Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $L$ be an atomistic ortholattice (i.e. every element can be written as a join of atoms) with top and bottom elements 0 and 1, and let $M$ be a distributive atomic sub-ortholattice of $L$.

Is $M$ generated by its atoms, in the sense that every element in $M$ can be written as a join of the atoms in $M$?

share|improve this question

1 Answer 1

up vote 3 down vote accepted

No. Let $L$ be the power set of ${\bf N}$ ordered by inclusion. It is an atomic Boolean algebra. Let a subset of ${\bf N}$ belong to $M$ if its characteristic function is periodic. Then $M$ is a nonatomic Boolean subalgebra of $L$.

share|improve this answer
    
what if I impose the additional condition that $M$ is atomic? –  Carmen Jan 14 '13 at 13:07
    
Not sure what you mean ... to me, "atomic" means that every element is a join of atoms. –  Nik Weaver Jan 14 '13 at 13:19
1  
Also, when you edit a question to add conditions you should include a comment that you've done so. –  Nik Weaver Jan 14 '13 at 13:22
    
Yes, good idea about the comment. I understand atomic to mean that $M$ has atoms, while atomistic means that every element is a join of atoms. –  Carmen Jan 14 '13 at 13:56
    
Having one atom is not going to help you, you can modify the example to take all subsets of $\mathbb N$ whose characteristic function is periodic on $\mathbb N-\{0\}$. This algebra has an atom, namely $\{0\}$, but only two of its elements are joins of atoms. (The general form of Nik’s answer is that every Boolean algebra is a subalgebra of a powerset algebra, by the way.) –  Emil Jeřábek Jan 14 '13 at 16:20

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.