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Smallest Filter Heyting Algebras

Lemma: Let $A$ be the subset of a Heyting Algebra $H$ . Then the smallest filter which contains $A$ is

F = {a \in H : a \geq a_{1} ⊓ a_{2} \dots a_{k} where a_{1}, a_{2} \dots a_{k} \in A}

Proof: $F$ contains $\overline{a} = a_{1} ⊓ \dots ⊓ a_{k}$ , because for any $a, b \in F$ we have $a ⊓ b \in F$ . Hence Any $a \geq \overline{a} \in F$ .

Thus, if $\overline{F}$ is the smallest filter $\overline{F} \subseteq F$ . But any Filter containing $A$ must be a superset of $F$ Hence $F$ is the smallest filter which contains $A$ .

Heyting Algebra

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