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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 $M$?

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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 $L$.

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  • $\begingroup$ what if I impose the additional condition that $M$ is atomic? $\endgroup$
    – Carmen
    Commented Jan 14, 2013 at 13:07
  • $\begingroup$ Not sure what you mean ... to me, "atomic" means that every element is a join of atoms. $\endgroup$
    – Nik Weaver
    Commented Jan 14, 2013 at 13:19
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    $\begingroup$ Also, when you edit a question to add conditions you should include a comment that you've done so. $\endgroup$
    – Nik Weaver
    Commented Jan 14, 2013 at 13:22
  • $\begingroup$ 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. $\endgroup$
    – Carmen
    Commented Jan 14, 2013 at 13:56
  • $\begingroup$ 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.) $\endgroup$
    – Emil Jeřábek
    Commented Jan 14, 2013 at 16:20

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