Proof. Suppose N is an open neighborhood of x in S, where S is a semicomputable compact set in $(X,\mathcal {T},(I_{i} ))$ , and $f:T_{k} \rightarrow N$ a homeomorphism. In view of Theorem 3.3, it only remains to consider the case when $f(0,\dots ,0)\neq x$ . But then $x\in f(\langle 0,1\rangle e_{i} )$ for some $i\in \{1,\dots ,k \}$ , and since $\langle 0,1\rangle e_{i}$ is open in $T_{k} $ , $f(\langle 0,1\rangle e_{i} )$ is open in N and so $f(\langle 0,1\rangle e_{i} )$ is an open neighborhood of x in S which is clearly homeomorphic to $\langle 0,1\rangle $ and therefore to $\mathbb {R}$ . By Theorem 3.2 S is c.e. at x.

Hence $T_{k} $ has local computable type.

Proof. The set $I_i^{\prime } = (J_{\phi (i)})/K$ is obviously open in $X/K$ for each $i \in \mathbb {N}$ , since $J_{\phi (i)}$ is K-saturated ( $\phi (i) \in \Omega $ implies $K \subseteq J_{\phi (i)}$ or $J_{\phi (i)} \cap K = \emptyset $ ).

Suppose V is open in $X/K$ , take $y \in V$ and let $U \subseteq X$ be the inverse image of V under the canonical quotient map. If $y=K$ , we know that $K \subseteq U$ . Since U is open and K is compact, we can find $j \in \mathbb {N}$ such that $K \subseteq J_j \subseteq U$ . Then $j \in \Omega $ , so $j = \phi (i)$ for some $i \in \mathbb {N}$ and we have

$$\begin{align*}K/K \subseteq (J_j)/K \subseteq U/K \text,\end{align*}$$

which gives us

$$\begin{align*}y \in I_i^{\prime} \subseteq V \text. \end{align*}$$

On the other hand, suppose $y = \{x\}$ for $x \in X \setminus K$ . We begin by fixing $J_j$ such that $x \in J_j \subseteq U$ and apply Theorem 2.1 to find $k, l \in \mathbb {N}$ such that $\{x\} \subseteq J_k$ , $K \subseteq J_l$ , $(k, j) \in C$ and $(k, l) \in D$ . Again, we have $k \in \Omega $ , so $k = \phi (i)$ for some $i \in \mathbb {N}$ and $\{x\} \in I_i^{\prime } \subseteq V$ . Therefore, $(I_i^{\prime })_{i\in \mathbb {N}}$ is a basis for $X/K$ .

Characteristic relations for this space can be defined as follows:

$$ \begin{align*}\mathcal{C} := \lbrace (i, j) \mid (\phi(i), \phi(j)) \in C \rbrace \text,\end{align*} $$

$$ \begin{align*}\mathcal{D} := \lbrace (i, j) \mid (\phi(i), \phi(j)) \in D \rbrace \text.\end{align*} $$

If $(i, j) \in \mathcal {C}$ , then $J_{\phi (i)} \subseteq J_{\phi (j)}$ and obviously $I_i^{\prime }=(J_{\phi (i)})/K \subseteq I_j^{\prime }=(J_{\phi (j)})/K$ . Similarly, if $(i, j) \in \mathcal {D}$ , then $J_{\phi (i)} \cap J_{\phi (j)} = \emptyset $ . Since $J_{\phi (i)}$ and $J_{\phi (j)}$ are K-saturated, we have $I_i^{\prime } \cap I_j^{\prime } = \emptyset $ . This proves (CR1) and (CR2).

To prove (CR3), suppose $y \in I_i^{\prime } \cap I_j^{\prime }$ . If $y=K$ , then we have $K \subseteq J_{\phi (i)} \cap J_{\phi (j)}$ , and by Theorem 2.1 we can find $l \in \mathbb {N}$ such that $K \subseteq J_l$ and $(l, \phi (i)), (l, \phi (j)) \in C$ . Then $l \in \Omega $ , so there exists $k \in \mathbb {N}$ such that $\phi (k)=l$ and we have $K \in I_k^{\prime }$ and $(k, i), (k, j) \in \mathcal {C}$ . If $y=\{x\}$ , $x \in X \setminus K$ , we can (again by Theorem 2.1) find ${l, m \in \mathbb {N}}$ such that $x \in J_l$ , $K \subseteq J_m$ , $(l, \phi (i)), (l, \phi (j)) \in C$ and $(l, m) \in D$ . Again $l \in \Omega $ , so there exists k such that $\phi (k)=l$ and we have

$$ \begin{align*}\{x\} \in I_k^{\prime} \quad\text{ and }\quad (k, i), (k, j) \in \mathcal{C} \text.\end{align*} $$

Finally, we prove (CR4). Let $y_1\text , y_2 \in X/K$ , $y_1 \neq y_2$ . If $y_1=\{x_1\}$ and  $y_2=\{x_2\}$ , we can find $k, l, m \in \mathbb {N}$ such that $x_1 \in J_k$ , $x_2 \in J_l$ , $K \subseteq J_m$ and ${(k, l), (k, m), (l, m) \in D}$ . If one of the points is K, for example $y_1=\{x\}$ and $y_2=K$ , we can, again, find $k, l \in \mathbb {N}$ such that $x \in J_k$ , $K\subseteq J_l$ and $(k, l) \in D$ . In either case, there exist $i, j \in \mathbb {N}$ such that $\phi (i)=k$ , $\phi (j)=l$ and we have

$$ \begin{align*}y_1 \in I_i^{\prime}, \quad\quad y_2 \in I_j^{\prime} \quad\text{ and }\quad (i, j) \in \mathcal{D}\text.\end{align*} $$

This concludes the proof that $(X/K, \mathcal {T}_K, (I_i)^{\prime })$ is a computable topological space.

Proof. We have

$$\begin{align*}J_j^{\prime} = \bigcup_{i \in [j]}I_i^{\prime} = \bigcup_{i \in [j]}(J_{\phi(i)})/K = \big(\bigcup_{i \in [j]}J_{\phi(i)}\big)/K = \big(\bigcup_{i \in [j]}\bigcup_{k \in [\phi(i)]}I_k\big)/K \text.\end{align*}$$

Since the function $j \mapsto \bigcup _{i \in [j]}[\phi (i)]$ is computable (see e.g., the proof of [Reference Iljazović15, Lemma 33]), there exists a recursive function $\xi : \mathbb {N} \to \mathbb {N}$ such that $\bigcup _{i \in [j]}[\phi (i)] = [\xi (j)]$ , $\forall j \in \mathbb {N}$ . Now we have $J_j^{\prime } = (J_{\xi (j)})/K$ . By the remark above, we know that $\xi (j) \in \Omega $ , $\forall j \in \mathbb {N}$ .

Proof. This is an immediate consequence of the following:

$$\begin{align*}S/K \subseteq J_j^{\prime} \ \Longleftrightarrow\ S/K \subseteq (J_{\xi(j)})/K \ \Longleftrightarrow\ S \subseteq J_{\xi(j)} \text.\end{align*}$$

The first equivalence and

$$\begin{align*}S \subseteq J_{\xi(j)} \ \Longrightarrow\ S/K \subseteq (J_{\xi(j)})/K\end{align*}$$

are obvious.

Suppose $S/K \subseteq (J_{\xi (j)})/K$ and take $x \in S$ . If $x \in S \cap K$ , we know that $S/K$ , and therefore also $(J_{\xi (j)})/K$ , contains K. Since $\xi (j) \in \Omega $ , this implies $K \subseteq J_{\xi (j)}$ and therefore $x \in J_{\xi (j)}$ . Alternatively, if $x \in S \setminus K$ , then $\{x\} \in S/K \subseteq (J_{\xi (j)})/K$ , which holds only if $x \in J_{\xi (j)}$ .

Proof. Suppose $I_i \cap S \neq \emptyset $ . Since $\operatorname {Int}_S K = \emptyset $ , there exists $x \in (S \cap I_i) \setminus K$ . We have $x \in I_i = J_{\langle i \rangle }$ and $x \notin K$ . By Theorem 2.1, we can find $k, l \in \mathbb {N}$ such that $x \in J_k$ , $K \subseteq J_l$ , $(k, \langle i \rangle ) \in C$ and $(k, l) \in D$ . Now, we know that $k \in \Omega $ , therefore $k = \phi (j)$ for some $j \in \mathbb {N}$ , and we have $\{x\} \in I_j^{\prime } \cap S/K$ .

Conversely, suppose that for some $i \in \mathbb {N}$ there exists $j \in \mathbb {N}$ such that $I_j^{\prime } \cap S/K \neq \emptyset $ and $(\phi (j), \langle i \rangle )\in C$ . Then

$$\begin{align*}\emptyset \neq I_j^{\prime} \cap S/K = (J_{\phi(j)})/K \cap S/K \subseteq (J_{\langle i \rangle})/K \cap S/K \text.\end{align*}$$

Since S is K-saturated, $J_{\langle i \rangle } \cap S$ cannot be empty – otherwise $(J_{\langle i \rangle })/K \cap S/K$ would also be empty. Therefore, $I_i \cap S = J_{\langle i \rangle } \cap S \neq \emptyset $ .

This proves

$$ \begin{align*}\{i \in \mathbb{N} \mid I_i \cap S \neq \emptyset \} = \{ i \in \mathbb{N} \mid (\exists j \in \mathbb{N})(I_j^{\prime} \cap S/K \neq \emptyset \land (\phi(j), \langle i \rangle)\in C)\}\text.\end{align*} $$

The set on the right side is c.e. because $S/K$ is c.e. This shows that S is c.e.

Proof. Fix $a, b \in \mathbb {N}$ such that $S \subseteq J_a$ , $T \subseteq J_b$ and $J_a \cap J_b = \emptyset $ . Then we have

$$\begin{align*}S \subseteq J_j \Leftrightarrow S \cup T \subseteq J_j \cup J_b \quad\text{ and }\quad T \subseteq J_j \Leftrightarrow S \cup T \subseteq J_j \cup J_a \text,\end{align*}$$

which implies that S and T are semicomputable.

Proof. Let $(X, \mathcal {T}, (I_i))$ be a computable topological space and let $f : A \to X$ be an embedding such that $f(A)$ and $f(B \cup C)$ are semicomputable. Denote $f(A)$ by S, $f(B)$ by K and $f(C)$ by T.

From the previous lemma we know that K and T are semicomputable.

Let $(X/K, \mathcal {T}_K, (I_i^{\prime }))$ be the computable quotient space obtained by identifying points in K. By Lemma 5.4, we know that $S/K$ and $T/K$ are semicomputable sets. Furthermore, since $K \subseteq S$ and $K \cap T = \emptyset $ , it holds that $S/K \cong A/B$ and $T/K \cong C/B \cong C$ . Given that $(A/B, C)$ has computable type, we can conclude that $S/K$ is computable. Since $\operatorname {Int}_SK = f(\operatorname {Int}_AB) = \emptyset $ , by Lemma 5.5, this implies that S is computable in $(X, \mathcal {T}, (I_i))$ .

Proof. Let $(X,\mathcal {T},(I_{i} ))$ be a computable topological space, S a semicomputable compact set in this space and $x_{0} \in S$ a point which has an open neighborhood N is S such that N is homeomorphic to $\mathbb {R}/K$ . We want to prove that S is c.e. at $x_{0} $ . Let $h:\mathbb {R}/K\rightarrow N$ be a homeomorphism.

If $K=\emptyset $ , then $\mathbb {R}/K$ is homeomorphic to $\mathbb {R}$ and the claim follows from Theorem 3.2. Suppose $K\neq \emptyset $ . Then $\mathbb {R}\setminus K$ has at least two connected components. If $\mathbb {R}\setminus K$ has exactly two connected components, then they must be of the form $\langle -\infty , a\rangle $ and $\langle b,\infty \rangle $ , where $a,b\in \mathbb {R}$ , $a\leq b$ , and $K=[a,b]$ (we use a well known fact that an open connected subset of $\mathbb {R}$ must be an open interval). Let $g:\mathbb {R}\rightarrow \mathbb {R}$ be defined by

$$ \begin{align*}g(t) =\begin{cases} t &\text{ if } t\leq a \text, \\ a &\text{ if } a\leq t\leq b \text, \\ t-b+a &\text{ if } b\leq t\text. \end{cases}\end{align*} $$

It is easy to conclude that g is a quotient map which collapses $[a,b]$ to a. Using the notation from Corollary 4.2 we have that $\mathbb {R}_{g}=\mathbb {R}/K$ . By the same corollary g induces a homeomorphism $\mathbb {R}_{g}\rightarrow \mathbb {R}$ , hence $\mathbb {R}/K$ and $\mathbb {R}$ are homeomorphic and thus S is c.e. at $x_{0}$ .

Let us consider now the case when $\mathbb {R}\setminus K$ has at least three connected components.

We may assume that h maps K to $x_{0} $ . Otherwise, $x_{0} =h(a)$ for some ${a\in (\mathbb {R} /K)\setminus \{K\}}$ . The canonical quotient map $q:\mathbb {R}\rightarrow \mathbb {R}/K$ restricts to the map $\mathbb {R}\setminus K\rightarrow (\mathbb {R} /K)\setminus \{K\}$ which is easily seen to be quotient and bijective, hence it is a homeomorphism. We conclude that $(\mathbb {R} /K)\setminus \{K\}$ is an open set in $\mathbb {R}/K$ which is homeomorphic to an open set in $\mathbb {R}$ . Thus $h((\mathbb {R} /K)\setminus \{K\})$ is an open neighborhood of $x_{0} $ in N (and therefore also in S) which is homeomorphic to an open subset of $\mathbb {R}$ . Now the fact that each open subset of $\mathbb {R} $ has local computable type implies that S is c.e. at $x_{0} $ .

We have

$$ \begin{align*}K=[a_{1} ,b_{1} ]\cup [a_{2} ,b_{2} ]\cup \dots \cup [a_{k} ,b_{k} ],\end{align*} $$

where $k\geq 2$ and $a_{i} \le b_{i} <a_{i+1} \le b_{i+1}$ for each $i\in \{1,\dots ,k-1\}$ . Choose $\varepsilon>0$ such that $b_{i} +\varepsilon <a_{i+1}-\varepsilon $ for each $i\in \{1,\dots ,k-1\}$ . Let

$$ \begin{align*}U=\bigcup_{i=1}^{k} \langle a_{i} -\varepsilon ,b_{i} +\varepsilon \rangle .\end{align*} $$

Then U is open and K-saturated and so $q(U)$ is an open neighborhood of K in $\mathbb {R}/K$ . We have that q restricts to the map $U\rightarrow q(U)$ which is also quotient and collapses the set K to the point K. It follows from Corollary 4.2 that

(10) $$ \begin{align} U/K\cong q(U). \end{align} $$

On the other hand, let $\varphi :U\rightarrow T_{2k}$ (recall the definition of $T_{k} $ from Section 3) be defined in the following way: if $t\in \langle a_{i} -\varepsilon ,b_{i} +\varepsilon \rangle $ , let

$$ \begin{align*}\varphi (t) =\begin{cases} \frac{a_{i} -t}{\varepsilon }e_{2i-1} &\text{ if } a_{i} -\varepsilon <t\leq a_{i} \text, \\ (0,\dots ,0) &\text{ if } a_{i} \leq t\leq b_{i} \text, \\ \frac{t-b_{i} }{\varepsilon }e_{2i} &\text{ if } b_{i} \leq t<b_{i} +\varepsilon \text. \end{cases}\end{align*} $$

It is easy to see that $\varphi $ is a quotient map which collapses K to $(0,\dots ,0)$ . It follows from Corollary 4.2 that $U/K\cong T_{2k}$ . This, together with (10), gives that $q(U)\cong T_{2k}$ . Since $h:\mathbb {R}/K\rightarrow N$ is a homeomorphism which maps K to $x_{0} $ and $q(U)$ is an open neighborhood of K in $\mathbb {R}/K$ , we have that $x_{0} $ has an open neighborhood in S which is homeomorphic to $T_{2k} $ . So S is c.e. at $x_{0} $ by Corollary 3.4.

Proof. Let $i \in \mathcal {I}$ . Since $Q_i$ is open in $X \setminus Y$ , it is also open in X and we have

$$ \begin{align*}\partial Q_i \subseteq X \setminus Q_i = Y \cup \bigcup_{j \neq i}Q_j \text.\end{align*} $$

Take $x \in \partial Q_i$ . If $x \in Q_j$ for some $j \neq i$ , $Q_j$ is an open neighborhood of x and must intersect $Q_i$ , which is a contradiction. Therefore, $\partial Q_i \cap Q_j = \emptyset $ for all $j \neq i$ , and $\partial Q_i \subseteq Y$ .

Proof. Since K is compact, there is $m> 0$ such that $K \subseteq \langle -m, m\rangle ^n$ . Then $\mathbb {R}^n \setminus \langle -m, m\rangle ^n$ is an unbounded subset of $\mathbb {R}^n \setminus K$ and, since it is connected, it must be a subset of a connected component (which must be unbounded). Any other component is then contained in $[-m, m]^n$ , i.e., it is bounded.

Proof. Let $\mathbb {R}^n \setminus K = P \cup Q$ be a separation of $\mathbb {R}^n \setminus K$ into disjoint open sets (possibly $Q=\emptyset $ ). Suppose

$$ \begin{align*}[-m, m]^n \setminus L \subseteq U_1 \cup \dots \cup U_n \cup V_1 \cup \dots \cup V_n \text.\end{align*} $$

Since $U_1, \ldots , U_n$ , $V_1, \ldots , V_n$ are K-saturated and $K \cap L = \emptyset $ , at least one set must contain K. Suppose $K \subseteq U_1$ and let

$$ \begin{align*}U_1^{\prime} = U_1 \cup Q \text,\quad\quad V_1^{\prime} = V_1 \setminus \overline{Q}\end{align*} $$

and

$$ \begin{align*}U_i^{\prime} = U_i, \quad V_i^{\prime} = V_i, \quad\quad \forall i \in \{2, \dots, n\} \text.\end{align*} $$

Obviously, $U_1^{\prime }, \ldots , U_n^{\prime }$ , $V_1^{\prime }, \ldots , V_n^{\prime }$ are open, $U_i^{\prime } \cap V_i^{\prime } = \emptyset $ and $V_i^{\prime } \cap B_i^{-m} = \emptyset $ for all $i \in \{1, \ldots , n\}$ . Furthermore, since $\partial [-m, m]^n \subseteq P$ and $Q \cap P = \emptyset $ , we have ${U_1^{\prime } \cap B_1^m = \emptyset }$ , so $U_i^{\prime } \cap B_i^m = \emptyset $ holds for all $i \in \{1, \ldots , n\}$ .

It follows from Lemma 7.2 that $\partial Q \subseteq K$ , therefore $\partial Q \subseteq U_1 \subseteq U_1^{\prime }$ and

$$ \begin{align*} U_1^{\prime} \cup V_1^{\prime} = (U_1 \cup Q) \cup (V_1 \setminus \overline{Q}) = (U_1 \cup \overline{Q}) \cup (V_1 \setminus \overline{Q}) \supseteq U_1 \cup V_1 \text. \end{align*} $$

Now, we have

$$ \begin{align*} [-m, m]^n \setminus L \subseteq U_1 \cup V_1 \cup \dots \cup U_n \cup V_n \subseteq U_1^{\prime} \cup V_1^{\prime} \cup \dots \cup U_n^{\prime} \cup V_n^{\prime} \text. \end{align*} $$

Figure 6

Open sets with properties as in Corollaries 7.5 and 7.6. K is in black, L is the white rectangle.

Since $L \subseteq Q \subseteq U_1^{\prime }$ , this implies

$$ \begin{align*} [-m, m]^n \subseteq U_1^{\prime} \cup \dots \cup U_n^{\prime} \cup V_1^{\prime} \cup \dots \cup V_n^{\prime} \text, \end{align*} $$

which contradicts Corollary 7.5.

Proof. Let U be the set of all $x_{0} \in P$ with the following property: for each $\varepsilon>0$ there exist compact sets $F_{1} $ ,…, $F_{n}$ , $G_{1}$ ,…, $G_{n}$ and H such that (I)–(V) hold.

Claim 1. If $x_{0} \in P$ and $r>0$ are such that $\overline {B}(x_{0} ,r)\subseteq P$ , then

$$ \begin{align*} B(x_{0} ,r)\cap U = \emptyset \quad\text{or}\quad B(x_{0} ,r)\subseteq U. \end{align*} $$

Proof of Claim 1.

Let $x_{0} \in P$ and $r>0$ be such that

(11) $$ \begin{align} \overline{B}(x_{0} ,r)\subseteq P~\mbox{ and }~B(x_{0} ,r)\cap U\neq\emptyset \text. \end{align} $$

Fix $x_{1} \in B(x_{0} ,r)\cap U$ and $r_{1}>0 $ such that

(12) $$ \begin{align} B(x_{1} ,r_{1} )\subseteq B(x_{0} ,r). \end{align} $$

Since $x_{1} \in U$ , there exist compact sets $F_{1} $ ,…, $F_{n}$ , $G_{1}$ ,…, $G_{n}$ and H such that

(13) $$ \begin{align} x_{1} \in H,\quad \mathop{\mathrm{diam}} H<r_{1} \end{align} $$

and such that (II)–(V) hold.

To prove that $ B(x_{0} ,r)\subseteq U$ , we take $x\in B(x_{0} ,r)$ and $\varepsilon>0$ and we proceed to construct sets with properties (I)–(V).

Let S be the sphere in $\mathbb {R}^{n} $ with radius r centered at $x_{0} $ , i.e., $S=\{z\in \mathbb {R}^{n} \mid d(z,x_{0} )=r\}$ . For $T\subseteq S$ let

$$ \begin{align*}\widetilde{T}=\bigcup_{z\in T}\overline{zx},\end{align*} $$

where $\overline {zx}$ is the line segment in $\mathbb {R}^{n} $ with endpoints z and x (see Figure 7a). Note that, since $S\subseteq \overline {B}(x_{0} ,r)$ , $x\in \overline {B}(x_{0} ,r)$ and $\overline {B}(x_{0} ,r)$ is convex, we have $\widetilde {T}\subseteq \overline {B}(x_{0} ,r)$ .

Figure 7

Sets used in the proof of Lemma 7.7 for $n=2$ . K is in black.

It is not hard to see that for all $z_{1} ,z_{2} \in S$ the following implication holds:

(14) $$ \begin{align} z_{1} \neq z_{2}~\Rightarrow~ \overline{z_{1} x}\cap \overline{z_{2} x}=\{x\}. \end{align} $$

Furthermore, if T is compact, then $\widetilde {T}$ is also compact. Namely, the function $T\times [0,1]\rightarrow \widetilde {T}$ , $(z,\lambda )\mapsto (1-\lambda )z+\lambda x$ , is a continuous surjection and $T\times [0,1]$ is compact.

Choose $\delta>0$ such that

(15) $$ \begin{align} \overline{B}(x,\delta )\subseteq B(x_{0} ,r)~\mbox{ and }~2\delta <\varepsilon. \end{align} $$

For $i\in \{1,\dots ,n\}$ let

$$\begin{align*}F_{i}^{\prime}=\widetilde{F_{i} \cap S}\setminus B(x,\delta )~\mbox{ and }~G_{i}^{\prime}=\widetilde{G_{i} \cap S}\setminus B(x,\delta ) \text,\end{align*}$$

and, furthermore

$$\begin{align*}F_{i}^{\prime\prime}=(F_{i} \setminus B(x_{0} ,r))\cup F_{i}^{\prime}~\mbox{ and }~G_{i}^{\prime\prime} =(G_{i} \setminus B(x_{0} ,r))\cup G_{i}^{\prime} \end{align*}$$

(see Figure 7b). As unions of compact sets, these sets are compact.

We claim that $F_1^{\prime \prime }, \dots , F_n^{\prime \prime }$ , $G_1^{\prime \prime }, \dots , G_n^{\prime \prime }$ and $\overline B(x, \delta )$ satisfy (I)–(V).

• • (I): By (15) we have $\mathop {\mathrm {diam}} \overline {B}(x,\delta )\leq 2\delta <\varepsilon $ . Hence

  $$ \begin{align*} x\in \overline{B}(x,\delta ) \quad\text{and}\quad \mathop{\mathrm{diam}} \overline{B}(x,\delta )<\varepsilon. \end{align*} $$

• • (II): Obviously $S\cap B(x_{0} ,r)=\emptyset $ . It follows from (13) and (12) that

  (16) $$ \begin{align} H\subseteq B(x_{0} ,r) \end{align} $$
  and thus $H\cap S=\emptyset $ . Furthermore, it follows from (11) and $S\subseteq \overline {B}(x_{0} ,r)$ that $S\subseteq [-m,m]^{n}$ . From this, (II) and $H\cap S=\emptyset $ it follows that
  $$ \begin{align*}S\subseteq F_{1} \cup \dots \cup F_{n} \cup G_{1} \cup \dots \cup G_{n} .\end{align*} $$

  So

  (17) $$ \begin{align} S= (F_{1}\cap S) \cup \dots \cup (F_{n}\cap S) \cup (G_{1}\cap S) \cup \dots \cup (G_{n}\cap S). \end{align} $$

  It is not hard to check that

  $$ \begin{align*}\overline{B}(x_{0} ,r)=\bigcup_{z\in S}\overline{zx}.\end{align*} $$

  This and (17) imply

  $$ \begin{align*}\overline{B}(x_{0} ,r)=\widetilde{F_{1}\cap S}\cup \dots \cup \widetilde{F_{n}\cap S} \cup \widetilde{G_{1}\cap S} \cup \dots \cup \widetilde{G_{n}\cap S}.\end{align*} $$

  From this and (15) it follows that

  (18) $$ \begin{align} \overline{B}(x_{0} ,r)=F_{1}^{\prime}\cup \dots \cup F_{n}^{\prime} \cup G_{1}^{\prime} \cup \dots \cup G_{n}^{\prime} \cup \overline{B}(x,\delta ). \end{align} $$

  By (16) and (II) we have

  $$ \begin{align*}[-m,m]^{n} =F_{1} \cup \dots \cup F_{n} \cup G_{1} \cup \dots \cup G_{n} \cup \overline{B}(x_{0} ,r),\end{align*} $$
  so
  $$ \begin{align*} [-m,m]^{n} =(F_{1}\setminus B(x_{0} ,r)) \cup \dots \cup (F_{n}\setminus B(x_{0} ,r)) \cup \\ \cup (G_{1}\setminus B(x_{0} ,r)) \cup \dots \cup (G_{n}\setminus B(x_{0} ,r)) \cup \overline{B}(x_{0} ,r) \text. \end{align*} $$

  This and (18) imply

  $$ \begin{align*} [-m,m]^{n} =F_{1}^{\prime\prime} \cup \dots \cup F_{n}^{\prime\prime}\cup G_{1}^{\prime\prime}\cup \dots \cup G_{n}^{\prime\prime}\cup \overline{B}(x,\delta ). \end{align*} $$

• • (III): Let $i\in \{1,\dots ,n\}$ . By (III) we have $(F_{i} \cap S)\cap (G_{i} \cap S)=\emptyset $ and now (14) implies that $\widetilde {F_{i} \cap S}\cap \widetilde {G_{i} \cap S}=\{x\}$ , which, together with the definitions of $F_{i}^{\prime }$ and $G_{i}^{\prime }$ , gives that

  (19) $$ \begin{align} F_{i}^{\prime}\cap G_{i}^{\prime}=\emptyset \text. \end{align} $$

  We claim that

  (20) $$ \begin{align} F_{i}^{\prime\prime}\cap G_{i}^{\prime\prime}=\emptyset \text. \end{align} $$

  Obviously $G_i^{\prime } \subseteq \overline {B}(x_0, r)$ . Using this and (19) we get

  $$ \begin{align*} (F_{i} \setminus B(x_{0} ,r))\cap G_{i}^{\prime} = (F_{i} \cap S)\cap G_{i}^{\prime}\subseteq F_{i}^{\prime}\cap G_{i}^{\prime}=\emptyset \text. \end{align*} $$

  So $(F_{i} \setminus B(x_{0} ,r))\cap G_{i}^{\prime }=\emptyset $ . In the same way we get $F_{i}^{\prime }\cap (G_{i} \setminus B(x_{0} ,r))=\emptyset $ , which together with (III) and (19) gives (20).

• • (IV): Let $i\in \{1,\dots ,n\}$ . We have $(F_{i} \setminus B(x_{0} ,r ))\cap B_{i}^{m} =\emptyset $ by (IV) and $F_{i}^{\prime } \cap B_{i}^{m} =\emptyset $ since $F_{i}^{\prime }\subseteq \overline {B}(x_{0} ,r)$ and $\overline {B}(x_{0} ,r)\subseteq \langle -m,m\rangle ^{n} $ by (11). So $F_{i}^{\prime \prime }\cap B_{i}^{m} =\emptyset $ . In the same way we get $G_{i}^{\prime \prime }\cap B_{i}^{-m} =\emptyset $ . Hence

  $$ \begin{align*} F^{\prime\prime}_{i} \cap B_{i}^{m} =\emptyset ~\mbox{ and }~G^{\prime\prime}_{i} \cap B_{i}^{-m} =\emptyset\mbox{ for each }i\in \{1,\dots ,n\}. \end{align*} $$

• • (V): By (11) we have $\overline {B}(x_{0} ,r)\cap K=\emptyset $ . Using (V) we conclude that

  $$ \begin{align*} K\subseteq F^{\prime\prime}_{1} ~\mbox{ and }~K\cap (F^{\prime\prime}_{2} \cup \dots \cup F^{\prime\prime}_{n} \cup G^{\prime\prime}_{2} \cup \dots \cup G^{\prime\prime}_{n} )=\emptyset. \end{align*} $$

To summarize, this shows that $x\in U$ and thus we have proved $B(x_0, r) \subseteq U$ , which concludes the proof of Claim 1.

Proof of Claim 2.

Since P is an outer connected component of $\langle -m,m\rangle ^{n} \setminus K$ , there exists $0 < r < m$ such that $\langle -m,m\rangle ^{n} \setminus [-r,r] ^{n} \subseteq P$ and

(21) $$ \begin{align} K\subseteq \langle -r,r\rangle ^{n}. \end{align} $$

Choose $\lambda $ such that $r < \lambda < m$ and let $x_0 =(\lambda ,\dots ,\lambda ) \in \mathbb {R}^n$ . Clearly $x_{0} \in P$ . We claim that $x_{0} \in U$ .

Let $\varepsilon>0$ . Choose $\delta>0$ so that $r<\lambda -\delta <\lambda <\lambda +\delta <m$ and

$$ \begin{align*}\mathop{\mathrm{diam}} [\lambda -\delta,\lambda +\delta ]^{n} <\varepsilon\text.\end{align*} $$

Let $H=[\lambda -\delta ,\lambda +\delta ]^{n}$ . Let

$$ \begin{align*} F_{1} &=\{(x_{1} ,\dots ,x_{n} )\in [-m,m]^{n} \mid x_{1} \leq \lambda -\delta \} \text, \\ G_{1} &=\{(x_{1} ,\dots ,x_{n} )\in [-m,m]^{n} \mid x_{1} \geq \lambda +\delta \} \end{align*} $$

and for $i\in \{2,\dots ,n\}$ let

$$ \begin{align*} F_{i} &=\{(x_{1} ,\dots ,x_{n} )\in [-m,m]^{n} \mid x_{i} \leq \lambda -\delta\mbox{ and }x_{1} \in [\lambda -\delta ,\lambda +\delta ] \} \text, \\ G_{i} &=\{(x_{1} ,\dots ,x_{n} )\in [-m,m]^{n} \mid x_{i} \geq \lambda +\delta\mbox{ and }x_{1} \in [\lambda -\delta ,\lambda +\delta ] \} \end{align*} $$

(see Figure 7c). Then $F_{1} $ , …, $F_{n} $ , $G_{1} $ , …, $G_{n} $ and H are compact sets and it is easy to check that (I)–(V) hold ((V) follows from (21)). So $x_{0} \in U$ , which concludes the proof of Claim 2.

Claim 1 implies that both U and $P\setminus U$ are open. Indeed, take $x_0 \in P$ . Since connected components of an open set in $\mathbb {R}^{n} $ are also open, we have that P is an open set in $\mathbb {R}^{n} $ . Thus we may choose $r>0$ such that $\overline {B}(x_{0} ,r)\subseteq P$ . If $x_0 \in U$ , we have $B(x_{0} ,r)\cap U\neq \emptyset $ . Claim 1 now implies that $B(x_{0} ,r)\subseteq U$ , so U is open. Similar argument follows for $x_0 \in P \setminus U$ .

Since P is connected and $U \neq \emptyset $ by Claim 2, we conclude that $U=P$ . This proves the statement of the lemma.

Proof. Let $m> 0$ be such that $K \subseteq \langle -m, m\rangle ^n$ .

Let $Q = \mathbb {R}^n \setminus (P \cup K)$ . Then $L \subseteq Q$ . Since $\mathbb {R}^n$ is normal, we can find compact sets $M_1$ and $M_2$ such that

$$\begin{align*}L \subseteq \operatorname{Int} M_1 \subseteq M_1 \subseteq \operatorname{Int} M_2 \subseteq M_2 \subseteq Q \end{align*}$$

(see Figure 8a). Note that, if Q is empty, then L must also be empty and we can take $M_1 = M_2 = \emptyset $ .

Figure 8

Compact subsets of $[-(m+2), m+2]^n$ used in the proof of Lemma 7.8.

Consider the set $\langle -(m+2), m+2 \rangle ^n \setminus M_1$ . This set is open and f-saturated (contains K), so its image under f is open in N and, consequently, open in S. Therefore,

(22) $$ \begin{align} S \setminus f(\langle -(m+2), m+2 \rangle^n \setminus M_1) \end{align} $$

is closed in a compact set S, which means it is itself compact. Since

$$\begin{align*}[-(m+1), m+1]^n \setminus \operatorname{Int} M_2 \subseteq \langle -(m+2), m+2 \rangle^n \setminus M_1 \end{align*}$$

and both sets are f-saturated, $f([-(m+1),m+1]^{n} \setminus \operatorname {Int} M_2)$ is a compact set which is disjoint with (22). By Theorem 2.1, there exists $m_0 \in \mathbb {N}$ such that

$$\begin{align*}S \setminus f(\langle -(m+2), (m+2) \rangle^n \setminus M_1) \subseteq J_{m_0}\end{align*}$$

and

$$\begin{align*}J_{m_0} \cap f([-(m+1), m+1]^n \setminus \operatorname{Int} M_2) = \emptyset \text.\end{align*}$$

Let $S^{\prime } := S \setminus J_{m_0}$ . This set is semicomputable and

(23) $$ \begin{align} f([-(m+1), (m+1)]^n \setminus \operatorname{Int} M_2) \subseteq S^{\prime} \subseteq f([-(m+2), m+2]^n \setminus \operatorname{Int} M_1) \text. \end{align} $$

For $i \in \{ 1, \ldots , n\}$ let

$$ \begin{align*} C_i &= \{ (x_1, \dots, x_n) \in [-(m+2), m+2]^n \mid x_i \leq m \} \setminus (\operatorname{Int} M_1)\text, \\ D_i &= \{ (x_1, \dots, x_n) \in [-(m+2), m+2]^n \mid x_i \geq -m \} \setminus (\operatorname{Int} M_1) \end{align*} $$

(see Figure 8b and 8c). These sets are compact, f-saturated and

$$\begin{align*}B_i^{-(m+1)} \cap D_i = B_i^{m+1} \cap C_i = \emptyset \text, \quad\quad \forall i \in \{ 1, \ldots, n\} \text,\end{align*}$$

which implies $f(B_i^{-(m+1)})$ , $f(B_i^{m+1})$ , $f(C_i)$ and $f(D_i)$ , $i \in \{1, \dots , n\}$ are compact in $(X, \mathcal {T})$ and

$$\begin{align*}f(B_i^{-(m+1)}) \cap f(D_i) = f(B_i^{m+1}) \cap f(C_i) = \emptyset \text, \quad\quad \forall i \in \{ 1, \ldots, n\} \text.\end{align*}$$

By Theorem 2.1, there exist $c_1, \ldots c_n$ , $d_1, \ldots , d_n$ such that

(24) $$ \begin{align} &f(C_i) \subseteq J_{c_i}\text,\quad f(D_i) \subseteq J_{d_i} \quad \text{and} \end{align} $$

(25) $$ \begin{align} &J_{c_i} \cap f(B_i^{m+1}) = J_{d_i} \cap f(B_i^{-(m+1)}) = \emptyset \text. \end{align} $$

Consider the boundary $\partial P$ of P in $\mathbb {R}^n$ . Since P is open in $\mathbb {R}^n \setminus K$ , it is also open in $\mathbb {R}^n$ . If $\partial P = \emptyset $ , this would imply that P is both open and closed in $\mathbb {R}^n$ , which is impossible since $P \neq \emptyset $ and $P \neq \mathbb {R}^n$ . Therefore, $\partial P \neq \emptyset $ . By Lemma 7.2, we know that $\partial P \subseteq K$ . Now we have

(26) $$ \begin{align} \emptyset \neq f(\partial P) \subseteq f(K) = \{x_0\} \text, \end{align} $$

so $f(\partial P) = \{x_0\}$ . Now,

(27) $$ \begin{align} f(P) \cup \{x_0\} = f(P) \cup f(\partial P) = f(\overline P) \text, \end{align} $$

where $\overline P = P \cup \partial P$ is the closure of P in $\mathbb {R}^n$ .

Let us denote

$$\begin{align*}N^{\prime} := f\left(\langle-m, m\rangle^n \cap P\right) \cup \{x_0\} \text.\end{align*}$$

First, note that $\langle -m, m\rangle ^n \cap P = \left ( \langle -m, m\rangle ^n \setminus L \right ) \cap P$ , since $P \cap L = \emptyset $ . Next, since f is injective outside K and $\left ( \langle -m, m\rangle ^n \setminus L\right ) \cap P$ does not intersect K, we have

(28) $$ \begin{align} f( \left(\langle-m, m\rangle^n \setminus L \right) \cap P) = f(\langle-m, m\rangle^n \setminus L) \cap f(P) \text, \end{align} $$

so

(29) $$ \begin{align} N^{\prime} = \left( f(\langle-m, m\rangle^n \setminus L) \cap f(P) \right) \cup \{x_0\} \text. \end{align} $$

Finally, since $x_0 \in f(K) \subseteq f(\langle -m, m\rangle ^n \setminus L)$ , we have

(30) $$ \begin{align} N^{\prime} = f(\langle -m, m\rangle^n \setminus L) \cap \left( f(P) \cup \{x_0\} \right) \text. \end{align} $$

Since $\langle -m, m\rangle ^n \setminus L$ is open in $\mathbb {R}^n \setminus L$ , its image under f is open in N. So $N^{\prime }$ is an open set in $f(P) \cup \{x_0\}$ which clearly contains $x_0$ , i.e., $N^{\prime }$ is a neighborhood of $x_0$ in $f(P) \cup \{x_0\}$ .

Suppose $I_{i_0} \cap N^{\prime } \neq \emptyset $ . Then $f^{-1}(I_{i_0})$ obviously intersects $\langle -m, m\rangle ^n$ . If $x_0 \in I_{i_0} \cap N^{\prime }$ , $f^{-1}(I_{i_0})$ is an open set which intersects $\partial P$ , so it must also intersect P. Therefore, we have

$$ \begin{align*} f^{-1}(I_{i_0}) \cap (\langle -m, m \rangle^n \cap P) \neq \emptyset \text. \end{align*} $$

Note that this is an open set. Take $x \in \mathbb {R}^n$ and $\varepsilon> 0$ such that

(31) $$ \begin{align} B(x, \varepsilon) \subseteq f^{-1}(I_{i_0}) \cap (\langle -m, m \rangle^n \cap P) \text. \end{align} $$

For this choice of x and $\varepsilon $ , let $F_1, \ldots , F_n$ , $G_1, \ldots , G_n$ , H be sets as in Lemma 7.7. Note that these sets are f-saturated. We have $H \subseteq B(x, \varepsilon ) \subseteq f^{-1}(I_{i_0})$ , so

(32) $$ \begin{align} f(H) \subseteq I_{i_0} = J_{\langle i_0 \rangle} \text. \end{align} $$

Let

$$ \begin{align*} F_i^{\prime} &= (F_i \setminus (\operatorname{Int} M_1)) \cup \{ (x_1, \dots, x_n) \in [-(m+2), m+2]^n \mid x_i \leq -m \} \text, \\ G_i^{\prime} &= (G_i \setminus (\operatorname{Int} M_1)) \cup \{ (x_1, \dots, x_n) \in [-(m+2), m+2]^n \mid x_i \geq m \} \end{align*} $$

for all $i \in \{ 1, \dots , n\}$ .

These sets are compact and f-saturated. Using (II), (III), and (IV) we get

(33) $$ \begin{align} F_i^{\prime} \subseteq C_i, \quad G_i^{\prime} \subseteq D_i \quad\text{and}\quad F_i^{\prime} \cap G_i^{\prime} = \emptyset \text, \qquad \forall i \in \{1, \ldots n\}\text. \end{align} $$

Furthermore, (II) implies

(34) $$ \begin{align} F_1^{\prime} \cup \cdots \cup F_n^{\prime} \cup G_1^{\prime} \cup \cdots \cup G_n^{\prime} \cup H = [-(m+2), m+2]^n \setminus (\operatorname{Int} M_1) \text, \end{align} $$

so

(35) $$ \begin{align} \bigcup_{i = 1}^n f(F_i^{\prime}) \cup \bigcup_{i = 1}^n f(G_i^{\prime}) \cup f(H) = f([-(m+2), m+2]^n \setminus (\operatorname{Int} M_1)) \text. \end{align} $$

From (33) and (24), we get

(36) $$ \begin{align} f(F_i^{\prime}) \subseteq J_{c_i}, \qquad f(G_i^{\prime}) \subseteq J_{d_i} \quad\text{and}\quad f(F_i^{\prime}) \cap f(G_i^{\prime}) = \emptyset \text. \end{align} $$

The sets $f(F_1^{\prime }), \ldots , f(F_n^{\prime })$ , $f(G_1^{\prime }), \ldots , f(G_n^{\prime })$ , $f(H)$ are non-empty compact sets in $(X, \mathcal {T})$ . By Theorem 2.1, (32) and (36) there exist numbers $f_1, \ldots , f_n$ , $g_1, \ldots , g_n$ , $h \in \mathbb {N}$ such that $f(F_i^{\prime }) \subseteq J_{f_i}$ , $f(G_i^{\prime }) \subseteq J_{g_i}$ for each $i \in \{1, \dots , n\}$ , $f(H) \subseteq J_h$ and

1. (i) $(f_i, c_i), (g_i, d_i) \in C$ for all $i \in \{1, \ldots , n\}$ ;

2. (ii) $(h,\langle i_0\rangle ) \in C$ ;

3. (iii) $(f_i, g_i) \in D$ for all $i \in \{1, \ldots , n\}$ .

Additionally, from (23) and (35) we get

1. (iv) $S^{\prime } \subseteq J_{f_1} \cup \dots \cup J_{f_n} \cup J_{g_1} \cup \dots \cup J_{g_n} \cup J_h$ .

Relations (i)–(iii) are clearly c.e. Moreover, it is possible to construct a c.e. function $\mathbb {N}^{2n+1} \to \mathbb {N}$ such that, if $(i_1, \dots , i_{2n+1}) \mapsto k$ , then $[k] = [i_1] \cup \dots \cup [i_{2k+1}]$ . Using this, we can conclude that (iv) is c.e. too.

Denote by $\Omega $ the set of all natural numbers $i_0$ for which there exist $f_1, \ldots , f_n$ , $g_1, \ldots , g_n$ , $h \in \mathbb {N}$ such that (i)–(iv) holds. Then $\Omega $ is a c.e. set.

So far, we have proved

$$ \begin{align*}I_{i_0} \cap N^{\prime} \neq \emptyset \ \implies\ i_0 \in \Omega \text.\end{align*} $$

We now prove

$$ \begin{align*}I_{i_0} \in \Omega \ \implies\ I_{i_0} \cap S^{\prime} \neq \emptyset\text.\end{align*} $$

Suppose $i_0 \in \Omega $ and $I_{i_0} \cap S^{\prime } = \emptyset $ . Let $f_1, \ldots , f_n$ , $g_1, \ldots , g_n$ , h be natural numbers such that (i)–(iv) holds. Since $I_{i_0} \cap S^{\prime } = \emptyset $ and $J_h \subseteq I_{i_0}$ by (ii), we have $J_h \cap S^{\prime } = \emptyset $ . From this, (iv) and (23), we know that

$$ \begin{align*}f([-(m+1), m+1]^n \setminus (\operatorname{Int} M_3)) \subseteq J_{f_1} \cup \dots \cup J_{f_n} \cup J_{g_1} \cup \dots \cup J_{g_n} \text,\end{align*} $$

therefore

$$ \begin{align*}[-(m+1), m+1]^n \setminus M_3 \subseteq f^{-1}(J_{f_1}) \cup \dots \cup f^{-1}(J_{f_n}) \cup f^{-1}(J_{g_1}) \cup \dots \cup f^{-1}(J_{g_n}) \text.\end{align*} $$

The sets $f^{-1}(J_{f_i})$ , $f^{-1}(J_{g_i})$ are open in $\mathbb {R}^n \setminus L$ and therefore open in $\mathbb {R}^n$ . They are also K-saturated. From (i) and (25) it follows that

$$ \begin{align*}f^{-1}(J_{f_i}) \cap B_i^{m+1} = f^{-1}(J_{g_i}) \cap B_i^{-(m+1)} = \emptyset \text, \quad \forall i \in \{1, \ldots, n\} \text.\end{align*} $$

This contradicts Corollary 7.6. Therefore $I_{i_0} \cap S^{\prime } \neq \emptyset $ .

Proof. If P is the unbounded connected component of $\mathbb {R}^{n} \setminus K$ , then the claim follows from Lemma 7.8.

Suppose P is bounded. Since K is compact, there exists $r>0$ such that ${K\subseteq \langle -r,r\rangle ^{n} }$ . Since $\mathbb {R}^{n} \setminus \langle -r,r\rangle ^{n} $ is connected, this set is contained in an unbounded connected component of $\mathbb {R}^{n} \setminus K$ , thus $(\mathbb {R}^{n} \setminus \langle -r,r\rangle ^{n})\cap P=\emptyset $ , which implies that $P\subseteq \langle -r,r\rangle ^{n} $ .

Choose $p\in P$ . It is a well known fact that there exists a homeomorphism $[-r,r]^{n} \rightarrow B^{n} $ which maps the boundary $\partial [-r,r]^{n}$ to $S^{n-1}$ . So $[-r,r]^{n} /\partial [-r,r]^{n}$ is homeomorphic to $B^{n} /S^{n-1}$ , hence to $S^{n} $ . Let $h:[-r,r]^{n} /\partial [-r,r]^{n}\rightarrow S^{n} $ be a homeomorphism. Then h induces a homeomorphism

$$ \begin{align*}([-r,r]^{n} /\partial [-r,r]^{n})\setminus \{p\}\rightarrow S^{n} \setminus \{h(p)\}.\end{align*} $$

It is also known that $S^{n} \setminus \{x\}$ is homeomorphic to $\mathbb {R}^{n} $ for each $x\in S^{n} $ . So there exists a homeomorphism

(37) $$ \begin{align} ([-r,r]^{n} /\partial [-r,r]^{n})\setminus \{p\}\rightarrow \mathbb{R}^{n}. \end{align} $$

The canonical quotient map $[-r,r]^{n} \rightarrow [-r,r]^{n} /\partial [-r,r]^{n}$ induces a map

(38) $$ \begin{align} [-r,r]^{n}\setminus \{p\}\rightarrow ([-r,r]^{n} /\partial [-r,r]^{n})\setminus \{p\} \end{align} $$

which is quotient (this follows easily from the fact that $[-r,r]^{n}\setminus \{p\}$ is open in $[-r,r]^{n}$ ). The composition of maps (37) and (38) gives the quotient map

$$ \begin{align*}g:[-r,r]^{n}\setminus \{p\}\rightarrow \mathbb{R}^{n} .\end{align*} $$

Note that g collapses $\partial [-r,r]^{n}$ to some point $y_{0} \in \mathbb {R}^{n} $ . It follows that g restricts to the map

$$ \begin{align*}h:\langle -r,r\rangle ^{n}\setminus \{p\}\rightarrow \mathbb{R}^{n} \setminus \{y_{0} \}\end{align*} $$

which is quotient and bijective. Hence h is a homeomorphism.

Note that $K\subseteq \langle -r,r\rangle ^{n}\setminus \{p\}$ . Since P is an open connected subset of $\mathbb {R}^{n} $ , it is easy to conclude that $P\setminus \{p\}$ is also connected (see e.g., [Reference Christenson and Voxman10, Corollary 2.F.4]). Clearly $P\setminus \{p\}\subseteq (\langle -r,r\rangle ^{n}\setminus \{p\})\setminus K$ .

Now we have that $P\setminus \{p\}$ is a connected component of $(\langle -r,r\rangle ^{n}\setminus \{p\})\setminus K$ . Otherwise, there would exist a connected set $Q\subseteq (\langle -r,r\rangle ^{n}\setminus \{p\})\setminus K$ such that $P\setminus \{p\} \subset Q$ (where $\subset $ means the strict inclusion). We would have that $Q\cup \{p\}$ is a connected set since $Q\cup \{p\}=Q\cup P$ (the union of two connected sets with nonempty intersection) and furthermore $Q\cup \{p\}\subseteq \mathbb {R}^{n} \setminus K$ and $P\subset Q\cup \{p\}$ , which is impossible since P is a connected component of $\mathbb {R}^{n} \setminus K$ .

Since h is a homeomorphism, $h(P\setminus \{p\})$ is a connected component of ${(\mathbb {R}^{n} \setminus \{y_{0} \})\setminus h(K)}$ . We claim that $h(P\setminus \{p\})$ is an unbounded set in $\mathbb {R}^{n} $ . It suffices to prove that the set $(\mathbb {R}^{n}\setminus \{y_{0} \})\setminus h(P\setminus \{p\})$ is bounded.

We have

$$ \begin{align*} (\mathbb{R}^{n}\setminus \{y_{0} \})\setminus h(P\setminus \{p\})=h((\langle -r,r\rangle ^{n}\setminus \{p\})\setminus (P\setminus \{p\})) \\ = h(\langle -r,r\rangle ^{n}\setminus P)\subseteq g([-r,r]^{n} \setminus P) \end{align*} $$

and $g([-r,r]^{n} \setminus P)$ is a bounded set since g is continuous and $[-r,r]^{n} \setminus P$ is compact.

So $h(P\setminus \{p\})$ is an unbounded connected component of $(\mathbb {R}^{n} \setminus \{y_{0} \})\setminus h(K)$ . We claim that $h(P\setminus \{p\})$ is an unbounded connected component of $\mathbb {R}^{n}\setminus h(K)$ .

Suppose that Q is a connected component of $\mathbb {R}^{n}\setminus h(K)$ such that $h(P\setminus \{p\})\subset Q$ . Then $y_{0} \in Q$ (otherwise Q would be a connected subset of $(\mathbb {R}^{n} \setminus \{y_{0} \})\setminus h(K)$ which contains $h(P\setminus \{p\})$ as a proper subset, impossible). Since Q is open, we have that $Q\setminus \{y_{0} \}$ is connected and now

$$ \begin{align*}h(P\setminus \{p\})\subseteq Q\setminus \{y_{0} \}\subseteq (\mathbb{R}^{n} \setminus \{y_{0} \})\setminus h(K)\end{align*} $$

implies that $h(P\setminus \{p\})= Q\setminus \{y_{0} \}$ . So $Q=h(P\setminus \{p\})\cup \{y_{0} \}$ . The set $\{y_{0} \}$ is clearly closed in Q. We now show that $\{y_{0} \}$ is also open in Q (which contradicts the fact that Q is connected).

Choose $s\in \langle 0,r\rangle $ so that $K\subseteq \langle -s,s\rangle ^{n} $ . As before we conclude that $P\subseteq \langle -s,s\rangle ^{n} $ . The set $[-r,r]^{n} \setminus [-s,s]^{n} $ is open in $[-r,r]^{n}\setminus \{p\}$ and contains $\partial [-r,r]^{n}$ , so ${g([-r,r]^{n} \setminus [-s,s]^{n} )}$ is an open set in $\mathbb {R}^{n} $ which is clearly disjoint with $g(P\setminus \{p\})$ , i.e., with $h(P\setminus \{p\})$ . Thus $g([-r,r]^{n} \setminus [-s,s]^{n} )$ is an open neighborhood of $y_{0} $ in $\mathbb {R}^{n} $ which is disjoint with $h(P\setminus \{p\})$ . This implies that $\{y_{0}\}$ is open in $h(P\setminus \{p\})\cup \{y_{0} \}$ , i.e., in Q.

We conclude that $h(P\setminus \{p\})$ is an unbounded connected component of ${\mathbb {R}^{n}\setminus h(K)}$ . Clearly $y_{0} \in \mathbb {R}^{n} \setminus (h(P\setminus \{p\})\cup h(K))$ .

The quotient map $f:\mathbb {R}^{n} \rightarrow N$ (which collapses K to $x_{0} $ ) restricts to the quotient map

$$ \begin{align*}f^{\prime}:\langle -r,r\rangle ^{n} \setminus \{p\}\rightarrow f(\langle -r,r\rangle ^{n} \setminus \{p\})\end{align*} $$

which collapses K to $x_{0} $ . Since $h^{-1} :\mathbb {R}^{n} \setminus \{y_{0} \}\rightarrow \langle -r,r\rangle ^{n} \setminus \{p\}$ is a homeomorphism, the composition

$$ \begin{align*}f^{\prime}\circ h^{-1} :\mathbb{R}^{n} \setminus \{y_{0} \}\rightarrow f(\langle -r,r\rangle ^{n} \setminus \{p\})\end{align*} $$

is a quotient map which collapses $h(K)$ to $x_{0} $ . The fact that f is quotient and collapses K to $x_{0} $ implies that $f(\langle -r,r\rangle ^{n} \setminus \{p\})$ is an open set in N, therefore $f(\langle -r,r\rangle ^{n} \setminus \{p\})$ is an open neighborhood of $x_{0} $ in S.

By Lemma 7.8 there exists a neighborhood $N^{\prime }$ of $x_{0} $ in

$$ \begin{align*}(f^{\prime}\circ h^{-1} )(h(P\setminus \{p\}))\cup \{x_{0}\}\end{align*} $$

such that $N^{\prime }$ is c.e. up to S. However, $(f^{\prime }\circ h^{-1} )(h(P\setminus \{p\}))=f^{\prime }(P\setminus \{p\})=f(P\setminus \{p\})$ , so $N^{\prime }$ is a neighborhood of $x_{0} $ in $f(P\setminus \{p\})\cup \{x_{0} \}$ .

Finally, $f(P\setminus \{p\})$ is an open set in N (since f is quotient) and if we choose an open set U in N such that $f(p)\notin U$ and $x_{0} \in U$ (which we certainly can because N is Hausdorff as a subspace of $(X,\mathcal {T})$ ), then we have

$$ \begin{align*}f(P\setminus \{p\})\cup \{x_{0} \}=(f(P)\cup \{x_{0} \})\cap (f(P\setminus \{p\})\cup U),\end{align*} $$

which shows that $f(P\setminus \{p\})\cup \{x_{0} \}$ is open in $f(P)\cup \{x_{0} \}$ . Therefore $N^{\prime }$ is a neighborhood of $x_{0} $ in $f(P)\cup \{x_{0} \}$ . This, together with the fact that $N^{\prime }$ is c.e. up to S, completes the proof of the lemma.

Proof of Theorem 7.1.

Suppose $(X,\mathcal {T},(I_{i} ))$ is a computable topological space, S is a semicomputable compact set in this space and $x_{0} \in S$ a point which has an open neighborhood N in S such that N is homeomorphic to $\mathbb {R}^{n} /K$ where $n \geq 2$ . We want to prove that S is c.e. at $x_{0} $ .

Let $h:\mathbb {R}^{n} /K\rightarrow N$ be a homeomorphism. Let $f=h\circ q$ , where $q:\mathbb {R}^{n} \rightarrow \mathbb {R}^{n} /K$ is the canonical quotient map. Then

$$ \begin{align*}f:\mathbb{R}^{n} \rightarrow N\end{align*} $$

is a quotient map which collapses K to some $y_{0} $ .

As in the proof of Theorem 6.1 we see that we may assume $y_{0} =x_{0} $ (i.e., ${h(K)=x_{0}} $ ). So f is a quotient map which collapses K to $x_{0} $ .

Let $P_{0} ,\dots ,P_{k} $ be all connected components of $\mathbb {R}^{n} \setminus K$ . By Lemma 7.9 for each $i\in \{0,\dots ,k \}$ there exists a neighborhood $N_{i}^{\prime }$ of $x_{0} $ in $f(P_{i} )\cup \{x_{0} \}$ such that $N_{i}^{\prime }$ is c.e. up to S. By the definitions of a neighborhood and a relative topology, for each $i\in \{0,\dots ,k \}$ there exists $U_{i}\in \mathcal {T} $ such that

$$ \begin{align*}x_{0} \in U_{i} \cap (f(P_{i} )\cup \{x_{0} \})\subseteq N_{i}^{\prime}.\end{align*} $$

We have $N=f(\mathbb {R}^{n} )=f(P_{0} \cup \dots \cup P_{k} \cup K)=f(P_{0} )\cup \dots \cup f(P_{k} )\cup \{x_{0} \}$ , hence