Proof Let $(M,d)$ be a right-c.e. Polish space, and let $(\alpha _i)_{i\in \omega }$ be its sequence of special points. By $\tau $ we denote the metric topology of $(M,d)$ . As usual, the base $\beta $ of $\tau $ contains basic open balls

$$\begin{align*}B(\alpha_i,q) = \{ x\in M\,\colon d(\alpha_i,x) < q\},\ \ \ i\in\omega,\ q\in\mathbb{Q}^{+}. \end{align*}$$

For $i\in \omega $ and $q\in \mathbb {Q}^{+}$ , we put $\nu (i,q) = B(\alpha _i,q)$ .

We prove that the tuple $(M,\tau ,\beta ,\nu )$ is a computable topological space. It is sufficient to establish the following: for any $i,j\in \omega $ and $q,r\in \mathbb {Q}^{+}$ , we can (uniformly) effectively enumerate a set $X \subseteq \omega \times \mathbb {Q}^{+}$ such that

(1) $$ \begin{align} B(\alpha_i,q) \cap B(\alpha_j, r) = \bigcup \{B(\alpha_k, t) \,\colon (k,t) \in X \}. \end{align} $$

Our set X is defined as follows: X contains all pairs $(k,t)$ such that

$$\begin{align*}d(\alpha_i,\alpha_k) < q - t \text{ and } d(\alpha_j,\alpha_k) < r-t. \end{align*}$$

Since the space $(M,d)$ is right-c.e., it is not hard to see that the set X is c.e., uniformly in $i,j,q,r$ . If $(k,t)\in X$ , then by using the triangle inequality, we can easily show that $B(\alpha _k,t)$ is a subset of $B(\alpha _i,q) \cap B(\alpha _j,r)$ .

Let x be an arbitrary point from $U = B(\alpha _i,q) \cap B(\alpha _j, r)$ . Choose positive rationals $\epsilon $ and $\delta $ such that $\epsilon < q - d(\alpha _i, x)$ and $\delta < r - d(\alpha _j, x)$ . Since U is open, we can find $k\in \omega $ and $t \in \mathbb {Q}^+$ such that $x\in B(\alpha _k, t) \subseteq U$ and $t < \min (\epsilon /2, \delta /2)$ . Then we have

$$\begin{align*}d(\alpha_i,\alpha_k) \leq d(\alpha_i,x) + d(\alpha_k,x) < (q - \epsilon) + t < q - \epsilon/2 < q - t. \end{align*}$$

Therefore, $(k,t)$ belongs to X, and the set X satisfies (1). Hence, $(M,\tau ,\beta ,\nu )$ is a computable topological space.

Notation 2.8. For a basic open B, write $B^c$ for the basic closed ball with the same centre and radius as B.

Proof $(1) \Rightarrow (2).$ Suppose $V = \bigcup _{i \in W} B_i$ . Note that (C2) implies that

$$ \begin{align*}f^{-1}(V) = \bigcup \{D \in X: (D, E) \in F \, \& \, \exists i \in W \, E \subseteq_{form} B_i\},\end{align*} $$

and thus the name of $f^{-1}(B_i)$ can be listed using only positive information about $W,$ with all possible uniformity.

$(2) \Rightarrow (3).$ Note that $B \in N^{f(x)}$ if and only if $f^{-1}(B)$ contains a basic open set in $N^x$ .

$(3) \Rightarrow (1).$ Define a collection F of pairs $(D,E)$ of (indices of) basic open sets in $X \times Y$ as follows. Fix a basic open E in Y. Enumerate all basic open D in X, and for each such D, enumerate all finite collections $D, A_1,\ldots , A_k$ of basic open sets (in X) such that $D \subseteq _{form} \cap _{i \leq k} A_i$ (meaning that D is formally contained in each $A_i$ ). Feed these finite collections to $\Phi $ and wait for some E to be enumerated in the output. When E is enumerated (if ever), put $(D,E)$ into F.

We claim that F defined above satisfies (C1) and (C2). We check (C1). If $(D,E) \in F$ then $f(D) \subseteq E$ . Indeed, suppose $d \in D$ . There exists a sequence $D, A_1, \ldots , A_k$ such that $\Phi ^{\{D, A_1,\ldots , A_k\}}$ enumerates E. Recall $D \subseteq _{form} \cap _i A_i$ implies $D \subseteq \cap _i A_i$ , thus for any $d \in D$ the sequence listed by $\Phi ^{N^d}$ will contain E, and therefore $f(D) \subseteq E$ . We now check (C2). Fix $x \in X$ and a basic open $E \ni f(x)$ . We must show that for some basic open $D \ni x$ , $(D, E) \in F$ . By assumption, $\Phi ^{N^x}$ enumerates $N^{f(x)}$ that contains E. Suppose E is listed with use $A_1, \ldots , A_k$ . Since the $A_i$ all contain x, there exists a basic open $D \ni x$ that is formally included into their intersection. Since the operator uses only positive information about its oracle, it will list E on input $\{D, A_1, \ldots , A_k\}$ as well, and thus $(D,E)$ will be enumerated into F.

Fact 2.14. An isolated point in a $\Pi ^0_1$ -class is computable.

Proof The implication $(2)\rightarrow (1)$ is obvious.

Assume $(1)$ . We prove $(2)$ . Take a finite collection $(B_i)$ of basic open sets and assume it is a cover. We must argue that eventually we will be able to effectively recognise that it is indeed a cover. The idea is that there exists an $\epsilon = 2^{-n}$ so small that every $\epsilon $ -cover of M is formally contained in this given cover. (This will be the Lebesgue number of the cover, in particular.) This will also be true for the $\epsilon $ -cover that will be given to us according to the definition of computable compactness. Since formal inclusion is c.e., we will be able to recognise that this formal inclusion has occurred. Noting that formal inclusion does imply set-theoretic inclusion, so if some $\epsilon $ cover is formally included in some other finite collection of basic open balls, then this other collection must also be a cover. Thus, if we succeed, it will show that $(B_i)$ is equivalent to saying that, for some n, every ball in the $2^{-n}$ -cover given to us by the definition of computable compactness (and indeed, any other $2^{-n}$ -cover) is formally included in one of the $B_i$ . This is, of course, a $\Sigma ^0_1$ -property.

It remains to prove that such an $\epsilon $ exists. We argue non-computably. Let $c_i$ be the center of $B_i$ , and $r_i$ be its the radius. Define for every i, a function $f_i (x) = r_i - d(x,c_i)$ if x is in the ball $B_i$ , and $0$ otherwise. This function is continuous. Now take the supremum of the finite family $(f_i)$ to define a new continuous g; $g(x)=\sup _i f_i(x)$ . If $(B_i)$ indeed was a cover, then the function g would be strictly positive, because each x is inside one of the $B_i$ .

Let v be its infimum that is achieved somewhere, by compactness. Take a rational $\epsilon =2^{-m}$ less than $v/2$ . Then for every point y, we have $ \epsilon < r_i - d(y, c_i)$ ; that is

$$ \begin{align*}d(y, c_i)+ \epsilon< r_i,\end{align*} $$

equivalently, $B(y, \epsilon ) \subset _{form} B_i$ . This inclusion will still hold if we replace $\epsilon $ with an even smaller $\epsilon '$ . Thus, in particular, every basic open $\epsilon '$ -ball is formally included in one of the $B_i$ . Consequently, $(1)$ implies $(2)$ .

Proof $(2) \rightarrow (1)$ . The supremums and infimums of computable functions are uniformly computable (see Proposition 3.5(1)), and so is for every computable f and g. The proof then proceeds by induction on the complexity of continuous formulae (with parameters).

1. (1) The distance function is computable by our assumption, and so are the constant functions $1$ and $0$ .

2. (2) If $f(x, \bar {y}), g(x, \bar {y})$ are computable, then so are , and $\dfrac {1}{2} f(x, \bar {y})$ (see Proposition 3.5(3)).

The latter also includes the case when there are no parameters, which means that the function is just the constant function (a computable real).

$(1) \rightarrow (2)$ . Let $(x_i)$ be the computable dense sequence. There is a formula of continuous logic saying that

which we can computably evaluate with precision $2^{-n}$ . If we discover that

$$ \begin{align*}U_n(x_0, ..., x_m) < 2^{-n},\end{align*} $$

then every point is at distance at most $2^{-n+1}$ from one of the $x_i$ , by the triangle inequality. Since we can effectively list such formulae, and for every n there exists an m for which the formula holds up to $2^{-n}$ (by compactness), we conclude that we can effectively produce at least one $2^{-n}$ -cover for every n.

Proof The open set $M \setminus B_i^c$ is c.e. Indeed, we just list all the basic open balls that are formally disjoint from $B_i^c$ via a standard argument.Footnote ⁷ Thus, the union of the complements, which is the complement of the intersection $B_i^c \cap B_j^c$ , is also c.e. open. It covers the space if, and only if, the intersection is empty. By computable compactness of M, this is c.e. The case of finitely many balls is similar.

Proof Obviously, $**$ -computably compactness implies computable compactness. To this end, we assume computable compactness of M. We will use the equivalence of computable compactness and $*$ -computable compactness throughout the rest of the proof without explicit reference.

We need to show that, for every $\epsilon>0$ , there exists a finite basic open $\epsilon $ -cover K of the space. Fix a finite $\epsilon /2$ -cover of the space by basic open balls, and replace each ball in the cover with an $\epsilon $ -ball with the same centre. Let S be this new $\epsilon $ -cover. Recall that $B^c$ denotes the basic closed ball with the same centre as B. For each basic open $B_1, \ldots , B_k \in S$ , (exactly) one of the possibilities is realized:

1. (a) $ \bigcap _{i \leq k} B_i^c = \emptyset $ , or

2. (b) $\bigcap _{i\leq k} B_i \neq \emptyset $ , or

3. (c) $ \bigcap _{i \leq k} B_i^c \neq \emptyset $ but $\bigcap _{i\leq k} B_i = \emptyset $ .

Note that there are only finitely many conditions like that in total.

If we shrink the radii of all $B \in S$ by a $\delta <\epsilon /2$ (but keep the same centres), then the conditions of the form $(a)$ will still hold, and the smaller balls will still cover the space because the $\epsilon $ -balls do. If $\delta $ is small enough, then the conditions of the form $(b)$ will still be satisfied, since there are only finitely many conditions like that involved.Footnote ⁸ The third alternative $(c)$ must be witnessed only by points y such that, for some $B_i = B(c,r)$ , $d(y,c) =r$ . This means that, after the shrinking by $\epsilon /2>\delta >0$ so small that the alternative $(b)$ still holds for each tuple of balls, we completely exclude the third alternative for the new cover.

This argument shows that such a cover exists. Since the conditions are c.e. by Lemma 3.11, it remains to search for a cover such that each finite collection of basic balls in the cover satisfies either $(a)$ or $(b)$ .

Proof of Theorem 3.21

Assume that, in M, we can compute $(\epsilon _n)_{n \in \omega }$ and the maximal number $D(n)$ corresponding to $\epsilon _n$ . In N, search for $D(n)$ -many special points that are $\epsilon _n$ -far from each other. The $\epsilon _n$ -balls around these points will give a finite cover of the space. We can slightly enlarge the radii to make sure that they are rational.

Proof of $\mathrm{(i)} \leftrightarrow \mathrm{(v)}$ in Theorem 1.1

The proof of Theorem 3.21 above essentially shows that $\mathrm{(v)}$ implies $\mathrm{(i)}$ . Thus, we assume computable compactness and prove $\mathrm{(v)}$ .

The reader perhaps wonders why we did not use $\epsilon _n = 2^{-n}$ in $\mathrm{(v)}$ . To clarify this subtlety, we shall attempt to show that, in a computably compact M, the following invariant $D(M,2^{-n})$ is uniformly computable in n:

The issue that we will face will clarify why we need to adjust our $\epsilon _n$ ’s. Also, it will be easy to modify this naive attempt and obtain a procedure that actually works for some $\epsilon _n \leq 2^{-n}$ .

We describe our attempt. Given $\bar {x} \in M^m$ , we can calculate $\inf _{i<j \leq m} d(x_i, x_j) $ and then

$$ \begin{align*}\sup_{\bar{x} \in M^m} \, \inf_{i<j \leq m} d(x_i, x_j).\end{align*} $$

If this supremum is $< 2^{-n}$ , then m is too large, i.e., $m>D(M,2^{-n})$ . Note that this is a c.e. event. On the other hand, by searching through all possible m-tuples we can bound the maximal number of such points from below. The issue is that the supremum could be exactly equal to $2^{-n}$ , so we may end up with a pair of integers $n_0, n_0+1$ each of which can potentially be equal to the invariant $D(M,2^{-n})$ ; here $n_0+1$ corresponds to the situation when there are $n_0+1$ points at distance exactly $2^{-n}$ from one another.

In this case we shall wait long enough so that any $(n_0+2)$ -tuple has at least one pair of points at distance $> 2^{-n}$ , and for some small $\xi _n < 2^{-n-1}$ there exist $n_0 +1$ points at distance $2^{-n}- \xi _n$ . Then

$$ \begin{align*}D(M,2^{-n}- \xi_n)= n_0+1.\end{align*} $$

This allows us to compute a computable sequence of rationals

$$ \begin{align*}\epsilon_n = 2^{-n}- \xi_n,\end{align*} $$

where clearly $\epsilon _n \leq 2^{-n}$ , such that $D(M,\epsilon _n)$ is a computable sequence of natural numbers. Indeed, for this $\epsilon _n$ there is an $n_0+1$ tuple $\bar {y}_n$ of points that satisfy the desired properties. We can assume these points are special.

Proof This is because $A \setminus f^{-1}(C) = f^{-1}(B\setminus C)$ is c.e. open. To see why, recall that f is computable if, and only if, it is effectively open. If a basic open D is enumerated into $B \setminus C$ , then we will list all pairs $(D', D")$ in the continuous name of f such that $D"$ is formally included in D. Since every $x \in D$ is contained in such a $D"$ (by surjectivity), this will give an enumeration of $f^{-1}(D)$ . Putting these enumerations together for all such D in $B \setminus C$ , we will list its preimage.

Fact 3.25 (Folklore).

Suppose $P = \{p\}$ is effectively closed singleton in a computably compact space X. Then the only point p of P is (uniformly) computable.Footnote ¹⁰

Proof Given n, wait for a basic open ball D of radius $2^{-n}$ and finitely many basic open $B_1, \ldots , B_n \in X \setminus P$ such that $D, B_1, \ldots , B_n$ cover X. Then $p \in D$ .

Proof of Lemma 3.27

Suppose C possesses such a computable sequence $(\alpha _i)_{i \in \mathbb {N}}$ . Then the density of the sequence in C implies that $B_i \cap C \neq \emptyset $ iff $\exists j \alpha _j \in B_i$ , which is a uniformly $\Sigma ^0_1$ statement.

Now suppose C is a computably enumerable closed subset of M. Our goal is to construct a uniformly computable (finite or infinite) sequence of points $(\alpha _i)_{i \in I}$ that is dense in C. The proof below does not have to be non-uniform, but for notational convenience we split it into two cases, namely, when C is finite or infinite.

If C is finite, then it clearly contains only computable points. To see why, assume it is not empty (in this case there is nothing to prove) and let x be any point in C. Take a ball small enough so that $\{x\}\in B\cap C$ . To get a $2^{-n}$ -approximation to x, wait for a basic open $B'$ of radius $< 2^{-n}$ so that $B'\cap C \neq \emptyset $ and additionally $B'$ is formally contained in B.

Without loss of generality, we assume C is infinite. We uniformly approximate a computable sequence by stages. Before we describe stage s, recall that two basic open balls U and W are formally s-disjoint if $r(U) + r(W) < d(cntr(U), cntr(W))$ and this can be seen after calculating the radii and the distance with precision $2^{-s}$ . Then U and W are formally disjoint if they are formally s-disjoint for some s.

At stage $0$ , search for a basic open ball $B_{0, 0}$ of radius ${<}1$ such that $B_{0,0} \cap C \neq \emptyset $ . If such a ball is never found then do nothing. If it is every found, go to the next stage.

At stage $s>1$ first check whether there exists a basic open ball with index $<s$ which is formally s-disjoint from $B_{0, s-1}, \ldots , B_{s-1, s-1}$ . If such a basic open B exists, then choose the first fund $B_{s,s} \subseteq _{form} B$ and $B_{i, s} \subseteq _{form} B_{i, s-1}$ , $i < s$ such that $B_{j, s} \cap C \neq \emptyset $ , the $B_{j, s}$ are pairwise formally disjoint and $r(B_{j, s}) < 2^{-s}$ , $j = 0, \ldots , s$ . Otherwise, if no such B exists, fix the first found pairwise formally disjoint $B_{0,s}, \ldots , B_{s,s}$ that intersect C, have radii $ < 2^{-s}$ , and such that $B_{i, s} \subseteq _{form} B_{i, s-1}$ for $i < s$ (note there is no further restriction on $B_{s,s}$ ). This ends the construction.

Let $\alpha _i$ be the unique point of the Polish space such that $\{\alpha _i\} = \bigcap _{j \geq i} B_{i,j}.$ Since the construction is uniform and the radii of balls are rapidly shrinking, the points $\alpha _i$ form a uniformly computable sequence. Since each of the $B_{i, j}$ ( $j =i, i+1, \ldots $ ) intersects C and C is closed, each $\alpha _i \in C$ . It remains to check that $(\alpha _i)_{i \in \mathbb {N}}$ is dense in C. Let $\overline {(\alpha _i)}_{i \in \mathbb {N}}$ be the completion of $(\alpha _i)_{i \in \mathbb {N}}$ .

Suppose $c \in C$ . We claim that $c \in \overline {(\alpha _i)}_{i \in \mathbb {N}}$ . Assume $c \notin \overline {(\alpha _i)}_{i \in \mathbb {N}}$ , and there is a ball U centred in c which is outside $\overline {(\alpha _i)}_{i \in \mathbb {N}}$ . There will be a basic open ball $B' \ni c$ of radius at most $2^{-n}$ and which is formally contained in U with precision $2^{-n}$ :

$$ \begin{align*}d(cntr(U), cntr(B')) + r(B') < r(U) +2^{-n}.\end{align*} $$

Then at every stage $s>n+4$ the balls $ B_{i,s-1}$ , $i = 0, \ldots , s-1$ will be formally s-disjoint from B, as will be readily witnessed by the metric. At some late enough stage $s'$ we will get a confirmation that $B \cap C \neq \emptyset $ . There exist only finitely many basic balls that have their index smaller than the index of B. Therefore, eventually B will be used to define $B_{t,t} \subseteq _{form} B$ , contradicting the assumption that $U \cap \overline {(\alpha _i)}_{i \in \mathbb {N}} = \emptyset .$

Proof Assume (1). It is clear that C is c.e. (by Lemma 3.27). To list its complement, fix $x \in M \setminus C$ . Let $\delta = \inf _{c \in C} d(x, c)$ . Then any $\delta /4$ -cover of C must be formally disjoint from any ball centred in y with $d(y,x)<\delta /4$ . For every n, fix a finite $2^{-n}$ -cover $\mathcal {K}$ of C. It follows that $M \setminus C$ is equal to the union of the (uniformly) effectively open sets $U_n$ , where

$$ \begin{align*}\{B: B \mbox{ basic open and } B \mbox{ is formally disjoint from every ball in }\mathcal{K}\}.\end{align*} $$

It follows that (2) holds.

Now assume (2). C is computably metrized by Lemma 3.27; let $(y_j)$ be the computable dense subsequence. Fix $\epsilon = 2^{-n}$ . We need to find an $\epsilon $ -cover of C by basic open balls.Footnote ¹¹ Regardless of whether the balls involved are basic or not, as long as their centres and radii are computable, the relation of formal containment remains c.e.

If a finite collection $\mathcal {K}$ of basic open (in C) balls formally contains a cover $\mathcal {K}'$ by basic open (in M) balls, then clearly $\mathcal {K}$ is a cover of C. We claim that this condition is also necessary (for $\mathcal {K}$ to cover C).

From the proof of Lemma 3.3 we know that, for a given cover $\mathcal {K}$ of C by (basic or not) open balls there is a small enough $\epsilon $ such that every $\epsilon $ -cover of C will be formally contained in at least one ball of $\mathcal {K}$ .

Take $\delta = \epsilon /4$ . Fix a finite $\delta $ -cover $\mathcal {K}'$ of C by balls that are centred in special points of M, not C. Every $B' \in \mathcal {K}'$ intersects C at some point x, and by the choice of $\delta $ , $d(x, cntr(B')) +\delta < \epsilon $ , thus $B(\epsilon , x) \supset _{form} B'$ . By transitivity of formal inclusion,Footnote ¹² we have that $B'$ must be formally contained in some ball in $\mathcal {K}$ .

By computable compactness of M and computability of C, we can produce at least one $\delta $ -cover $\mathcal {K}'$ of C by basic open balls of M, uniformly in $\delta $ . (To see why, replace every basic open ball in the c.e open name of $M \setminus C$ by the effective union of balls of radii at most $\delta $ that are formally contained in it. This gives a new c.e. enumeration of the complement of C but with balls of radii at most $\delta $ . Then take the c.e. collection of all basic $\delta $ -balls that intersect C. Together these sets of balls cover M. Initiate the combined enumeration of these two c.e. sets and wait until at some finite stage we discover that we have a cover of M.)

Since formal inclusion is c.e., this gives a procedure of listing covers of C by basic balls (in C).

Proof Lemma 3.20 says that each such $D^c$ is c.e. closed. If $x \in M \setminus D^c$ , then x is inside an open ball C that is formally disjoint from $D^c$ , and such balls can be computably enumeratedFootnote ¹⁴ . Thus, the c.e. union of such open balls formally disjoint from $D^c$ makes up the complement of $D^c$ .

Proof Let $(x_i)$ is a computable dense sequence in X. Then $(f(x_i))$ is dense in $f(X)$ . (Every $\alpha = \lim _j x_j$ for some subsequence $(x_j)$ and, by continuity, $f(\alpha ) = \lim _j f(x_j)$ , so $f(X)\subseteq cl(f(x_i))$ . Suppose $\xi \in cl(f(x_i))$ , say $\xi = \lim _j f(x_j)$ . By compactness, $(x_j)$ has a convergent subsequence $(x_{j_k})$ , so let $z = \lim _k x_{j_k} \in X$ be its limit. Then $f(z) = \lim _k f(x_{j_k}) = \lim _j f(x_j) = \xi $ .)

Given a cover $B_j$ of $f(X)$ by basic open (in $f(X)$ ) balls of radius $2^{-n}$ centred in $f(x_j)$ , calculate the c.e. names of each $B_j$ in Y and begin enumeration of $f^{-1}(C)$ for each such open basic C; note that it could be that some of these $f^{-1}(C)$ will be undefined. At some stage the preimages must cover the whole X. We can see which $B_j$ included the basic open (in Y) balls whose images were sufficient to cover X. This gives a way of producing at least one $2^{-n}$ -cover of $f(X)$ uniformly in n; now apply Lemma 3.3.

Proof Computable compactness of Y follows from the corollary above. Given a (not necessarily) special point $y \in Y$ , act computably relative to y. The set $Y \setminus \{y\}$ is effectively open relative to y. Indeed, for every $z \neq y$ there must exist formally disjoint basic open $B \ni y$ and $D \ni z$ . Thus, it is sufficient to list, effectively in y, all basic open balls formally disjoint from some ball in the name of y.

Since f is computable, $f^{-1}(Y \setminus \{y\})$ is effectively open relative to y. Since f is bijective, we have that

$$ \begin{align*}X \setminus f^{-1}(Y \setminus \{y\}) = \{f^{-1}(y)\}.\end{align*} $$

To list a basic open D into the name of $x =f^{-1}(y)$ , wait for finitely many basic open balls $B_1,\ldots , B_k$ in $X \setminus C_0 = f^{-1}(Y \setminus \{y\}) $ such that $D, B_1, \ldots , B_k$ cover X. Note that, for each $D\ni x$ , such a finite collection must exist by compactness. Since X is computably compact, the process described above is uniformly computable in y, and thus $f^{-1}$ is computable.

Proof Let N be the computably compact induced computable metrization of $f(C)$ that exists because of Lemma 3.31 and which is furthermore computably compact by Corollary 3.32 and Proposition 3.29. The map $f: C \rightarrow f(C)$ can be viewed as a computable map from C to the induced computable metrication on C, as follows. When $(B, C)$ is enumerated in the name of f, find a basic open ball D in $f(C)$ that formally contains C. The basic open balls in C are balls with centres that are special in $f(C)$ but are computable in M, but formal containment is still c.e. So we enumerate $(B, D)$ into the new name of f.

Another way to view this is to replace an $\epsilon $ -approximation $x_i \in M$ to $f(y)$ by a $2\epsilon $ -approximation $c_i \in C$ to $f(y)$ ; it must exist.

To compute $f^{-1}$ , apply the previous theorem.

Proof We know that $X \times Y$ is computably compact. Suppose f is computable. Then $f(x)=y$ is clearly a $\Pi ^0_1$ property.

Now assume $graph(f)$ is effectively closed. The subspace $\{x\} \times Y$ is effectively closed relative to x, thus $graph(f) \cap \{x\} \times Y $ is an effectively closed singleton relative to x. By Fact 3.25 relativized to x, given x we can compute $(x,f(x))$ and, thus $f(x)$ .

It remains to note that $graph(f)$ is actually c.e. closed for a computable f, because if a basic open B (in $X \times Y$ ) intersects the graph then we will eventually recognise it.

Proof Note that H being a (computable) product of computably compact spaces is itself computably compact by Proposition 3.5.

If the space has a computably compact presentation, then its image under the canonical embedding will also be computably compact (Proposition 3.29 and Lemma 3.31), and thus the image will be computable closed by Corollary 3.31. On the other hand, if a closed subset of H homeomorphic to the space is a computable closed subset, then it gives a computably compact homeomorphic presentation of the space by Proposition 3.29 because H, being a (computable) product of computably compact spaces, is itself computably compact.

Proof We can list the basic open balls $B(x_i, r)$ for which $D(x_i)>r$ , and they must cover $M \setminus C$ . But we can also list the balls $B(x_j, q)$ such that $D(x_i) <r$ , and these are exactly the basic open balls that intersect C.

Fact 3.39 (Folklore).

For a c.e. closed subset C of H, C is computable iff C is a computable point in $\mathcal {C}(H)$ .

Proof If C is computable then it is computably compact by Proposition 3.29. If $(B_i)$ is a finite $2^{-n}$ cover of C and $x_i$ is a special point in $B_i$ , then every point of C is at most $2^{-n+1}$ -far from one of the $x_i$ .

On the other hand, assume $c \in B_i \cap C$ ; indeed c is contained in $B_i$ together with an $\epsilon $ -ball. Thus, there is $\epsilon $ so small that any D that is $\epsilon $ -close to C in $\mathcal {C}(H)$ contains a special point in $B_i$ . This will be eventually recognised, and thus such $B_i$ can be computably enumerated, making C c.e. closed.

The first proof

Fix a fully $\cap $ -decidable system of covers $(K_n)$ of M that exists by Theorem 3.3. By Remark 3.18, we can use finite basic computable closed covers throughout.

We follow the standard classical topological proof very closely, and we use $\cap $ -decidability of $\bigcup _{n} K_n$ throughout. Suppose $K_0 = \{D_1, \ldots , D_{k}\}$ , and fix $D_i \in K_0$ . Then $K_1$ contains finitely many balls that cover $D_i$ ; denote them $D_{i,j}$ . These can be computed and indeed, these $D_{i,j}$ are exactly the closed balls in $K_1$ that intersect $D_i$ , because the rest are disjoint from it. Define $\hat {D}_{\langle i,j\rangle } = D_i \cap D_{i, j}$ .

We can computably proceed by recursion and define $\hat {D}_{\sigma }$ , which is a finite non-empty intersection of basic closed balls, where $\sigma $ ranges over a computably branching tree T with no dead ends (observe that $\sigma \neq \tau $ does not necessarily imply that $\hat {D}_\sigma \neq \hat {D}_\tau $ ). Equip the set of all infinite paths through $[T]$ with the standard (longest common prefix) ultrametric; then $[T]$ is computably metrized space in which the dense sequence is given by $\sigma 1^\omega $ , $\sigma \in T$ . We identify $\sigma $ with the basic clopen ball of $[T]$ consisting of all strings with prefix $\sigma $ .

Also recall that, by the construction of $(K_n)$ , without loss of generality we can assume that basic open balls intersect whenever the respective closed balls intersect, and thus we can always calculate a special point $x_\sigma $ in each $\hat {D}_\sigma $ (see Remark 3.18). We could view $x_\sigma $ to be an $\epsilon $ -approximation to any path in $[T]$ extending $\sigma $ , where $\epsilon =2^{-n+1}$ for the largest n such that a ball from $K_n$ is mentioned in $\hat {D}_\sigma $ . So we define $f(\sigma ) = x_{\sigma }$ with precision $2^{-n+1}$ .

For an infinite path $\xi \in [T]$ , $\hat {D}_{\xi \upharpoonright n} \subseteq \hat {D}_{\xi \upharpoonright m}$ whenever $ m\leq n $ are prefixes of $\xi $ , and since the diameter of $\hat {D}_{\xi \upharpoonright n}$ is at most $2^{-n+1}$ and it is non-empty, we conclude that

$$ \begin{align*}\bigcap_{n} \hat{D}_{\xi \upharpoonright n} = \{\alpha\}\end{align*} $$

for some $\alpha \in M$ . So we set $f(\xi ) = \alpha \in M$ . It is routine to show that the procedure above defines a computable and surjective $f: [T] \rightarrow M$ . It is not difficult to see that $[T]$ is a computable image of $2^{\omega }$ ; as we promised in the preliminaries, we include a proof of this well-known fact.

In combination with f defined above, this gives a computable surjection of $2^{\omega }$ onto M.

The second proof

Recall that we assumed that M is computably compact. Without loss of generality, we can assume it is a computable closed subset of H (see Remark 3.37). Recall that there is a computable map from $2^{\omega }$ onto $[0,1]$ ; for instance, map every infinite sequence $\sigma $ in $2^{\omega }$ into the binary expansion $\sum _{i \in \omega } 2^{-i-1} \sigma (i)$ . Also, the famous Hilbert’s curve computably maps $[0,1]$ onto $H = [0,1]^{\omega }$ (see, e.g., [Reference Sagan130] for a primitive recursive version due to Schoenberg). (See [Reference Bagaviev, Batyrshin, Bazhenov, Bushtets, Dorzhieva, Kornev, Melnikov, Ng and Koh5] for a detailed explanation of computability of this particular construction. We also cite [Reference Couch, Daniel and McNicholl32] for more about computability of space-filling curves.) This gives a computable surjective $f: 2^{\omega } \rightarrow H \supseteq M$ , and we know from Lemma 3.24 that $f^{-1}(M)$ is a $\Pi ^0_1$ -class which, unfortunately, does not have to be computable in general. Recall that, in the beginning of the subsection, we assumed that M (thus, P) is non-empty. If it were a computable $\Pi ^0_1$ class, then we would be able to computably surjectively map $2^{\omega }$ onto it by Claim 1. But first, we must be in the position to apply the claim. The lemma below helps.

Furthermore, $\tilde {f}$ can be chosen so that $\tilde {f}$ agrees with f on $f^{-1}(P)$ and $\tilde {f}(2^{\omega } \setminus C) \subseteq K \setminus P$ ; we will also include the verification of these properties in our proof below. Also, we will see that the simultaneous construction of $\tilde {f}$ and C is uniform. The proof below is somewhat informal: after all, we have already proven the theorem. We hope that the elementary formal details that are missing should be easy to reconstruct.

Proof This is done as follows. We use a c.e. formal open name of f and computable compactness of K and computability of P throughout. In particular, we will use that f is uniformly computably continuous. This is done by listing $2^{-n}$ -covers $K_n$ of K and using the formal name of f to find a cover S of $2^{\omega }$ such that, for every $\sigma \in S$ there is a $D \in K_n$ so that $(\sigma , D)$ is in the name.

Since P is computable and thus is a computably compact subspace (see Proposition 3.29), we can assume that we know which open sets in $K_n$ intersect P and which do not. To see why, create two lists: one is the list of all finite covers of P, and the second list includes finite collections of basic open balls in K that are formally disjoint from some cover from the first list. For every $\epsilon $ , there is a finite $\epsilon $ -cover $K' \cup K"$ of K, where $K'$ is from the first list and $K"$ is from the second list.Footnote ¹⁷ The finite set of balls $K"$ can be empty, but $K'$ is never empty. If $\epsilon = 2^{-n}$ then we can set $K_n = K' \cup K"$ .

The construction of C and $\tilde {f}$ proceeds as follows.

Suppose at a stage we see that, for some basic open set $\sigma $ in $2^{\omega }$ , $f(\sigma ) \subseteq B$ for some $\epsilon $ -ball B such that $B \cap P = \emptyset $ . Then declare $\sigma $ outside of the closed set C that we build, and let (the name of) $\tilde {f}$ copy (the name of) f.

Now suppose, using the same notation and premises, $B \cap P \neq \emptyset $ . It still possible that $f(\sigma ) \cap P = \emptyset $ (because all we know is that $f(\sigma ) \subseteq B$ , so it can miss P), but we do not know that yet. In this case declare that $\sigma $ intersects C and proceed to the next stage. (As we made sure above, for any such B and $\sigma $ that we use at the stage we can decide whether B intersects P or not.)

At the next stage we will find a refined cover of K and $2^{\omega }$ and, in particular, of $\sigma $ . If at least one such $\sigma '$ refining (extending) $\sigma $ has the property that $f(\sigma ') \subset B'$ , where $B' \cap P \neq \emptyset $ (here $B'$ is taken from the more refined cover of K and $B'$ ) then we let (the name of) $\tilde {f}$ copy (the name of) f.

Of course, it is entirely possible that we discover that for all of the finitely many extensions $\sigma '$ of $\sigma $ , the respective ball $B'$ (such that $f(\sigma ') \subseteq B'$ ) does not intersect P. However, we have already declared that $\sigma \cap C \neq \emptyset $ .

In this case, go to the previous stage and find a computable point $x \in B \cap P$ (that can be uniformly extracted from the proof of Lemma 3.27) and declare

$$ \begin{align*}\tilde{f}(\xi) = x\end{align*} $$

for every $\xi \in B$ . In this case we also say that the clopen ball B is declared artificially in C. (This is also the only case when $\tilde {f}$ does not agree with f; we set f and $\tilde {f}$ equal in all other cases.)

By construction, $\tilde {f}$ is well-defined on all of $2^{\omega }$ and is computable. We need to argue that it additionally has the properties claimed in the statement of the lemma.

If $\xi \in B$ then clearly $\tilde {f}(\xi ) \in P$ . If $\xi \in f^{-1}(P)$ then at no stage it can be declared in a ball which is out or artificially in C, so it must be that $\tilde {f}$ and f agree on $\xi $ .

If $\xi $ is such that it is never in any ball that is artificially in or out, it means that for every $2^{-n}$ there must be a point $\alpha _n \in P$ that is $2^{-n}$ -close to $\tilde {f}(\xi )$ , and we see that $\tilde {\xi } = \lim _n \alpha _n \in P$ because P is compact and thus closed.

Finally, $f(2^{\omega } \setminus C) \subseteq K \setminus P$ follows from the fact that if a basic clopen $\sigma $ is declared out of C, we let $\tilde {f}$ follow f in its definition within $\sigma $ .

It follows that M is a computable surjective image of a computable $\Pi ^0_1$ -class. It remains to map $2^{\omega }$ to such class surjectively using Claim 1. This finishes the second proof.

Sketch.

Define a computable function by stages on the Cantor set (linear elsewhere) so that, while an interval is not yet declared out of C, f keeps getting closer to $0$ , say, from below. If the interval is declared out, freeze the function at this interval and, thus, keep it away from zero. It shall approach zero but only at points that correspond to paths in C.

Proof

1. (1) Fix an uncountable $\Pi ^0_1$ -class P without computable points and apply Theorem 4.1; for example, some $\Pi _1^0$ class consisting of only Martin-Löf random reals (e.g., [Reference Downey and Hirschfeldt36]).

2. (2) By Theorem 3.40, there is a computable surjection $g: 2^{\omega } \rightarrow [0,1]$ , and by Lemma 3.41 the pre-image of the effectively closed $X = f^{-1}Z_f$ is also effectively closed in $2^\omega $ , that is, it is a $\Pi ^0_1$ class. Using this argument, we easily see that if $Z_f$ is nonempty then it has a low point, by the Low Basis Theorem.

3. (3) This is because isolated members of effectively closed sets are computable.

4. (4) If $Z_f$ is infinite and countable, and thus it has no perfect kernel. In particular, there must be infinitely many isolated points; apply (3).

Proof In view of Theorem 3.35, it is sufficient to take $C_1 = graph(f)$ for f in (2) of the theorem above, and $C_2 = [0,1]$ (the “x-axis”) which is clearly computable too.

Sketch.

This is somewhat similar to the proof of Theorem 4.2(1). Fix a $\Pi ^0_1$ class without computable members and define f on the Cantor set (and linearly elsewhere), but this time make its value approach a left-c.e. non-computable real $\alpha $ . Note that, if a point x is computable, then it has to be either on a segment of the Cantor set that was declared out at some stage, or it is on a linear segment connecting two such points that have been declared out. Observe also that each linear segment connects computable points. In either case, we can wait for the point to be listed in the effectively open complement of the homeomorphic image of the $\Pi ^0_1$ class, go to the stage of the construction where that happened, and compute the index of the image.