It doesn't violate any laws per say, due to the general lawlessness of Foldable, but violates aesthetics. If we restrict it to NonEmpty then we get a lovely semigroup homomorphism!
If we loosen it to lists, then the identity element of the list monoid gets mapped to an absorbing element in the Set semigroup under intersection, which just feels... off.

On Sun, Dec 6, 2020 at 10:50 AM Sven Panne <svenpanne@gmail.com> wrote:
Am So., 6. Dez. 2020 um 07:20 Uhr schrieb Reed Mullanix <reedmullanix@gmail.com>:
[...]
  intersections :: Ord a => NonEmpty (Set a) -> Set a
  intersections (s :| ss) = Foldable.foldl' intersection s ss
[...]

Why NonEmpty? I would expect "intersections [] = Set.empty", because the result contains all the elements which are in all sets, i.e. none. That's at least my intuition, is there some law which this would violate?