Proof. It follows from [Reference Bezhanishvili6, Lemma 7]Footnote ¹ that $Y_n$ is a closed upset in Y. So, [Reference Esakia, Bezhanishvili and Holliday17, Lemma 3.4.11] yields that $Y_n$ is an Esakia space. It is then clear that $Y_n \in \mathsf {ERS}_n$ . Since $Y_n$ is a closed upset of Y, the inclusion $e \colon Y_n \hookrightarrow Y$ is an $\mathsf {ERS}$ -morphism. Let $f \colon Z \to Y$ be an $\mathsf {ERS}$ -morphism with $Z \in \mathsf {ERS}_n$ . We show that there is a unique continuous p-morphism $g \colon Z \to Y_n$ such that $e \circ g = f$ . If $z \in Z$ , then $d(z) \le n$ , and so ${\uparrow } f(z) = f[{\uparrow } z]$ has cardinality smaller or equal to n. Thus, $f[Z] \subseteq Y_n$ . Take $g \colon Z \to Y_n$ to be the restriction of f. Then g is the unique continuous p-morphism such that $e \circ g = f$ . It follows from [Reference Mac Lane28, Theorem IV.1.2] that $(-)_n$ is a well-defined functor that is right adjoint to the inclusion $\mathsf {ERS}_n \hookrightarrow \mathsf {ERS}$ .

Proof. As the elements of $\mathsf {CC}(X)$ are closed chains, it is clear that $\mathsf {CC}(X) \subseteq \mathsf {V}(X)$ . To show that $\mathsf {CC}(X)$ is closed, let $F \in \mathsf {V}(X) \setminus \mathsf {CC}(X)$ . Our goal is to exhibit an open neighborhood of F in $\mathsf {V}(X)$ that is disjoint from $\mathsf {CC}(X)$ . Since $F \in \mathsf {V}(X) \setminus \mathsf {CC}(X)$ , it is a nonempty closed subset of X that is not totally ordered with respect to the order on X. Thus, there are $x_1,x_2 \in F$ such that $x_1 \nleq x_2$ and $x_2 \nleq x_1$ . Since X is a Priestley space, there are clopen upsets $U_1,U_2$ of X such that $x_1 \in U_1 \setminus U_2$ and $x_2 \in U_2 \setminus U_1$ . Consider $\mathcal {V} := \Diamond (U_1 \setminus U_2) \cap \Diamond (U_2 \setminus U_1)$ , which is a clopen subset of $\mathsf {V}(X)$ because $U_1 \setminus U_2$ and $U_2 \setminus U_1$ are clopen in X. Moreover, $F \in \mathcal {V}$ because $x_1 \in F \cap (U_1 \setminus U_2)$ and $x_2 \in F \cap (U_2 \setminus U_1)$ . It remains to show that $\mathcal {V}$ is disjoint from $\mathsf {CC}(X)$ . Assume that there is $C \in \mathsf {CC}(X) \cap \mathcal {V}$ . Then $C \in \Diamond (U_1 \setminus U_2) \cap \Diamond (U_2\setminus U_1)$ , and so there are $y_1,y_2 \in X$ such that $y_1 \in C \cap (U_1 \setminus U_2)$ and $y_2 \in C \cap (U_2 \setminus U_1)$ . Since C is a chain, $y_1 \le y_2$ or $y_2 \le y_1$ . If $y_1 \le y_2$ , it follows that $y_2 \in U_1$ because $y_1 \in U_1$ and $U_1$ is an upset. This contradicts that $y_2 \in U_2 \setminus U_1$ . If $y_2 \le y_1$ , we also obtain a contradiction with a similar argument. Therefore, $\mathcal {V}$ is an open neighborhood of F in $\mathsf {V}(X)$ that is disjoint from $\mathsf {CC}(X)$ . As F was an arbitrary element of $\mathsf {V}(X) \setminus \mathsf {CC}(X)$ , we have shown that $\mathsf {CC}(X)$ is a closed subset of $\mathsf {V}(X)$ .

Proof. (1) and (2) are straightforward consequences of the definitions of $\Box A$ and $\Diamond A$ .

(3) and (4) follow immediately from the definition of the topology on $\mathsf {CC}(X)$ .

To verify (5), observe that it follows from (6) that every open subset of $\mathsf {CC}(X)$ is a union of subsets of the form $\Box V_1 \cap \dots \cap \Box V_m \cap \Diamond V_1' \cap \dots \cap \Diamond V_n'$ , where each $V_i$ and $V_j'$ is a clopen subset of X. By (1), we obtain that $\Box V_1 \cap \dots \cap \Box V_m = \Box V$ , where $V := V_1 \cap \dots \cap V_m$ is clopen in X. Finally, it follows from the definitions of $\Box A$ and $\Diamond B$ that $\Box A \cap \Diamond B = \Box A \cap \Diamond (A \cap B)$ for every $A,B \subseteq X$ , and hence

$$\begin{align*}\Box V \cap \Diamond V_1' \cap \dots \cap \Diamond V_n' = \Box V \cap \Diamond (V \cap V_1') \cap \dots \cap \Diamond (V \cap V_n'). \end{align*}$$

This yields the claim because each $V \cap V_j'$ is a clopen subset of X contained in V.

Proof. If $C_2 = {\uparrow } x \cap C_1$ , then $C_2$ is an upset of $C_1$ , and hence $C_1 \unlhd C_2$ . To show the other implication, assume $C_1 \unlhd C_2$ and let x be the least element of $C_2$ , which exists by Proposition 3.6. Since $C_1 \unlhd C_2$ , it follows that $C_2 \subseteq C_1$ , and hence $C_2 \subseteq {\uparrow } x \cap C_1$ . Conversely, if $y \in {\uparrow } x \cap C_1$ , then $y \in C_2$ because $C_2$ is an upset in $C_1$ and $x \in C_2$ . Thus, ${\uparrow } x \cap C_1 \subseteq C_2$ .

Proof. It is an immediate consequence of its definition that $\unlhd $ is reflexive and antisymmetric. To show that $\unlhd $ is transitive, let $C_1,C_2,C_3 \in \mathsf {CC}(X)$ with $C_1 \unlhd C_2$ and $C_2 \unlhd C_3$ . Then $C_3 \subseteq C_2 \subseteq C_1$ and $C_3$ is an upset in $C_2$ , which is an upset in $C_1$ . Thus, $C_3$ is an upset in $C_1$ , and hence $C_1 \unlhd C_3$ . Thus, $\unlhd $ is a partial order. Consider $C,C_1,C_2 \in \mathsf {CC}(X)$ with $C \unlhd C_1,C_2$ . We need to show that $C_1 \unlhd C_2$ or $C_2 \unlhd C_1$ . By Lemma 3.7, there are $x_1,x_2 \in C$ such that $C_1 = {\uparrow } x_1 \cap C$ and $C_2 = {\uparrow } x_2 \cap C$ . Since C is a chain, $x_1 \le x_2$ or $x_2 \le x_1$ . Therefore, $C_1 \unlhd C_2$ or $C_2 \unlhd C_1$ . This proves that $(\mathsf {CC}(X), \unlhd )$ is a root system.

Notation 3.9. To avoid confusion, subsets of $\mathsf {CC}(X)$ will be denoted with calligraphic capital letters and we will write ${\Uparrow } \mathcal {A}$ and ${\Downarrow } \mathcal {A}$ to denote the downset and upset in $(\mathsf {CC}(X), \unlhd )$ generated by a subset $\mathcal {A}$ of $\mathsf {CC}(X)$ . As usual, if $C \in \mathsf {CC}(X)$ , we will simply write ${\Uparrow } C$ and ${\Downarrow } C$ instead of ${\Uparrow } \{C\}$ and ${\Downarrow } \{C\}$ .

Proof. (1). Let $C_1,C_2 \in \mathsf {CC}(X)$ such that $C_1 \unlhd C_2$ and $C_1 \in \Box A$ . Then $C_2 \subseteq C_1$ and $C_1 \subseteq A$ . Thus, $C_2 \subseteq A$ , and hence $C_2 \in \Box A$ . This shows that $\Box A$ is an upset.

(2). It follows from (1) that $\Box (X \setminus A)$ is an upset. By Lemma 3.4(2), we have that $\Diamond A = \mathsf {CC}(X) \setminus \Box (X \setminus A)$ , and so $\Diamond A$ is a downset because it is the complement of an upset.

(3). The left-to-right inclusion is an immediate consequence of the fact that taking the downset of a subset preserves inclusions. To prove the other inclusion, let

$$\begin{align*}C \in {\Downarrow} (\Box A \cap \Diamond B_1) \cap \dots \cap {\Downarrow} (\Box A \cap \Diamond B_1). \end{align*}$$

Then there are $C_1, \dots , C_n \in \mathsf {CC}(X)$ such that $C \unlhd C_i$ and $C_i \in \Box A \cap \Diamond B_i$ for each i. By Theorem 3.8, $\mathsf {CC}(X)$ is a root system, and hence ${\Uparrow } C$ is totally ordered with respect to $\unlhd $ . Since $C_1,\dots ,C_n \in {\Uparrow } C$ , there is j such that $C_j \unlhd C_i$ for every i. By (2), $\Diamond B_i$ is a downset, and so $C_j \in \Diamond B_i$ for every i because $C_i \in \Diamond B_i$ and $C_j \unlhd C_i$ . Then $C_j \in \Box A \cap \Diamond B_1 \cap \dots \cap \Diamond B_n$ . Therefore, $C \in {\Downarrow } (\Box A \cap \Diamond B_1 \cap \dots \cap \Diamond B_n)$ since $C \unlhd C_j$ .

(4). Let $C_1,C_2 \in \mathsf {CC}(X)$ with $C_1 \unlhd C_2$ and $C_2 \in \Box D$ . We want to show that $C_1 \in \Box D$ . Since $C_1 \unlhd C_2$ , there is $x \in C_1$ such that $C_2 = {\uparrow } x \cap C_1$ by Lemma 3.7. Let $y \in C_1$ . Then $x \le y$ or $y \le x$ because $x,y \in C_1$ and $C_1$ is a chain. If $x \le y$ , we have $y \in {\uparrow } x \cap C_1 = C_2 \subseteq D$ . Otherwise, $y \le x \in C_2 \subseteq D$ , and hence $y \in D$ because D is a downset. In either case, $y \in D$ . This shows that $C_1 \subseteq D$ , and so we have proved that $C_1 \in \Box D$ .

(5). Since U is an upset of X, we have that $X \setminus U$ is a downset, and so $\Box (X \setminus U)$ is a downset of $\mathsf {CC}(X)$ by (4). Lemma 3.4(2) implies that $\Diamond U = \mathsf {CC}(X) \setminus \Box (X \setminus U)$ . Thus, $\Diamond U$ is an upset because it is the complement of a downset.

(6). To show the left-to-right inclusion, assume that $C \in {\Downarrow } (\Box A \cap \Diamond (U \cap D))$ . Then there is $K \in \mathsf {CC}(X)$ such that $C \unlhd K$ and $K \in \Box A \cap \Diamond (U \cap D)$ . Since $\Diamond (U \cap D)$ is a downset by (2), we have that $C \in \Diamond (U \cap D)$ . It then remains to show that $C \in \Box (A \cup D)$ . From $K \in \Box A \cap \Diamond (U \cap D)$ it follows that $K \subseteq A$ and $K \cap U \cap D \neq \varnothing $ . Let $x \in K \cap U \cap D$ . Since $C \unlhd K$ and $x \in K$ , we obtain ${\uparrow } x \cap C \subseteq K$ . So, ${\uparrow } x \cap C \subseteq A$ because $K \subseteq A$ . We also have that ${\downarrow } x \cap C \subseteq D$ because D is a downset and $x \in D$ . That C is a chain implies $C = ({\uparrow } x \cap C) \cup ({\downarrow } x \cap C)$ . Therefore, $C = ({\uparrow } x \cap C) \cup ({\downarrow } x \cap C) \subseteq A \cup D$ , and hence $C \in \Box (A \cup D)$ . This shows that $C \in \Box (A \cup D) \cap \Diamond (U \cap D)$ .

We now prove the other inclusion. Suppose that $C \in \Box (A \cup D) \cap \Diamond (U \cap D)$ . Then $C \subseteq A \cup D$ and $C \cap U \cap D \neq \varnothing $ . Take $x \in C \cap U \cap D$ and let $K = {\uparrow } x \cap C$ . Thus, $K \in \mathsf {CC}(X)$ and $C \unlhd K$ . We show that $K \in \Box A \cap \Diamond (U \cap D)$ . Since $x \in K \cap U \cap D$ , we have that $K \in \Diamond (U \cap D)$ . By (1), $\Box (A\cup D)$ is an upset. Then $C \unlhd K$ implies $K \in \Box (A \cup D)$ , and so $K \subseteq A \cup D$ . Because $x \in U$ and U is an upset, we have $K \subseteq {\uparrow } x \subseteq U$ . Then the hypothesis that $U \cap D \subseteq A$ yields

$$\begin{align*}K \subseteq (A \cup D) \cap U \subseteq A \cup (U \cap D) = A, \end{align*}$$

which implies that $K \in \Box A$ . Therefore, $K \in \Box A \cap \Diamond (U \cap D)$ , and so $C \in {\Downarrow } (\Box A \cap \Diamond (U \cap D))$ because $C \unlhd K$ .

(7). It immediately follows from (6) by taking $U=X$ .

Proof. Corollary 3.3 and Theorem 3.8 yield that $\mathsf {CC}(X)$ is a Stone space and a root system. It remains to show that it is an Esakia space. We first prove that $\mathsf {CC}(X)$ is a Priestley space. Let $C_1,C_2 \in \mathsf {CC}(X)$ with $C_1 \ntrianglelefteq C_2$ . First, consider the case in which $C_2 \nsubseteq C_1$ . Then there is $x \in C_2 \setminus C_1$ . Since X is a Stone space and $C_1$ is closed, we can find a clopen subset V of X such that $C_1 \subseteq V$ and $x \notin V$ . Thus, $C_1 \in \Box V$ and $C_2 \notin \Box V$ . By Lemma 3.10(1), $\Box V$ is a clopen upset of $\mathsf {CC}(X)$ containing $C_1$ and not $C_2$ . Let us now assume that $C_2 \subseteq C_1$ . By Proposition 3.6, there exist the greatest elements $x_1$ and $x_2$ of $C_1$ and $C_2$ , respectively. Since $C_2 \subseteq C_1$ , it must be that $x_2 < x_1$ or $x_1 = x_2$ . If $x_2 < x_1$ , then there is a clopen upset U of X such that $x_1 \in U$ and $x_2 \notin U$ because X is a Priestley space. Then $x_1 \in C_1 \cap U$ and $C_2 \cap U = \varnothing $ because $x_2$ is the greatest element of $C_2$ and U is an upset. By Lemma 3.10(5), $\Diamond U$ is a clopen upset of $\mathsf {CC}(X)$ such that $C_1 \in \Diamond U$ and $C_2 \notin \Diamond U$ . It then remains to consider the case in which $x_1=x_2$ . Since $C_2 \subseteq C_1$ and $C_1 \ntrianglelefteq C_2$ , there is $y \in C_1 \setminus C_2$ such that ${\downarrow } y \cap C_2 \neq \varnothing $ . Then $x_2 \in {\uparrow } y \cap C_2$ because $y \in C_1$ and $x_2$ coincides with the greatest element of $C_1$ . Thus, ${\downarrow } y \cap C_2$ and ${\uparrow } y \cap C_2$ are nonempty closed chains in X. Let $z_1$ and $z_2$ be the greatest and least elements of ${\downarrow } y \cap C_2$ and ${\uparrow } y \cap C_2$ , respectively. Since $y \notin C_2$ , we have $y \nleq z_1$ and $z_2 \nleq y$ . Because X is a Priestley space, there is a clopen downset D such that $z_1 \in D$ and $y \notin D$ and there is a clopen upset U such that $z_2 \in U$ and $y \notin U$ . Then $C_1 \notin \Box (U \cup D) \cap \Diamond D$ because $y \in C_1$ and $y \notin U \cup D$ . Since $C_2$ is a chain and $z_1,z_2$ are the greatest and least elements of ${\downarrow } y \cap C_2$ and ${\uparrow } y \cap C_2$ , we have that ${C_2 \subseteq ({\uparrow } y \cap C_2) \cup ({\downarrow } y \cap C_2) \subseteq {\uparrow } z_2 \cup {\downarrow } z_1}$ . So, $C_2 \in \Box (U \cup D) \cap \Diamond D$ because $C_2 \subseteq {\uparrow } z_2 \cup {\downarrow } z_1 \subseteq U \cup D$ and $z_1 \in C_2 \cap D$ . By Lemma 3.10(7), $\mathcal {U} := \mathsf {CC}(X) \setminus (\Box (U \cup D) \cap \Diamond D)$ is a clopen upset of $\mathsf {CC}(X)$ such that $C_1 \in \mathcal {U}$ and $C_2 \notin \mathcal {U}$ . We have proved that $\mathsf {CC}(X)$ is a Priestley space.

To prove that $\mathsf {CC}(X)$ is an Esakia space, we need to show that the downset of any clopen subset of $\mathsf {CC}(X)$ is clopen. By Lemma 3.4(5), any clopen of $\mathsf {CC}(X)$ can be written as a finite union of clopens of the form $\Box V \cap \Diamond W_1 \cap \cdots \cap \Diamond W_n$ with $V,W_1, \dots , W_n$ clopen subsets of X. Since taking the downset commutes with unions, it is enough to show that ${\Downarrow } (\Box V \cap \Diamond W_1 \cap \dots \cap \Diamond W_n)$ is clopen for every $V,W_1, \dots , W_n$ clopens of X. Lemma 3.10(3) implies that it is sufficient to show that ${\Downarrow } (\Box V \cap \Diamond W)$ is clopen for any $V,W$ clopens of X. By Lemma 3.11, if V and W are clopens, then the clopen $V \cap W$ can be written as

(1) $$ \begin{align} V \cap W=(U_1 \cap D_1) \cup \dots \cup (U_m \cap D_m), \end{align} $$

where $U_i$ and $D_i$ are, respectively, a clopen upset and a clopen downset of X for each i. We obtain

$$ \begin{align*} \Box V \cap \Diamond W &= \Box V \cap \Diamond (V \cap W) = \Box V \cap \Diamond ((U_1 \cap D_1) \cup \dots \cup (U_1 \cap D_m))\\ &= \Box V \cap (\Diamond (U_1 \cap D_1) \cup \dots \cup \Diamond (U_m \cap D_m))\\ &= (\Box V \cap \Diamond (U_1 \cap D_1)) \cup \dots \cup (\Box V \cap \Diamond (U_m \cap D_m)), \end{align*} $$

where the first equality follows from the fact that $\Box A \cap \Diamond B = \Box A \cap \Diamond (A \cap B)$ for every $A,B \subseteq X$ , the second from Equation (1), the third from Lemma 3.4(1), and the fourth is straightforward. Therefore,

$$ \begin{align*} {\Downarrow} (\Box V \cap \Diamond W) = {\Downarrow} (\Box V \cap \Diamond (U_1 \cap D_1)) \cup \dots \cup {\Downarrow} (\Box V \cap \Diamond (U_m \cap D_m)). \end{align*} $$

Since $U_i \cap D_i \subseteq V$ for every i, Lemma 3.10(6) yields that

$$\begin{align*}{\Downarrow} (\Box V \cap \Diamond (U_i \cap D_i))= \Box (V \cup D_i) \cap \Diamond (U_i \cap D_i), \end{align*}$$

which is a clopen subset of $\mathsf {CC}(X)$ . Thus, ${\Downarrow } (\Box V \cap \Diamond W)$ is clopen. We have shown that the downset of any clopen of $\mathsf {CC}(X)$ is clopen, and hence that $\mathsf {CC}(X)$ is an Esakia space.

Proof. (1). Let U be an upset of X and $C \in \mathsf {CC}(X)$ . Since $C \subseteq {\uparrow } m(C)$ , we have that $m(C) \in U$ iff $C \subseteq U$ . Thus, $m^{-1}[U]=\Box U$ .

(2). Let D be a downset of X and $C \in \mathsf {CC}(X)$ . Since $m(C) \in {\downarrow } x$ for every $x \in C$ , we have that $m(C) \in D$ iff $C \cap D \neq \varnothing $ . Thus, $m^{-1}[D]=\Diamond D$ .

Proof. We first show that m is continuous. Since X is a Stone space, it is enough to prove that $m^{-1}[V]$ is clopen for each clopen subset V of X. Since $m^{-1}$ commutes with unions and intersections, Lemma 3.11 implies that it is sufficient to show that $m^{-1}[U]$ and $m^{-1}[D]$ are clopen in $\mathsf {CC}(X)$ for each U clopen upset and D clopen downset of X. By Lemma 3.13, if U is a clopen upset and D is a clopen downset, then $m^{-1}[U] = \Box U$ and $m^{-1}[D] = \Diamond D$ , which are clopen in $\mathsf {CC}(X)$ . Thus, m is continuous.

It remains to show that m is order preserving. Let $C_1,C_2 \in \mathsf {CC}(X)$ with $C_1 \unlhd C_2$ . Then there is $x \in C_1$ such that $C_2 = {\uparrow } x \cap C_1$ . Thus, $m(C_2) = x \in C_1$ , and hence $m(C_1) \le m(C_2)$ . Therefore, m is order preserving.

Proof. Define $g \colon Y \to \mathsf {CC}(X)$ by mapping $y \in Y$ to $f[{\uparrow } y]$ . Since Y is an Esakia root system, ${\uparrow } y$ is a closed chain in Y for every $y \in Y$ . By Lemma 3.15, $f[{\uparrow } y] \in \mathsf {CC}(X)$ because f is continuous and order preserving. So, g is well defined.

We show that g is continuous. By Lemma 3.4(5), the clopen subsets of $\mathsf {CC}(X)$ of the form $\Box V$ and $\Diamond V$ , with V clopen in X, form a subbasis for the topology on $\mathsf {CC}(X)$ . So, it is sufficient to show that $g^{-1}[\Box V]$ and $g^{-1}[\Diamond V]$ are clopen for every clopen subset V of X. The definitions of g and $\Diamond V$ imply that for every $y \in Y$ we have

$$ \begin{align*} y \in g^{-1}[\Diamond V] & \iff g(y) \in \Diamond V \iff f[{\uparrow} y] \in \Diamond V \iff f[{\uparrow} y] \cap V \neq \varnothing\\ & \iff {\uparrow} y \cap f^{-1}[V] \neq \varnothing \iff y \in {\downarrow} f^{-1}[V]. \end{align*} $$

Thus,

(2) $$ \begin{align} g^{-1}[\Diamond V] = {\downarrow} f^{-1}[V]. \end{align} $$

The continuity of f yields that $f^{-1}[V]$ is clopen, and hence $g^{-1}[\Diamond V]={\downarrow } f^{-1}[V]$ is clopen because Y is an Esakia space. We also have

$$ \begin{align*} g^{-1}[\Box V] &= g^{-1}[\mathsf{CC}(X) \setminus \Diamond(X \setminus V)] = Y \setminus g^{-1}[\Diamond(X \setminus V)]\\ &= Y \setminus {\downarrow} f^{-1}[X \setminus V] = Y \setminus {\downarrow} (Y \setminus f^{-1}[V]), \end{align*} $$

where the first equality follows from Lemma 3.4(2), the third from Equation (2), and the remaining are consequences of the fact that preimages commute with complements. Since V is clopen, f is continuous, and Y is an Esakia space, we obtain that $g^{-1}[\Box V]$ is clopen. This shows that g is continuous.

To show that g is a p-morphism, we first need to prove that

(3) $$ \begin{align} f[{\uparrow} y_2] = {\uparrow} f(y_2) \cap f[{\uparrow} y_1], \text{ for every } y_1,y_2 \in Y \text{ with } y_1 \le y_2. \end{align} $$

From $y_1 \le y_2$ it follows that ${\uparrow } y_2 \subseteq {\uparrow } y_1$ , and so $f[{\uparrow } y_2] \subseteq {\uparrow } f(y_2) \cap f[{\uparrow } y_1]$ because f is order preserving. If $x \in {\uparrow } f(y_2) \cap f[{\uparrow } y_1]$ , then $f(y_2) \le x$ and there is $y_3 \in {\uparrow } y_1$ such that $f(y_3) = x$ . Since ${\uparrow } y_1$ is a chain and $y_2,y_3 \in {\uparrow } y_1$ , we have $y_2 \le y_3$ or $y_3 \le y_2$ . If $y_2 \le y_3$ , then $y_3 \in {\uparrow } y_2$ , and so $x=f(y_3) \in f[{\uparrow } y_2]$ . If $y_3 \le y_2$ , then $x=f(y_3) \le f(y_2) \le x$ , and hence $x=f(y_2) \in f[{\uparrow } y_2]$ . In either case, $x \in f[{\uparrow } y_2]$ . Thus, Equation (3) holds. It immediately follows that $y_1 \le y_2$ implies $f[{\uparrow } y_1] \unlhd f[{\uparrow } y_2]$ , and hence $g(y_1) \unlhd g(y_2)$ . Therefore, g is order preserving. Let $y \in Y$ and $C \in \mathsf {CC}(X)$ be such that $g(y) \unlhd C$ . Then $C={\uparrow } x \cap g(y)$ for some $x \in g(y) = f[{\uparrow } y]$ . Thus, $x=f(z)$ with $y \le z$ , and so $C={\uparrow } f(z) \cap f[{\uparrow } y]$ . By Equation (3), $C= f[{\uparrow } z] = g(z)$ . This shows that g is a p-morphism.

Since f is order preserving, $m g(y)=m(f[{\uparrow } y])=f(y)$ for every $y \in Y$ . Thus, $m \circ g=f$ . It remains to show the uniqueness of g. Let $h \colon Y \to \mathsf {CC}(X)$ be a continuous p-morphism such that $m \circ h = f$ . We show that $h(y)=g(y)$ for every $y \in Y$ . By the definition of m, we have that $h(y)$ is a closed chain in X whose last element is $f(y)$ . To prove $h(y) \subseteq g(y)$ , consider $x \in h(y)$ and let $C = {\uparrow } x \cap h(y)$ . Then $C \in \mathsf {CC}(X)$ and $h(y) \unlhd C$ . Since h is a p-morphism, there is $z \in {\uparrow } y$ such that $C=h(z)$ . Therefore, $x=m(C)=mh(z)=f(z)$ , and so $x \in f[{\uparrow } y]=g(y)$ . Thus, $h(y) \subseteq g(y)$ . To show that $g(y) \subseteq h(y)$ , let $x \in g(y)$ . By the definition of g, we have $x \in f[{\uparrow } y]$ , and so $x=f(z)$ with $y \le z$ . Since h is order preserving, $h(y) \unlhd h(z)$ , which implies $h(z) \subseteq h(y)$ . From $f=m \circ h$ it follows that $x=f(z)=mh(z)$ . Thus, x is the least element of $h(z)$ , and hence $x \in h(z) \subseteq h(y)$ . Thus, $g(y) \subseteq h(y)$ . We have shown that $g=h$ . Therefore, g is the unique continuous p-morphism such that $m \circ g=f$ .

Proof. As we recalled in Section 2, $(-)^* \colon \mathsf {Pries} \to \mathsf {DL}$ is a dual equivalence that restricts to a dual equivalence between $\mathsf {ERS}$ and $\mathsf {HA}$ . It then follows from Theorem 3.16 that for every $H \in \mathsf {GA}$ and lattice homomorphism $f \colon X^* \to H$ there is a unique Heyting homomorphism $g \colon \mathsf {CC}(X)^* \to H$ such that $g \circ m^* = f$ . Therefore, the Gödel algebra $\mathsf {CC}(X)^*$ is free over the distributive lattice $X^*$ via the map $m^* \colon X^* \to \mathsf {CC}(X)^*$ . Since $\sigma _L \colon L \to X^*$ is a lattice isomorphism, it is straightforward to verify that $\mathsf {CC}(X)^*$ is free over L via the map $m^* \circ \sigma _L$ . It remains to observe that ${m^* \circ \sigma _L=e}$ . Since $\sigma _L(a)$ is a clopen upset of X and $m^*$ is the inverse image under m, Lemma 3.13(1) yields that $m^*(\sigma _L(a))=m^{-1}[\sigma _L(a)]= \Box \sigma _L(a) = e(a)$ .

Proof. By Proposition 3.21, the distributive lattice $({\mathsf {2}}^S)^*$ is free over S via the map q. Let $X=(({\mathsf {2}}^S)^*)_*$ be the double dual of ${\mathsf {2}}^S$ . Then it is straightforward to verify that the Gödel algebra $\mathsf {CC}(X)^*$ is free over S via the map $e \circ q$ , where $q \colon S \to ({\mathsf {2}}^S)^*$ and $e \colon ({\mathsf {2}}^S)^* \to \mathsf {CC}(X)^*$ are the maps appearing in Proposition 3.21 and Theorem 3.19. To show that $\mathsf {CC}({\mathsf {2}}^S)^*$ is free over S via r, it is then sufficient to exhibit an isomorphism of Gödel algebras $\varphi \colon \mathsf {CC}(X)^* \to \mathsf {CC}({\mathsf {2}}^S)^*$ such that $\varphi \circ (e \circ q) = r$ . Let $\varepsilon \colon {\mathsf {2}}^S \to X$ be the isomorphism of Priestley spaces described in Section 2, where we omitted the subscript $2^S$ from $\varepsilon $ for ease of readability. Define $\varphi $ to be $\mathsf {CC}(\varepsilon )^*$ . Since $\varepsilon $ is an isomorphism in $\mathsf {Pries}$ , and $\mathsf {CC}$ and $(-)^*$ are functors, it follows that $\varphi $ is an isomorphism in $\mathsf {ERS}$ . It then remains to show that $\varphi \circ (e \circ q) = r$ . The definitions of e and q imply that $eq(s)=\Box \sigma q(s) = \Box \sigma (U_s)$ for every $s \in S$ , where we omitted the subscript $({\mathsf {2}}^S)^*$ from the isomorphism $\sigma _{({\mathsf {2}}^S)^*} \colon ({\mathsf {2}}^S)^* \to X^*$ . If $s \in S$ , then

$$ \begin{align*} \varphi (eq(s)) & = \varphi(\Box \sigma q(s)) = \mathsf{CC}(\varepsilon)^{-1}(\Box \sigma q(s)) = \{C \in \mathsf{CC}({\mathsf{2}}^S) \mid \varepsilon[C] \in \Box \sigma q(s) \}\\ & = \{C \in \mathsf{CC}({\mathsf{2}}^S) \mid \varepsilon[C] \subseteq \sigma q(s) \} = \{C \in \mathsf{CC}({\mathsf{2}}^S) \mid C \subseteq \varepsilon^{-1} [\sigma q(s)] \}\\ & = \{C \in \mathsf{CC}({\mathsf{2}}^S) \mid C \subseteq \varepsilon^* \sigma q(s)] \} = \{C \in \mathsf{CC}({\mathsf{2}}^S) \mid C \subseteq q(s) \}\\ & = \Box q(s) = r(s). \end{align*} $$

In the above display, the first three, the sixth, and the last equalities follow from the definitions of e, $\varphi $ , $\mathsf {CC}(\varepsilon )$ , $\varepsilon ^*$ , and r, respectively. The fourth and the eighth equalities are a consequence of the definition of $\Box A$ for a subset A of a Priestley space. The fifth equality is straightforward, and the seventh is an instance of the triangle identity $\varepsilon ^* \circ \sigma =\text {id}_{({\mathsf {2}}^S)^*}$ , where $\text {id}_{({\mathsf {2}}^S)^*}$ is the identity on $({\mathsf {2}}^S)^*$ (see, e.g., [Reference Mac Lane28, Theorem IV.1.1(ii)]). Therefore, $\varphi \circ (e \circ q) = r$ , and this concludes the proof.

Proof. Note that, since every finite subset of a Stone space is closed, every finite chain of X is closed in X. Lemma 3.7 yields that ${\Uparrow } C = \{{\uparrow } x \cap C \mid x \in C\}$ , which is a set in bijection with C. Thus, C has depth n iff it has size n.

Proof. Since $\prod _i Y_i$ is a Priestley space, Theorem 3.12 yields that $\mathsf {CC}(\prod _i Y_i)$ is an Esakia root system. Since closed upsets of Esakia spaces equipped with the subspace topology and the restriction of the order are Esakia spaces (see, e.g., [Reference Esakia, Bezhanishvili and Holliday17, Lemma 3.4.11]), it is sufficient to show that $\bigotimes _i Y_i$ is a closed upset of $\mathsf {CC}(\prod _i Y_i)$ .

To prove that $\bigotimes _i Y_i$ is an upset of $\mathsf {CC}(\prod _i Y_i)$ , let $C_1 \in \bigotimes _i Y_i$ and $C_2 \in \mathsf {CC}(\prod _i Y_i)$ such that $C_1 \unlhd C_2$ . We show that $C_2 \in \bigotimes _i Y_i$ , which means that ${\uparrow } \pi _i[C_2] = \pi _i[C_2]$ for every $i \in I$ . Let $i \in I$ and $y \in {\uparrow } \pi _i[C_2]$ . Since $C_2 \subseteq C_1$ and $\pi _i[C_1]$ is an upset of $Y_i$ , we obtain that $y \in {\uparrow } \pi _i[C_2] \subseteq {\uparrow } \pi _i[C_1] = \pi _i[C_1]$ . So, there is $x \in C_1$ such that $\pi _i(x)=y$ . Let $m(C_2)$ be the least element of $C_2$ . Since $x,m(C_2) \in C_1$ and $C_1$ is a chain, we have that $x \le m(C_2)$ or $m(C_2) \le x$ . If $x \le m(C_2)$ , then $y=\pi _i(x) \le \pi _i(m(C_2))$ . We have that $\pi _i(m(C_2)) \le y$ because $y \in {\uparrow } \pi _i[C_2]$ and $\pi _i$ is order preserving. So, $y=\pi _i(m(C_2)) \in \pi _i[C_2]$ . If $m(C_2) \le x$ , then $x \in C_2$ because $C_1 \unlhd C_2$ . Thus, ${y=\pi _i(x) \in \pi _i[C_2]}$ . In either case, $y \in \pi _i[C_2]$ . This shows that $\pi _i[C_2]$ is an upset of $Y_i$ for each $i \in I$ , and hence $C_2 \in \mathsf {CC}(\prod _i Y_i)$ .

It remains to show that $\bigotimes _i Y_i$ is a closed subset of $\mathsf {CC}(\prod _i Y_i)$ . Let $C \in \mathsf {CC}(\prod _i Y_i)$ be such that $C \notin \bigotimes _i Y_i$ . We show that C is contained in an open subset of $\mathsf {CC}(\prod _i Y_i)$ disjoint from $\bigotimes _i Y_i$ . Since $C \notin \bigotimes _i Y_i$ , there is $i \in I$ such that $\pi _i[C]$ is not an upset of $Y_i$ . Then $\pi _i[C] \ne {\uparrow } \pi _i[C]$ . Because $\pi _i[C] \ne {\uparrow } \pi _i[C]$ , there is $x \in {\uparrow } \pi _i[C]$ that is not in $\pi _i[C]$ . From $x \notin \pi _i[C]$ it follows that ${\uparrow } x \cap ({\downarrow } x \cap \pi _i[C]) = \varnothing $ and ${\downarrow } x \cap ({\uparrow } x \cap \pi _i[C])=\varnothing $ . Since ${\downarrow } x \cap \pi _i[C]$ and ${\uparrow } x \cap \pi _i[C]$ are closed subsets of $Y_i$ , two applications of Lemma 4.5 yield a clopen upset U disjoint from ${\downarrow } x \cap \pi _i[C]$ such that $x \in U$ and a clopen downset D disjoint from ${\uparrow } x \cap \pi _i[C]$ such that $x \in D$ . Then

(4) $$ \begin{align} {\downarrow} x \cap \pi_i[C] \subseteq {\downarrow} (U \cap D), \ {\downarrow} x \cap \pi_i[C] \subseteq Y_i \setminus (U \cap D), \text{ and } {\uparrow} x \cap \pi_i[C] \subseteq U \setminus D, \end{align} $$

where the first inclusion holds because $x \in U \cap D$ , the second because ${\downarrow } x \cap \pi _i[C] \subseteq Y_i \setminus U$ , and the third follows from $x \in U$ and ${\uparrow } x \cap \pi _i[C] \subseteq Y_i \setminus D$ . Since ${\uparrow } \pi _i[C] \subseteq {\uparrow } \pi _i(m(C))$ and $Y_i$ is a root system, we have that ${\uparrow } \pi _i[C]$ is a chain. It then follows from $x \in {\uparrow } \pi _i[C]$ and $\pi _i[C] \subseteq {\uparrow } \pi _i[C]$ that $\pi _i[C]=({\downarrow } x \cap \pi _i[C]) \cup ({\uparrow } x \cap \pi _i[C])$ . The inclusions in (4) imply that

(5) $$ \begin{align} \pi_i[C] = ({\downarrow} x \cap \pi_i[C]) \cup ({\uparrow} x \cap \pi_i[C]) \subseteq ({\downarrow}(U \cap D) \setminus (U \cap D)) \cup (U \setminus D). \end{align} $$

Since $x \in {\uparrow } \pi _i[C]$ , we have that $\pi _i(m(C)) \le x$ , and hence $\pi _i(m(C)) \in {\downarrow } x \cap \pi _i[C]$ . It then follows from the inclusions in (4) that $\pi _i(m(C)) \in {\downarrow }(U \cap D)$ and $\pi _i(m(C)) \notin U \cap D$ . Thus,

(6) $$ \begin{align} \pi_i[C] \cap ({\downarrow}(U \cap D) \setminus (U \cap D)) \neq \varnothing. \end{align} $$

Then (5) and (6) imply

$$\begin{align*}C \in \mathcal{U} := \Box \pi_i^{-1}[({\downarrow}(U \cap D) \setminus (U \cap D)) \cup (U \setminus D)] \cap \Diamond \pi_i^{-1}[{\downarrow}(U \cap D) \setminus (U \cap D)]. \end{align*}$$

We show that $\mathcal {U}$ is a clopen of $\mathsf {CC}(\prod _i Y_i)$ disjoint from $\bigotimes _i Y_i$ . Since $Y_i$ is an Esakia space, U and D are clopen, and $\pi _i$ is continuous, we have that $\mathcal {U}$ is clopen. To prove that each $K \in \mathcal {U}$ is not in $\bigotimes _i Y_i$ , assume that $K \in \mathcal {U} \cap \bigotimes _i Y_i$ . Then $K \in \mathcal {U} \subseteq \Diamond \pi _i^{-1}[{\downarrow }(U \cap D) \setminus (U \cap D)]$ , and hence there is $y \in K$ such that $\pi _i(y) \in {\downarrow } (U \cap D) \setminus (U \cap D)$ . This implies that there is $z \in Y_i$ such that

$$\begin{align*}z \in {\uparrow} \pi_i(y) \cap U \cap D \subseteq {\uparrow} \pi_i (m(K)) \cap U \cap D \subseteq \pi_i[K] \cap U \cap D, \end{align*}$$

where ${\uparrow } \pi _i (m (K)) \subseteq {\uparrow } \pi _i[K] = \pi _i[K]$ because $K \in \bigotimes _i Y_i$ . However,

$$\begin{align*}K \in \mathcal{U} \subseteq \Box \pi_i^{-1}[({\downarrow}(U \cap D) \setminus (U \cap D)) \cup (U \setminus D)] \subseteq \Box \pi_i^{-1}[Y_i \setminus (U \cap D)], \end{align*}$$

and so $\pi _i[K] \subseteq Y_i \setminus (U \cap D)$ . This contradicts the existence of $z \in \pi _i[K] \cap U \cap D$ . Therefore, $\mathcal {U}$ is a clopen subset of $\mathsf {CC}(\prod _i Y_i)$ containing C that is disjoint from $\bigotimes _i Y_i$ . This shows that $\bigotimes _i Y_i$ is closed in $\mathsf {CC}(\prod _i Y_i)$ , and concludes the proof that $\bigotimes _i Y_i$ is an Esakia root system.

Proof. We first show that $p_i \colon \bigotimes _i Y_i \to Y_i$ is a continuous p-morphism for every $i \in I$ . We have that $p_i = \pi _i \circ m$ , and so it is a continuous map because every projection $\pi _i$ is continuous and m is continuous by Lemma 3.14. We show that $p_i$ is a p-morphism. Both $\pi _i$ and m are order preserving, so $p_i$ is also order preserving. Let $C \in \bigotimes _i Y_i$ and $y \in Y_i$ such that $p_i(C) \le y$ . Then $\pi _i(m(C)) \le y$ , and hence $y \in {\uparrow } \pi _i(m(C)) \subseteq {\uparrow } \pi _i[C]$ . Since $C \in \bigotimes _i Y_i$ , we have that ${\uparrow } \pi _i[C]=\pi _i[C]$ . Thus, $y \in \pi _i[C]$ . So, there is $x \in C$ such that $\pi _i(x)=y$ . Let $K = C \cap {\uparrow } x$ . Then $C \unlhd K$ , and hence $K \in \bigotimes _i Y_i$ because $\bigotimes _i Y_i$ is an upset in $\mathsf {CC}(\prod _i Y_i)$ as shown in the proof of Theorem 4.6. Therefore, $y=\pi _i(x)=\pi _i(m(K)) = p_i(K)$ . This shows that $p_i$ is a p-morphism.

It remains to verify the universal property of products. Let Z be an Esakia root system and $f_i \colon Z \to Y_i$ a continuous p-morphism for each $i \in I$ . By Proposition 4.1, there is a map $q \colon Z \to \prod _i Y_i$ sending $z \in Z$ to $(f_i(z))_{i \in I}$ that is continuous and order preserving. Thus, Theorem 3.16 yields a continuous p-morphism $g \colon Z \to \mathsf {CC}(\prod _i Y_i)$ given by $g(z) = q[{\uparrow } z]= \{ (f_i(w))_{i \in I} \mid w \in {\uparrow } z \}$ for every $z \in Z$ . Since each $f_i$ is a p-morphism, we have that $f_i[{\uparrow } z]$ is an upset of $Y_i$ for every $z \in Z$ . So, ${\uparrow } \pi _i[g(z)] = {\uparrow } f_i[{\uparrow } z]=f_i[{\uparrow } z] = \pi _i[g(z)]$ . Therefore, $g(z) \in \bigotimes _i Y_i$ , and hence g restricts to a continuous p-morphism $g \colon Z \to \bigotimes _i Y_i$ . Moreover, $p_i g(z)=\pi _i m g(z)=\pi _i ((f_i(z))_{i \in I})=f_i(z)$ for every $z \in Z$ . Thus, $p_i \circ g = f_i$ for each $i \in I$ . We now show that g is the unique map with such properties. Suppose that $h \colon Z \to \bigotimes _i Y_i$ is a continuous p-morphism such that $p_i \circ h = f_i$ for every $i \in I$ . Since $\bigotimes _i Y_i$ is an upset of $\mathsf {CC}(\prod _i Y_i)$ , it follows that $h \colon Z \to \mathsf {CC}(\prod _i Y_i)$ is a continuous p-morphism. We also have that $\pi _i m(h(z)) = p_i(h(z)) = f_i(z)$ for all $i \in I$ and $z \in Z$ . Thus, $m(h(z))=(f_i(z))_{i \in I} = m(g(z))$ for each $z \in Z$ . Then $h \colon Z \to \mathsf {CC}(\prod _i Y_i)$ is a continuous p-morphism such that $m \circ h=m \circ g$ . It follows from Theorem 3.16 that $h=g$ .

Proof. Since $(-)^* \colon \mathsf {ERS} \to \mathsf {GA}$ is a dual equivalence of categories, it sends products into coproducts. Thus, Theorem 4.7 yields that $(\bigotimes _i Y_i)^*$ together with the maps $p_i^*$ for each $i \in I$ is the coproduct of $\{ Y_i^* \mid i \in I\}$ in $\mathsf {GA}$ . Since $\sigma _{G_i} \colon G_i \to Y_i^*$ is an isomorphism of Gödel algebras for each $i \in I$ , it is straightforward to verify that $(\bigotimes _i Y_i)^*$ together with the maps $p_i^* \circ \sigma _{G_i}$ is the coproduct of $\{ G_i \mid i \in I\}$ in $\mathsf {GA}$ . It remains to observe that $p_i^* \circ \sigma _{G_i}$ maps $a \in G_i$ to $\Box \pi _i^{-1}[\sigma _{G_i}(a)] \in (\bigotimes _i Y_i)^*$ . Since $p_i=\pi _i \circ m$ , we obtain that $p_i^*(\sigma _{G_i}(a))=m^{-1}[\pi _i^{-1}[\sigma _{G_i}(a)]]$ . Because $\sigma _{G_i}(a)$ is a clopen upset of $Y_i$ , Lemma 3.13(1) yields that $p_i^*(\sigma _{G_i}(a))=\Box \pi _i^{-1}[\sigma _{G_i}(a)]$ .

Proof. Let $Y_i=(G_i)_*$ be the Esakia root system dual to $G_i$ . Theorem 4.8 implies that the depth of the coproduct of $\{G_i \mid i \in I\}$ coincides with the depth of $\bigotimes _i Y_i$ . Since $d(G_i)=d(Y_i)$ for every $i \in I$ , it is sufficient to show that $d(\bigotimes _i Y_i)=1 + \sum _{i \in I} (d(Y_i)-1)$ . We first prove the following technical fact.

Claim 4.10. Let $w_i \in \max Y_i$ for every $i \in I$ . Let also $i_1, \dots , i_n \in I$ and $z_{i_j} \in Y_{i_j}$ such that $z_{i_j} \le w_{i_j}$ for each $j=1, \dots , n$ . Define $C_j \subseteq \prod _i Y_i$ for each $j=1, \dots , n$ as follows:

$$ \begin{align*} (y_i)_{i \in I} \in C_j \iff \begin{cases} y_i=z_i &\text{if } i \in \{i_1, \dots, i_{j-1}\},\\ y_i \in {\uparrow} z_{i_j} &\text{if } i=i_j,\\ y_i=w_i &\text{if } i \notin \{i_1, \dots, i_j\}. \end{cases} \end{align*} $$

Then $C := C_1 \cup \dots \cup C_n$ is a point of depth $1 + \sum _{j=1}^n (d(z_{i_j})-1)$ in $\bigotimes _i Y_i$ , where $d(z_{i_j})$ is the depth of $z_{i_j}$ in $Y_{i_j}$ .

Proof of the Claim.

For every j, we have

$$\begin{align*}C_j=\bigcap \{ \pi_i^{-1}[z_i] \mid i \in \{i_1, \dots, i_{j-1}\}\} \cap \pi_{i_j}^{-1}[{\uparrow} z_{i_j}] \cap \bigcap \{ \pi_i^{-1}[w_i] \mid i \notin \{i_1, \dots, i_j\}\}, \end{align*}$$

and hence $C_j$ is a closed subset of $\prod _i Y_i$ because $\{ z_i\}$ , ${\uparrow } z_{i_j}$ , and $\{w_i\}$ are all closed and $\pi _i$ is continuous for every $i \in I$ . Since $Y_{i_j}$ is a root system, ${\uparrow } z_{i_j}$ is a chain, and hence $C_j \in \mathsf {CC}(\prod _i Y_i)$ . If $1 \le j < n$ , then the least element of $C_j$ is the greatest element of $C_{j+1}$ . Therefore, $C = C_1 \cup \dots \cup C_n$ is a chain in $\prod _i Y_i$ . Since $C_1, \dots , C_n$ are closed in $\prod _i Y_i$ , we obtain that $C \in \mathsf {CC}(\prod _i Y_i)$ . By the definition of $C_1, \dots , C_n$ , it follows that $\pi _i[C]={\uparrow } z_i$ if $i \in \{i_1, \dots , i_n\}$ , and $\pi _i[C]=\{w_i\}$ , otherwise. Therefore, for every $i \in I$ the set $\pi _i[C]$ is an upset of $Y_i$ , and hence $C \in \bigotimes _i Y_i$ . By Proposition 3.25, the depth of C in $\bigotimes _i Y_i$ coincides with its size. If $d(z_{i_j}) = \infty $ for some j, then C is infinite, and so has depth $\infty =1 + \sum _{j=1}^n (d(z_{i_j})-1)$ in $\bigotimes _i Y_i$ . Otherwise, C has size $1+\sum _{j=1}^n (d(z_{i_j})-1)$ . Indeed, each $C_j$ has size $d(z_{i_j})$ and the least element of $C_j$ coincides with the greatest element of $C_{j+1}$ for every $j=1, \dots , n-1$ . This shows that C is an element of $\bigotimes _i Y_i$ of depth $1 + \sum _{j=1}^n (d(z_{i_j})-1)$ .

We first consider the case in which there is $k \in I$ with $d(Y_k)=\infty $ . Then there is an element of $Y_k$ of infinite depth or there are elements of $Y_k$ of arbitrary large finite depth. Suppose first that there is $z_k \in Y_k$ such that $d(z_k)=\infty $ . Since every $G_i$ is nontrivial, all the $Y_i$ are nonempty. Then $\max Y_i$ and $\max ({\uparrow } z_k)$ are nonempty by [Reference Priestley30, Proposition 2.6]. So, we can pick $w_i \in \max Y_i$ for every $i \in I$ so that $z_k \le w_k$ . By Claim 4.10, there is $C \in \bigotimes _i Y_i$ of infinite depth. Thus, $d(\bigotimes _i Y_i)=\infty $ . Suppose now that for each $n \in \mathbb {N}$ there is $z_k \in Y_k$ such that $d(z_k) \ge n$ . By arguing as in the previous case, we obtain that there are elements of $\bigotimes _i Y_i$ of arbitrarily large finite depth, and hence $d(\bigotimes _i Y_i)=\infty $ . So, we can assume that $d(Y_i) \ne \infty $ for every $i \in I$ . Suppose there are infinitely many $i \in I$ such that $d(Y_i)>1$ . Then for each $n \in \mathbb {N}$ we can find $i_1, \dots , i_n \in I$ and $z_{i_j} \in Y_{i_j}$ such that $d(z_{i_j}) \ge 2$ . Claim 4.10 allows us to construct $C \in \bigotimes _i Y_i$ of depth $1 + \sum _{j=1}^n (d(z_{i_j})-1)> n$ . It follows that $d(\bigotimes _i Y_i)=\infty $ . The last case to consider is when $d(Y_i) \ne \infty $ for every $i \in I$ and $\{i \in I \mid d(Y_i)> 1\}$ is finite. Let $\{i \in I \mid d(Y_i)> 1\} = \{ i_1, \dots , i_n\}$ and $z_{i_j} \in Y_{i_j}$ such that $d(z_{i_j})=d(Y_{i_j})$ . Then Claim 4.10 implies that there is $C \in \bigotimes _i Y_i$ of depth $1 + \sum _{j=1}^n (d(z_{i_j})-1)$ , and hence

$$\begin{align*}d(\textstyle{\bigotimes_i} Y_i) \ge 1 + \sum_{j=1}^n (d(z_{i_j})-1) = 1 + \sum_{j=1}^n (d(Y_{i_j})-1). \end{align*}$$

We now show that $d(\bigotimes _i Y_i) \le 1 + \sum _{j=1}^n (d(Y_{i_j})-1)$ . Let $K \in \bigotimes _i Y_i$ . We prove that the size of K is smaller or equal to $1+\sum _{j=1}^n (d(Y_{i_j})-1)$ . If $i \notin \{i_1, \dots , i_n \}$ , then $\pi _i[K]=\{w_i\}$ with $w_i \in \max Y_i$ because $d(Y_i)=1$ . If $j=1, \dots , n$ , let $\pi _{i_j}[K]={\uparrow } z_{i_j}$ for some $z_{i_j} \in Y_{i_j}$ . Thus, every $\pi _i[K]$ is finite, and it is a singleton for all but finitely many $i \in I$ . Then K is a finite chain because $K \subseteq \prod _i \pi _i[K]$ . Let

$$\begin{align*}W=\{ (j,y) \mid j \in \{1, \dots, n\} \text{ and } y \in {\uparrow} z_{i_j} \setminus \{z_{i_j}\}\}. \end{align*}$$

For every $(j,y) \in W$ we have that $\pi _{i_j}^{-1}[y] \cap K$ is a finite nonempty chain because K is a finite chain and $y \in {\uparrow } z_{i_j} = \pi _{i_j}[K]$ . Recall that $m(K)$ denotes the least element of K. Define $f \colon W \to K \setminus m(K)$ by mapping $(j,y) \in W$ to the least element of $\pi _{i_j}^{-1}[y] \cap K$ , which belongs to $K \setminus m(K)$ because $\pi _{i_j}(m(K))=z_{i_j}$ and $y \neq z_{i_j}$ . We show that f is onto. Let $\overline {y}=(y_i)_{i \in I} \in K \setminus m(K)$ . Then there exists $j \in \{1, \dots , n\}$ such that the predecessor of $\overline {y}$ in K differs from $\overline {y}$ in the $i_j$ -th component. So, $y_{i_j}\hspace{-1pt} \neq\hspace{-1pt} z_{i_j}$ , and hence $(j,y_{i_j})\hspace{-1pt} \in\hspace{-1pt} W$ . From the definition of f it follows that $f(j,y_{i_j})\hspace{-1pt} =\hspace{-1pt} \overline {y}$ . Thus, $f \colon W \to K \setminus m(K)$ is onto. Then the size of $K \setminus m(K)$ is smaller or equal to the cardinality of W, which is $\sum _{j=1}^n (d(z_{i_j})-1)$ . Since $d(z_{i_j}) \le d(Y_{i_j})$ for every j, we obtain that the size of K is smaller or equal to $1 + \sum _{j=1}^n (d(Y_{i_j})-1)$ for every $K \in \bigotimes _i Y_i$ . By Proposition 3.25, the size of K coincides with its depth in $\bigotimes _i Y_i$ , so we get $d(\bigotimes _i Y_i) \le 1 + \sum _{j=1}^n (d(Y_{i_j})-1)$ . We assumed that $d(Y_i) \ne \infty $ for every $i \in I$ and that $\{i \in I \mid d(Y_i)> 1\}$ is finite, so $1 + \sum _{j=1}^n (d(Y_{i_j})-1) = 1 + \sum _{i \in I} (d(Y_i)-1)$ . It follows that $d(\bigotimes _i Y_i) \le 1 + \sum _{i \in I} (d(Y_i)-1)$ . This concludes the proof that $d(\bigotimes _i Y_i)=1+\sum _{i \in I} (d(Y_i)-1)$ . By what we observed at the beginning of the proof, it follows that the coproduct of $\{G_i \mid i \in I\}$ has depth $1 + \sum _{i \in I} (d(G_i)-1)$ .

Proof. Let $Y_i=(G_i)_*$ be the Esakia root system dual to $G_i$ for each $i \in I$ . By Theorem 4.8, the coproduct of $\{ G_i \mid i \in I\}$ in $\mathsf {GA}$ is dual to $\bigotimes _i Y_i$ , while Theorem 4.12 yields that the coproduct of $\{ G_i \mid i \in I\}$ in $\mathsf {GA}_n$ is dual to $(\bigotimes _i Y_i)_n$ . Thus, the two coproducts are isomorphic iff $\bigotimes _i Y_i$ and $(\bigotimes _i Y_i)_n$ are isomorphic Esakia root systems, which happens exactly when $d(\bigotimes _i Y_i) \le n$ . Theorem 4.9 yields that $d(\bigotimes _i Y_i)=1 + \sum _{i \in I} (d(G_i)-1)$ . Therefore, the two coproducts coincide when $1 + \sum _{i \in I} (d(G_i)-1) \le n$ , which is equivalent to $\sum _{i \in I} (d(G_i)-1) \le n-1$ .

Proof. Let $L^\partial $ be the order dual of L, and $X^\partial $ the Priestley space that is order dual to X and is equipped with the same topology. It is straightforward to check that the Priestley dual of $L^\partial $ is $X^\partial $ , and that X is co-Esakia iff $X^\partial $ is Esakia. By Esakia duality, $L^\partial $ is a Heyting algebra iff $X^\partial $ is an Esakia space, so L is a co-Heyting algebra iff X is a co-Esakia space. It also follows that L is a bi-Heyting algebra iff X is a bi-Esakia space.

Proof. (1). We prove the claim by induction on n. The case $n=0$ is clear. We assume that the claim is true for $n-1$ with $n \ge 1$ and show that it is true for n. We first prove the left-to-right implication. Let . The definition of implies that there is $i \le n$ such that

Then there is $y \le x$ with . By the induction hypothesis, there exists a chain K in X such that $K \subseteq {\downarrow } y$ and $K \cap A_j \neq \varnothing $ for every $j \neq i$ . Then $C=K \cup \{y\}$ is a chain such that $C \subseteq {\downarrow } x$ and $C \cap A_i \neq \varnothing $ for every i. To prove the other implication, suppose there is a chain C in X such that $C \subseteq {\downarrow } x$ and $C \cap A_i \neq \varnothing $ for every i. Choose an element $a_i \in C \cap A_i$ for every i. Since C is a chain, $K:= \{ a_1, \dots , a_n\}$ is a chain such that $K \cap A_i \neq \varnothing $ for every i. Let $a_j$ be the greatest element of K. Since $K \subseteq {\downarrow } a_j$ , the induction hypothesis yields that . Thus, . It follows that

(2). We again argue by induction on n. When $n=0$ , the set is clearly clopen. We assume that the claim is true for $n-1$ with $n \ge 1$ and show that it is true for n. By the induction hypothesis, is clopen for every i. Since each $A_i$ is clopen, is clopen for every i. Thus, is clopen because it is the upset of a clopen subset and X is a co-Esakia space. So,

is clopen since it is a finite union of clopens.

Proof. To show the left-to-right inclusion, assume that

$$\begin{align*}C \in {\Uparrow } (\Box V \cap \Diamond W_1 \cap \dots \cap \Diamond W_n). \end{align*}$$

Then there is $K \in \mathsf {CC}(X)$ such that $K \unlhd C$ , $K \in \Box V$ , and $K \in \Diamond W_i$ for every $i=1, \dots , n$ . Let $I \subseteq \{1, \dots , n\}$ be such that $i \in I$ iff $C \notin \Diamond W_i$ . Then $C \in \Diamond W_j$ for every $j \in I^c$ . Thus, $C \in \bigcap \{ \Diamond W_j \mid j \in I^c\}$ . It remains to show that . By Lemma 3.10(1), $\Box V$ is an upset, and hence $C \in \Box V$ because $K \unlhd C$ and $K \in \Box V$ . Thus, $C \subseteq V$ . Since $K \in \Diamond W_i$ for each $i= 1, \dots , n$ and $C \notin \Diamond W_i$ for every $i \in I$ , it follows that $(K \setminus C) \cap W_i \ne \varnothing $ for every $i \in I$ . We have that $K \unlhd C$ implies $K \setminus C \subseteq {\downarrow } x$ for every $x \in C$ . So, for every $x \in C$ the set $K \setminus C$ is a chain in X such that $K \setminus C \subseteq {\downarrow } x$ and $(K \setminus C) \cap W_i \ne \varnothing $ for every $i \in I$ . Then Lemma 5.3(1) implies that . Therefore, , and hence . We have then found $I \subseteq \{1, \dots , n\}$ such that . It follows that C belongs to the right-hand side of the claimed equality.