Observation 3.6. The formula $\theta $ in the previous theorem can be seen as defining a tree with respect to X. So over $\mathbf {ACA}_0$ being an ill founded tree is a universal $\Sigma ^1_1$ formula and being a well-founded tree is a universal $\Pi ^1_1$ formula.

Observation 4.2. One may wonder if all countable spaces are second countable. This is not the case. A simple procedure to construct countable but not second countable spaces is to take any downwards directed family $\mathcal {F}\subseteq\mathcal {P}(\mathbb {N}_{>0})$ that does not have a countable cofinal sequence, and define on $\mathbb {N}$ the topology in which every $n\in \mathbb {N}_{>0}$ is isolated and the basic open neighborhoods of $0$ are the ones of the form $\{0\}\cup S$ where $S\in \mathcal {F}$ .

To find such a directed subset of $\mathcal {P}(\mathbb {N}_{>0})$ one can use the fact that there exists a family $\mathcal {B}$ of size $2^{\aleph _0}$ of infinite subsets of $\mathbb {N}_{>0}$ such that the intersection of any two members is finite. Let $\mathcal {F}$ be the family of all finite intersections of the complements of elements in $\mathcal {B}$ . We have that $\mathcal {F}$ is downwards directed, but it does not have a cofinal sequence.

Observation 4.5. The coded open collections will be closed under finite intersections and countable unions. Given $A_0$ and $A_1$ open collections with codes $f_0$ and $f_1$ , the partial function $h:\mathbb {N}\times \mathbb {N}\times X\rightarrow \mathbb {N}$ given by $h(x,i,j)=k(x,f_0(i),f_1(j))$ will code $A_0\cap A_1$ . Given a sequence of coded open collections $(A_i)_{i\in \mathbb {N}}$ with codes $(f_i)_{i\in \mathbb {N}}$ the partial function $F:\mathbb {N}\times \mathbb {N}\rightarrow \mathbb {N}$ given by $F(i,n)=f_i(n)$ is the code for the open collection $\bigcup _{i\in \mathbb {N}} A_i$ . Similarly, coded closed collections are closed under finite unions and countable intersections.

Observation 4.8. Since over $\mathbf {ACA}_0$ every CSC space has a function k we will often omit it from the definition of CSC space. Similarly, we won’t distinguish homeomorphisms from effective homeomorphisms over $\mathbf {ACA}_0$ .

Proof. Informally, we will construct a real a that is the diagonal of the reals $(q\cdot d(x,y))_{q\in \mathbb {Q}_{>0},x,y\in X}$ . Let $(t_n)_{n\in \mathbb {N}}$ be an enumeration of all triples $(x,y,q)\in X\times X\times \mathbb {Q}_{>0}$ . We define a real $a=(a_i)_{i\in \mathbb {N}}$ recursively as follows. Set $a_0=\frac {1}{2}$ . At step $i+1$ let $t_i=(x,y,q)$ and let $q\cdot d(x,y)=(b_j)_{j\in \mathbb {N}}$ . If $|b_{2i+2}-a_{2i}|<2^{-2i-1}$ then:

1. (1) $b_{2i+2}\geq a_{2i}$ , in which case we define $a_{2i+1}=a_{2i+2}=b_{2i+2}- 2^{-2i-1}$ .

2. (2) $b_{2i+2}< a_{2i}$ , in which case we define $a_{2i+1}=a_{2i+2}=b_{2i+2}+ 2^{-2i-1}$ .

If instead $|b_{i+1}-a_i|\geq 2^{-2i-1}$ then we define $a_{i+1}=a_i$ . By induction, the triangle inequality, and using the fact that

$$ \begin{align*} \forall i \in\mathbb{N}\;(|a_{2i}-a_{2i+1}|=|a_{2i}-a_{2i+1}|\leq 2^{-2i-1}) \end{align*} $$

we have that $(a_i)_{i\in \mathbb {N}}$ is a real number. Since $a=(a_{i})_{i\in \mathbb {N}}$ is recursive relative to $(X,d)$ it exists by $\Delta ^0_1$ comprehension. By construction of a, we have that for all $x,y\in X$ , $q\in \mathbb {Q}\, a\neq _{\mathbb {R}} q\cdot d(x,y)$ . In particular we have that for all $q\in \mathbb {Q}$ and all $x\in X$ the ball $B(x,q\cdot a)$ is clopen. Since the clopen balls are uniformly $\Delta ^0_1$ definable relative to $(X,d)$ , we have that the collection $(B(x,q\cdot a))_{q\in \mathbb {Q}_{>0},x\in X}$ exists by $\Delta ^0_1$ comprehension. Let $(q_i)_{i\in \mathbb {N}}$ be an enumeration of $\mathbb {Q}_{>0}$ and let $k:X\times (\mathbb {Q}_{>0} \times X)^2\rightarrow \mathbb {Q}_{>0} \times X$ be the function given by

$$ \begin{align*} k(x,(y,p),(z,r)) =(x,q_j), \end{align*} $$

where

$$ \begin{align*} j=\min \{ i\in \mathbb{N} : p\neq q_i\neq r\wedge d(x,y) <_{\mathbb{R}} (p-q_i)\cdot a \wedge d(x,z) <_{\mathbb{R}} (r-q_i)\cdot a\} \end{align*} $$

or rather j is the least such that $q_j\cdot a <_{\mathbb {R}} p\cdot a-d(x,y)$ and $q_j\cdot a <_{\mathbb {R}} r\cdot a-d(x,z)$ . By construction of a, we also have that

$$ \begin{align*} j=\min \{ i\in \mathbb{N} : p\neq q_i\neq r\wedge d(x,y) \leq_{\mathbb{R}} (p-q_i)\cdot a \wedge d(x,z) \leq_{\mathbb{R}} (r-q_i)\cdot a\}. \end{align*} $$

So j can be searched effectively. By the triangle inequality, we have

$$ \begin{align*} B(x,q_j\cdot a)\subseteq B(y,p\cdot a)\cap B(z,r\cdot a). \end{align*} $$

By construction of a, we have that k is recursive and so it exists by $\Delta ^0_1$ comprehension.

Observation 4.14. If L is well-ordered, its order topology coincides with its upper limit topology since $]a,b]\ {=}\ ]a, b+1[$ where $b+1$ is either the successor of b or $+\infty $ if b is maximal. However, we will see later that over $\mathbf {RCA}_0$ , the two topologies are not, in general, effectively homeomorphic.

Observation 4.15. Let $(L,<_L)$ be a linear order, and let $S\subseteq L$ be a set that is downwards closed with respect to $<_L$ then S is open in the upper limit topology since $S=\bigcup _{s\in S}\;]-\infty ,s]$ .

Proof. Let $f:(X,(U_i)_{i\in \mathbb {N}},k)\rightarrow (Y,(V_i)_{i\in \mathbb {N}},k')$ be an effectively continuous injection and let $H^Y_0$ and $H^Y_1$ witness that $(Y,(V_i)_{i\in \mathbb {N}},k')$ is $eT_2$ . Let $v:X\times \mathbb {N}\rightarrow \mathbb {N}$ verify that f is continuous, meaning that for all $x\in X$ and $i\in \mathbb {N}$ such that $f(x)\in V_i$ we have that

$$ \begin{align*} x\in U_{v(x,i)}\subseteq f^{-1}(V_i). \end{align*} $$

For all $x,y\in X$ we define

$$ \begin{align*} H^X_0(x,y)=v(x,(H^Y_0(f(x),f(y))) \end{align*} $$

$$ \begin{align*} H^X_1(x,y)=v(y,(H^Y_1(f(x),f(y))). \end{align*} $$

The functions $H^X_0$ and $H^X_1$ exist by $\Delta ^0_1$ comprehension and witness that X is $eT_2$ .

Proof. Assume there is an injective function $f:\mathbb {N}\rightarrow \mathbb {N}$ whose range does not exist as a set. Let $<_f$ be the order on $\mathbb {N}\cup \{\infty \}$ given by:

• • $\forall n\in \mathbb {N}\; n<_f\infty $ .

• • $\forall n,m\in \mathbb {N}\; n<_f m\leftrightarrow f(n)<f(m)$ .

Seeking a contradiction, assume that $(\mathbb {N}\cup \{\infty \},<_f)$ with the order topology is $eT_2$ and let $H_0$ and $H_1$ be the functions that witness it. We have that n is the $<_f$ successor of m if and only if $m<_f n$ and $H_0(m,n)=(y,n)$ for some $y<_f n$ and $H_1(m,n)=(m,x)$ for some $x>_f m$ . Since the image of f is unbounded, every $n\in \mathbb {N}$ has an $<_f$ successor and the successor function exists since it is computable relative to $H_0$ and $H_1$ . We can therefore define recursively g where $g(0)$ the $<_f$ -minimal element and $g(n+1)$ is the $<_f$ -successor of $g(n)$ . So $f\circ g$ is strictly increasing and has the same image as f, which implies that $\text {rng}(f)$ exists, contradicting our initial assumption.

Observation 5.8. The space constructed in the proof above is a well-order with the order topology. While the upper limit topology and the order topology coincide on a well-order, they may not be effectively homeomorphic. This is because the upper limit topology is always $eT_2$ , and arithmetic comprehension is equivalent to the order topology of any well-order is $eT_2$ . However, it is straightforward to show that over $\mathbf {RCA}_0$ , the identity on a well-order is an effective homeomorphism between the upper limit topology and the order topology if and only if the successor function for the order exists.

Proof. $(1\rightarrow 2)$ follows from the fact that the successor function on a well-order is arithmetically definable.

$(2\rightarrow 1)$ Let $f:\mathbb {N}\rightarrow \mathbb {N}$ be an injection, $(X,<_f)$ be as in the proof of Theorem 5.7, and let $\infty _X$ denote the maximal element of X. We show that $\text {rng}(f)$ exists, which by Proposition 3.2 implies arithmetic comprehension. By assumption, X with the upper limit topology is effectively homeomorphic to some linear order $(L,<_L)$ with the order topology. Since X is $eT_2$ , we also have that L, with the order topology, is $eT_2$ . As in the proof of Theorem 5.7, we have that the successor and predecessor functions for L will exist. Since X has one limit point and doesn’t contain an infinite closed discrete subspace, we have by induction that L will be order isomorphic to $\mathbb {N}+n$ or its dual for some $n\in \mathbb {N}$ or to $\mathbb {N}+1+\mathbb {N}^*$ . The order topology of all these orders is effectively homeomorphic to that of $\mathbb {N}+1$ so without loss of generality we may assume $L=\mathbb {N}+1$ . Let $g:X\rightarrow \mathbb {N}+1$ be an effective homeomorphism and let $v$ be the function which witnesses that g is effectively open. Let $\infty _{\mathbb {N}}$ denote the maximal element of $\mathbb {N}+1$ , we have that

$$ \begin{align*} \forall n\in\mathbb{N}\;(]v(\infty_X,n),\infty_{\mathbb{N}}]_{<_{\mathbb{N}}}\subseteq g(\, ] n,\infty_X ]_{<_f})), \end{align*} $$

which implies that $\forall n\; \forall m\;(m\leq _f n\rightarrow g^{-1}(m)\leq _{\mathbb {N}} v(\infty _X,n))$ . So in particular we have that n is the $<_f$ successor of m if and only if

$$ \begin{align*} \forall y\leq_{\mathbb{N}} v(\infty_X,n)\;(g^{-1}(y)\leq_{\mathbb{N}} m\vee g^{-1}(y)\geq_{\mathbb{N}} n). \end{align*} $$

The $<_f$ successor function is $\Delta ^0_1$ definable, so it exists by $\Delta ^0_1$ comprehension. Using the successor function as in the proof of Theorem 5.7, we get that the range of f exists.

Observation 5.12. All $T_3$ spaces are $T_2$ .

Observation 5.15. Subspaces of regular CSC spaces are regular and subspaces of uniformly effectively regular CSC space are uniformly effectively regular.

Proof. Let $f:X\rightarrow Y$ be an effective embedding and $v_0$ verify that f is continuous and $v_1$ verify that f is open with respect to its image (see Definition 4.7). Let $R^Y_0, R^Y_1$ be the functions given by uniform effective regularity of Y. We define

$$ \begin{align*} R^X_0(x,i)=v_0(x,R^Y_0(f(x),v_1(f(x),i))). \end{align*} $$

We have that

$$ \begin{align*} x\in U_{R^X_0(x,i)}\subseteq f^{-1}(V_{R^Y_0(f(x),v_1(f(x),i))})\subseteq U_i. \end{align*} $$

We define for $y\notin U_i$

$$ \begin{align*} R^X_1(x,i,y)=v_0(y,R^Y_1(f(x),v_1(f(x),i),f(y))). \end{align*} $$

Since $y\notin U_i$ we have $f(y)\notin V_{v_1(f(x),i)}$ so $R^Y_1(f(x),v(f(x),i),f(y))$ is well defined. By definition of $R^Y_1$ we have

$$ \begin{align*} f(y)\in V_{R^Y_1(f(x),v_1(f(x),i),f(y))}\subseteq Y\setminus V_{R^Y_0(f(x),v_1(f(x),i))}\subseteq Y\setminus f(U_{R^X_0(x,i)}). \end{align*} $$

In particular

$$ \begin{align*} y\in U_{R^X_1(x,i,y)}\subseteq f^{-1}(V_{R^Y_1(f(x),v_1(f(x),i),f(y))})\subseteq f^{-1}(Y\setminus f(U_{R^X_0(x,i)}))=X\setminus U_{R^X_0(x,i).} \end{align*} $$

This means that the functions $R^X_0$ and $R^X_1$ have the desired properties, and so X is uniformly effectively regular.

Proof. Let $(X,(U_i)_{i\in \mathbb {N}},k)$ be a $uT_3$ CSC space and let $x,y\in X$ be distinct points. Since X is $T_0$ , let i be the least number such that $x\in U_i\leftrightarrow y\notin U_i$ . If $x\in U_i$ then we let $H_0(x,y)=R_0(x,i)$ and $H_1(x,y)=R_1(x,i,y)$ otherwise if $y\in U_i$ we set $H_0(x,y)=R_1(x,i,y)$ and $H_1(x,y)=R_0(x,i)$ .

Proof. Follows from Theorem 5.7 and Proposition 5.

Proof. Let $(I_e)_{e\in \mathbb {N}}$ be the standard indexing of the open intervals of $(L,<_L)$ and $H_0$ , $H_1$ be the functions that witness that the order topology on L is $eT_2$ . Given $y<_L x<_L z$ we define

$$ \begin{align*} R_0(x,(y,z))=k(x,H_0(x,y),H_0(x,z)) \end{align*} $$

and

$$ \begin{align*} R_1(w,x,(y,z)) &=\begin{cases} &\!\!\!\!\!\!\!\!\text{ index for: } I_{H_1(x,y)}\cup\ ]-\infty,y] \text{ if } w\leq_L y \\ &\!\!\!\!\!\!\!\!\text{ index for: } I_{H_1(x,z)}\cup[z,+\infty [ \text{ if } z\leq_L w. \end{cases} \end{align*} $$

The functions $R_0$ and $R_1$ witness that the order topology on L is $uT_3$ .

Proof. We show that arithmetic comprehension proves that all regular CSC spaces are uniformly regular. Given a regular CSC space $(X,(U_i)_{i\in \mathbb {N}},k)$ , for each $(x,j)\in X\times \mathbb {N}$ such that $x\in U_j$ we define

$$ \begin{align*} R_0(x,j)=\min\{s\in\mathbb{N}: x\in U_s\subseteq \overline{U_s}\subseteq U_j\} \end{align*} $$

and for all $y\in X\setminus U_j$ we define

$$ \begin{align*} R_1(x,j,y)=\min\{s\in \mathbb{N}: y\in U_s\wedge U_s\cap U_{R_0(x,j)}=\emptyset\}. \end{align*} $$

The $R_0$ and $R_1$ have the desired properties and are arithmetically definable. So $\mathbf {ACA}_0$ proves that every regular CSC space is uniformly regular.

We show the converse. Let A be a set. We show that the Turing jump of A exists, which will imply arithmetic comprehension. We define

$$ \begin{align*} X^e=\{-\infty\}\cup((\mathbb{Z}\cup\{+\infty\})\times\{t\in\mathbb{N}:\Phi^A_e(e){\downarrow}_{\leq t}\}). \end{align*} $$

We define on $X^e$ the following topology:

$$ \begin{align*} U_0=\{-\infty\}\cup (\mathbb{Z}\times\{t\in\mathbb{N}:\Phi^A_e(e){\downarrow}_{\leq t}\}). \end{align*} $$

For all $(m,n)\in (\mathbb {Z}\times \{t\in \mathbb {N}:\Phi ^A_e(e){\downarrow }_{\leq t}\})$ we define

$$ \begin{align*} U_{4(m,n)+1}=\{(m,n)\} \end{align*} $$

$$ \begin{align*} U_{4(m,n)+2}=\{(+\infty,n)\}\cup\{(l,n):l\geq m\} \end{align*} $$

$$ \begin{align*} U_{4(m,n)+3}=X^e\setminus\{(l,s):(s=+\infty)\vee(l\geq m\wedge s\leq n)\} \end{align*} $$

$$ \begin{align*} U_{4(m,n)+4}=\begin{cases} X^e \quad \quad\quad \text{ if } \quad \quad \quad \neg\Phi^A_e(e){\downarrow}_{\leq n}\\ \{-\infty\}\cup\{(l,s):l\leq m \wedge \Phi^A_e(e){\downarrow}_{\leq s}\}\quad \text{ if }\quad \Phi^A_e(e){\downarrow}_{\leq n}. \end{cases} \end{align*} $$

We define $H_0$ and $H_1$ on $X^e$ by:

• • $H_0(-\infty ,(m,n))=4(m,n)+3\;$ and $\;H_1(-\infty ,(m,n))=4(m,n)+1.$

• • $H_0(-\infty ,(+\infty ,n))=4(m,n)+3\;$ and $\;H_1(-\infty ,(+\infty ,n))=4(m,n)+2.$

• • $H_0((+\infty ,l),(+\infty ,n))=4(0,l)+2\;$ and $\;H_1((+\infty ,l),(+\infty ,n))= 4(0,n)+2.$

• • $H_0((+\infty ,l),(m,n))=4(m+1,l)+2\;$ and $\;H_1((+\infty ,l),(m,n))= 4(m,n)+1.$

• • $H_0((p,l),(m,n))=4(p,l)+1\;$ and $\;H_1((p,l),(m,n))=4(m,n)+1.$

Given $a=4(i,j)+r$ and $b=4(p,l)+s$ let $c = \max \{|i|,j, |p|,l\}+1$ and define:

• • $k((m,n),a,b)=4(m,n)+1.$

• • $k((\infty ,n),a,b)= 4(c,n)+2.$

• • $k(-\infty ,a,b)=\begin {cases}4(-c,c)+4 \text { if } \Phi ^A_e(e){\downarrow }_{\leq c}\\ 4(-c,c)+3 \text { otherwise.} \end {cases}$

It is straightforward to show that the functions $H_0$ , $H_1$ , and k have the desired properties and that $X^e$ is an $eT_2$ CSC space. To show that $X^e$ is regular, we first observe that every basic open set, except perhaps $U_0$ and $U_{4(m,n)+3}$ , is clopen and $-\infty $ is the only non-isolated point of $U_0$ and $U_{4(m,n)+3}$ . If $\Phi ^A_e(e){\uparrow } $ then $X^e= \{-\infty \}$ otherwise let t be the least number such that $\Phi ^A_e(e){\downarrow }_{\leq t}$ then the sequence $(U_{4(-n,t)+4})_{n\in \mathbb {N}}$ is a basis of clopen neighborhoods of $-\infty $ . So every point of $X^e$ has a clopen neighborhood, so $X^e$ is regular.

The CSC space $X=\coprod _{e\in \mathbb {N}} X^e$ will be regular and $eT_2$ so by assumption X is also $uT_3$ . Let $R_0$ and $R_1$ be functions that witness the uniform regularity of X. For $e\in \mathbb {N}$ we have $\Phi ^A_e(e){\downarrow }_{\leq n}$ if and only if for all $m\in \mathbb {N}\, U_{4(m,n)+4}\subseteq U_0$ in the space $X^e$ . For all e there is some $m\in \mathbb {N}$ such that $R_0((e,-\infty ),(e,0))=(e,4(m,n)+4)$ , where by $(e,-\infty )$ we mean $-\infty $ of the space $X^e$ , and $(e,0)$ is the index for $U_0$ in $X^e$ . So by construction of $X^e$ we have $\Phi ^A_e(e){\downarrow }_{\leq n}$ if for some $m\leq R_0((e,-\infty ),(e,0))$ we have $R_0((e,-\infty ),(e,0))=(e,4(m,n)+4)$ . In particular, $A'\leq _T R_0$ , which means the Turing jump of A exists by $\Delta ^0_1$ comprehension.

Observation 5.21. We observe that the space in the previous proof is scattered. So, arithmetic comprehension is equivalent to all $T_3$ scattered spaces being $uT_3$ . A CSC space is scattered if every subspace has an isolated point, we will look at $T_3$ scattered CSC space in the next part. We also point out that the space in the previous proof is effectively regular in the sense of Dorais (Definition 5.13). This shows that effective regularity is a strictly weaker property than uniform regularity over $\mathbf {RCA}_0$ .

Proof. Fix a set $A\subseteq \mathbb {N}$ , we show that $A'$ exists. Let $(X,(U_i)_{i\in \mathbb {N}},k)$ be as in the proof of Proposition 5.20. Let $(L,<_L)$ be a linear order, $f:X\rightarrow L$ an effective embedding, and $v_0$ and $v_1$ be functions that witness, respectively, that f is effectively continuous and effectively open. For each e, we define $m_e$ to be the unique number such that

$$ \begin{align*} v_0((-\infty,e),v_1((-\infty,e),0))=4(m_e,n)+r=s_e, \end{align*} $$

where $r\in \{3,4\}$ and $n\in \mathbb {N}$ , or $s_e=0$ . We have that the map $e\mapsto m_e$ is recursive relative to $v_0$ and $ v_1$ , so it exists by $\Delta ^0_1$ comprehension. Assume that $\Phi ^A_e(e){\downarrow }$ , we have that $v_1((-\infty ,e),0)$ will be the index of some interval $]a,b[_{<_L}$ . By construction we have $U_{s_e}\subseteq f^{-1}(]a,b[_{<_L})\subseteq U_0$ . We also have that $f^{-1}([a,b]_{<_L})$ is a closed set contained in $U_0$ plus at most two other points. If $\Phi ^A_e(e){\downarrow }$ and either $r=3$ or $s_e=0$ , then the closure of $U_{s_e}$ contains infinitely many points which are not in $U_0$ . But the closure of $U_{s_e}$ is contained in $U_0$ plus at most two other points. So if $\Phi ^A_e(e){\downarrow }$ then $r=4$ . By construction of X, we have $\Phi ^A_e(e){\downarrow }$ and $U_{4(m_e,n)+4}\subseteq U_0$ if and only if $\Phi ^A_e(e){\downarrow }_{\leq m_e}$ . So we have that $\Phi ^A_e(e){\downarrow } \leftrightarrow \Phi ^A_e(e){\downarrow }_{\leq m_e}$ , which means that $A'$ exists since it is computable from the map $e\mapsto m_e$ . Since every set has a Turing jump by Theorem 3.2, we have arithmetic comprehension.

Proof. Let A be a set, we show that $A'$ exists. Define

$$ \begin{align*} X^e=\{-\infty\}\cup((\mathbb{Z}\times\{t\in\mathbb{N}:\Phi^A_e(e){\downarrow}_{\leq t}\})\cup\{(n,+\infty):n\in\mathbb{N}\} \end{align*} $$

with basic open sets:

$$ \begin{align*} U_0=\{-\infty\}\cup (\mathbb{Z}\times\{t\in\mathbb{N}:\Phi^A_e(e){\downarrow}_{\leq t}\}) \end{align*} $$

and for all $(m,n)\in (\mathbb {Z}\times \{t\in \mathbb {N}:\Phi ^A_e(e){\downarrow }_{\leq t}\}),$

$$ \begin{align*} U_{3(m,n)+1}=\{(m,n)\} \end{align*} $$

$$ \begin{align*} U_{3(m,n)+2}=\{(+\infty,n)\}\cup\{(l,n):l\geq m\} \end{align*} $$

$$ \begin{align*} U_{3(m,n)+3}=\begin{cases} U_0 \quad \quad\quad \text{ if } \quad \quad \quad \neg\Phi^A_e(e){\downarrow}_{\leq n}\\ \{-\infty\}\cup\{(l,s):l\leq m \wedge \Phi^A_e(e){\downarrow}_{\leq s}\}\quad \text{ if }\quad \Phi^A_e(e){\downarrow}_{\leq n}. \end{cases} \end{align*} $$

Given $a=4(i,j)+r$ and $b=4(p,l)+s$ let $c =\max \{|i|,j, |p|,l\}+1$ and define:

• • $k((m,n),a,b)=3(m,n)+1.$

• • $k((\infty ,n),a,b)= 3(c,n)+2.$

• • $k(-\infty ,a,b)=3(-c,c)+3.$

Take $X=\coprod _{e\in \mathbb {N}} X^e$ , we have that it is $T_3$ and scattered. We note that for each e the subspace $X^e\times \{e\}$ is clopen in X. By assumption, there is an effectively continuous bijection $f:X\rightarrow W$ where W has the order topology of a well-order $<_W$ . Let $(I_i)_{i\in \mathbb {N}}$ be the standard indexing of the basic open intervals of $(W,<_W)$ and $v$ be a function that witnesses that f is effectively continuous. We have that all intervals of a well-order are clopen. Define for each $e\in \mathbb {N,}$

$$ \begin{align*} i_e=\begin{cases} \text{ index for: } ]f((e,+\infty,0)),+\infty[_{<_W} &\text{if}\quad f((e,+\infty,0))<_W f((e,-\infty)) \\ \text{ index for: } ]-\infty,f((e,+\infty,0))[_{<_W} &\text{if}\quad f((e,-\infty)) <_W f((e,+\infty,0)). \end{cases} \end{align*} $$

We have that $f^{-1}(I_{i_e})$ will be a clopen neighborhood of $(e,-\infty )$ . So, in particular, we have that $v((e,-\infty ),i_e)$ will be an index for a basic open set of $(e,-\infty )$ that is contained in a clopen set that does not contain $(e,+\infty ,0)$ . But by construction of X we have that either $\Phi ^A_e(e){\downarrow }_{\leq v((e,-\infty ),i_e)}$ or $\Phi ^A_e(e){\uparrow }$ . So we have that $A'$ is computable relative to $v,(i_e)_{e\in \mathbb {N}},$ and X, and so it exists by $\Delta ^0_1$ comprehension.

Observation 6.3. For any sequence of sets $(A_i)_{i\in \mathbb {N}}$ the algebra generated by them is $\Delta ^0_1$ definable and so are functions $Int$ and $Comp$ that compute, respectively, intersections and complements. A similar construction is done in [Reference Dorais6, Definition 2.11]. So for any zero-dimensional CSC space $(X,(U_i)_{i\in \mathbb {N}},k)$ there is a CSC space with an algebra of clopen sets $(X,(V_i)_{i\in \mathbb {N}},Int,Comp)$ such that $(U_i)_{i\in \mathbb {N}}$ is a subsequence of $(V_i)_{i\in \mathbb {N}}$ and every $V_i$ is open in $(X,(U_i)_{i\in \mathbb {N}},k)$ . This means that the identity $X\rightarrow X$ is a homeomorphism. It will not, in general, be an effective homeomorphism. What is needed to ensure that the identity is an effective homeomorphism is a function that encodes complements in some way.

Observation 6.5. $\mathbf {RCA}_0$ proves that any $T_0$ effectively zero-dimensional CSC space is $eT_2$ .

Observation 6.6. Let $(L,<_L)$ be a linear order, the partial function

$$ \begin{align*} G(z,(x,y))=\begin{cases} (-\infty,x)\quad \quad\ \mathrm{if}\ \quad \quad z\leq x \\ (y,+\infty)\quad \quad\ \mathrm{if}\ \quad \quad z> y \end{cases} \end{align*} $$

witnesses that the upper limit topology is effectively zero-dimensional.

Proof. Let $(X,(U_i)_{i\in \mathbb {N}},k)$ be an effectively zero-dimensional CSC space and G be a function that witnesses that X is effectively zero-dimensional. For all $x\in X$ and $i\in \mathbb {N}$ such that $x\in U_i$ we define $R_0(x,i)=i$ and for all $y\notin U_i$ we define $R_1(x,i,y)=G(y,i)$ . It is easy to verify that $R_0$ and $R_1$ witness that X is uniformly regular.

Proof. We prove the case for one pair $(C_0,C_1)$ of disjoint closed collections coded by $f_0$ and $f_1$ , respectively. The general case will follow from the fact that the construction can be carried out uniformly. To construct D we will define recursively two increasing sequences of closed sets $(C^n_0)_{n\in \mathbb {N}}$ and $ (C^n_1)_{n\in \mathbb {N}}$ such that $C^0_0=C_0$ and $C^0_1=C_1$ and for all n the sets $C^n_0$ and $C^n_1$ are disjoint. We also define two increasing sequences of open sets $(A^n_0)_{n\in \mathbb {N}}$ and $(A^n_1)_{n\in \mathbb {N}}$ such that for all $n\in \mathbb {N}$ and $j\leq 1$ we have $A^n_j\subseteq C^n_j$ . We would like that $\bigcup _{n\in \mathbb {N}}A^n_0\cup A^n_1 =X$ and in such case we define $D=\bigcup _{n\in \mathbb {N}}A^n_0$ which by construction will be clopen and $C_0\subseteq D\subseteq X\setminus C_1$ . At every step, we will pick an $x\notin A^n_0\cup A^n_1$ and find a small enough neighborhood $U_{l}$ such that for some $j\leq 1$ we have $x\in U_l\subseteq X\setminus C^n_j$ . We then set $A^{n+1}_{1-j}=A^n_{1-j}\cup U_{R_0(x,l)}$ and $C^{n+1}_{1-j}=C^n_{1-j}\cup \bigcap _{y\notin U_l}(X\setminus U_{R_1(x,l,y)})$ .

To be able to construct the sequences $(C^n_0)_{n\in \mathbb {N}},(C^n_1)_{n\in \mathbb {N}},(A^n_0)_{n\in \mathbb {N}},$ and $(A^n_1)_{n\in \mathbb {N}}$ we will want them to be coded as finite objects. We will code $A^n_j$ with a finite set $F^n_j$ of indices such that $A^n_j=\bigcup _{i\in F^n_j} U_i$ . We will code $C^n_j$ as a finite set of pairs $E^n_j$ such that

$$ \begin{align*}C^n_j=\bigcap_{i\in\mathbb{N}}(X \setminus U_{f_j(i)})\cup \bigcup_{(x,l)\in E^n_j}\left(\bigcap_{y\notin U_l}(X\setminus U_{R_1(x,l,y}))\right). \end{align*} $$

We note that, in general, $C^n_j$ may not exist as a set.

Let $X=(x_n)_{n\in \mathbb {N}}$ and let $F^0_0=F^0_1=E^0_0=E^0_1=\emptyset $ . Given $C^n_0,C^n_1$ and $A^n_0,A^n_1$ , if $A^n_0\sqcup A^n_1=X$ then we are done. Otherwise, let m be the least number such that $x_m\in X\setminus (A^n_0\cup A^n_1)$ . Let $(i,j)$ be the least pair such that $j\leq 1$ and

$$ \begin{align*} (\exists s\leq i\;x_m\in U_{f_j(s)})\wedge \forall (z,l)\in E^n_j\; \exists y\leq i\; x_m \in U_{R_1(z,l,y)}. \end{align*} $$

The formula above states that in i steps, we verify that $x_m$ is not in $C^n_j$ . Since $C^n_0$ and $C^n_1$ are disjoint we have that the search will eventually terminate. Let s be the least number such that $x_m\in U_{f_j(s)}$ and define

$$ \begin{align*} I=\{R_1(z,l,y):(z,l)\in E^n_j \wedge y=\min\{ w\leq i :x_m \in U_{R_1(z,l,w)}\}\}\cup\{f_j(s)\}, \end{align*} $$

which will be a finite set by $\mathbf {B}\Sigma ^0_1$ and will be coded. We then define

$$ \begin{align*} F^{n+1}_{1-j}=F^n_{1-j}\cup\{R_0(x_m,K(x_m,I))\}\quad \text{ and }\quad F^{n+1}_j=F^n_j \end{align*} $$

and

$$ \begin{align*} E^{n+1}_{1-j}=E^n_{1-j}\cup\{(x_m,K(x_m,I))\}\quad \text{ and }\quad E^{n+1}_j=E^n_j. \end{align*} $$

We now verify that the sets $A^{n+1}_j$ , $C^{n+1}_j$ , $A^{n+1}_{1-j}$ , and $C^{n+1}_{1-j}$ have the wanted properties. By construction $A^{n+1}_j=A^n_j\subseteq C^n_j=C^{n+1}_j$ . We have that

$$ \begin{align*}A^{n+1}_{1-j}=A^n_{1-j}\cup U_{R_0(x_m,K(x_m,I))}\subseteq C^{n+1}_{1-j}\cup\bigcap_{y\notin U_{K(x_m,I)}} X\setminus U_{R_1(x_m,K(x_m,I),y)} =C^{n+1}_{1-j} \end{align*} $$

since $A^n_{1-j}\subseteq C^n_{1-j}$ and by definition of $R_0$ and $R_1$ we have

$$ \begin{align*} U_{R_0(x_m,K(x_m,I))}\subseteq \bigcap_{y\notin U_{K(x_m,I)}} X\setminus U_{R_1(x_m,K(x_m,I),y)}. \end{align*} $$

We prove that $C^{n+1}_j\cap C^{n+1}_{1-j}=\emptyset $ . We have by construction that

$$ \begin{align*} C^{n+1}_j=C^n_j\quad \text{ and }\quad C^{n+1}_{1-j}=C^n_{1-j}\cup \bigcap_{y\notin U_{K(x_m,I)}} X\setminus U_{R_1(x_m,K(x_m,I),y)}. \end{align*} $$

By definition of the functions $R_1$ and K we have

$$ \begin{align*} \bigcap_{y\notin U_{K(x_m,I)}} X\setminus U_{R_1(x_m,K(x_m,I),y)}\subseteq U_{K(x_m,I)}\subseteq \bigcap_{i\in I}U_i. \end{align*} $$

It therefore suffices to show that $C^n_j\cap \bigcap _{i\in I}U_i=\emptyset $ . Let $r\in C^n_j$ using the definition of $C^n_j$ we have the following cases.

Case 1: $r\in \bigcap _{i\in \mathbb {N}}X\setminus U_{f_j(i)}$ so $r\notin U_{f_j(s)}$ and $r\notin \bigcap _{i\in I} U_i$ since $f_j(s)\in I$ .

Case 2: There is $(z,l)\in E^n_j$ such that $r\in \bigcap _{y\notin U_l} X\setminus U_{R_1(z,l,y)}$ . By definition of I there is some y such that $R_1(z,l,y)\in I$ and so $r\notin U_{R_1(z,l,y)}$ .

Therefore, the sequences $(C^n_0)_{n\in \mathbb {N}},(C^n_1)_{n\in \mathbb {N}},(A^n_0)_{n\in \mathbb {N}},$ and $(A^n_1)_{n\in \mathbb {N}}$ have the desired properties. Since the sequences $(F^n_0)_{n\in \mathbb {N}}$ , $(F^n_1)_{n\in \mathbb {N}}$ , $(E^n_0)_{n\in \mathbb {N}}$ , and $(E^n_1)_{n\in \mathbb {N}}$ are defined recursively we have that they exist by $\Delta ^0_1$ comprehension. As observed before we have that the set

$$ \begin{align*}D=\bigcup_{n\in\mathbb{N}}A^n_0=X\setminus \bigcup_{n\in\mathbb{N}}A^n_1 \end{align*} $$

will be the wanted clopen set. Both D and its complement are $\Sigma ^0_1$ so D exists by $\Delta ^0_1$ comprehension.

Observation 6.10. Applying the previous result to the space $(\mathbb {N},(\{i\})_{i\in \mathbb {N}},k)$ , where $k(i,i,i)=i$ , gives us the classic recursion theory result that any pair of disjoint $\Pi ^0_1$ sets can be separated by a recursive set.

Proof. Let $R_0$ and $R_1$ be the functions that witness that X is $uT_3$ . For each pair $(x,i)$ , where $x\in U_i$ , let $C_0^{(i,x)}$ be the closure of the point x; which by regularity is contained in $U_{i}$ , and $C^{(x,i)}_1=X\setminus U_i$ . We point out that the closure of x may not be a set, but it will be a coded closed collection. By Theorem 6.9 we have that there exists a sequence of clopen sets $(D_{(x,i)})_{x\in U_i}$ such that for all $x\in U_i$ we have $x\in D_{(x,i)}\subseteq X\setminus (X\setminus U_i)=U_i$ . We define $k'$ . Given $x\in D_{(y,j)}\cap D_{(z,l)}$ , by the construction carried out in Theorem 6.9 we have that

$$ \begin{align*}D_{(y,j)}=\bigcup_{n\in\mathbb{N}}A^n_{0,(y,j)}=\bigcup_{n\in\mathbb{N}}\bigcup_{s\in F^n_{0,(y,j)}} U_s\quad \text{and}\quad D_{(z,j)}=\bigcup_{n\in\mathbb{N}}A^n_{0,(z,l)}=\bigcup_{n\in\mathbb{N}}\bigcup_{t\in F^n_{0,(z,l)}} U_t. \end{align*} $$

Let $m=\min \{n\in \mathbb {N}: x\in A^n_{0,(y,j)}\cap A^n_{0,(z,l)}\}$ and define

$$ \begin{align*} s=\min \{s'\in F^m_{0,(y,j)}: x\in U_{s'}\} \quad \text{and}\quad t=\min \{t'\in F^m_{0,(z,l)}: x\in U_{t'}\} \end{align*} $$

and let $k'(x,(y,i),(z,j))=(x,k(x,s,t))$ . It is straightforward, using $R_0$ , to show that $(X,(D_{(x,i)})_{x\in U_i},k')$ is a CSC space with a basis of clopen sets and it is effectively homeomorphic to $(X,(U_i)_{i\in \mathbb {N}},k)$ . We define a function G witnessing that X is effectively zero-dimensional. Let $x\in X$ and $i\in \mathbb {N}$ be such that $x\in U_i$ . Observing the construction carried out in Theorem 6.9 we have that $X\setminus D_{(x,i)}=\bigcup _{n\in \mathbb {N}} A^n_1=\bigcup _{n\in \mathbb {N}}\bigcup _{j\in F^n_1} U_j$ . Given $y\notin D_{(x,i)}$ let $m=\min \{n\in \mathbb {N}: y\in A^n_1\}$ and define $G(y,(x,i))=(y,\min \{j\in F^n_1: y\in U_j\})$ . We have that G witnesses that $(X,(D_{(x,i)})_{x\in U_i},k')$ is effectively zero-dimensional and exists by $\Delta ^0_1$ comprehension.

Proof. Let $(V_i)_{i\in \mathbb {N}}$ be the sequence of sets given by $V_{2j}=U_i$ and $V_{2i+1}=X\setminus U_i$ . We then define $A_m=\bigcap _{i\in m}\bigcup _{j\in i} V_i$ . We define $Int(n,m)$ to be the number that codes the union of m and n. For $m\in \mathbb {N}$ coding the set

$$ \begin{align*} \{\{a_{0,0},\dots,a_{0,n_0}\},\dots,\{a_{r,0},\dots,a_{r,n_r}\}\}, \end{align*} $$

then $Comp(m)$ will code the set of all sets of the form

$$ \begin{align*} \{a_{0,f(0)}^c,a_{1,f(1)}^c,\dots,a^c_{r,f(r)}\}, \end{align*} $$

where $f:r\rightarrow \max \{n_j:j\leq r\}$ is a coded function such that for all $j\leq r $ we have $f(j)\leq n_j$ and where $a^c=a-1$ if a is odd and $a^c=a+1$ if a is even. One observes that $A_{Comp(m)}=X\setminus A_m$ .

Let $Id:(X,(U_i)_{i\in \mathbb {N}},k)\rightarrow (X,(A_m)_{m\in \mathbb {N}},Int,Comp)$ be the identity function from X to X, we show that it’s an effective homeomorphism. The function $(x,i)\mapsto \{\{2i\}\}$ witnesses that $Id$ is effectively continuous. We need to show that there is a function $v$ such that for any $(x,m)$ such that $x\in A_m$ then $x\in U_{v(x,m)}\subseteq A_m$ . We define $v$ on the complexity of the set m codes. If $m=\{\{2i\}\}$ then we set $v(x,m)=i$ . If $m=\{\{2i+1\}\}$ then $v(x,m)=G(x,i)$ . For $m=\{\{a_0,\dots ,a_n\}\}$ where $a_0<a_1<\dots <a_n$ then we define $v(x,m)=v(x,\{\{ a_0\}\})$ . Finally for any $m=\{m_0,\dots ,m_r\}$ then let $I=\{v(x,\{m_i\}): i\leq r\}$ we define

$$ \begin{align*} v(x,m)=K(x,I). \end{align*} $$

The function $v$ is defined primitively recursively, and so it exists by $\Delta ^0_1$ comprehension, and it verifies that the identity is effectively open between $(X,(U_i)_{i\in \mathbb {N}},k)$ and $(X,(A_m)_{m\in \mathbb {N}},Int,Comp)$ .

Proof. The case in which X is finite is trivial, so we assume that X is infinite and let $(x_i)_{i\in \mathbb {N}}$ be the elements of X enumerated increasingly. $(3\rightarrow 4)$ is trivial.

$(1\rightarrow 2)$ follows from Corollary 6.11 and Proposition 6.12. $(2\rightarrow 1)$ follows from the fact that any CSC space with an algebra of clopen sets is effectively zero-dimensional, which by Proposition 6.7 is $uT_3$ .

$(4\rightarrow 2)$ We have already seen in Proposition 4.10 that all countable metric spaces have a CSC structure that consists of clopen balls. In particular given a metric space $(X,d)$ there exists an $a\in \mathbb {R}_{>0}$ such that $(X, (B(x,q\cdot a) )_{q\in \mathbb {Q}_{>0},x\in X},k)$ is a CSC space with a basis of clopen sets. We define $G:X\times (X\times \mathbb {Q}_{>0})\rightarrow (X\times \mathbb {Q}_{>0})$ which will send $(x,(y,q))$ , where $x\notin B(y,q\cdot a)$ , to $(x,\frac {1}{n})$ where n is the least n such that $\frac {1}{2^n}\cdot a < d(x,y)- q\cdot a$ . Since the CSC structure of X is effectively zero-dimensional, by Proposition 6.12, it is effectively homeomorphic to a space with an algebra of clopen sets.

$(2\rightarrow 3)$ Assume that X has an algebra of clopen sets. We will show that there is an embedding f from X to the rational points of the Cantor space, which we will denote as $2^{\mathbb {N}}\cap \mathbb {Q}$ . We observe that $2^{\mathbb {N}}\cap \mathbb {Q}$ is a CSC space with an algebra of clopen sets of the form $[[\tau ]]=\{g\in 2^{\mathbb {N}}\cap \mathbb {Q}:\tau \sqsubseteq g\}$ , and it effectively embeds into $[0,1]\cap \mathbb {Q}$ .

We construct recursively the following things:

• • A total surjection $\phi :\mathbb {N}\rightarrow \mathbb {N}$ such that for all $n\in \mathbb {N}$ and for all $i<j<2n$ we have

  $$ \begin{align*}\{m<2n:x_i\in U_{\phi(m)}\}\neq \{m< 2n:x_j\in U_{\phi(m)}\}. \end{align*} $$

  The function $\phi $ will define a sequence of basic open sets $(U_{\phi (n)})_{n\in \mathbb {N}}$ . For $\tau \in 2^{<\mathbb {N}}$ we write

  $$ \begin{align*} N_\tau=\{x\in X:\forall m<|\tau|\;(x\in U_{\phi(m)}\leftrightarrow \tau(m)=1)\}. \end{align*} $$

  Since we are assuming that X has an algebra of sets, we have that for every sequence $\tau $ that

  $$ \begin{align*} N_\tau=\left(\bigcap_{i< |\tau|\wedge \tau(i)=1}U_{\phi(i)}\right)\cap\left(\bigcap_{i< |\tau|\wedge \tau(i)=0}U_{Comp(\phi(i))}\right). \end{align*} $$

  So there is some j such that $N_\tau =U_j$ . Furthermore, there is a uniformly effective procedure to find an index j from $\tau $ . In general, $N_\tau $ might be empty, but this will not affect the construction.

• • $f(x_0),\dots ,f(x_n)\in 2^{<\mathbb {N}}\cap \mathbb {Q}$ .

• • A function $r_n:2^{\leq 2n}\rightarrow 2^{\leq 2n}$ which is monotone, level preserving and for all $\sigma \in \text {rng}(r_n)$ there is an $i\leq n$ such that $f(x_i)\in N_{\sigma }$ . We also require that for all $i\leq n$ we have $f(x_i)\in [[r_n(seq_{2n}(x_i))]]$ where $seq_{2n}(x_i)$ is a sequence of length $2n$ such that

  $$ \begin{align*}seq_{2n}(x_i)(j)=\begin{cases} 1 \quad \text{ if }\quad x\in U_{\phi(j)} \\ 0 \quad \text{ if }\quad x\notin U_{\phi(j)}. \end{cases} \end{align*} $$

  This ensures us that for all $\tau $ of length $n\, f(N_\tau )=f(X)\cap [[r_n(\tau )]]$ and that f will be injective. Finally, we will require for all $m<n$ that $r_m\subseteq r_n$ .

Assume we have $f(x_0),\dots ,f(x_{n}),r_n,\phi |_{< 2n}$ and that for all $\sigma \in 2^{=2n}\, [[r_n(\sigma )]]$ contains a unique $f(x_i)$ .

Step 1: We set $\phi (2n)=\min \mathbb {N}\setminus \text {rng}(\phi |_{<2n})$ , this ensures us that $\phi $ will be surjective. Given $\tau \in 2^{<\mathbb {N}}$ of length $2n-1$ there exists exactly one $j\leq n$ such that $f(x_j)\in [[r_n(\tau )]]$ . Let $i\leq 1$ be such that $f(x_j)\in [[r_n(\tau )^\frown (i)]]$ , we define $r_n'(\tau ^\frown (0))=r_n'(\tau ^\frown (1))=r_n(\tau )^\frown (i)$ .

Step 2: There is, by construction of $r^{\prime }_n$ , exactly one $j\leq n$ such that $r^{\prime }_n(seq_{2n+1}(x_{n+1}))=r^{\prime }_n(seq_{2n+1}(x_j))$ . We define

$$ \begin{align*} \phi(2n+1)=\min \{ l\in\mathbb{N}\setminus\text{rng}(\phi|_{<2n+1}):x_{n+1}\in U_l\leftrightarrow x_j\notin U_l\}. \end{align*} $$

That is, the first index for a basic open set not in the range of $\phi |_{<2n+1}$ that separates $x_{n+1}$ and $x_j$ . The search for such l will eventually terminate since we are assuming that X is $T_0$ . Assume that $x_j\in U_{\phi (2n+1)}$ and $x_{n+1}\notin U_{\phi (2n+1)}$ , the proof for the other case is analogous. Let $\rho $ be the unique sequence of length $2n+2$ such that $f(x_j)\in [[\rho ]]$ and let $i=\rho (2n+1)$ . For all $\sigma \in 2^{=2n+1}$ such that

$$ \begin{align*} r^{\prime }_n(\sigma )= r^{\prime }_n(seq_{2n+1}(x_{n+1}))=r^{\prime }_n(seq_{2n+1}(x_j)) \end{align*} $$

we define

$$ \begin{align*} r_{n+1}(\sigma^\frown(0))=r^{\prime}_n(\sigma)^\frown (1-i) \end{align*} $$

$$ \begin{align*} r_{n+1}(\sigma^\frown(1))=r^{\prime}_n(\sigma)^\frown (i)=\rho. \end{align*} $$

For all other sequences of length $2n+2$ we define $r_{n+1}$ as in step $1$ .

In either case, we are ensured that for all $j\leq n$ , we have

$$ \begin{align*} f(x_j)\in [[r_{n+1}(seq_{2n+2}(x_j))]]. \end{align*} $$

Step 3: We define $f(x_{n+1})$ to be the least Cantor rational in the open set $[[r_{n+1}(seq_{2n+2}(x_{n+1}))]]$ . This ends the construction.

Given the functions $f,\phi $ and $r=\bigcup _{n\in \mathbb {N}}r_n$ we show that f is an effective homeomorphism with its range.

We first show that the range of f exists. By definition $y\in \text {rng}(f)\leftrightarrow \exists x\; f(y)=x$ , and so the range is $\Sigma ^0_1$ definable relative to f. We also have that $y\in \text {rng}(f)$ if and only if for all $n\in \mathbb {N}$ there is a $\sigma \in 2^{=n}$ such that $y\in [[r_n(\sigma )]]$ , which is a $\Pi ^0_1$ definition. So we have that $\text {rng}(f)$ is $\Delta ^0_1$ definable and therefore exists by $\Delta ^0_1$ comprehension. We note that $\Pi ^0_1$ definition of $\text {rng}(f)$ tells us it is a closed subspace of $2^{\mathbb {N}}\cap \mathbb {Q}$ .

Let $x_j\in X$ be a point and $U_{\phi (i)}$ be an open set containing it. We wish to effectively find an open neighborhood of $f(x)$ that is contained in $f(U_{\phi (i)})$ . Let $n>\max \{i,j\}$ , we have that $seq_{2n}(x_j)(i)=1$ and so

$$ \begin{align*} x_j\in N_{seq_{2n}(x_j)}\subseteq U_{\phi(i)}. \end{align*} $$

Which by construction implies

$$ \begin{align*} f( x_j)\in f(N_{seq_{2n}(x_j)})=f(X)\cap [[r_n(seq_{2n}(x_j))]]\subseteq f(U_{\phi(i)}). \end{align*} $$

This implies that $f:X\rightarrow f(X)$ is effectively open.

Let $[[\tau ]]$ be a basic open set of the Cantor space and $f(x_j)\in [[\tau ]]$ . By construction we have that $\tau \in \text {rng}(r)$ , since r is level preserving we can effectively find $\rho =r^{-1}(\tau )$ . We have that $x\in N_{\rho }$ and $f(N_\rho )\subseteq[[\tau ]]$ by construction. This shows that f is effectively continuous. Therefore f is an effective embedding of X into $2^{<\mathbb {N}}\cap \mathbb {Q}$ .

Proof. By Corollary 6.11 and Proposition 6.12 any $uT_3$ CSC space has an algebra of clopen sets, so by Theorem 6.14, any $uT_3$ space without isolated points will be homeomorphic to $\mathbb {Q}$ .

Observation 6.16. Since over $\mathbf {ACA}_0$ all $T_3$ CSC spaces are $uT_3$ we have that $\mathbf {ACA}_0$ proves that every $T_3$ space is homeomorphic to a subset of $\mathbb {Q}$ and is homeomorphic to a space with an algebra of clopen sets.

Proof. By Theorem 6.13, $(2)$ – $(4)$ are equivalent over $\mathbf {RCA}_0$ . By Proposition 6.7, all effectively zero-dimensional CSC spaces, and therefore all CSC spaces with an algebra of clopen sets, are uniformly regular. By Proposition 5.16, $(4)$ is equivalent to every regular $eT_2$ CSC space is uniformly regular, which is equivalent to arithmetic comprehension over $\mathbf {RCA}_0$ by Theorem 5.20.

Proof. Let $(X,(U_i)_{i\in \mathbb {N}})$ be a uniformly regular CSC space by Corollary 6.11 and Proposition 6.12 we can assume without loss of generality that X has an algebra of clopen sets. Let $U[x]=\{i\in \mathbb {N}:x\in U_i\}$ we define d by

$$ \begin{align*} d(x,y)=\sum_{k\in U[x]\triangle U[y]} \frac{1}{2^k}. \end{align*} $$

The function d is symmetric, and the triangle inequality follows from the following property of the symmetric difference:

$$ \begin{align*} \forall A,B,C\;(A\triangle C)\subseteq ((A\triangle B)\cup(B\triangle C)). \end{align*} $$

By definition of d we have

$$ \begin{align*}B\left(x,\frac{1}{2^k}\right)=N_{U^k[x]}=\{y\in X: \forall i<k\;(x\in U_i\leftrightarrow y\in U_i)\}. \end{align*} $$

So we have that $B(x,\frac {1}{2^k})$ is the Boolean combination of clopen sets and is, therefore, clopen. We show that the identity $Id:(X,(U_i)_{i\in \mathbb {N}}, Int, Comp)\rightarrow (X,(B(x,\frac {1}{2^k}))_{k\in \mathbb {N}},k)$ is an effective homeomorphism. For $x\in X$ and $i\in \mathbb {N}$ such that $x\in U_i$ then we have that $x\in B(x,\frac {1}{2^i})\subseteq U_i$ and so the map $(x,i)\mapsto i$ verifies that the identity is effectively open. The map that sends $(x,k)$ to the code for the open set

$$ \begin{align*} \left(\bigcap_{i< k\wedge x\in U_i} U_i\right)\cap \left(\bigcap_{i< k\wedge x\notin U_i} X\setminus U_{i}\right) \end{align*} $$

is computable relative to the functions $Comp$ and $Int$ and it verifies that the identity is effectively continuous.

Proof. $(1\rightarrow 2)$ is Corollary 6.11, $(2\rightarrow 3)$ is Proposition 6.12, and $(3\rightarrow 4)$ is Proposition 6.19. $(4\rightarrow 1)$ follows from the fact that the fact that the CSC structure on a pseudometric space is effectively zero-dimensional.

Observation 7.3 (Dorais [Reference Dorais6, Example 5.4])

Over $\mathbf {ACA}_0$ a CSC space is compact if and only if it is sequentially compact.

Observation 7.4. $\mathbf {RCA}_0$ proves that the union of two effectively compact sets is effectively compact. In particular, given $C_0,C_1$ the covering relations of $K_0,K_1$ , respectively, then $C_0\cap C_1$ is the covering relation for $K_0\cup K_1$ .

Proof. $(1\rightarrow 2)$ since the covering relation for a CSC space is arithmetically definable. $(2\rightarrow 3)$ is trivial. For $(3\rightarrow 1)$ consider the CSC space $(X,(U_i)_{i\in \mathbb {N}},k)$ where $X=\mathbb {N}\cup \{\infty \}$ :

$$ \begin{align*} U_{2i}=\{i\} \end{align*} $$

$$ \begin{align*} U_{2(n,s)+1}= \{\infty\}\cup\{j\in\mathbb{N}: j\geq n\wedge \forall e\in s\; \Phi^A_e(e){\downarrow}_{\leq j}\rightarrow \exists r<j \Phi^A_e(e){\downarrow}_{\leq r}\} \end{align*} $$

that is all the $j\geq n$ such that for all $e\in s$ the Turing machine of index e does not halt for the first time at step j. We define k that sends $(n,a,b)\mapsto 2n$ and $(\infty ,2(i,s)+1,2(j,t)+1)$ to $2(\max \{i,j\},r)+1$ where r is the code for the union of s and t. X is $uT_3$ follows from the fact that the basis consists of clopen sets and all points except $\infty $ are isolated. By assumption, X is effectively compact and, therefore, has a covering relation C. We have that

$$ \begin{align*} \{2(0,\{e\})+1\}\in C\leftrightarrow \Phi ^A_e(e){\uparrow } \end{align*} $$

and so $A'\leq _T C$ .

Proof. Let $H_0,H_1$ the functions that witness that X is $eT_2$ . Define $p:\mathbb {N}\times X \rightarrow \mathbb {N}$ such that for any $n\in \mathbb {N}$ and $x\notin K_n$ we have

$$ \begin{align*} p(n,x)=\min\{F \text{ code for a finite subset of }K_n: \{H_0(x,y):y\in F\}\in C_n\}. \end{align*} $$

We observe that p is computable relative to $(K_n,C_n)_{n\in \mathbb {N}}$ and so it exists by $\Delta ^0_1$ comprehension. For each $x\notin K_n$ we define $s(n,x)=K(x,\{H_1(x,y):y\in p(n,x)\})$ . We have that by construction

$$ \begin{align*} x\in U_{s(n,x)}\subseteq \bigcap_{y\in p(n,x)} U_{H_1(x,y)}\subseteq X\setminus \bigcup_{y\in p(n,x)}U_{H_0(x,y)}\subseteq X\setminus K_n. \end{align*} $$

We have that s exists by $\Delta ^0_1$ comprehension and, for each n, we have that $s(n,\cdot )$ is a code for the closed set $K_n$ , so $(K_n)_{n\in \mathbb {N}}$ is uniformly effectively closed.

Proof. Let $(X,(U_i)_{i\in \mathbb {N}},k)$ be an effectively compact $eT_2$ CSC space. Let C be the covering relation of X and $H_0$ and $H_1$ be functions that witness that X is $eT_2$ . We define $R_0$ and $R_1$ on X. For any $x\in U_i$ we have that $\{U_i\}\cup \{U_{H_1(x,y)}:y\notin U_i\}$ is a covering of X. Let $F\subseteq X\setminus U_i$ be the least finite set such that $\{i\}\cup \{H_1(x,y):y\in F\}\in C$ . Define

$$ \begin{align*} R_0(x,i)=K(x,\{H_0(x,y):y\in F\}) \end{align*} $$

and for each $z\notin U_i$ we define

$$ \begin{align*} R_1(x,i,z)=\min\{y\in F: z \in U_{H_1(x,y)}\}. \end{align*} $$

By definition, we have

$$ \begin{align*} x\in U_{R_0(x,i)}\subseteq\bigcap_{y\in F} U_{H_0(x,y)}\subseteq X\setminus \bigcup_{y\in F} U_{H_1(x,y)}=X\setminus\bigcup_{z\notin U_i} U_{R_1(x,i,z)}\subseteq U_i. \end{align*} $$

So the functions $R_0$ and $R_1$ witness that X is $uT_3$ and are recursive relative to the sets and functions $C,(X,(U_i)_{i\in \mathbb {N}},k)$ , $H_0$ , and $H_1$ so they exist by $\Delta ^0_1$ comprehension.

Observation 7.13. $\mathbf {RCA}_0$ proves that every linear order with its order topology has a covering relation, so $\mathbf {RCA}_0$ proves that every well-order with maximal element is effectively compact. Using the same proof, we can show that the upper limit topology on a well-order with a maximal element is effectively compact. We will see later that over $\mathbf {ACA}_0$ all compact $T_2$ CSC spaces will be homeomorphic to a well-order with maximum element.

Proof. $(1\rightarrow 2)$ follows from the fact that over $\mathbf {ACA}_0$ compact is equivalent to strongly compact.

$(2\rightarrow 1)$ Let $f:\mathbb {N}\rightarrow \mathbb {N}$ be an injection, we wish to prove that $\text {rng}(f)$ exists as a set. Define

$$ \begin{align*} T =\{ &\sigma: \forall m,n<|\sigma |\;((f(n)=m\leftrightarrow\sigma(m)=n+1) \\ &\quad \wedge(\sigma(m)>0\leftrightarrow f(\sigma(m)-1)=m))\}. \end{align*} $$

That is, the tree T contains all sequences that approximate the inverse of the function g. The same construction is used in [Reference Simpson32, Theorem III.7.2] to prove that Kőnig’s lemma implies arithmetic comprehension. We have that T has the following properties:

1. (1) Every $\sigma \in T$ has at most two immediate $\sqsubseteq$ -successors.

2. (2) T has at most one branch, and the unique branch of T can compute $\text {rng}(f)$ .

We prove that the Kleene–Brouwer order on T is of order type $\omega +\omega ^*$ , that is, T is infinite and for every $\sigma \in T$ either $]-\infty ,\sigma [_{<_{\text {KB}}}$ is finite and we say that $\sigma $ is in the $\omega $ part or $]\sigma ,+\infty [_{<_{\text {KB}}}$ is finite and we say that $\sigma $ is in the $\omega ^*$ part. Given $\sigma \in T$ , we consider the following cases.

Case 1: If $\forall j<|\sigma |\;(\exists n\; f(n)=j\rightarrow \sigma (j)=n+1)$ then we have that there is no element to the right of $\sigma $ in T and so the only elements that are $<_{\text {KB}}$ greater than $\sigma $ are its initial segments. This means that in $]\sigma ,+\infty [_{<_{\text {KB}}}$ has $|\sigma |$ many elements and so $\sigma $ is in the $\omega ^*$ part.

Case 2: If there is a $j<|\sigma |$ and $n\in \mathbb {N}$ such that $f(n)=j$ and $ \sigma (j)=0$ then we have that $\sigma $ and any sequence to the left of $\sigma $ in T cannot have an extension of length more than n in T. This in particular means that $]-\infty ,\sigma ]_{<_{\text {KB}}}$ has at most $2^n$ elements and so $\sigma $ is in the $\omega $ part.

We note that by the fact that $(T,<_{\text {KB}})$ has order type $\omega +\omega ^*$ we have that every element of T which isn’t the maximum has a successor and every element other than the minimum will have a predecessor. In particular, we have that the upper limit topology on $(T,<_{\text {KB}})$ is $eT_2$ and discrete so it cannot be strongly compact. T does however have a covering relation, we prove that it is compact. Assume that T is not compact, and let $(]a_n,b_n[)_{n\in \mathbb {N}}$ be an infinite covering of T which does not admit a finite subcovering. We have that every interval in $(T,<_{\text {KB}})$ will be either finite or cofinite, so for every $n\in \mathbb {N}$ we have that $]a_n,b_n[$ will be finite. We show that both the $\omega $ part and the $\omega ^*$ part of $(T,<_{\text {KB}})$ will be $\Sigma ^0_1$ definable. We have that x is in the $\omega $ part if there exists a finite sequence $\sigma $ such that $a_{\sigma (0)}=-\infty $ and for all $j<|\sigma |-1$ we have $b_j\leq _{\text {KB}} a_{j+1}$ and $x\in\ ]a_{\sigma (|\sigma |-1)},b_{\sigma (|\sigma |-1)}[$ . That is, there exists a finite sequence of intervals of the covering whose union is an initial segment of $(T,<_{\text {KB}})$ containing x. In a similar fashion we may give a $\Sigma ^0_1$ definition of the $\omega ^*$ part of $(T,<_{\text {KB}})$ . By $\Delta ^0_1$ comprehension, we have that both the $\omega $ and $\omega ^*$ parts of T exist. So the Kleene–Brouwer order of T is not complete. By Lemma 7.16 it will have a branch which will compute $\text {rng}(f)$ .

Proof. $(1\rightarrow 2)$ follows from the fact that being compact is a $\Pi ^1_1$ definable property. We first show that $(2)$ implies arithmetic comprehension. Let A be a set, we define for each $e\in \mathbb {N}$ the space $X^e=\{t\in \mathbb {N}: \Phi ^A_e(e){\downarrow }_{\leq t}\}$ with the discrete topology. We observe that $X^e$ is compact if $X^e$ is finite which holds if and only if $\Phi ^A_{e}(e){\uparrow }$ . So $\mathbb {N}\setminus A'=\{e\in \mathbb {N}:X^e \text { is compact}\}$ . So $(2)$ implies that the Turing jump of every set exists which implies arithmetic comprehension. It therefore suffices to show $(2\rightarrow 1)$ over $\mathbf {ACA}_0$ .

We show that $(2)$ implies that for any sequence of trees $(T_i)_{i\in \mathbb {N}}$ the collection $ \{i\in \mathbb {N}: T_i \text { is well-founded}\,\}$ exists, which by Theorem 3.8 is equivalent to $\Pi ^1_1$ comprehension. It suffices to prove that a tree T is well-founded if and only if T with the upper limit topology with respect to the Kleene–Brouwer ordering is compact. If T is well-founded, then $(T,<_{\text {KB}})$ is well-ordered, which implies that its order topology is equal to its upper limit topology. By Proposition 7.12 and the fact that $\emptyset \in T$ is $<_{\text {KB}}$ -maximum, we have that $(T,<_{\text {KB}})$ with the upper limit topology is compact. If T is not well-founded then $(T,<_{\text {KB}})$ has an infinite descending chain $(x_j)_{j\in \mathbb {N}}$ . The space T admits a partition of basic open sets

$$ \begin{align*} \{T\setminus{\uparrow}\{x_j:j\in\mathbb{N}\},]x_0,+\infty[\,\}\cup\{\,]x_{j+1},x_j]:j\in\mathbb{N}\} \end{align*} $$

so T is not compact.

Observation 7.19. In the previous proof, we saw that for every tree, there is a canonically defined CSC space which is compact if and only if the tree is well-founded. This implies that being a compact CSC space is a universal $\Pi ^1_1$ formula over $\mathbf {ACA}_0$ , so in particular, compactness for CSC space cannot be expressed by a $\Sigma ^1_1$ formula. Furthermore, Shafer proved that over $\mathbf {WKL}_0$ , every linear order is complete if and only if it is compact [Reference Shafer29, Corollary 4.2]. This means that being a complete linear order is also a $\Pi ^1_1$ universal formula over $\mathbf {ACA}_0$ .