ) are computably equivalent. For each i ∈N the set D i is open and connected. So (X, d, α) is effectively locally connected.

Proposition 2

Let (X, d, α)be a computable metric space and let (D i )be a sequence of open connected subsets of (X, d)which is computably equivalent to (I i ). Then the sets

$$S=\{(n,i)\in \mathbf{N}^{2}\mid \alpha_{n} \in D_{i} \}~\text{ and }~T=\{(i,j)\in \mathbf{N}^{2}\mid D_{i} \cap D_{j} \neq\emptyset \}$$

arec.e.

Proof

Let \(\mathcal {C}\subseteq \mathbf {N}^{2}\) be a c.e. set such that \((I_{i} )\preceq _{\mathcal {C}}(D_{i} )\). Let n, i ∈N. Then

$$\alpha_{n} \in D_{i} \Leftrightarrow \exists j\in \mathbf{N}\text{ such that }\alpha_{n} \in I_{j} \text{ and }(j,i)\in \mathcal{C}. $$

It follows from Proposition 1 that S is c.e.

Let i, j ∈N. The set D i ∩ D j is open and therefore, if it is nonempty, it contains α n for some n ∈N. Hence

$$D_{i} \cap D_{j} \neq\emptyset \Leftrightarrow \exists n\in \mathbf{N}\text{ such that }\alpha_{n} \in D_{i} \text{ and }\alpha_{n} \in D_{j}. $$

The fact that S is c.e. now implies that T is c.e. □

Proposition 3

Let (X, d, α)be a computable metric space and let (D i )be a sequence of connected open sets in (X, d)which is computably equivalent to (I i ). Then there exists a c.e. subset \(\mathcal {F}\) of N 3 such that:

  1. (i)

    if \((l,i,k)\in \mathcal {F}\) , then D l ⊆ D i and diamD l < 2−k ;

  2. (ii)

    if i, k ∈N and x ∈ D i , then there exists l ∈N such that x ∈ D l and \((l,i,k)\in \mathcal {F}\) .

Proof

Let \(\mathcal {C},\mathcal {C}^{\prime }\subseteq \mathbf {N}^{2}\) be c.e. sets such that \((D_{i} )\preceq _{\mathcal {C}}(I_{i} )\) and \((I_{i} )\preceq _{\mathcal {C}^{\prime }}(D_{i} )\).

Suppose i, k ∈N and x ∈ D i . There exists j ∈N such that x ∈ I j and \((j,i)\in \mathcal {C}^{\prime }\). It is easy to conclude (similarly as in Example 1) that there exists p ∈N such that x ∈ I p , I p ⊆ F I j and ρ p < 2−(k+1). Finally, there exists l ∈N such that x ∈ D l and \((l,p)\in \mathcal {C}\).

We have the following conclusion: if i, k ∈N and x ∈ D i , then there exists l ∈N such that \((l,i,k)\in \mathcal {F}\), where

$$\mathcal{F}=\{(l,i,k)\in \mathbf{N}^{3}\mid (\exists p,j\in \mathbf{N})~(l,p)\in \mathcal{C},~\rho_{p} <2^{-(k+1)},~I_{p} \subseteq_{F}I_{j} ,~(j,i)\in \mathcal{C}^{\prime}\}.$$

Suppose \((l,i,k)\in \mathcal {F}\). Then there exist p, j ∈N such that \((l,p)\in \mathcal {C}\), ρ p < 2−(k+1), I p ⊆ F I j , and \((j,i)\in \mathcal {C}^{\prime }\). It follows D l ⊆ I p and since diamI p ≤ 2ρ p < 2−k we have diamD l < 2−k. Furthermore, D l ⊆ I p , I p ⊆ I j and I j ⊆ D i , and so D l ⊆ D i .

It is easy to conclude that \(\mathcal {F}\) is a c.e. set and the claim of the proposition follows. □

4 Co-c.e. Sets with Effectively Disconnected Complements

In this section we examine connected components of c.e. open sets.

Proposition 4

Let (X, d, α)be an effectively locally connected computable metric space and let U be a c.e. open set in this space. Let

$${\Delta} =\{(i,j)\in \mathbf{N}^{2}\mid \alpha_{i}~\text{and}~\alpha_{j}\textrm{ lie in the same component of }~U\}. $$

Then Δisc.e.

Proof

Let (D i ) be a sequence of connected open sets computably equivalent to (I i ). It is easy to conclude that there exists a c.e. set A ⊆N such that

$$ U=\underset{i\in A}{\bigcup} D_{i}. $$
(5)

Let ∼ be the binary relation on U defined in the following way: x ∼ y if there exists a finite sequence l 0,…, l m in A such that

$$ x\in D_{l_{0} },~~ y \in D_{l_{m} }, $$
(6)

and

$$ D_{l_{k} }\cap D_{l_{k+1}}\neq\emptyset \text{ for each }k\in \{0,{\dots} ,m-1\}. $$
(7)

We have that ∼ is an equivalence relation on U (reflexivity follows from (5); symmetry and transitivity are obvious).

Suppose x ∼ y. Then there exist l 0,…, l m ∈ A such that (6) and (7) hold. It follows from (7) that the set

$$P=D_{l_{0} }\cup {\dots} \cup D_{l_{m} } $$

is connected (in general, if two connected sets have a nonempty intersection, then their union is connected). Moreover, this set is contained in U and, by (6), it contains x and y. Note also that P is open and P ⊆ [x], where [x] is the equivalence class of x.

So, if x ∈ U and y ∈ [x], there exists a connected open set P such that x, y ∈ P ⊆ [x]. We conclude that [x] is an open and connected subset of (X, d). Since {[x]∣x ∈ U} is a partition of U, the sets [x] for x ∈ U are exactly the connected components of U.

So x ∼ y if and only if x and y lie in the same component of U. From this and the definition of Δ it follows

$$ {\Delta} =\{(i,j)\in \mathbf{N}^{2}\mid \alpha_{i}\sim \alpha_{j}\}. $$
(8)

Using this fact, we now prove that Δ is c.e.

Let i, j ∈N. By (8), we have (i, j) ∈ Δ if and only if there exists l ∈N such that:

$$ \alpha_{i}\in D_{(l)_{0} },~~~ \alpha_{j} \in D_{(l)_{\overline{l}}}, $$
(9)
$$ D_{(l)_{k}}\cap D_{(l)_{k+1}}\neq\emptyset \text{ for each }k\in \mathbf{N}\text{ such that }0\leq k<\overline{l} $$
(10)

and

$$ (l)_{0},{\dots} , (l)_{\overline{l}}\in A . $$
(11)

Let Ω be the set of all (i, j, l) ∈N 3 for which (9), (10) and (11) hold. It is easy to conclude that Ω is c.e. (see the proof of Proposition 24 in [5]). For all i, j ∈N we have

$$(i,j)\in {\Delta} \Longleftrightarrow \text{ there exists }l\in \mathbf{N}\text{ such that }(i,j,l)\in {\Omega} $$

and it follows that Δ is c.e. □

Remark 1

Suppose (X, d, α) is effectively locally connected in the sense of [1]. To prove that (X, d, α) is effectively locally connected (in the sense of our definition), we proceed in the following way. The set {(i, j) ∈N 2∣α j ∈ I i } is c.e. and therefore {(i, j) ∈N 2∣α j ∈ I i } = f(N) for some computable function f : N →N 2. Let f 1, f 2 : N →N be the component functions of f .

Let k ∈N. Let i = f 1(k) and j = f 2(k). We define D k to be the connected component of the point α j in the set I i . Since (X, d) is locally connected, a connected component of an open set is open. Hence D k is an open connected set. For each i ∈N we have \(I_{i} =\bigcup\limits_{f_{1} (k)=i} D_{k}\) and we conclude that (D k ) ≼ (I i ).

Now we outline the proof that (I i ) ≼ (D k ). Using the definition of effective local connectedness from [1], it is not hard to conclude that we can effectively, for each i ∈N, find a sequence (U i, j ) j∈N of open sets, each of which is a finite union of rational open balls, such that \(I_{i}=\bigcup\limits_{j\in \mathbf {N}}U_{i,j}\) and each U i, j is contained in some open connected subset of I i (although U i, j need not necessarily be connected). In a similar way as in the proof of Proposition 4 we get that, for i, j 1, j 2 ∈N, the sets \(U_{i,j_{1} }\) and \(U_{i,j_{2} }\) lie in the same component of I i if and only if there exist l 0,…, l n ∈N such that l 0 = j 1, l n = j 2 and \(U_{i,l_{p} }\cap U_{i,l_{p+1}}\neq \emptyset \) for each p ∈{0,…, n − 1}, and, consequently, the set \(\{(i,j_{1} ,j_{2})\in \mathbf {N}^{3}\mid U_{i,j_{1} }\) and \(U_{i,j_{2} }\) lie in the same component of I i } is c.e.

Let k ∈N. Let i = f 1(k) and j = f 2(k). We can effectively find some l ∈N such that α j ∈ U i, l and then D k is the union of all \(U_{i,l^{\prime }}\) such that U i, l and \(U_{i,l^{\prime }}\) lie in the same component of I i . It follows (I i ) ≼ (D k ).

Lemma 1

Let (X, d, α)be a computable metric space and let S be a co-c.e. closed set in this space. Let (D i )and \(\mathcal {F}\) be as in Proposition 3.

  1. (i)

    Suppose i ∈N is such that D i intersects two different components of X ∖ S and such that the interior of D i ∩ S is empty. Then for each k ∈N there exists l ∈N such that \((l,i,k)\in \mathcal {F}\) and such that D l intersects two different components of X ∖ S .

  2. (ii)

    Suppose U is an open connected set in (X, d)which intersects two different components of X ∖ S and such that the interior of U ∩ S is empty. Then there exists l ∈N such that D l ⊆ U and such that D l intersects two different components of X ∖ S .

Proof

(i) Let k ∈N. By (i) and (ii) of Lemma 3 we have

$$ D_{i} =\underset{(l,i,k)\in \mathcal{F}}\bigcup D_{l}. $$
(12)

Suppose that there exists no l ∈N such that \((l,i,k)\in \mathcal {F}\) and such that D l intersects two different components of X ∖ S.

For each component K of X ∖ S let

$$[K]=\{l\in \mathbf{N}\mid (l,i,k)\in \mathcal{F}\text{ and }D_{l} \cap K\neq\emptyset \}. $$

Let K 1 and K 2 be two different components of X ∖ S and let l 1 ∈ [K 1] and l 2 ∈ [K 2]. We claim that

$$ D_{l_{1}} \cap D_{l_{2}} =\emptyset . $$
(13)

Suppose the opposite. Then \(D_{l_{1}} \cap D_{l_{2}} \) is a nonempty open set and since \(D_{l_{1}} \subseteq K_{1} \cup S\) and \(D_{l_{2}} \subseteq K_{2} \cup S\) we have

$$D_{l_{1}} \cap D_{l_{2}}\subseteq S. $$

This, together with \(D_{l_{1}} \cap D_{l_{2}}\subseteq D_{i} \) implies that the interior of D i ∩ S is nonempty which contradicts the assumption of the lemma.

Hence (13) holds. This and (12) imply that D i is the disjoint union of the sets

$$\underset{l\in [K]}{\bigcup} D_{l}, $$

where K is a component of X ∖ S. These sets are obviously open and at least two of them are nonempty (since D i intersects two different components of X ∖ S). But this is impossible since D i is connected.

Hence there exists l ∈N such that \((l,i,k)\in \mathcal {F}\) and such that D l intersects two different components of X ∖ S.

In the same way we prove claim (ii). □

Let (X, d, α) be a computable metric space and let V be a disconnected open set in (X, d). We say that V is effectively disconnected in (X, d, α) if there exists a c.e. subset A of N such that for each component K of V there exists a unique number i ∈ A such that α i ∈ K.

For example, each disconnected open set which has finitely many components is effectively disconnected.

Proposition 5

Let (X, d, α)be an effectively locally connected computable metric space and let U be a c.e. open set in (X, d, α)which is effectively disconnected. Then the set

$${\Gamma} =\{(i,j)\in \mathbf{N}^{2}\mid \alpha_{i}~\text{and}~\alpha_{j}~\textrm{lie in different components of}~U\}$$

isc.e.

Proof

Since U is effectively disconnected, there exists a c.e. subset A of N such that for each component K of U there exists a unique number i ∈ A such that α i ∈ K.

Let Δ be the set from Proposition 4 and let i, j ∈N. Then

$$(i,j)\in {\Gamma}~\Leftrightarrow~\exists~ a,b\in \mathbf{N}\text{ such that }a\neq b,~a,b\in A,~(i,a)\in {\Delta},~ (j,b)\in {\Delta}$$

and the claim follows from the fact that Δ is c.e. □

It is easy to prove the following simple fact from the theory of computable functions.

Lemma 2

Let S ⊆N and T ⊆N 3 be computably enumerable sets such that for all i ∈ S and k ∈N there exists l ∈ S such that (l, i, k) ∈ T.Then for each i 0 ∈ S there exists a computable function f : N →N such that f(0) = i 0 , f(N) ⊆ S and (f(k + 1), f(k), k + 1) ∈ T for each k ∈N .

Let (X, d, α) be a computable metric space and let S be a co-c.e. set in (X, d, α). Then, as mentioned in the introduction, S need not contain a computable point.

Suppose X ∖ S is effectively disconnected. If the ambient space is Euclidean space, then S contains a computable point [5]. In general, however, S need not contain a computable point (even if (X, d) is connected), as the following example shows.

Example 2

Let f : [0,1] →R be a computable function (in the sense of [13]) which has zero-points, but none of them is computable [14]. We may assume that 0 ≤ f(x) < 1 for each x ∈ [0,1]. Then it is not hard to prove that the set

$$X=\{(x,y)\mid x\in [0,1],~f(x)\leq y\leq 1\}\cup \{(x,y)\mid x\in [0,1],~-1\leq y\leq -f(x)\} $$

is c.e. closed in R 2 (where we consider R 2 as a computable metric space in the way which is described in Section 2), see figure. Since R 2 is complete, there exists a computable sequence α in R 2 such that X = Cl({α i ∣i ∈N}).

Let d be the Euclidean metric on X. Then (X, d, α) is a computable metric space.

figure a

Since α is computable in R 2, its component sequences are computable and it is easy to conclude the following: if (x, y) is a computable point in (X, d, α), then x and y are computable numbers.

Let S = f −1({0}) ×{0}. Then S ⊆ X, moreover

$$S=X\cap (\mathbf{R}\times \{0\}).$$

It is easy to conclude that S is a co-c.e. set in (X, d, α). Clearly X ∖ S has two connected components, thus X ∖ S is effectively disconnected. But it is immediate from the definition of S that S does not contain a point which is computable in (X, d, α). Note that the metric space (X, d) is connected.

We will see that the assumption that (X, d, α) is effectively locally connected assures that S has a computable point if (X, d) is connected (Corollary 1). This is a consequence of the following theorem.

Theorem 1

Let (X, d, α)be an effectively locally connected computable metric space. Let S be a co-c.e. set in (X, d, α)such that X ∖ S is effectively disconnected.

  1. (i)

    Suppose (X, d)is a complete metric space and U is an open connected set in (X, d)which intersects two different components of X ∖ S . Then U ∩ S contains a computable point.

  2. (ii)

    Suppose that each point x ∈ S lies in the boundary of at least two different components of X ∖ S . Then S is a computable set in (X, d, α).

Proof

Let (D i ) be a sequence of open connected sets in (X, d) which is computably equivalent to (I i ). Let \(\mathcal {F}\subseteq \mathbf {N}^{3}\) be a c.e. set as in Proposition 3.

Let

$${\Gamma} =\{(i,j)\in \mathbf{N}^{2}\mid \alpha_{i}~\text{and}~\alpha_{j}~\textrm{lie in different components of}~X\setminus S\}. $$

By Proposition 5 the set Γ is c.e.

Let

$${\Omega}=\{i\in \mathbf{N}\mid D_{i}\text{ intersects two}\text{ different components of }X\setminus S\}. $$

Then

$$i\in {\Omega} ~\Leftrightarrow~\exists a,b\in \mathbf{N}\text{ such that }\alpha_{a},\alpha_{b}\in D_{i}\text{ and }(a,b)\in {\Gamma}, $$

and it follows from Proposition 2 that Ω is c.e.

(i) If Int(U ∩ S)≠∅, then there exists a ∈N such that α a ∈ U ∩ S and this is the desired computable point.

Suppose Int(U ∩ S) = ∅. By Lemma 1(ii) there exists P ∈N such that D P ⊆ U and D P intersects two different components of X ∖ S.

By Lemma 1 for all i ∈ Ω and k ∈N there exists l ∈ Ω such that \((l,i,k)\in \mathcal {F}\). By Lemma 2 there exists a computable sequence (i k ) k∈N such that i 0 = P, i k ∈ Ω and \((i_{k+1} ,i_{k},k+1)\in \mathcal {F}\) for each k ∈N.

In general, if V is a connected set in (X, d) which intersects two different components of X ∖ S, then V cannot be contained in X ∖ S, hence V ∩ S≠∅. Therefore for each k ∈N we have

$$ D_{i_{k} }\cap S\neq\emptyset. $$
(14)

Furthermore,

$$D_{i_{k+1}}\subseteq D_{i_{k} }\text{ and } {\text{diam}} D_{i_{k+1}}<2^{-(k+1)} $$

for each k ∈N. It follows \({\text {Cl}}(D_{i_{k+1}})\subseteq {\text {Cl}}(D_{i_{k} })\) and \({\text {diam}} {\text {Cl}}(D_{i_{k+1}})={\text {diam}} D_{i_{k+1}}<2^{-(k+1)}\) for each k ∈N. Since (X, d) is a complete metric space, the Cantor intersection theorem implies that

$$\underset{k\in \mathbf{N}}{\bigcap} {\text{Cl}}(D_{i_{k}})=\{x\},$$

where x ∈ X. It is easy to conclude, using Proposition 2, that there exists a computable function f : N →N such that \(\alpha _{f(k)}\in D_{i_{k} }\) for each k ∈N. For each k ∈N, k ≥ 1, we have

$$d(x,\alpha_{f(k)})<{\text{diam}}(D_{i_{k}})<2^{-k} $$

and it follows that x is a computable point. Each open neighborhood of x in (X, d) contains some \(D_{i_{k} }\) and therefore, by (14), it intersects S. It follows that x ∈Cl(S), hence x ∈ S.

(ii) By the assumption of the theorem S is a co-c.e. set. So we only need to prove that S is a c.e. set. Let \(\mathcal {C}\subseteq \mathbf {N}^{2}\) be a c.e. set such that \((D_{i} )\preceq _{\mathcal {C}}(I_{i} )\).

Let i ∈N. Suppose that I i ∩ S≠∅. Then there exists x ∈ I i ∩ S, and, since x ∈ I i , there exists j ∈N such that

$$x\in D_{j}~\text{ and }~ (j,i)\in \mathcal{C}. $$

Since x ∈ S, x lies in the boundary of two different components of X ∖ S. We have that D j is an open neighborhood of x and therefore D j intersects two different components of X ∖ S, hence j ∈ Ω.

On the other hand, suppose j ∈N is such that \((j,i)\in \mathcal {C}\) and j ∈ Ω. Since D j intersects two different components of X ∖ S, we have D j ∩ S≠∅, which, together with D j ⊆ I i , gives I i ∩ S≠∅.

So

$$I_{i} \cap S\neq\emptyset \Leftrightarrow \text{ there exists }j\in \mathbf{N}\text{ such that }(j,i)\in \mathcal{C}\text{ and }j\in {\Omega} .$$

It follows that the set {i ∈N∣I i ∩ S≠∅} is c.e. Thus S is c.e. and we have that S is computable. □

Corollary 1

Let (X, d, α)be an effectively locally connected computable metric space and let S be a co-c.e. set in (X, d, α)such that X ∖ S is effectively disconnected. Additionally, let us assume that (X, d)is connected and complete. Then S contains a computable point.

Proof

We take U = X and apply Theorem 1(i). □

Let us consider the computable metric space (X, d, α) and the set S constructed in Example 2. We have that S is co-c.e. and X ∖ S is effectively disconnected. Also, (X, d) is complete (moreover, it is compact). If we take U = X, then U is an open and connected set in (X, d). However, the set U ∩ S = S does not contain a computable point. It follows from Theorem 1 that the computable metric space (X, d, α) is not effectively locally connected. On the other hand, using the definition of X, it is not hard to check that the metric space (X, d) is locally connected (and also connected).

Let (X, d, α) be an effectively locally connected computable metric space and let S be a co-c.e. set in (X, d, α) such that X ∖ S is effectively disconnected. Although Theorem 1(i) gives a sufficient condition that S contains a computable point, the set S in general need not contain a computable point. Let us consider the following example.

Example 3

Let b be a left computable number which is not computable and such that 0 < b < 1. It is easy to construct a computable sequence of rational numbers β such that ([0, b], d ′, β) is a computable metric space, where d ′ is the Euclidean metric on [0, b], and to conclude that {b} is a co-c.e. set in such a space (see Example 3.2 in [7]). Furthermore, it is easy to construct a computable sequence of rational numbers γ such that {γ i ∣i ∈N} = Q ∩ [1,2] and a computable sequence of rational numbers α such that {α i ∣i ∈N} = {β i ∣i ∈N}∪{γ i ∣i ∈N}. Let X = [0, b] ∪ [1,2] and let d be the Euclidean metric on X. Then (X, d, α) is a computable metric space in which the set S = {b} is co-c.e. Furthermore, X ∖ S has precisely two components, so X ∖ S is effectively disconnected. Since all open balls in (X, d) with sufficiently small radius are connected, the computable metric space (X, d, α) if effectively locally connected (Example 1). However, S does not contain a computable point (the fact that b is not a computable number implies that b is not a computable point in (X, d, α)).

Now we give another sufficient condition that S contains a computable point. We show that S contains a computable point x if there exists a connected set A which intersects two different components of X ∖ S. Moreover, such x can be found sufficiently close to A.

Theorem 2

Let (X, d, α)be an effectively locally connected computable metric space and let S be a co-c.e. set in (X, d, α)such that X ∖ S is effectively disconnected. Let us assume that the metric space (X, d)is complete and that A is a connected set in (X, d)which intersects at least two different components of X ∖ S . Then for each ε > 0, there exists a computable point x 0 ∈ S such that

$$d(x_{0},A)<\varepsilon.$$

Proof

Let (D i ) be a sequence of open connected sets which is computably equivalent to (I i ) and let ε > 0. It is easy to conclude, using Proposition 3, that {D i ∣i ∈N, diamD i < ε} is an open cover of (X, d). Let

$$A^{\prime}=\{i\in \mathbf{N}\mid D_{i} \cap A\neq\emptyset \text{ and }{\text{diam}} D_{i} <\varepsilon \}.$$

We have

$$ A\subseteq \underset{i\in A^{\prime}}{\bigcup} D_{i}. $$
(15)

Case 1

There exists i ∈ A ′ such that D i intersects two different components of X ∖ S. Since (X, d) is complete and D i is open and connected, Theorem 1(i) implies that D i ∩ S contains a computable point x 0. We have D i ∩ A≠∅ and diamD i < ε and it follows

$$d(x_{0},A)<\varepsilon.$$

Case 2

For each i ∈ A ′ the set D i intersects at most one component of X ∖ S. Now, if there exists i ∈ A ′ such that D i does not intersect any component of X ∖ S, then D i ⊆ S; since D i is open and nonempty, there exists n ∈N such that α n ∈ D i and, as in Case 1, we conclude that d(α n , A) < ε, meaning that α n is the desired point.

We may therefore suppose that D i intersects precisely one component of X ∖ S for each i ∈ A ′. Let \(\mathcal {K}\) be the set of all components of X ∖ S. For \(K\in \mathcal {K}\) let

$$A^{\prime}_{K}=\left\{i\in A^{\prime}\mid D_{i}\cap K\neq\emptyset\right\}. $$

We have

$$ A^{\prime}=\underset{K\in \mathcal{K}}{\bigcup} A^{\prime}_{K}, $$
(16)

and for each \(K\in \mathcal {K}\) and \(i\in A^{\prime }_{K}\) we have

$$ D_{i}\subseteq K\cup S. $$
(17)

For \(K\in \mathcal {K}\) let

$$U_{K}=\underset{i\in A^{\prime}_{K}}{\bigcup} D_{i}.$$

It follows from (15) and (16) that

$$ A\subseteq \underset{K\in \mathcal{K}}{\bigcup} U_{K}. $$
(18)

Note that for each \(K\in \mathcal {K}\) we have K ∩ A ⊆ U K ∩ A. By the assumption of the theorem, A intersects two components of X ∖ S and therefore at least two of the sets U K , \(K\in \mathcal {K}\), intersect A. Since A is connected and (18) holds, the sets U K , \(K\in \mathcal {K}\), (which are clearly open) cannot be mutually disjoint. Hence there exist \(K_{1} ,K_{2} \in \mathcal {K}\) such that K 1≠K 2 and

$$U_{K_{1} }\cap U_{K_{2} }\neq\emptyset .$$

It follows that there exist \(i\in A^{\prime }_{K_{1} }\) and \(j\in A^{\prime }_{K_{2} }\) such that D i ∩ D j ≠∅. By (17) we have

$$D_{i} \cap D_{j} \subseteq (K_{1} \cup S)\cap (K_{2} \cup S)\subseteq S. $$

So D i ∩ D j is an open nonempty set contained in S, and if we choose some n ∈N such that α n ∈ D i ∩ D j , we have d(α n , A) < ε and α n is the desired point.

□

In view of Theorem 2, it is natural to ask the following question: can we find a computable point x 0 ∈ S such that x 0 ∈ A? Of course, it is reasonable to assume that A itself has some nice computability property, for example that A is a computable compact set. The next example shows that even under this assumption such a point need not exist.

Example 4

Let f : [0,1] →R be a nonnegative computable function which has zero-points, but none of them is computable. Using the fact that f is computable, it is not hard to prove that the graph Γ(f) of the function f if a computable compact set in R 2. Also, Γ(−f) is a computable compact set and therefore

$$A={\Gamma} (f)\cup {\Gamma} (-f)$$

is a computable compact set in R 2. Let S = R ×{0}. Then S is a co-c.e. set in R 2 such that R 2 ∖ S is effectively totally disconnected. The computable metric space R 2 is clearly effectively locally connected. Furthermore, the set A intersects two components of R 2 ∖ S. However, the intersection S ∩ A does not contain a computable point. Namely, S ∩ A = f −1({0}) ×{0}.