[proofplan] We prove the result in the generality of an arbitrary [topological space](/page/Topological%20Space) $P$, so it applies to a forcing poset with its forcing topology. The main device is the regularisation map $r: \mathcal{P}(P) \to \operatorname{RO}(P)$ defined by $r(A)=\operatorname{int}(\overline{A})$; we prove directly that this operator sends arbitrary subsets to regular open sets and fixes exactly the regular open sets, so no extra separation or semiregularity hypothesis is needed. This gives arbitrary joins by regularising unions, while finite meets are ordinary intersections and complements are regularised set-theoretic complements. We then prove the finite distributive law directly from the topological definitions and obtain arbitrary meets by De Morgan duality, giving a complete Boolean algebra. [/proofplan]

[step:Show that regularisation produces regular open sets]Let $\mathcal{T}$ denote the topology on $P$, and for every subset $A \subset P$ define \begin{align*} r(A) := \operatorname{int}(\overline{A}), \end{align*} where $\overline{A}$ is the closure of $A$ in $P$ and $\operatorname{int}(\overline{A})$ is its interior in $P$. We claim that $r(A) \in \operatorname{RO}(P)$ for every $A \subset P$. Since $r(A)$ is open and $r(A) \subset \overline{A}$, monotonicity of closure gives \begin{align*} \overline{r(A)} \subset \overline{A}. \end{align*} Taking interiors gives \begin{align*} \operatorname{int}(\overline{r(A)}) \subset \operatorname{int}(\overline{A}) = r(A). \end{align*} Conversely, since $r(A) \subset \overline{r(A)}$ and $r(A)$ is open, we have \begin{align*} r(A) \subset \operatorname{int}(\overline{r(A)}). \end{align*} Thus \begin{align*} r(A)=\operatorname{int}(\overline{r(A)}), \end{align*} so $r(A)$ is regular open.[/step]

[guided]The operation $r(A)=\operatorname{int}(\overline{A})$ is the mechanism that repairs an arbitrary subset into a regular [open set](/page/Open%20Set). We verify this directly, because it is the structural reason arbitrary joins exist. Let $A \subset P$ be any subset. Define \begin{align*} r(A) := \operatorname{int}(\overline{A}). \end{align*} By definition, $r(A)$ is open. To prove that $r(A)$ is regular open, we must prove \begin{align*} r(A)=\operatorname{int}(\overline{r(A)}). \end{align*} First, since $r(A) \subset \overline{A}$, monotonicity of closure implies \begin{align*} \overline{r(A)} \subset \overline{A}. \end{align*} Taking interiors preserves inclusion, so \begin{align*} \operatorname{int}(\overline{r(A)}) \subset \operatorname{int}(\overline{A}) = r(A). \end{align*} For the reverse inclusion, use that $r(A)$ is open and $r(A) \subset \overline{r(A)}$. Because $\operatorname{int}(\overline{r(A)})$ is the largest open subset of $\overline{r(A)}$, the open set $r(A)$ is contained in it: \begin{align*} r(A) \subset \operatorname{int}(\overline{r(A)}). \end{align*} Combining both inclusions gives \begin{align*} r(A)=\operatorname{int}(\overline{r(A)}), \end{align*} which is exactly the definition of regular openness.[/guided]

[step:Verify that complements of regular open sets are regular open] Let $U \in \operatorname{RO}(P)$, and define \begin{align*} \neg U := \operatorname{int}(P \setminus U). \end{align*} Since $U$ is open, $P \setminus U$ is closed, and hence \begin{align*} \overline{\neg U} \subset P \setminus U. \end{align*} Therefore \begin{align*} \operatorname{int}(\overline{\neg U}) \subset \operatorname{int}(P \setminus U)=\neg U. \end{align*} The reverse inclusion follows because $\neg U$ is open and $\neg U \subset \overline{\neg U}$. Hence \begin{align*} \neg U = \operatorname{int}(\overline{\neg U}), \end{align*} so $\neg U \in \operatorname{RO}(P)$. Also, \begin{align*} U \cap \neg U = \varnothing. \end{align*} For every open set $O \subset P$, the identity \begin{align*} \operatorname{int}(P \setminus O)=P \setminus \overline{O} \end{align*} holds: a point has an open neighbourhood contained in $P \setminus O$ exactly when it is not in the closure of $O$. Applying this with $O=U$ gives \begin{align*} \neg U=P \setminus \overline{U}. \end{align*} Since $U$ is regular open, we also get the involution identity \begin{align*} \neg\neg U=\operatorname{int}(P \setminus (P \setminus \overline{U}))=\operatorname{int}(\overline{U})=U. \end{align*} If $U,V \in \operatorname{RO}(P)$ and $U \subset V$, then $P \setminus V \subset P \setminus U$, and monotonicity of interior gives \begin{align*} \neg V=\operatorname{int}(P \setminus V) \subset \operatorname{int}(P \setminus U)=\neg U. \end{align*} Thus the complement operation is order reversing on $\operatorname{RO}(P)$. Finally, \begin{align*} r(U \cup \neg U)=\operatorname{int}(\overline{U \cup (P \setminus \overline{U})})=P, \end{align*} because every non-empty open subset of $P$ either meets $U$ or is contained in $P \setminus \overline{U}$, so $U \cup (P \setminus \overline{U})$ is dense in $P$. [/step]

[step:Show that finite meets are ordinary intersections] Let $U,V \in \operatorname{RO}(P)$. We prove that $U \cap V$ is regular open. Since $U \cap V$ is open, we have \begin{align*} U \cap V \subset \operatorname{int}(\overline{U \cap V}). \end{align*} For the opposite inclusion, monotonicity of closure gives \begin{align*} \overline{U \cap V} \subset \overline{U} \cap \overline{V}. \end{align*} Taking interiors gives \begin{align*} \operatorname{int}(\overline{U \cap V}) \subset \operatorname{int}(\overline{U} \cap \overline{V}). \end{align*} Since $\operatorname{int}(A \cap B) \subset \operatorname{int}(A) \cap \operatorname{int}(B)$ for all subsets $A,B \subset P$, we obtain \begin{align*} \operatorname{int}(\overline{U \cap V}) \subset \operatorname{int}(\overline{U}) \cap \operatorname{int}(\overline{V}) = U \cap V. \end{align*} Therefore \begin{align*} U \cap V = \operatorname{int}(\overline{U \cap V}), \end{align*} so $U \cap V \in \operatorname{RO}(P)$. Thus the finite meet operation is \begin{align*} U \wedge V = U \cap V. \end{align*} [/step]

[step:Identify arbitrary joins by regularising unions] Let $(U_i)_{i \in I}$ be an indexed family in $\operatorname{RO}(P)$, and define \begin{align*} W := r\left(\bigcup_{i \in I} U_i\right) = \operatorname{int}\left(\overline{\bigcup_{i \in I} U_i}\right). \end{align*} By the first step, $W \in \operatorname{RO}(P)$. For each $i \in I$, \begin{align*} U_i \subset \bigcup_{j \in I} U_j \subset W, \end{align*} where the last inclusion holds because $U_i$ is open and contained in the closure of the union. Hence $W$ is an upper bound of the family. Now let $Z \in \operatorname{RO}(P)$ be any upper bound, so $U_i \subset Z$ for every $i \in I$. Then \begin{align*} \bigcup_{i \in I} U_i \subset Z. \end{align*} Taking closures and then interiors gives \begin{align*} W=\operatorname{int}\left(\overline{\bigcup_{i \in I} U_i}\right) \subset \operatorname{int}(\overline{Z})=Z, \end{align*} because $Z$ is regular open. Thus $W$ is the least regular open upper bound, and therefore \begin{align*} \bigvee_{i \in I} U_i = \operatorname{int}\left(\overline{\bigcup_{i \in I} U_i}\right). \end{align*} [/step]

[step:Verify the finite Boolean join and distributive laws]For $U,V \in \operatorname{RO}(P)$, the finite join obtained from the arbitrary join formula is \begin{align*} U \vee V=r(U \cup V)=\operatorname{int}(\overline{U \cup V}). \end{align*} This is the least regular open set containing $U \cup V$, by the preceding step applied to the two-element family indexed by $\{1,2\}$. We verify distributivity. Let $U,V,W \in \operatorname{RO}(P)$. Define \begin{align*} A := (U \cap V) \cup (U \cap W). \end{align*} The right-hand side of the desired identity is \begin{align*} (U \cap V) \vee (U \cap W)=r(A)=\operatorname{int}(\overline{A}). \end{align*} Since $A \subset U$ and $A \subset V \cup W$, monotonicity of closure and regular openness of $U$ give \begin{align*} r(A) \subset \operatorname{int}(\overline{U}) \cap \operatorname{int}(\overline{V \cup W})=U \cap (V \vee W). \end{align*} For the reverse inclusion, take $x \in U \cap (V \vee W)$. Because $U \cap (V \vee W)$ is open, it is enough to prove that this open set is contained in $\overline{A}$, since then \begin{align*} U \cap (V \vee W) \subset \operatorname{int}(\overline{A})=r(A). \end{align*} Let $O \in \mathcal{T}$ be an open neighbourhood of $x$. Since $x \in U$ and $U$ is open, $O \cap U$ is an open neighbourhood of $x$. Since $x \in V \vee W=\operatorname{int}(\overline{V \cup W}) \subset \overline{V \cup W}$, the non-empty open set $O \cap U$ meets $V \cup W$. Choose $y \in O \cap U \cap (V \cup W)$. Then $y \in O \cap A$, so every open neighbourhood $O$ of $x$ meets $A$. Hence $x \in \overline{A}$. Therefore \begin{align*} U \cap (V \vee W) \subset r(A)=(U \cap V) \vee (U \cap W). \end{align*} Combining both inclusions gives \begin{align*} U \cap (V \vee W)=(U \cap V) \vee (U \cap W). \end{align*} Commutativity, associativity, idempotence, and absorption for $\wedge$ follow from the corresponding set-theoretic laws for intersection. Commutativity, associativity, idempotence, and absorption for $\vee$ follow from uniqueness of least upper bounds in the inclusion order. The complement identities $U \cap \neg U=\varnothing$, $U \vee \neg U=P$, and $\neg\neg U=U$ were proved above. Together with the distributive law just proved, these are precisely the finite Boolean algebra laws for $\operatorname{RO}(P)$.[/step]

[guided]The distributive law requires a proof because $V \vee W$ is not the raw union $V \cup W$; it is the regularised union $\operatorname{int}(\overline{V \cup W})$. We prove the identity directly from the definitions: \begin{align*} U \cap (V \vee W)=(U \cap V) \vee (U \cap W). \end{align*} Define the subset \begin{align*} A := (U \cap V) \cup (U \cap W). \end{align*} Then the join on the right is \begin{align*} r(A)=\operatorname{int}(\overline{A}). \end{align*} Because $A \subset U$ and $A \subset V \cup W$, monotonicity of closure gives $\overline{A} \subset \overline{U}$ and $\overline{A} \subset \overline{V \cup W}$. Taking interiors and using that $U$ is regular open yields \begin{align*} r(A)=\operatorname{int}(\overline{A}) \subset \operatorname{int}(\overline{U}) \cap \operatorname{int}(\overline{V \cup W})=U \cap (V \vee W). \end{align*} This proves one inclusion. For the reverse inclusion, fix $x \in U \cap (V \vee W)$. The strategy is to show that $x$ lies in $\overline{A}$; because $U \cap (V \vee W)$ is itself open, this will put the whole open set inside $\operatorname{int}(\overline{A})$. Let $O \in \mathcal{T}$ be any open neighbourhood of $x$. Since $x \in U$ and $U$ is open, the set $O \cap U$ is an open neighbourhood of $x$. Since \begin{align*} x \in V \vee W=\operatorname{int}(\overline{V \cup W}) \subset \overline{V \cup W}, \end{align*} this non-empty open neighbourhood $O \cap U$ must meet $V \cup W$. Choose \begin{align*} y \in O \cap U \cap (V \cup W). \end{align*} Then either $y \in U \cap V$ or $y \in U \cap W$, so $y \in A$. Since also $y \in O$, every open neighbourhood $O$ of $x$ meets $A$, which is exactly the statement $x \in \overline{A}$. Thus \begin{align*} U \cap (V \vee W) \subset \operatorname{int}(\overline{A})=r(A)=(U \cap V) \vee (U \cap W). \end{align*} Combining the two inclusions proves distributivity.[/guided]

[step:Obtain arbitrary meets by De Morgan duality]Let $(U_i)_{i \in I}$ be an indexed family in $\operatorname{RO}(P)$. Define \begin{align*} M := \neg\left(\bigvee_{i \in I} \neg U_i\right). \end{align*} The previous steps show that each $\neg U_i$ is regular open, that the displayed join exists in $\operatorname{RO}(P)$, and that the complement of a regular open set is regular open. Hence $M \in \operatorname{RO}(P)$. For every $i \in I$, since $\neg U_i \subset \bigvee_{j \in I} \neg U_j$, the order-reversal property proved above gives \begin{align*} M=\neg\left(\bigvee_{j \in I} \neg U_j\right) \subset \neg\neg U_i = U_i. \end{align*} Thus $M$ is a lower bound. If $Y \in \operatorname{RO}(P)$ is any lower bound, so $Y \subset U_i$ for every $i \in I$, then the order-reversal property gives \begin{align*} \neg U_i \subset \neg Y \end{align*} for every $i \in I$. Hence \begin{align*} \bigvee_{i \in I} \neg U_i \subset \neg Y. \end{align*} Applying the order-reversal property once more, and using the involution identity on $Y$, gives \begin{align*} Y=\neg\neg Y \subset \neg\left(\bigvee_{i \in I} \neg U_i\right)=M. \end{align*} Therefore $M$ is the greatest lower bound of the family. Since arbitrary joins and arbitrary meets exist, and since the finite Boolean operations above satisfy the complement, meet, join, and distributive laws inside $\operatorname{RO}(P)$, the regular open sets form a complete Boolean algebra with the stated operations.[/step]

[guided]The only subtle point in constructing arbitrary meets is that the complement operation is not the raw set-theoretic complement. It is the regular-open complement $U \mapsto \neg U=\operatorname{int}(P \setminus U)$, so De Morgan duality must use the two identities already proved: the map is order reversing and satisfies $\neg\neg U=U$ for every $U \in \operatorname{RO}(P)$. Let $(U_i)_{i \in I}$ be an indexed family in $\operatorname{RO}(P)$. Define \begin{align*} M := \neg\left(\bigvee_{i \in I} \neg U_i\right). \end{align*} Each $\neg U_i$ belongs to $\operatorname{RO}(P)$, the join exists by the join step, and the complement of a regular open set is regular open; hence $M \in \operatorname{RO}(P)$. We first prove that $M$ is a lower bound. Fix $i \in I$. Since $\neg U_i$ is one member of the family whose join is $\bigvee_{j \in I} \neg U_j$, we have \begin{align*} \neg U_i \subset \bigvee_{j \in I} \neg U_j. \end{align*} Because complement reverses order on regular open sets, applying $\neg$ to this inclusion gives \begin{align*} \neg\left(\bigvee_{j \in I} \neg U_j\right) \subset \neg\neg U_i. \end{align*} Using $\neg\neg U_i=U_i$, this becomes \begin{align*} M \subset U_i. \end{align*} Since $i$ was arbitrary, $M$ is a lower bound for the whole family. Now let $Y \in \operatorname{RO}(P)$ be any other lower bound, meaning $Y \subset U_i$ for every $i \in I$. Applying order reversal to each inclusion gives \begin{align*} \neg U_i \subset \neg Y \end{align*} for every $i \in I$. Thus $\neg Y$ is a regular open upper bound of the family $(\neg U_i)_{i \in I}$. Since $\bigvee_{i \in I} \neg U_i$ is the least such upper bound, we get \begin{align*} \bigvee_{i \in I} \neg U_i \subset \neg Y. \end{align*} Apply order reversal again and use $\neg\neg Y=Y$: \begin{align*} Y=\neg\neg Y \subset \neg\left(\bigvee_{i \in I} \neg U_i\right)=M. \end{align*} Therefore every lower bound $Y$ is contained in $M$, so $M$ is the greatest lower bound.[/guided]