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

[Aryth1]

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!

 

[Aryth1]

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.