Proof. First observe that $(Cl (\mathfrak {S}), \cap , \cup , ^C, \emptyset , S)$ is a Boolean algebra via Stone duality. We now prove that the defining equations of $\mathsf {VA}$ involving hold, i.e., (1)–(4) in Definition 2.6. For (1), let $A \in C(\mathfrak {S})$ , then by ( $\alpha $ 1).

We now prove (2), i.e., for all $A,B,C \in Cl (\mathfrak {S})$ . Since

if then $f(A,x) \subseteq B$ and $f(B,x) \subseteq A$ , thus by ( $\alpha $ 2) we get that $f(A,x) = f(B,x)$ . Now, coincides with

$$ \begin{align*}&(\{x \in S: f(A,x) \subseteq C\} \cap \{x \in S: f(B,x) \subseteq C\})\\ &\quad \cup (\{x \in S: f(A,x) \subseteq C\}^C \cap \{x \in S: f(B,x) \subseteq C\}^C).\end{align*} $$

Since $f(A,x) = f(B,x)$ , the claim follows from the fact that either $x \in \{y \in S: f(A,y) \subseteq C\}$ or $x \in \{y \in S: f(A,y) \subseteq C\}^C$ .

For (3), we verify that for all $A,B,C \in C (\mathfrak {S})$

It suffices to verify the right-to-left inclusion. Let $x \in S$ ; by ( $\alpha $ 3) either $f(A \cup B,x) \subseteq A$ , $f(A \cup B,x) \subseteq B$ , or $f(A \cup B,x) = f(A,x) \cup f(B,x)$ . In the first case we get that , while in the second one it follows that . Finally, suppose $f(A \cup B,x) = f(A,x) \cup f(B,x)$ . We show that then , or equivalently, that

Indeed, by definition, the latter is equivalent to:

$$ \begin{align*}f(A \cup B,x) \subseteq C \;\mbox{ iff }\; (f(A,x) \subseteq C \mbox{ and } f(B,x) \subseteq C),\end{align*} $$

which clearly holds given the hypothesis $f(A \cup B, x) = f(A,x) \cup f(B,x)$ .

Lastly, it is straightforward from the definition of that for all $A,B,C \in C (\mathfrak {S})$ , which proves (4). We have shown that $\mathscr{C} (\mathfrak {S}) \in \mathsf {VA}$ .

Proof. (1) implies (2) follows directly from the definition of $f_C$ . We prove (2) implies (1) by contraposition. Suppose , and let ; then F is a Boolean filter of ${\mathbf{C}}$ . Indeed, it is an upset because of the fact that is order-preserving on the right and X is upwards closed, it is closed under meet since distributes over meets on the right and X is closed under meets, and $1 \in F$ given that . Moreover, since $b \notin F$ given that , by the Boolean Ultrafilter Theorem we obtain that there is an ultrafilter Y of the Boolean reduct of ${\mathbf{C}}$ such that , $b \notin Y$ ; therefore $Y \in f_C(\mathfrak {s}(a), X)$ but $b \notin Y$ .

Finally, the fact that (2) and (3) are equivalent follows from the fact that $\mathfrak {s}(b) = \{X \in Ul({\mathbf{C}}): b \in X\}$ . This concludes the proof.

Proof. Since $\mathscr{S}({\mathbf{C}})$ is a Stone space by Stone duality, we only have to prove that $f_C$ satisfies the properties ( $\alpha $ 1)–( $\alpha $ 5) in Definition 3.1. Note once again that by Stone duality, every clopen in $\mathscr{S}({\mathbf{C}})$ is of the kind $\mathfrak {s}(a)$ for some $a \in C$ .

For ( $\alpha $ 1), let $a \in C, X \in Ul({\mathbf{C}})$ , we have to verify that $f_C(\mathfrak {s}(a), X) \subseteq \mathfrak {s}(a)$ . Suppose $Y \in f_C(\mathfrak {s}(a), X)$ , which by definition is equivalent to implies $c \in Y$ , thus since by the first axiom of $\mathsf {VA}$ : , we get that $a \in Y$ and then $Y \in \mathfrak {s}(a) = \{Z \in Ul({\mathbf{C}}): a \in Z\}$ .

To prove ( $\alpha $ 2), we show that for $a, b \in C, X \in Ul({\mathbf{C}})$ , $f_C(\mathfrak {s}(a), X) \subseteq \mathfrak {s}(b)$ and $f_C(\mathfrak {s}(b), X) \subseteq \mathfrak {s}(a)$ imply $f_C(\mathfrak {s}(a), X) = f_C(\mathfrak {s}(b), X)$ . Suppose then $f_C(\mathfrak {s}(a), X) \subseteq \mathfrak {s}(b)$ and $f_C(\mathfrak {s}(b), X) \subseteq \mathfrak {s}(A)$ , which by Lemma 3.4 is equivalent to assuming that and . Then also , and by the property (2) of $\mathsf {V}$ -algebras, we get that:

We prove $f_C(\mathfrak {s}(a), X) \subseteq f_C(\mathfrak {s}(b), X)$ , the reader shall observe that the other inclusion is proved symmetrically. Assume $Y \in f_C(\mathfrak {s}(a), X)$ , i.e., implies $c \in Y$ , we show that then $Y \in f_C(\mathfrak {s}(b), X)$ . Indeed, suppose , then since also and X is a Boolean filter, we get that ; therefore $c \in Y$ , which shows $Y \in f_C(\mathfrak {s}(b), X)$ .

For ( $\alpha $ 3), we verify that, given $a,b \in C, X \in Ul({\mathbf{C}})$ , either $f_C(\mathfrak {s}(a) \cup \mathfrak {s}(b), X) \subseteq \mathfrak {s}(a)$ , or $f_C(\mathfrak {s}(a) \cup \mathfrak {s}(b), X) \subseteq \mathfrak {s}(b)$ , or $f_C(\mathfrak {s}(a) \cup \mathfrak {s}(b), X) = f_C(\mathfrak {s}(a), X) \cup f_C(\mathfrak {s}(b), X)$ . By property (3) of $\mathsf {V}$ -algebras, for all $a,b,c \in C$ :

Since X is an ultrafilter, either , or . Using again Lemma 3.4, the first two cases correspond to, respectively, $f_C(\mathfrak {s}(a) \cup \mathfrak {s}(b), X) \subseteq \mathfrak {s}(a)$ and $f_C(\mathfrak {s}(a) \cup \mathfrak {s}(b), X) \subseteq \mathfrak {s}(b)$ . Consider then the remaining case where . Then by the definition of $\leftrightarrow $ and the fact that $\to $ distributes over $\land $ we get that also the following elements are in X:

We show that $f_C(\mathfrak {s}(a) \cup \mathfrak {s}(b), X) = f_C(\mathfrak {s}(a), X) \cup f_C(\mathfrak {s}(b), X)$ . First, consider $Y \in f_C(\mathfrak {s}(a), X) \cup f_C(\mathfrak {s}(b), X)$ ; without loss of generality, we assume $Y \in f_C(\mathfrak {s}(a), X)$ , i.e., implies $c \in Y$ . We prove $Y \in f_C(\mathfrak {s}(a) \cup \mathfrak {s}(b), X)$ , i.e., implies $c \in Y$ . Indeed, if , given that also , we get and then $c \in Y$ .

For the other inclusion, suppose that $Y \notin f_C(\mathfrak {s}(a), X) \cup f_C(\mathfrak {s}(b), X)$ , we prove that $Y \notin f_C(\mathfrak {s}(a) \cup \mathfrak {s}(b), X)$ . By assumption, there is $c \in C$ such that but $c \notin Y$ ; in more detail, there must be $c', c" \in C$ such that but $c',c" \notin Y$ , then one can take $c = c' \lor c"$ . Now, , and also ; thus is in X too. But $c \notin Y$ , thus $Y \notin f_C(\mathfrak {s}(a) \cup \mathfrak {s}(b), X)$ .

( $\alpha $ 4) requires that for all $a, b \in C$ , is clopen. Using the definition of and Lemma 3.4 we get that $\mathfrak {s}$ is also a homomorphism with respect to :

which implies that is clopen.

Lastly, to verify ( $\alpha $ 5) we show that each $f_C(\mathfrak {s}(a), X)$ is closed. We start by proving that

$$ \begin{align*}f_C(\mathfrak{s}(a), X) = \bigcap \{\mathfrak{s}(b): f_C(\mathfrak{s}(a), X) \subseteq \mathfrak{s}(b)\}.\end{align*} $$

While the left-to-right inclusion is clear, we verify the converse. Let $Y \in \bigcap \{\mathfrak {s}(b): f_C(\mathfrak {s}(a), X) \subseteq \mathfrak {s}(b)\}$ , then Y is such that $f_C(\mathfrak {s}(a), X) \subseteq \mathfrak {s}(b)$ implies $Y \in \mathfrak {s}(b)$ for any $b \in C$ ; once again using Lemma 3.4, the latter translates to saying that implies $b \in Y$ for any $b \in C$ , which is exactly the definition of $Y \in f_C(\mathfrak {s}(a),X)$ . Thus, $f_C(\mathfrak {s}(a), X) = \bigcap \{\mathfrak {s}(b): f_C(\mathfrak {s}(a), X) \subseteq \mathfrak {s}(b)\}$ , which means that $f_C(\mathfrak {s}(a), X)$ is an intersection of clopen sets, and therefore it is closed.here

Proof. The isomorphism is given by the Stone map $\mathfrak {s}$ which also preserves the operation , indeed as shown in the proof of Proposition 3.5 .

Proof. Following the definitions, and given that $\beta [A] = \mathfrak {s}(A)$ , we get:

where the last step is the tightness property discussed in Remark 3.2, which follows from the fact that $f(A,x) = \bigcap \{C \in Cl(S): f(A,x) \subseteq C\}$ . Indeed, by axiom $(\alpha $ 5), $f(A,x)$ is closed and hence can be written as the intersection of all the clopen sets that contain it.

Proof. $\mathscr{S}(h)$ is a continuous map from $\mathscr{A}({\mathbf{C}})$ to $\mathscr{A}({\mathbf{C}}')$ by Stone duality. Let us now show that for all $a' \in C', X,Y \in Ul({\mathbf{C}})$ ,

$$ \begin{align*}Y \in f_C(\mathscr{S}(h)^{-1}[\mathfrak{s}(a')], X) \,\mbox{ implies }\, \mathscr{S}(h)(Y) \in f_{C'}(\mathfrak{s}(a'), \mathscr{S}(h)(X)).\end{align*} $$

Suppose then that $Y \in f_C(\mathscr{S}(h)^{-1}[\mathfrak {s}(a')], X)$ . Direct calculations show that $\mathscr{S}(h)^{-1}[\mathfrak {s}(a')] = \mathfrak {s}(h(a))$ ; thus $Y \in f_C(\mathscr{S}(h)^{-1}[\mathfrak {s}(a')], X)$ if and only if $Y \in f_C(\mathfrak {s}(h(a')), X)$ if and only if, by definition of $f_C$ , implies $c \in Y$ for any $c \in C$ . Therefore, consider any $c' \in C'$ , $h(c') \in C$ , and we get that implies $h(c') \in Y$ ; equivalently, for any $c' \in C'$ , implies $c' \in h^{-1}(Y) = \mathscr{S}(h)(Y)$ . We have showed $\mathscr{S}(h)(Y) \in f_{C'}(\mathfrak {s}(a'), \mathscr{S}(h)(X))$ .

It is left to show that for all $a' \in C', X \in Ul({\mathbf{C}}), Y' \in Ul({\mathbf{C}}')$ , $Y'\in f_{C'}(\mathfrak {s}(a'), \mathscr{S}(h)(X))$ implies that there exists $Y \in Ul(C)$ such that $\mathscr{S}(h)(Y) = Y'$ and $Y \in f_C(\mathscr{S}(h)^{-1}[\mathfrak {s}(a')], X)$ . Assume $Y' \in f_{C'}(\mathfrak {s}(a'), \mathscr{S}(h)(X))$ , i.e., for all $c' \in C'$ , implies $c' \in Y'$ . Equivalently, for all $c' \in C'$ ,

(8)

We want to find $Y \in Ul(C)$ such that $\mathscr{S}(h)(Y) = Y'$ and $Y \in f_C(\mathscr{S}(h)^{-1}[\mathfrak {s}(a')], X)$ , equivalently $Y \in f_C(\mathfrak {s}(h(a')), X)$ , which means that for all $c \in C$ , implies $c \in Y$ . Consider the sets

F is easily seen to be a filter of the Boolean reduct of ${\mathbf{C}}$ , I is a Boolean ideal, and we can show that $I \cap F = \emptyset $ ; indeed, if $d \in I \cap F$ , we get $d \leq h(c')$ for some $c' \notin Y'$ , and . But then also , which by (8) above implies $c' \in Y'$ , a contradiction. Thus we can apply the Boolean Ultrafilter Theorem and obtain an ultrafilter $Y \in Ul(C)$ such that $F \subseteq Y$ and $Y \cap I = \emptyset $ . By construction, for all $c \in C$ , implies $c \in Y$ , since . We now prove that $\mathscr{S}(h)(Y) = Y'$ , which will conclude the proof. Consider $c' \notin Y'$ , then $h(c') \in I$ , and $h(c') \notin Y$ , and thus ${c' \notin h^{-1}[Y] = \mathscr{S}(h)(Y)}$ . This implies that $\mathscr{S}(h)(Y) \subseteq Y'$ . Since $\mathscr{S}(h)(Y)$ is an ultrafilter and therefore maximal, it follows that $\mathscr{S}(h)(Y) = Y'$ , completing the proof.

Proof. $\mathscr{S}^{\ -1}(\varphi )$ is a homomorphism on the Boolean reducts by Stone duality, thus we only need to show that for all $A,B \in Cl(S)$ , .

Now,

and

Let , i.e., $f(A, \varphi (x')) \subseteq B$ ; we show that , i.e., $f'(\varphi ^{-1}[A], x') \subseteq \varphi ^{-1}[B]$ . Indeed, if $y' \in f'(\varphi ^{-1}[A], x')$ , by Definition 3.8(2) we get $\varphi (y') \in f(A, \varphi (x'))$ and then $\varphi (y') \in B$ , that is, $y' \in \varphi ^{-1}[B]$ .

Finally, let , i.e., $f'(\varphi ^{-1}[A], x') \subseteq \varphi ^{-1}[B]$ , we verify that , i.e., $f(A, \varphi (x')) \subseteq B$ . If $y \in f(A, \varphi (x'))$ , by Definition 3.8(3) we get that there exists $y' \in S'$ such that $\varphi (y') = y$ and $y' \in f'(\varphi ^{-1}[A], x')$ ; then $y' \in \varphi ^{-1}[B]$ , that is, $\varphi (y') = y \in B$ . This shows the second inclusion and concludes the proof.

Proof. The dual equivalence between $\mathsf {V}$ -algebras and $\mathsf {AM}$ , or equivalently the categorical equivalence between $\mathsf {V}$ -algebras and the opposite category $\mathsf {AM}^{\mathrm {{op}}}$ (obtained by reversing the morphisms), is established as an extension of Stone duality. Note that via Propositions 3.9 and 3.10, we get that both $\mathscr{C}$ and $\mathscr{A}$ are full and faithful (with respect to the opposite category). It is then left to prove that $\mathscr{A}$ and $\mathscr{C}$ are adjoint functors. By Proposition 3.6, the composition $\mathscr{C} \circ \mathscr{A}$ is naturally isomorphic to the identity functor on $\mathsf {VA}$ , via the Stone map $\mathfrak {s}$ . Similarly, by Proposition 3.7 the composition $\mathscr{A} \circ \mathscr{C}$ is naturally isomorphic to the identity on $\mathsf {AM}$ , via the Stone homeomorphism $\beta $ . This completes the proof.

Proof. For the first point, let us start by recalling that $\mathscr{C}(\mathfrak {S}) \in \mathsf {VCA}$ if and only if

Moreover, by definition, . Let now $x \in A \cap B$ ; then in particular $x \in A$ and then if $\mathfrak {S}$ is a topological $\alpha _1$ -model, $f(A, x) =\{x\} \subseteq B$ , which shows the first inclusion. For the second one, let ; if $x \notin A^C$ , i.e., $x \in A$ , once again $f(A, x) =\{x\}$ , and then $x \in B$ given that .

For the second point, we recall that $\mathscr{C}(\mathfrak {S}) \in \mathsf {VCSA}$ if and only if , where

Consider any $x \in S$ ; if $\mathfrak {S}$ is a topological $\alpha _2$ -model, $f(A,x)$ contains at most one element. Thus either $f(A,x) = \emptyset $ , and then x is in both and , or $f(A,x)= \{z\}$ ; thus since either $z \in B$ or $z \in B^C$ , either or .

Proof. Suppose ${\mathbf{C}} \in \mathsf {VCA}$ , we show that for each $a \in C$ and $X \in Ul({\mathbf{C}})$ such that $a \in X$ , $f(\mathfrak {s}(a),X) = \{X\}$ . First, we show $X \in f(\mathfrak {s}(a),X)$ ; let , then . Now, $a, a \to b \in X$ implies $b \in X$ , which proves $X \in f(\mathfrak {s}(a),X)$ . We now verify that if $Y \in f(\mathfrak {s}(a),X)$ holds, then $Y=X$ . Indeed, let $b \in X$ ; since also $a \in X$ , $a \land b \in X$ , and , thus from $Y \in f(\mathfrak {s}(a),X)$ we get $b \in Y$ , which shows $X \subseteq Y$ . Since X is an ultrafilter and hence maximal, it follows that $X=Y$ , completing the proof of (1).

For the second point, suppose ${\mathbf{C}} \in \mathsf {VCSA}$ . We prove that $f_C(\mathfrak {s}(a), X)$ has at most one element for all $a \in C, X \in Ul({\mathbf{C}})$ . Suppose that both $Y \in f_C(\mathfrak {s}(a),X)$ and $Z \in f_C(\mathfrak {s}(a),X)$ hold; this means that implies $b \in Y \cap Z$ . Let $c \in Y$ ; since ${\mathbf{C}} \in \mathsf {VCSA}$ , and hence or , since X is a Boolean ultrafilter. If , then $c \in Y \cap Z \subseteq Z$ ; instead yields a contradiction, since it entails that $\neg c \in Y\cap Z \subseteq Y$ . Thus $c \in Z$ and $Y \subseteq Z$ . By the maximality of the ultrafilter Y, we have $Y=Z$ , which concludes the proof.

Proof. We have to show that $f_\sigma $ satisfies the conditions ( $\alpha $ 1)–( $\alpha $ 5) of Definition 3.1. ( $\alpha $ 1) is easily seen, since $f_\sigma (A, x) \subseteq A$ for all $A \in Cl(S)$ by definition.

For ( $\alpha $ 2), we proceed to prove that $f_\sigma (A, x) \subseteq B$ and $f_\sigma (B, x) \subseteq A$ implies $f_\sigma (A, x) = f_\sigma (B, x)$ . Let us then assume that $f_\sigma (A, x) \subseteq B$ and $f_\sigma (B, x) \subseteq A$ . It follows by the definition and Lemma 4.3 that either both sets are empty (and there is nothing else to prove), or they are both nonempty. Suppose we are in the latter case; thus $f_\sigma (A, x) = A \cap \Sigma (A, x) \subseteq B$ and $f_\sigma (B, x) = B \cap \Sigma (B, x) \subseteq A$ . Since

$$ \begin{align*}\emptyset \neq f_\sigma(B, x) = B \cap \Sigma(B, x) \subseteq A,\end{align*} $$

we get that $\Sigma (B, x) \cap A \neq \emptyset $ ; but $\Sigma (A, x)$ is the least element in $\sigma (x)$ that intersects A, thus $\Sigma (A, x) \subseteq \Sigma (B, x)$ . Analogously, $\Sigma (B, x) \subseteq \Sigma (A, x)$ and therefore $\Sigma (A, x) = \Sigma (B, x)$ . Thus

$$ \begin{align*}f_\sigma(A, x) &= A \cap \Sigma(A, x) = A \cap \Sigma(A, x) \cap B = A \cap \Sigma(B, x) \cap B\\ & = \Sigma(B, x) \cap B = f_\sigma(B,x).\end{align*} $$

Let us move to ( $\alpha $ 3). Given $A, B \in Cl(S), x \in S$ we show that either $f_\sigma (A \cup B, x) \subseteq A$ , or $f_\sigma (A \cup B, x) \subseteq B$ , or $f_\sigma (A \cup B, x) = f_\sigma (A, x) \cup f_\sigma (B, x)$ . In other words:

$$ \begin{align*}(A \cup B) \cap \Sigma(A \cup B,x) &\subseteq A \mbox{ or } (A \cup B) \cap \Sigma(A \cup B,x)\\& \subseteq B \mbox{ or } (A \cup B) \cap \Sigma(A \cup B,x) = (A \cap \Sigma(A,x)) \cup (B \cap \Sigma(B,x)).\end{align*} $$

Since $\sigma (x)$ is nested, we can assume without loss of generality that $\Sigma (A,x) \subseteq \Sigma (B,x)$ . We consider the following cases:

1. 1. $\emptyset \neq \Sigma (A,x) = \Sigma (B,x)$ ; then necessarily $\Sigma (A \cup B,x) = \mathrm {min} \{U \in \sigma (x): U \cap (A \cup B) \neq 0\} = \Sigma (A,x) = \Sigma (B,x)$ . Thus: $(A \cup B) \cap \Sigma (A \cup B,x) = (A \cap \Sigma (A \cup B,x)) \cup (B \cap \Sigma (A \cup B,x)) = (A \cap \Sigma (A,x)) \cup (B \cap \Sigma (B,x)).$

2. 2. $\emptyset \neq \Sigma (A,x) \subsetneq \Sigma (B,x)$ ; it follows that $\Sigma (A,x) \cap B = \emptyset $ and then $\emptyset \neq \Sigma (A \cup B,x) = \Sigma (A,x)$ . Hence $(A \cup B) \cap \Sigma (A \cup B,x) = (A \cap \Sigma (A \cup B,x)) \cup (B \cap \Sigma (A \cup B,x)) = A \cap \Sigma (A, x) \subseteq A.$

3. 3. $\emptyset = \Sigma (A,x) = \Sigma (B,x)$ ; in this case also $\Sigma (A \cup B,x) = \emptyset $ , and then trivially the last condition holds, i.e., $(A \cup B) \cap \Sigma (A \cup B,x) = (A \cap \Sigma (A,x)) \cup (B \cap \Sigma (B,x))$ .

4. 4. $\emptyset = \Sigma (A,x) \subsetneq \Sigma (B,x)$ ; then $\Sigma (A \cup B,x) = \Sigma (B,x)$ and $A \cap \Sigma (A \cup B,x) = \emptyset $ , hence

   $$ \begin{align*}(A \cup B) \cap \Sigma(A \cup B,x) &= (A \cap \Sigma(A \cup B,x)) \cup (B \cap \Sigma(A \cup B,x))\\& = B \cap \Sigma(B, x) \subseteq B.\end{align*} $$

This proves ( $\alpha $ 3). For ( $\alpha $ 4), the fact that is clopen follows from the fact that is clopen and and coincide, indeed:

Lastly, $f_\sigma (A, x) = A \cap \Sigma (A, x)$ is closed by (S5) in the definition, and ( $\alpha $ 5) holds.

Proof. For the first property, suppose $A \subseteq B$ and $A \cap f(B,x) \neq \emptyset $ . Notice that

$$ \begin{align*}B = B \cap (A \cup A^C) = (B \cap A) \cup (B \cap A^C).\end{align*} $$

In particular, since $A \subseteq B$ , $B = A \cup (B \cap A^C)$ and then $f(B,x) = f(A \cup (B \cap A^C), x)$ . Thus, by ( $\alpha 3$ ) we get the following cases:

1. (i) $f(B,x) \subseteq A$ , and therefore, since by ( $\alpha 1$ ) $f(A,x) \subseteq A \subseteq B$ , by ( $\alpha 2$ ) $f(B,x) = f(A,x)$ , which implies that $f(A,x) = A \cap f(B,x)$ ;

2. (ii) $f(B,x) \subseteq B \cap A^C \subseteq A^C$ , which yields a contradiction since $A \cap f(B,x) \neq \emptyset $ , and thus it is never the case;

3. (iii) $f(B,x) = f(A,x) \cup f(B \cap A^C, x)$ , which (together with ( $\alpha 1$ )) directly implies that $f(A,x) \subseteq A \cap f(B,x)$ . For the other inclusion, let $y \in A \cap f(B,x)$ ; then $y \in f(A,x) \cup f(B \cap A^C, x)$ . Since $f(B \cap A^C, x) \subseteq B \cap A^C \subseteq A^C$ , and $y \in A$ , it follows that $y \notin f(B \cap A^C, x)$ . Therefore, $y \in f(A,x)$ , and we have shown that $f(A,x) = A \cap f(B,x)$ .

Hence, in all the possible cases $f(A,x) = A \cap f(B,x)$ .

The second property is easier to show. Indeed, suppose $A \subseteq B$ and $f(B,x) = \emptyset $ . Then $f(B,x) = \emptyset \subseteq A$ and, using ( $\alpha 1$ ), $f(A,x) \subseteq A \subseteq B$ ; by ( $\alpha 2$ ) it follows that $f(A,x) = f(B,x) =\emptyset $ .

For (3), note that if $A \subseteq B$ and $\emptyset \neq f(B,x) \subseteq A$ , then in particular $A \cap f(B,x) \neq \emptyset $ , and then by (1) $f(A,x) = A \cap f(B,x)$ . This implies that $f(A,x) \subseteq f(B,x) \subseteq B$ (by ( $\alpha $ 1)). Hence by ( $\alpha $ 2) we get $f(A,x) = f(B,x)$ .

Proof. (1) We start by verifying that $<_x$ is a total preorder on $Cl(S)$ for each $x \in S$ . Reflexivity is straightforward from the definition, so let us prove transitivity. We assume that $A <_x B$ and $B <_x C$ , i.e.,

$$ \begin{align*}f(B,x) = \emptyset \;\mbox{ or }\;\; \emptyset \neq f(A,x) \subseteq f(A \cup B,x),\end{align*} $$

$$ \begin{align*}f(C,x) = \emptyset \;\mbox{ or }\;\; \emptyset \neq f(B,x) \subseteq f(B \cup C,x),\end{align*} $$

and we show that $A <_x C$ , that is, $f(C,x) = \emptyset \;\mbox { or }\;\; \emptyset \neq f(A,x) \subseteq f(A \cup C, x)$ . Suppose then that $f(C,x) \neq \emptyset $ ; thus since $B <_x C$ , $\emptyset \neq f(B,x) \subseteq f(B \cup C,x)$ , and since $A <_x B$ also $\emptyset \neq f(A,x) \subseteq f(A \cup B,x)$ . We need to show that $f(A,x) \subseteq f(A \cup C, x)$ . First, we observe that $(A \cup B) \cap f(A \cup (B \cup C), x) \neq \emptyset $ . Indeed, from $(\alpha 3)$ we get that either $f(A \cup (B \cup C), x) \subseteq A$ , $f(A \cup (B \cup C), x) \subseteq B \cup C$ , or $f(A \cup (B \cup C), x) =f(A,x) \cup f(B \cup C, x)$ . In the first case it is clear that $f(A \cup (B \cup C), x)$ intersects with $A \cup B$ . If $f(A \cup (B \cup C), x) \subseteq B \cup C$ , by Lemma 4.5(3) we get that $f(A \cup (B \cup C), x) = f(B \cup C, x) \supseteq f(B,x)$ (the last inclusion by hypothesis), and then again it intersects $A \cup B$ . In the last case, $f(A \cup (B \cup C), x) \supseteq f(B \cup C, x)$ which again contains $f(B,x)$ , and so in any case $f(A \cup (B \cup C), x)$ intersects $A \cup B$ . Therefore (using Lemma 4.5(1) for the last inclusion):

$$ \begin{align*}\emptyset \neq f(A, x) \subseteq f(A \cup B,x) \subseteq f(A \cup B \cup C, x),\end{align*} $$

which implies that $A \cap f(A \cup B \cup C, x) \neq \emptyset $ . Then also $(A \cup C) \cap f(A \cup B \cup C, x) \neq \emptyset $ , and again by Lemma 4.5,

$$ \begin{align*}f(A \cup C, x) &= (A \cup C) \cap f(A \cup B \cup C, x)\\& = (A \cap f(A \cup B \cup C, x)) \cup (C \cap f(A \cup B \cup C, x)).\end{align*} $$

Thus, since $A \cap f(A \cup B \cup C, x) \neq \emptyset $ , we get that $A \cap f(A \cup C, x)\neq \emptyset $ , and by a last application of Lemma 4.5 we obtain that $f(A,x) \subseteq f(A \cup C, x)$ which shows that $A <_x C$ ; hence $<_x$ is transitive for all $x \in S$ .

For the first point, it is left to show that $<_x$ is total. Consider $f(A \cup B,x)$ ; if $f(A \cup B,x) = \emptyset $ , by Lemma 4.5 also $f(A,x) = f(B,x) = \emptyset $ , and thus $A <_x B$ and $B <_x A$ . If $f(A \cup B,x) \neq \emptyset $ , then by $(\alpha 3)$ either $f(A \cup B,x) \subseteq A$ , which by Lemma 4.5(3) yields $f(A,x) = f(A \cup B,x)$ and then $A <_x B$ , or $f(A \cup B,x) \subseteq B$ , which analogously yields $B <_x A$ , or $f(A \cup B,x) = f(A,x) \cup f(B,x)$ , and therefore $A <_x B$ and $B <_x A$ . In any case, A and B are comparable with respect to $<_x$ , which is then a total preorder.

(2) Suppose that $A <_x B$ , i.e.,

$$ \begin{align*}f(B,x) = \emptyset \;\mbox{ or }\;\; \emptyset \neq f(A,x) \subseteq f(A \cup B,x).\end{align*} $$

Assume first that $f(B,x) = \emptyset $ , and let us consider two cases: whether $f(A \cup B,x)$ is empty or not. If $f(A \cup B,x) = \emptyset $ , by Lemma 4.5 also $f(A,x) = \emptyset $ and then $B \cap f(A,x) =\emptyset = f(B,x)$ . If $f(A \cup B,x) \neq \emptyset $ , from the fact that $f(B,x) = \emptyset $ , we get that $f(A \cup B,x) = f(A,x)$ using $(\alpha 3)$ and Lemma 4.5(3), and, necessarily, $B \cap f(A \cup B,x) = \emptyset $ by Lemma 4.5(1). Thus $B \cap f(A,x) = \emptyset = f(B,x)$ .

Assume now that $\emptyset \neq f(A,x) \subseteq f(A \cup B,x)$ . Then either $B \cap f(A,x) = \emptyset \subseteq f(B,x)$ or

$$ \begin{align*}\emptyset \neq B \cap f(A,x) \subseteq B \cap f(A \cup B,x)\end{align*} $$

and then by Lemma 4.5(1) $f(B,x) = B \cap f(A \cup B,x)$ thus

$$ \begin{align*}B \cap f(A,x) \subseteq B \cap f(A \cup B,x) = f(B,x).\end{align*} $$

This completes the proof of (2).

(3) Finally, assume $A \cap f(B,x) \neq \emptyset $ , which implies that in particular $f(B,x) \neq \emptyset $ . Consider $f(A \cup B,x)$ , then by $(\alpha 3)$ we have three cases:

1. (i) $f(A \cup B,x) \subseteq A$ ; then by Lemma 4.5(3), $f(A \cup B,x) = f(A,x)$ which is necessarily nonempty, since otherwise by Lemma 4.5(2) also $f(B,x)$ would be empty, which is a contradiction. Thus we have that $A <_x B$ .

2. (ii) $f(A \cup B,x) \subseteq B$ ; then again by Lemma 4.5(3), $f(A \cup B,x) = f(B,x)$ and then $A \cap f(A \cup B,x) = A \cap f(B,x) \neq \emptyset $ . By Lemma 4.5(1) this yields that $\emptyset \neq f(A, x) \subseteq f(A \cup B, x)$ and then $A <_x B$ .

3. (iii) $f(A \cup B,x) = f(A,x) \cup f(B,x)$ ; thus $f(A, x) \subseteq f(A \cup B, x)$ and $f(A, x)$ is nonempty (otherwise we would be in the case above), and once again $A <_x B$ .

We have shown that in every case $A <_x B$ , which concludes this proof.

Proof. We show that $\sigma _f$ satisfies the properties (S1)–(S4) of Definition 4.1. First, notice that $\sigma _f$ , defined as

$$ \begin{align*}\sigma_f(x) = \left\{\bigcup_{A <_x B} f(A,x) : B \in Cl(S)\right\} \end{align*} $$

is indeed a map from S to $\mathcal {P}(\mathcal {P}(S))$ . For (S1), note that for all $x \in S$ , $\sigma _f(x)$ is nested, since $<_x$ is total by Lemma 4.6.

Let now $\Sigma _f(A,x) = \emptyset $ if $A \cap \bigcup \sigma _f(x) = \emptyset $ , and $\Sigma _f(A,x) := \bigcap \{U \in \sigma _f(x): U \cap A \neq \emptyset \}$ otherwise. We prove (S3), i.e., that for all $x \in S$ and $A \in Cl(S)$ such that $A \cap \bigcup \sigma _f(x) \neq \emptyset $ , $\Sigma _f(A,x) \in \sigma _f(x)$ and $\Sigma _f(A,x) \cap A \neq \emptyset $ . Suppose then that $A \cap \bigcup \sigma _f(x) \neq \emptyset $ , hence we will show that

$$ \begin{align*}\Sigma_f(A,x) = \displaystyle\bigcup_{C <_x A} f(C,x).\end{align*} $$

Call $U_A = \bigcup _{C <_x A} f(C,x)$ . Then $U_A \in \sigma _f(x)$ by definition. We verify that $U_A \cap A \neq \emptyset $ . Observe that:

(12) $$ \begin{align} f(A,x) = \emptyset \;\; \mbox{ iff }\;\;\not\exists U\in \sigma_f(x): A \cap U \neq \emptyset. \end{align} $$

Indeed, the right-to-left direction is obvious since $f(A, x) \in U_A$ and $f(A,x) \subseteq A$ . We show the left-to-right one by contraposition. Note that if there is $U \in \sigma _f(x)$ such that $A \cap U \neq \emptyset $ , then there is a B such that $A \cap f(B,x) \neq \emptyset $ and so $A <_x B$ by Lemma 4.6; thus $f(A, x) \neq \emptyset $ by the definition of $<_x$ . Hence, since by hypothesis $A \cap \bigcup \sigma _f(x) \neq \emptyset $ , $\emptyset \neq f(A, x) \subseteq A$ (the last inclusion by $(\alpha 1)$ ), and since $A <_x A$ ( $<_x$ is reflexive by Lemma 4.6), then $A \cap U_A \neq \emptyset $ .

We now observe that $U_A$ is the smallest $U \in \sigma _f(x), U \cap A \neq \emptyset $ ; indeed, suppose $A \cap U_B = A \cap \bigcup _{C <_x B} f(C,x) \neq \emptyset $ . Then $A \cap f(C,x) \neq \emptyset $ for some $C <_x B$ , thus by Lemma 4.6 $A <_x C$ ; by the transitivity of $<_x$ , it holds that $A <_x B$ , thus $U_A \subseteq U_B$ . It follows that $U_A = \bigcap \{U \in \sigma _f(x): U \cap A \neq \emptyset \} = \Sigma _f(A, x) \in \sigma _f(x)$ , and also $\Sigma _f(A, x) \cap A \neq \emptyset $ .

We now proceed to prove (S4), that is, for all $x \in S$ , $A \in Cl(S)$ , $A \cap \Sigma _f(A,x)$ is closed. By what we have shown in the previous point, either $\Sigma _f(A,x) = \emptyset $ or $\Sigma _f(A,x) = U_A = \bigcup _{C <_x A} f(C,x)$ . In the first case, the empty set is a closed set in any topological space. In the second case, note that by Lemma 4.6 for all $C <_x A$ , $A \cap f(C,x) \subseteq f(A,x) = A \cap f(A,x)$ . Thus, since $A <_x A$ :

$$ \begin{align*}A \cap \displaystyle\bigcup_{C <_x A} f(C,x) = \displaystyle\bigcup_{C <_x A} A \cap f(C,x) = A \cap f(A,x)\end{align*} $$

and the latter set is closed, since $f(A,x)$ is closed by definition of a topological $\alpha $ -model, A is closed, and closed sets are closed under finite intersection.

Finally, for (S2), the fact that is clopen for all $A,B \in Cl(S)$ follows from the fact that , given that the latter is clopen by definition of a topological $\alpha $ -model. Let us verify the equality. It is helpful to first recall the definitions:

By (12) above, $f(A, x) = \emptyset $ if and only if there is no $U \in \sigma _f(x), U \cap A \neq \emptyset $ . If x is such, then $\Sigma _f(A,x) = \emptyset $ and therefore iff for any $B \in Cl(S)$ .

Consider now $x \in S$ such that $f(A, x) \neq \emptyset $ , and then iff $A \cap U_A \subseteq B$ . Hence, also since $f(A,x) \subseteq A \cap U_A$ . Conversely, suppose that , i.e., $f(A,x) \subseteq B$ . Then,

$$ \begin{align*}A \cap U_A = A \cap \displaystyle\bigcup_{C <_x A} f(C,x) \subseteq f(A,x) \subseteq B,\end{align*} $$

since if $C <_x A$ then $A \cap f(C,x) \subseteq f(A, x)$ by Lemma 4.6, and so . This completes the proof.

Proof. We verify that given any $A \in Cl(S), x,y \in S$ , $y \in f(A,x)$ if and only if $y \in f_{\sigma _{f}}(A,x)$ . Recall that

$$ \begin{align*}f_{\sigma_{f}}(A,x)& = \emptyset \mbox{ if } A \cap \bigcup\sigma_f(x) = \emptyset,\\f_{\sigma_{f}}(A,x)& = A \cap \displaystyle\bigcap\{U: U \in \sigma_f(X), U \cap A \neq \emptyset\} \mbox{ otherwise}.\end{align*} $$

First, note that $f(A,x) = \emptyset $ iff there is no $U \in \sigma _f(x)$ such that $A \cap U \neq \emptyset $ , as shown in the proof for (S3) of Proposition 4.7. Thus $f(A,x) = \emptyset $ iff $f_{\sigma _f} (A,x) = \emptyset $ .

Suppose then that $f(A, x) \neq \emptyset $ , and let $y \in f(A,x)$ . Since, again by the proof of (S3) in Proposition 4.7, we have that the smallest sphere intersecting A is $U_A = \bigcup _{C <_x A}f(C,x)$ , by definition $f_{\sigma _{f}}(A,x) = A \cap U_A$ . Given that $y \in f(A, x) \subseteq A$ , and since $A <_x A$ , we obtain that $y \in f_{\sigma _{f}}(A,x)$ .

Vice versa, suppose $y \in f_{\sigma _{f}}(A,x)$ . Since $f_{\sigma _{f}}(A,x) = A \cap \bigcup _{C <_x A}f(C,x)$ , and if $C <_x A$ then $A \cap f(C,x) \subseteq f(A, x)$ by Lemma 4.6, $y \in f_{\sigma _{f}}(A,x)$ implies $y \in f(A, x)$ , which concludes the proof.

Proof. We only need to show the two following properties:

1. 1. $y \in f_\sigma (\varphi ^{-1}[A'], x)$ implies $\varphi (y) \in f_{\sigma '}(A', \varphi (x))$ ;

2. 2. $y' \in f_{\sigma '}(A', \varphi (x))$ implies that there exists $y \in S$ such that $\varphi (y) = y'$ and $y \in f_\sigma (\varphi ^{-1}[A'], x)$ .

For (1), assume that $y \in f_\sigma (\varphi ^{-1}[A'], x)$ . Since

$$ \begin{align*}f_\sigma(\varphi^{-1}[A'], x) = \varphi^{-1}[A'] \cap \Sigma(\varphi^{-1}[A'], x)\end{align*} $$

we get that $\varphi ^{-1}[A'] \cap \bigcup \sigma (x) \neq \emptyset $ , and $y \in \varphi ^{-1}(A') \cap \Sigma (\varphi ^{-1}[A'], x)$ ; by definition of a sphere morphism, this implies $\varphi (y) \in \Sigma (A', \varphi (x))$ , and of course $\varphi (y) \in A'$ . Thus, since $f_{\sigma '}(A', \varphi (x)) = A' \cap \Sigma (A', \varphi (x))$ , we get $\varphi (y) \in f_{\sigma '}(A', \varphi (x))$ .

For (2) we proceed similarly; suppose $y' \in f_{\sigma '}(A', \varphi (x))$ . Thus $A' \cap \bigcup \sigma '(\varphi (x)) \neq \emptyset $ , and by definition of $f_{\sigma '}(A', \varphi (x))$ , $y' \in A' \cap \Sigma (A', \varphi (x))$ . Hence, by definition of a sphere morphism, there is a $y \in S$ such that $\varphi (y) = y'$ and $y \in \Sigma (\varphi ^{-1}[A'], x)$ ; this yields exactly $y \in f_\sigma (\varphi ^{-1}[A'], x)$ .

Proof. We need to show the following:

1. 1. for all $y \in S, A' \in Cl(S')$ such that $\varphi ^{-1}[A'] \cap \bigcup \sigma _f(x) \neq \emptyset $ , $y \in \varphi ^{-1}(A') \cap \Sigma _f(\varphi ^{-1}[A'], x)$ implies $\varphi (y) \in \Sigma _{f'}(A', \varphi (x))$ ;

2. 2. for all $y' \in S', A' \in Cl(S')$ such that $A' \cap \bigcup \sigma _{f'}(\varphi (x)) \neq \emptyset $ , $y' \in A' \cap \Sigma _f(A', \varphi (x))$ implies that there is $y \in S$ such that $\varphi (y) = y'$ and $y \in \Sigma _f(\varphi ^{-1}[A'], x)$ .

Let us show (1). Assume that $\varphi ^{-1}[A'] \cap \bigcup \sigma _f(x) \neq \emptyset $ , $y \in \varphi ^{-1}(A') \cap \Sigma _f(\varphi ^{-1}[A'], x)$ . Then we have that:

$$ \begin{align*}\Sigma_f(\varphi^{-1}[A'], x) = \bigcup_{C<_x \varphi^{-1}[A']}f(C,x).\end{align*} $$

Then $y \in \varphi ^{-1}(A') \cap f(C,x)$ for some $C<_x \varphi ^{-1}[A']$ ; thus $y \in f(\varphi ^{-1}(A'), x)$ by Lemma 4.6. Moreover by definition of an $\alpha $ -morphism, we get that $\varphi (y) \in f'(A', \varphi (x))$ . Thus

$$ \begin{align*}\varphi(y) \in A' \cap \bigcup_{C'<_x A'} f(C', \varphi(x)) = \Sigma_{f'}(A', \varphi(x)).\end{align*} $$

(2) can be verified by easy calculations in a similar fashion; we leave its proof to the reader.

Proof. In order to show that $\mathcal {F}$ establishes a categorical equivalence, it suffices to say that it is full, faithful and essentially surjective (see, for instance, [Reference Mac Lane20]). Fullness and faithfulness (i.e., injectivity and surjectivity on morphisms among corresponding pairs of objects) are obvious since $\mathcal {F}$ is the identity on morphisms. The fact that $\mathcal {F}$ is essentially surjective follows from Proposition 4.8.

Proof. (1) Suppose $(S, \tau , \sigma )$ is centered, i.e., $\{x\} \in \sigma (x)$ for all $x \in S$ . We need to show that if $x \in A$ , $f_\sigma (A,x) = \{x\}$ . Now, $f_\sigma (A,x) = A \cap \Sigma (A, x)$ . Recall that $\Sigma (A, x)$ is either empty or the intersection of all $U \in \sigma (x)$ that intersects with A, and $\Sigma (A, x)$ belongs to $\sigma (x)$ if nonempty. Thus if $x \in A$ , necessarily $\Sigma (A, x) = \{x\}$ , which implies the claim.

(2) Suppose now $(S, \tau , \sigma )$ is Stalnakerian. We want to show that for each $A \in Cl(S)$ and $x \in S$ , $f_\sigma (A,x)$ contains at most one element. By hypothesis, if $A \cap \bigcup \sigma (x) \neq \emptyset $ , then there exists $U \in \sigma (x)$ and $y \in S$ such that $A \cap U = \{y\}$ . Since $f_\sigma (A,x) = A \cap \Sigma (A, x)$ , either $f_\sigma (A,x) = \emptyset $ , or otherwise $A \cap \Sigma (A, x) = \{y\}$ .

Proof. (1) First, assume that for each $A \in Cl(S)$ and $x \in A$ , $f(A,x) = \{x\}$ ; we show that $\{x\} \in \sigma _f(x)$ for all $x \in S$ . Recall that

$$ \begin{align*}\sigma_f(x) = \left\{\bigcup_{B <_x A} f(B,x) : A \in Cl(S)\right\}.\end{align*} $$

Consider a clopen A such that $x \in A$ (since clopen sets are a base of the Stone space, and singletons are closed, there is surely at least one). Then $f(A,x) = \{x\}$ ; moreover, if $B <_x A$ , by definition of the preorder $\emptyset \neq f(B,x) \subseteq f(A \cup B,x)$ , and since $x \in A\subseteq A \cup B$ , by hypothesis $f(A \cup B,x) = \{x\}$ and thus $f(B,x) = \{x\}$ . Hence,

$$ \begin{align*}\{x\} = \bigcup_{B <_x A} f(B,x) \in \sigma_f(x).\end{align*} $$

(2) Now, assume that for each $A \in Cl(S)$ and $x \in S$ , $f(A,x)$ contains at most one element. We prove that if $A \cap \bigcup \sigma _f(x) \neq \emptyset $ , then there exists $U \in \sigma _f(x)$ and $y \in S$ such that $A \cap U = \{y\}$ . Assume $A \cap \bigcup \sigma _f(x) \neq \emptyset $ ; then $A \cap \Sigma _f(A, x)$ is nonempty, and as shown in the proof of Proposition 4.7, $\Sigma _f(A, x) = \bigcup _{B <_x A}f(B,x)$ , hence,

$$ \begin{align*}A \cap \bigcup_{B <_x A}f(B,x)\neq \emptyset.\end{align*} $$

In particular, using Lemma 4.6(2), $f(A, x)$ is nonempty, and thus it contains exactly one element, say $f(A, x) = \{y\}$ . Moreover, if $B <_x A$ , $A \cap f(B,x) \subseteq f(A, x) = \{y\} \subseteq A$ . Therefore, $A \cap \Sigma _f(A, x) = \{y\}$ and $\Sigma _f(A, x)$ is the desired element of $\sigma _f(x)$ .

Proof. We have to show that $f_\prec $ satisfies the conditions ( $\alpha $ 1)–( $\alpha $ 5) of Definition 3.1. For ( $\alpha $ 1), $f_\prec (A, x) \subseteq A$ for all $A \in Cl(S)$ by definition.

We proceed to prove ( $\alpha $ 2), i.e., for arbitrary $x \in S$ and $A, B \in Cl(S)$ , $f_\prec (A, x) \subseteq B$ and $f_\prec (B, x) \subseteq A$ implies $f_\prec (A, x) = f_\prec (B, x)$ . Let us then assume that $f_\prec (A, x) \subseteq B$ and $f_\prec (B, x) \subseteq A$ . It follows by the definition and Lemma 5.2 that either both sets are empty (and there is nothing else to prove), or they are both nonempty. Suppose we are in the latter case; thus $f_\prec (A, x) = min_{\prec _x}(A) \subseteq B$ and $f_\prec (B, x) = min_{\prec _x}(B) \subseteq A$ . Let $y \in f_\prec (A, x)$ and $z \in f_\prec (B, x)$ . Since $f_\prec (A, x) \subseteq B$ we get $y \in B$ , and then from $z \in f_\prec (B, x) = min_{\prec _x}(B)$ it follows that $z \prec _x y$ ; similarly we can show that $y \prec _x z$ . Since y and z were arbitrary elements of $f_\prec (A, x)$ and $f_\prec (B, x)$ , we have that $f_\prec (A, x)=f_\prec (B, x)$ .

For ( $\alpha $ 3) we have to prove that given $A, B \in Cl(S), x \in S$ , either $f_\prec (A \cup B, x) \subseteq A$ , or $f_\prec (A \cup B, x) \subseteq B$ , or $f_\prec (A \cup B, x) = f_\prec (A, x) \cup f_\prec (B, x)$ . In other words:

$$ \begin{align*}&min_{\prec_x}(A \cup B) \subseteq A \;\;\mbox{ or }\;\; min_{\prec_x}(A \cup B) \subseteq B \;\;\mbox{ or }\\ &min_{\prec_x}(A \cup B) = min_{\prec_x}(A) \cup min_{\prec_x}(B).\end{align*} $$

Assume that $min_{\prec _x}(A \cup B) \nsubseteq A$ and $min_{\prec _x}(A \cup B) \nsubseteq B$ , and notice that necessarily $min_{\prec _x}(A \cup B) \subseteq min_{\prec _x}(A) \cup min_{\prec _x}(B)$ . To prove the converse inclusion, let $w \in min_{\prec _x}(A) \cup min_{\prec _x}(B)$ , and suppose without loss of generality that $w \in min_{\prec _x}(A)$ . Since $min_{\prec _x}(A \cup B) \nsubseteq B$ , there is $z \in min_{\prec _x}(A \cup B)$ such that $z \in A$ . Since $w \in min_{\prec _x}(A)$ , $w \prec _x z \in min_{\prec _x}(A \cup B)$ , and given that $w \in A \subseteq A \cup B$ , it follows that $w \in min_{\prec _x}(A \cup B)$ . We can reason similarly if $w \in min_{\prec _x}(B)$ . Therefore, $min_{\prec _x}(A \cup B) = min_{\prec _x}(A) \cup min_{\prec _x}(B)$ .

Finally, ( $\alpha $ 4) and ( $\alpha $ 5) are direct consequences of (O3) and (O4) in the definition of topological order, together with the fact that $f_\prec (A, x) = min_{\prec _x}(A)$ .

Proof. We verify that for arbitrary $x \in S$ , $\prec ^f_x$ satisfies the properties (O1)–(O4) of Definition 5.1.

(O1) Notice first that $\prec ^f_x$ is indeed a binary relation over a subset $S_x$ of S. Precisely, by definition,

$$ \begin{align*} S_x = \bigcup\{f(A, x) \mid A \in Cl(S)\text{ and }f(A, x)\neq \emptyset\}. \end{align*} $$

Let us prove that $\prec _x^f$ is a total preorder, that is, it is reflexive, transitive, and total. If $S_x=\emptyset $ there is nothing to show, thus assume $S_x \neq \emptyset $ . First, $\prec _x^f$ is reflexive. Indeed, if $y \in S_x$ then $y \in f(A, x)$ for some $A \in Cl(S)$ , and since $A <_x A$ by Lemma 4.6, $y \prec _x^f y$ . Moreover, $\prec _x^f$ is transitive; in fact, assume $y \prec _x^f z$ and $z \prec _x^f w$ . This means that there are $A, B, C, D \in Cl(S)$ such that $y \in f(A, x)$ , $z \in f(B, x)$ , $z \in f(C, x)$ , $w \in f(D, x)$ , such that $A <_x B$ and $C <_x D$ . Now, $f(B, x) \subseteq B$ by ( $\alpha 1$ ), thus $z \in B \cap f(C, x)$ which is therefore nonempty; by Lemma 4.6(3) we get that $B <_x C$ . By transitivity of $<_x$ , it follows that $A <_x D$ and then $y \prec _x^f w$ .