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Are all atoms of the lattice of filters (ordered by set-theoretic inclusion) principal filters?

Note that atoms of this lattice are not ultrafilters (ultrafilters are atoms of the dual lattice).

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The answer is yes.

Suppose a non-principal filter $F$ on a set $U$ is an atom. Then there exists a set $M\ne U$ such that $M\in F$. Consider an element $x\in U$ such that $x\notin M$. Then the principal filter corresponding to the set $U\setminus\{x\}$ is strictly above $F$. So $F$ is not an atom.

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