Proof. We may assume that \(P\) has more than one element, since otherwise it is the one-element Boolean algebra, in which case we can take \(e\) to be the identity embedding, or it is the empty poset, in which case we can take \(e\) to be the empty embedding into the one-element Boolean algebra. In addition, since each poset can be embedded into a bounded poset by adjoining a top and/or bottom if necessary, which preserves all existing meets and joins, as well as the exactness of meets and joins, we may assume that \(P\) is bounded. We denote its minimum element by \(0\) and its maximum element by \(1\).

Where \(\leq\) is the order of \(P\), define \((X_P,\sqsubseteq)\) as follows:

1. \(X_P=\{(a,b)\in P^2\mid a\not\leq b\}\);

2. \((a,b)\sqsubseteq (c,d)\) iff \(a\leq c\) and \(b\geq d\).

Note that \(\sqsubseteq\) is a partial order whose top element is \((1,0)\). Given \(x\in X_P\), let \[{\Downarrow}x=\{y\in X_P\mid y\sqsubseteq x\} and \mathcal{D}(X_P)=\{U\subseteq X_P\midfor all x\in U, \mathord{\Downarrow}x\subseteq U\}.\] Note that \(\mathcal{D}(X_P)\) is the downset topology on \(X_P\). The interior and closure operations are defined for \(U\subseteq X_P\) by \[\Box U = \{x\in X_P \mid {\Downarrow} x\subseteq U\}and\Diamond U = \{x\in X_P \mid {\Downarrow} x\cap U\neq \varnothing\},\] respectively. Hence the regular open sets of the topology are the fixpoints of the \(\Box\Diamond\) operation, where \(\Box\Diamond U = \{x\in X_P\mid {\Downarrow}x \subseteq \Diamond U \}\). We denote the algebra of regular opens by \[\mathbf{B}P = (\{\Box\Diamond U\mid U\in \mathcal{D}(X_P)\}, \subseteq )\] and note that it is a complete Boolean algebra by a result of Tarski [11] (see, e.g., [15, Ch. 10]). Recall that in \(\mathbf{B}P\), meets are intersections, and joins are given by \(\bigsqcup \{U_i\mid i\in I\}=\Box\Diamond \bigcup \{U_i\mid i\in I\}\).

Now consider the map \(e : P \to \mathbf{B}P\) defined by \[e(a) = \begin{cases} \Box\Diamond {\Downarrow} (a,0) & if a\neq 0 \\ \varnothing &otherwise\end{cases}.\] We will prove that \(e\) is the desired embedding of \(P\) into \(\mathbf{B}P\). If \(P\) was obtained from an original poset \(P^{-}\) by adjoining a new top and/or bottom, then the restriction of \(e\) to \(P^{-}\) is the desired embedding of \(P^{-}\) into a complete Boolean algebra.

First, we prove that \(e\) is a poset embedding. To see that \(e\) is order-preserving, suppose \(a\leq b\). If \(a=0\), then \(e(a)=\varnothing\subseteq e(b)\). Otherwise, from \(a\leq b\) we have \((a,0)\sqsubseteq (b,0)\) and hence \(\Box\Diamond \mathord{\Downarrow}(a,0)\subseteq \Box\Diamond \mathord{\Downarrow}(b,0)\), so \(e(a)\subseteq e(b)\). Next, we prove that \(e\) is order-reflecting. Indeed, if \(a \not \leq b\), then \((a,b) \in X_P\) and \((a,b) \in \Box\Diamond{\Downarrow}(a,0)\) since for each \((x,y)\in X_P\), if \((x,y) \sqsubseteq (a,b)\), then \(x\leq a\) and \(y\geq b\), so \(x \leq a\) and \(y \geq 0\), and hence \((x,y) \sqsubseteq (a,0)\). On the other hand, we claim that \((a,b)\not\in e(b)\). If \(b=0\), this is immediate since \(e(0)=\varnothing\), so suppose \(b\neq 0\). We must show \((a,b) \notin \Box\Diamond{\Downarrow}(b,0)\), i.e., \({\Downarrow}(a,b) \not\subseteq \Diamond {\Downarrow}(b,0)\). In particular, \((a,b) \in {\Downarrow}(a,b)\), since \((a,b) \sqsubseteq (a,b)\), but we claim that \((a,b) \notin \Diamond {\Downarrow}(b,0)\). For if \((x,y) \sqsubseteq (a,b)\), then \(x \leq a\) and \(y \geq b\). Thus, \(x\not\leq b\), for otherwise \(x\leq y\), contradicting \((x,y)\in X_P\). Hence \((x,y)\not\sqsubseteq (b,0)\).

Now we prove for part [MainThm1] that \(e\) preserves finite meets. First observe that \[e(1) = \Box\Diamond {\Downarrow}(1,0) = \Box\Diamond X_P = X_P.\] For nonempty finite meets, suppose \(a_1\wedge \dots \wedge a_n\neq 0\). Then by definition, \(e(a_1\wedge \dots \wedge a_n) = \Box\Diamond {\Downarrow}(a_1\wedge \dots \wedge a_n,0)\). We claim that \[\Box\Diamond {\Downarrow}(a_1\wedge \dots \wedge a_n,0) = \Box\Diamond {\Downarrow} (a_1,0)\cap \dots \cap \Box\Diamond {\Downarrow}(a_n,0).\] First, where \(\mathord{\downarrow}x=\{y\in P\mid y\leq x\}\) for \(x\in P\), observe that since \({\downarrow}(a_1\wedge\dots\wedge a_n)= {\downarrow}a_1\cap \dots \cap {\downarrow}a_n\), it follows that \({\Downarrow}(a_1\wedge \dots\wedge a_n,0) = {\Downarrow}(a_1,0)\cap\dots\cap {\Downarrow}(a_n,0)\). Then since \(\Box\Diamond\) distributes over finite intersections, \[\Box\Diamond {\Downarrow}(a_1\wedge \dots \wedge a_n,0) = \Box\Diamond( {\Downarrow}(a_1,0)\cap\dots \cap {\Downarrow}(a_n,0)) = \Box\Diamond {\Downarrow}(a_1,0) \cap \dots \cap \Box\Diamond {\Downarrow}(a_n,0).\] Next, suppose \(a_1\wedge \dots \wedge a_n=0\), so \(e(a_1\wedge \dots \wedge a_n)=\varnothing\). We show that \(e(a_1) \cap \dots \cap e(a_n) = \varnothing\). Indeed, if \(a_i=0\) for some \(i\), then we are done by the definition of \(e\), so suppose \(a_i\neq 0\) for each \(i\). Then since \(a_1\wedge \dots \wedge a_n =0\), we have \({\Downarrow}(a_1,0)\cap \dots \cap {\Downarrow}(a_n,0)=\varnothing\), so \[\Box\Diamond {\Downarrow}(a_1,0) \cap \dots\cap \Box\Diamond {\Downarrow}(a_n,0) =\Box\Diamond ( {\Downarrow}(a_1,0)\cap \dots \cap {\Downarrow}(a_n,0)) = \Box\Diamond \varnothing =\varnothing.\]

For part [MainThm2], we assume that \(P\) is a meet semilattice. To show that \(e\) preserves all exact joins, consider any \(S\subseteq P\) that has an exact join. If \(\bigvee S=0\), then either \(S=\varnothing\) or \(S=\{0\}\). In either case, \(\bigsqcup \{e(s)\mid s\in S\}=\varnothing\), since \(e(0)=\varnothing\). Hence \(e(\bigvee S)=e(0)=\varnothing=\bigsqcup \{e(s)\mid s\in S\}\). Suppose \(\bigvee S\neq 0\). Without loss of generality, we can assume \(0\not\in S\). Then to show that \(e(\bigvee S)=\bigsqcup \{e(s)\mid s\in S\}\), we must show \[\Box\Diamond {\Downarrow}\left(\bigvee S,0\right) = \bigsqcup \{\Box\Diamond{\Downarrow}(s,0) \mid s\in S\}.\] By definition of \(\bigsqcup\) in \(\mathbf{B}P\), we have

\[\bigsqcup \{\Box\Diamond{\Downarrow}(s,0) \mid s\in S\} = \Box\Diamond\bigcup \{\Box\Diamond{\Downarrow}(s,0) \mid s\in S\} = \Box\bigcup \{\Diamond\Box\Diamond{\Downarrow}(s,0) \mid s\in S\}.\] Thus, it suffices to show \[\Box\Diamond {\Downarrow}\left(\bigvee S,0\right) = \Box\bigcup \{\Diamond\Box\Diamond{\Downarrow}(s,0) \mid s\in S\}.\label{KeyEq}\tag{1}\] The right-to-left inclusion is straightforward: for each \(s\in S\), we have \((s,0)\sqsubseteq (\bigvee S,0)\) and thus \(\Box\Diamond {\Downarrow} (s,0)\subseteq \Box\Diamond {\Downarrow} (\bigvee S,0)\). Since this holds for all \(s\in S\), we have \(\bigcup\{\Box\Diamond {\Downarrow} (s,0) \mid s\in S\}\subseteq \Box\Diamond {\Downarrow} \left(\bigvee S,0\right)\), which implies \(\Box\Diamond\bigcup\{\Box\Diamond {\Downarrow} (s,0) \mid s\in S\}\subseteq \Box\Diamond\Box\Diamond {\Downarrow} \left(\bigvee S,0\right)\) and hence \(\Box\bigcup \{\Diamond\Box\Diamond {\Downarrow} (s,0) \mid s\in S\}\subseteq \Box\Diamond {\Downarrow} \left(\bigvee S,0\right)\).

For the left-to-right inclusion in (1 ), suppose \((a,b)\) is in the left-hand side. To show that \((a,b)\) is in the right-hand side of (1 ), consider any \((a',b')\sqsubseteq (a,b)\). Then since \((a,b)\in \Box\Diamond {\Downarrow} (\bigvee S,0)\), there is an \((a'',b'')\sqsubseteq (a',b')\) such that \((a'',b'')\in {\Downarrow} (\bigvee S,0)\), i.e., \(a''\leq \bigvee S\). Since \(\bigvee S\) is an exact join, we have \(a''=a''\wedge\bigvee S= \bigvee_{s\in S} (a''\wedge s)\). Since \(a''\not\leq b''\), it follows that for some \(s\in S\), we have \(a''\wedge s\not\leq b''\). Now we claim that \((a''\wedge s, b'')\in \Box\Diamond{\Downarrow}(s,0)\). If not, then there is an \((a_s,b_s)\sqsubseteq (a''\wedge s, b'')\) such that \({\Downarrow}(a_s,b_s)\cap {\Downarrow}(s,0)=\varnothing\). Thus, \((a_s\wedge s, b_s )\not\in X_P\), which implies \(a_s\wedge s\leq b_s\). Since \((a_s,b_s)\sqsubseteq (a''\wedge s, b'')\), we have \(a_s\leq a''\wedge s\) and hence \(a_s\leq s\), which with \(a_s\wedge s\leq b_s\) implies \(a_s\leq b_s\), a contradiction. Hence \((a''\wedge s, b'')\in \Box\Diamond{\Downarrow}(s,0)\), which with \((a''\wedge s, b'')\sqsubseteq (a',b')\) implies \((a',b')\in \Diamond\Box\Diamond{\Downarrow}(s,0)\). Therefore, \((a,b)\) is in the right-hand side of (1 ). ◻

Proof. As in the proof of Theorem 3, we may assume without loss of generality that \(P\) is bounded. Let \(P^\partial\) be the dual poset of \(P\). Let \(e^\partial\) be the embedding of \(P^\partial\) into the complete Boolean algebra \(\mathbf{B}(P^\partial)\) constructed in the proof of Theorem 3. Then \(e^\partial\) (i) preserves finite meets and (ii) preserves all existing exact joins if \(P^\partial\) is a meet semilattice. Therefore, \(e^\partial\) is an embedding of \(P\) into the dual \((\mathbf{B} (P{^\partial}))^\partial\) of \(\mathbf{B}(P^\partial)\), which (a) preserves finite joins by (i) and (b) preserves all existing exact meets if \(P\) is a join semilattice, in which case \(P^\partial\) is a meet semilattice and (ii) applies. ◻

Proof. We continue from the end of the proof of Theorem 3 but now assume that \(P\) is a distributive lattice. To show that \(e\) preserves exact meets, consider any \(S\subseteq P\) that has an exact meet. First, suppose \(\bigwedge S\neq 0\). Then to show that \(e(\bigwedge S)= \bigsqcap \{e(s)\mid s\in S\}\), we must show that \[\Box\Diamond {\Downarrow}\left(\bigwedge S,0\right) = \bigcap \{\Box\Diamond{\Downarrow}(s,0) \mid s\in S\}.\] For the left-to-right inclusion, for each \(s\in S\), since \(\bigwedge S\leq s\), we have \((\bigwedge S, 0)\sqsubseteq (s,0)\) and hence \(\Box\Diamond{\Downarrow}(\bigwedge S, 0)\subseteq \Box\Diamond{\Downarrow}(s,0)\). Therefore, \(\Box\Diamond {\Downarrow}(\bigwedge S,0) \subseteq \bigcap \{\Box\Diamond{\Downarrow}(s,0) \mid s\in S\}\).

For the right-to-left inclusion, assume \((x,y)\in \bigcap \{\Box\Diamond {\Downarrow} (s,0)\mid s\in S\}\). To show \((x,y)\in \Box\Diamond {\Downarrow}(\bigwedge S,0)\), consider any \((x',y')\sqsubseteq (x,y)\). For contradiction, suppose \((x',y')\not\in \Diamond {\Downarrow}(\bigwedge S,0)\), i.e., \({\Downarrow}(x',y')\cap {\Downarrow}(\bigwedge S,0)=\varnothing\). Then \((x'\wedge \bigwedge S, y')\not\in X_P\), so \(x'\wedge \bigwedge S\leq y'\) and hence \[\begin{align} y'&=&y'\vee \left(x'\wedge \bigwedge S\right)\\ &=&(y'\vee x')\wedge \left(y'\vee \bigwedge S\right) \quad by distributivity\\ &=& (y'\vee x')\wedge \bigwedge_{s\in S} (y'\vee s) \quad since \bigwedge S is an exact meet. \end{align}\] Since \(x'\not\leq y'\), it follows that \(x'\not\leq \bigwedge_{s\in S} (y'\vee s)\), so there is some \(s\in S\) such that \(x'\not\leq y'\vee s\). Now \((x',y'\vee s)\sqsubseteq (x,y)\), which with \((x,y)\in \bigcap \{\Box\Diamond {\Downarrow} (s,0)\mid s\in S\}\) implies \((x',y'\vee s)\in \Diamond {\Downarrow}(s,0)\). Hence there is some \((x_s,y_s)\sqsubseteq (x',y'\vee s)\) such that \((x_s,y_s)\sqsubseteq (s,0)\). But then \(x_s\leq x'\wedge s \leq y'\vee s\leq y_s\), a contradiction.

Next suppose \(\bigwedge S=0\). Then \(e(\bigwedge S)=e(0)=\varnothing\subseteq \bigsqcap\{e(s)\mid s\in S\}\). For the reverse inclusion, we claim that \(\bigsqcap\{e(s)\mid s\in S\}=\varnothing\). If \(0\in S\), this is immediate, since \(e(0)=\varnothing\). So assume \(0\not\in S\). Further suppose for contradiction that \((x,y)\in \bigsqcap\{e(s)\mid s\in S\}\), so \((x,y)\in \bigcap \{\Box\Diamond {\Downarrow} (s,0)\mid s\in S\}\). Since \(\bigwedge S=0\), we have \(x\wedge \bigwedge S\leq y\). Then we derive a contradiction exactly as above but replacing \(x'\) by \(x\) and \(y'\) by \(y\). ◻

Proof. As noted in the Introduction, the left-to-right direction is clear from von Neumann’s [3] observation that complete Boolean algebras satisfy both JID and MID. For the right-to-left direction, if \(L\) satisfies both JID and MID, then every existing join and meet is exact. Since \(L\) is a lattice, all nonempty finite subsets have exact meets and joins. Thus, \(L\) is distributive. Now apply 7. ◻

Proof. It suffices to show that \(\mathscr{B}L\) is dense in \(\mathbf{B}L\). Consider a nonempty \(U\in \mathbf{B} L\), so there is some \((c,d)\in U\). Let \(e\) be the embedding of \(L\) into \(\mathbf{B}L\) used in the proofs of Theorems 3 and 7. Since \((c,d)\in X_L\) and hence \(c \not\leq d\), we have \(e(c) \not\subseteq e(d)\), so \(e(c) \wedge \lnot e(d) \ne \varnothing\). We claim that \(e(c)\wedge\neg e(d)\subseteq U\). Since \((c,d)\in U\in\mathbf{B}L\), we have \(\Box\Diamond \mathord{\Downarrow} (c,d)\subseteq U\), so it suffices to show that \(e(c)\wedge\neg e(d)\subseteq \Box\Diamond \mathord{\Downarrow} (c,d)\). Suppose \((w,v)\in e(c)\wedge\neg e(d)\). To show that \((w,v)\in\Box\Diamond\mathord{\Downarrow} (c,d)\), consider any \((w',v')\sqsubseteq (w,v)\), so \(w'\leq w\) and \(v\leq v'\). We must find a \((w'',v'')\sqsubseteq (w',v')\) such that \((w'',v'')\sqsubseteq (c,d)\). Since \((w,v)\in e(c)=\Box\Diamond\mathord{\Downarrow}(c,0)\), we have \((w',v')\in \Diamond\mathord{\Downarrow}(c,0)\), so there is some \((w'',v'')\sqsubseteq (w',v')\) such that \((w'',v'')\sqsubseteq (c,0)\). It follows that \(w''\leq c\wedge w'\) and \(w''\not\leq v''\geq v'\), so \(w''\not\leq v'\) and hence \(c\wedge w'\not\leq v'\). Since \((w,v)\in\neg e(d)=\neg\Box\Diamond\mathord{\Downarrow}(d,0)\), we have \(d\wedge w\leq v\), since otherwise \((d\wedge w,v)\sqsubseteq (w,v)\) and \((d\wedge w,v)\in \mathord{\Downarrow}(d,0)\subseteq \Box\Diamond\mathord{\Downarrow}(d,0)\). Then given \(d\wedge w\leq v\), \(w'\leq w\), and \(v\leq v'\), we have \(d\wedge w'\leq v'\). Toward showing that \(c\wedge w'\not\leq d\vee v'\), suppose for contradiction that \(c\wedge w'\leq d\vee v'\). Then \(c\wedge w'=(c\wedge w')\wedge (d\vee v')=(c\wedge w'\wedge d)\vee (c\wedge w'\wedge v')\), using distributivity for the second equation. But then since \(d\wedge w'\leq v'\) and \(c\wedge w'\wedge v'\leq v'\), it follows that \(c\wedge w'\leq v'\), contradicting our derivation of \(c\wedge w'\not\leq v'\) above. Thus, \(c \wedge w'\not\leq d\vee v'\), so \((c\wedge w',d\vee v')\in X_L\). Moreover, \((c\wedge w',d\vee v')\sqsubseteq (w',v')\) and \((c\wedge w',d\vee v')\sqsubseteq (c,d)\). Therefore, we may take \((w'',v'')=(c\wedge w',d\vee v')\). This completes the proof that \((w,v)\in\Box\Diamond \mathord{\Downarrow}(c,d)\) and hence that \(e(c)\wedge\neg e(d)\subseteq \Box\Diamond \mathord{\Downarrow}(c,d)\), which shows that \(\mathscr{B}L\) is dense in \(\mathbf{B}L\). ◻

Proof. We show that \(\{(a,b)\} \in \mathbf{B} M_3 \setminus \mathscr{B}M_3\). First, we show that \(\{(a,b)\} \in \mathbf{B} M_3\). For this, it suffices to show that \(\{(a,b)\}=\Box\Diamond\mathord{\Downarrow}(a,b)\). To see this, consider Figure 1, which shows both \(M_3\) (left) and \(X_{M_3}\) (right). Observe that \(\{(a,b)\}\) is a \(\sqsubseteq\)-downset. To see that it is a \(\Box\Diamond\)-fixpoint, consider any \((x,y)\not\in \{(a,b)\}\), so \(x\neq a\) or \(y\neq b\). We will show that in either case there is an \((x',y')\sqsubseteq (x,y)\) such that for all \((x'',y'')\sqsubseteq (x',y')\), we have \((x'',y'')\not\in \{(a,b)\}\). If \(x\neq a\), then either \(x=1\) or \(x\in \{b,c\}\). If \(x=1\), then let \((x',y')=(b,y)\) if \(y\neq b\) and \((x',y')=(c,b)\) otherwise. On the other hand, if \(x\in \{b,c\}\), then let \((x',y')=(x,y)\). Now assume \(x=a\), so \(y\neq b\). Then \(y=0\) or \(y=c\). In either case, let \((x',y')=(a,c)\).

Towards showing that \(\{(a,b)\}\not\in \mathscr{B}M_3\), we first claim that \(e(a) = \{(a, 0), (a, b), (a, c)\}\), where \(e\) is the embedding of \(M_3\) into \(\mathbf{B}M_3\) used in the proofs of Theorems 3 and 7, so \(e(a)=\Box\Diamond\mathord{\Downarrow}(a,0)\). Observe that \(\mathord{\Downarrow}(a,0)= \{(a, 0), (a, b), (a, c)\}\). We claim that \(\mathord{\Downarrow}(a,0)\) is a \(\Box\Diamond\)-fixpoint. Consider any \({(x,y)\not\in \mathord{\Downarrow}(a,0)}\), so \(x=1\) or \(x\in \{b,c\}\). Then by reasoning analogous to that in the previous paragraph, there is an \((x',y')\sqsubseteq (x,y)\) such that for all \((x'',y'')\sqsubseteq (x',y')\), we have \((x'',y'')\not\in \mathord{\Downarrow}(a,0)\).

Next, we show that \(\neg e(a)=\{(b, 0), (b, a), (b, c), (c, 0), (c, a), (c, b), (1, a)\}\). Call the right-hand side \(Y\). Clearly, for all \((x,y)\in Y\), \(\mathord{\Downarrow}(x,y)\cap e(a)=\varnothing\), so \((x,y)\in \neg e(a)\). Also observe that \[X_{M_3}\setminus Y=\{(a, 0), (a, b), (a, c),(1,b), (1,c), (1,0)\},\] and for every \((x,y)\in X_{M_3}\setminus Y\), there is some \((x',y')\in e(a)\) with \((x',y')\sqsubseteq (x,y)\), so \((X_{M_3}\setminus Y)\cap \neg e(a)=\varnothing\). This completes the proof that \(\neg e(a)=Y\).

By symmetric reasoning for \(b\) and \(c\), we can compute the \(e\)-images of all elements of \(M_3\setminus \{0,1\}\) and their negations:

Observe that \(e(a)\subseteq \neg e(b)\cap \neg e(c)\), \(e(b)\subseteq \neg e(a)\cap \neg e(c)\), \(e(c)\subseteq \neg e(a)\cap \neg e(b)\), and \(\neg e(a)\cap \neg e(b)\cap \neg e(c) =\varnothing\). Since every element of \(\mathscr{B}M_3\) is a join of meets, where each meet includes either \(e(x)\) or \(\neg e(x)\) for \(x\in \{a,b,c\}\), each atom of \(\mathscr{B}M_3\) is such a meet. It follows from the inclusions noted before that \(e(a)\), \(e(b)\), and \(e(c)\) are the only atoms.³ Then since \(\{(a,b)\}\) is a proper subset of \(e(a)\), it follows that \(\{(a,b)\}\not\in \mathscr{B}M_3\). ◻

Proof. The poset \(X_{N_5}\) is shown on the right of Figure 4. Direct calculation shows that each of the following sets is a \(\Box\Diamond\)-fixpoint of \(\mathcal{D}(X_{N_5})\):

Where \(e\) is the embedding of \(N_5\) into \(\mathbf{B}N_5\) used in the proofs of Theorems 3 and 7, we have \[e(a)=S_6, \;e(b)=S_3, \;and e(c)=S_4.\] We claim that there are no other \(\Box\Diamond\)-fixpoints besides \(S_1,\dots,S_8\). Each \(\Box\Diamond\)-fixpoint must be a \(\sqsubseteq\)-downset, so either it is \(\varnothing\) or it must contain at least one of the minimal points, which are \((b,c)\), \((a,b)\), and \((c,a)\). Considering the minimal points from left to right, we reason as follows:

1. Case 1

   : \(\mathrm{min}U=\{(b,c)\}\). Given that \((b,c)\in U\), we have \(S_3\subseteq U\), since for every \((x,y)\in S_3\) and \((x',y')\sqsubseteq (x,y)\), we have \((b,c)\sqsubseteq (x',y')\). Moreover, given that \((a,b)\not\in U\) and \((c,a)\not\in U\), we have \(U\subseteq S_3\), since for any \((x,y)\in X_{N_5}\setminus S_3\), we have \((a,b)\sqsubseteq (x,y)\) or \((c,a)\sqsubseteq (x,y)\). Therefore, \(U=S_3\).

2. Case 2: \(\mathrm{min}U=\{(a,b)\}\). Then \(U=S_2\), since for any \((x,y)\in X_{N_5}\setminus S_2\), we have either \((b,c)\sqsubseteq (x,y)\) or \((c,a)\sqsubseteq (x,y)\).

3. Case 3

   : \(\mathrm{min}U=\{(c,a)\}\). Then by reasoning symmetric to that in Case 1, \(U=S_4\).

4. Case 4

   : \(\mathrm{min}U=\{(b,c),(a,b)\}\). Given that \((b,c)\in U\) and \((a,b)\in U\), we have \((a,0)\in U\), since for any \((x',y')\sqsubseteq (a,0)\), we have either \((b,c)\sqsubseteq (x',y')\) or \((a,b)\sqsubseteq (x',y')\); it follows that \(S_6\subseteq U\). Moreover, given that \((c,a)\not\in U\), we have \(U\subseteq S_6\), since for any \((x,y)\in X_{N_5}\setminus S_6\), we have \((c,a)\sqsubseteq (x,y)\). Therefore, \(U=S_6\).

5. Case 5: \(\mathrm{min}U=\{(b,c),(c,a)\}\). Then \(S_7=S_3\cup S_4\subseteq U\), by reasoning as in Case 1 and Case 3. Moreover, \(U\subseteq S_7\), since for every \((x,y)\in X_{N_5}\setminus S_7\), we have \((a,b)\sqsubseteq (x,y)\). Therefore, \(U=S_7\).

6. Case 6: \(\mathrm{min}U=\{(a,b),(c,a)\}\). Then by reasoning symmetric to that in Case 4, \(U=S_5\).

7. Case 7: \(\mathrm{min}U=\{(b,c),(a,b),(c,a)\}\). Then clearly \(U=X_{N_5}\).

When ordered by inclusion, \(S_1,\dots,S_8\) form the \(8\)-element Boolean algebra \(\mathbf{B} N_5\). Then since \(e[N_5]\) has 5 elements, the Boolean subalgebra \(\mathscr{B}N_5\) of \(\mathbf{B}N_5\) generated by \(e[N_5]\) must be \(\mathbf{B} N_5\) itself. ◻