to include more general lattices, such as distributive lattices or Heyting algebras. However, in these cases maximal ideals are different from prime ideals, and the relation between PITs and MITs is not obvious. Indeed, it turns out that the MITs for distributive lattices and even for Heyting algebras are equivalent to the axiom of choice. On the other hand, it is known that the strong PIT for distributive lattices is equivalent to BPI (i.e. to the MIT and PIT for Boolean algebras). Hence this statement is strictly weaker than the axiom of choice. Furthermore, observe that Heyting algebras are not self dual, and thus using filters in place of ideals yields different theorems in this setting. Maybe surprisingly, the MIT for the duals of Heyting algebras is not stronger than BPI, which is in sharp contrast to the abovementioned MIT for Heyting algebras. Finally, prime ideal theorems do also exist for other (not order-theoretical) abstract algebras. For example, the MIT for rings implies the axiom of choice. This situation requires to replace the order-theoretic term "filter" by other concepts—for rings a "multiplicatively closed subset" is appropriate. The ultrafilter lemma A filter on a set is a nonempty collection of nonempty subsets of that is closed under finite intersection and under superset. An ultrafilter is a maximal filter. The ultrafilter lemma states that every filter on a set is a subset of some ultrafilter on . An ultrafilter that does not contain finite sets is called "non-principal". The ultrafilter lemma, and in particular the existence of non-principal ultrafilters (consider the filter of all sets with finite complements), can be proven using from Zorn's lemma. The ultrafilter lemma is equivalent to the Boolean prime ideal theorem, with the equivalence provable in ZF set theory without the axiom of choice. The idea behind the proof is that the subsets of any set form a Boolean algebra partially ordered by inclusion, and any Boolean algebra is representable as an algebra of sets by Stone's representation theorem. If the set is finite then the ultrafilter lemma can be proven from the axioms ZF. This is no longer true for infinite sets; an additional axiom be assumed. Zorn's lemma, the axiom of choice, and Tychonoff's theorem can all be used to prove the ultrafilter lemma. The ultrafilter lemma is strictly weaker than the axiom of choice. The ultrafilter lemma has many applications in topology. The ultrafilter