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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 $M$?

edited Jan 14 at 13:58

asked Jan 14 at 11:07

Katie
145●4

Nik Weaver · Accepted Answer · 2013-01-14 11:30:34Z UTC

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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 $L$.

answered Jan 14 at 11:30

Nik Weaver
4,810●1●8●21

what if I impose the additional condition that $M$ is atomic? – Katie Jan 14 at 13:07

Not sure what you mean ... to me, "atomic" means that every element is a join of atoms. – Nik Weaver Jan 14 at 13:19

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Also, when you edit a question to add conditions you should include a comment that you've done so. – Nik Weaver Jan 14 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. – Katie Jan 14 at 13:56

as a side question, I've added a comment in the edit summary box, but the comment doesn't seem to appear anywhere now, am I missing something? – Katie Jan 14 at 14:02

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distributive sublattices of atomistic ortholattices

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