202309121209

Extension of Proper Filter to Prime Filter in Heyting Algebras

Lemma: Let $F$ be a proper filter in $H$ and let $a \in / F$ . Then there exists a prime filter $G$ such that $F \subseteq G$ and $a \in / G$ .

Proof: Consider $F$ to be the set of filters that do not contain $a$ but contain $F$ with inclusion as the partial order.

Here the union of every chain is also in $F$ hence all chains have upper bounds. By Zorn’s Lemma, There is a Maximal element $G$ .

We need to prove that $G$ is prime.

Consider $G_{y} = {x : x \geq g ⊓ y}$ for some $g \in G$ (Pick $y$ and $g$ and find the smallest filter which contains both). If $c ⊔ b \in G$ . Then consider $G_{c}$ and $G_{b}$ .

If both of them contain $a$ Then there are $g_{1}, g_{2} \in G$ such that $g_{1} ⊓ b \leq a$ and $g_{2} ⊓ c \leq a$ . since $g_{1}, g_{2}, b ⊔ c \in G$ we have $(g_{1} ⊓ g_{2}) ⊓ (b ⊔ c) \in G$ .

We also have $a = a ⊔ a \geq (g_{1} ⊓ g_{2} ⊓ b) ⊔ (g_{1} ⊓ g_{2} ⊓ c) = (g_{1} ⊓ g_{2}) ⊓ (b ⊔ c) \in G$ which is a contradiction.

Thus one of $G_{b}$ and $G_{c} \in F$ and by primality of $G$ we have either $b, c \in G$ thus $G$ is prime.

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Extension of Proper Filter to Prime Filter in Heyting Algebras

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