# Boolean Algebra with decomposition property

1. Does there exist a (countable) Boolean algebra $(B,\bigcup, \bigcap, 0, 1)$ with the following property:

$\forall A\in B\setminus \{0\}$ there exists $A_1,A_2\in B$ such that $A_i\neq A$, $A_1 \bigcup A_2 = A$ and $A_1 \cap A_2=0$.

(e.g. in the uncountable case: take the Boolean algebra consisting of all infinite subets of the natural numbers)

1. Does such a property have a (well established) name?
• You must be leaving something out. Any Boolean algebra of sets has this property: Let $A_1=A$, $A_2=\emptyset$. – David C. Ullrich Jan 16 '18 at 14:40
• With $A_i \neq A$, this is impossible if $A = 0$, so the answer is no, for any Boolean algebra, whatever its cardinality. So you have to tune your question a little bit more... – amrsa Jan 16 '18 at 16:01
• Thank you both for your input. My property should be: For any non-zero element $A$, there exists a (proper) diamond $A,A_1,A_2,0$ in $B$. – Tom Q Jan 16 '18 at 16:59
• Your example for an uncountable case is wrong. The family of infinite subsets of any set is not a Boolean algebra. For example, two infinite subsets may have a finite intersection. – amrsa Jan 17 '18 at 10:48

What you're looking for is a Boolean algebra $\mathbf{B} = \langle B, \wedge, \vee,',0,1 \rangle$ without join-irreducible elements.
Definition. In any lattice (in particular, a Boolean algebra) an element $x$ is said to be join-irreducible if $x = a$ or $x = b$, whenever $x = a \vee b$.
In a lattice with $0$, an atom is an element $x$ such that $y = 0$, whenever $y<x$.
Lemma. If $\mathbf{B} = \langle B, \wedge, \vee,',0,1 \rangle$ is a Booelan algebra and $j \in B$ is a join-irreducible element, then $j$ is an atom.
Proof. Let $0 \leq x < j$. We ought to prove that $x=0$. We have $$j = x \vee j = (x \vee j) \wedge (x \vee x') = x \vee (j \wedge x').$$ Since $j$ is join-irreducible and $x < j$, it follows that $j = j \wedge x'$, that is, $j \leq x'$, whence $$x = x \wedge j \leq x \wedge x' = 0.$$