Does disjunction of two Boolean algebra cuts always produce their ideal sum? Let $(B, 0, 1, \leq, \wedge, \vee, \neg)$ be a Boolean algebra.
For a subset $A \subseteq B,$ denote by $L(A) = \{l \in B \mid (\forall a\in A) \, l \leq a\}$ the set of all lower bounds of $A,$ and similarly by $U(A) = \{u \in B \mid (\forall a \in A) \, a \leq u\}$ the set of all upper bounds of $A.$
A subset $C \subseteq B$ is called a cut, iff $L(U(C)) = C.$ These cuts appear in the Dedekind–MacNeille completion of the Boolean algebra.
A nonempty subset $I \subseteq B$ is called an ideal, iff $a \in I, b \leq a$ implies $b \in I$, and $a, b\in I$ implies $a \vee b \in I.$ It can be shown that if $I$ and $J$ are ideals, then $I \oplus J = \{c \in B \mid (\exists i \in I)(\exists j \in J) \, c \leq i \vee j\}$ is the smallest ideal, containing both $I$ and $J.$
It is easy to see that every cut is an ideal. Also, if $C$ and $C'$ are two cuts, then $C'' = L(U(C \cup C'))$ is also a cut. My intuition is that this should also be the smallest ideal, containing both $C$ and $C',$ but couldn't prove that. So, is $C'' = C \oplus C'$?
 A: This is false in general.  In terms of Stone duality, Boolean algebras correspond to Stone spaces, ideals correspond to open sets, and cuts correspond to regular open sets; your question is then whether the union of two regular open sets in a Stone space is regular.  This is false; for instance, consider $X=\mathbb{N}\cup \{\infty\}$ and let $U$ be the set of even natural numbers and $V$ be the set of odd natural numbers.  Then $U$ and $V$ are both regular open sets, but their union is $\mathbb{N}$, which is not regular (its regularization is all of $X$).
In terms of Boolean algebras, this translates into the following example: let $B$ be the set of finite or cofinite subsets of $\mathbb{N}$, let $C\subset B$ be the ideal of (finite) sets all of whose elements are even, and let $C'\subset B$ be the ideal of (finite) sets all of whose elements are odd.  Then $C$ and $C'$ are cuts, but $C\oplus C'$ is the ideal of all finite sets, while $C''$ is all of $B$ since $U(C\cup C')$ is just $\{\mathbb{N}\}$.
