MHB Is a Boolean Lattice Atomic if the Top Element is the Join of Atoms?

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A Boolean lattice L is atomic if the top element is the join of a set of atoms. The forward implication has been proven using Zorn's lemma, establishing that the maximal element in the set of joins of atoms must be the top. The reverse implication was resolved by leveraging the join continuity of Boolean lattices. It was shown that if the top element is the join of all atoms, then every element in L can be expressed as a join of atoms, confirming that L is atomic. This proof highlights the relationship between the structure of Boolean lattices and their atomic properties.
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My problem for this thread is:

Let $L$ be a Boolean lattice. Prove that $L$ is atomic if and only if the top element is the join of a set of atoms.

For the forward implication, I am already done. I used Zorn's lemma to show that the set, $\mathcal{F}$, of the elements in $L$ which are the joins of some set of atoms has a maximal element, and that that element must be the top.

The reverse implication is what is tripping me up. Any help is greatly appreciated!
 
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Well, I'm pretty sure I solved this a few days ago, but I forgot that posting here might help others or myself so here's the proof to the reverse implication. I used an earlier exercise that proved that Boolean lattices were join continuous. Turns out this problem also solved my next problem that needed to show that, in a complete atomic Boolean lattice, every element is the join of a set of atoms.

Let $L$ be a Boolean lattice where $\top$ is the join of a set of atoms and let $A$ be the set of all atoms from $L$. Then $\bigvee A = \top$. Let $y\in L$. Then $y\wedge \bigvee A = \bigvee\{y\wedge a: a\in A\} = y$ since $L$ is join continuous. If $y\wedge a = \bot$ for all $a\in A$, then it must be that $y = \bot$. So, if $y\neq \bot$, then there exists an $a\in A$ such that $y\wedge a = a$ and so $\downarrow y$ contains an atom from $L$. Since $y$ was arbitrary, $L$ is atomic.
 

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