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  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?
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    $\begingroup$ You must be leaving something out. Any Boolean algebra of sets has this property: Let $A_1=A$, $A_2=\emptyset$. $\endgroup$ – David C. Ullrich Jan 16 '18 at 14:40
  • $\begingroup$ 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... $\endgroup$ – amrsa Jan 16 '18 at 16:01
  • $\begingroup$ 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 $B$. $\endgroup$ – Tom Q Jan 16 '18 at 16:59
  • $\begingroup$ 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. $\endgroup$ – amrsa Jan 17 '18 at 10:48
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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 $$x = x \wedge j \leq x \wedge x' = 0.$$

So, rephrasing the first paragraph, you're looking for an atomless Boolean algebra.
The answer is yes, there are atomless Boolean algebras and in particular, countable ones. See, for example, this Wikipedia article.

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