Let $H$ be a Heyting algebra. It is a well-known result that there is a partially ordered set (Kripke frame) X such that there is an embedding of Heyting algebras $f: H \to \mathsf{Up}(X)$, where $\mathsf{Up}(X)$ consists of the up-sets of $X$. Specifically, take $X$ to consist of the prime filters of $H$, ordered by inclusion, and let $f(a) := \{ P \in X | a \in P \}$.


Is it possible to modify the result like this: Let $H$ be a **complete** Heyting algebra. Does there exists a **complete** embedding $g: H \to \mathsf{Up}(X)$ for some partially ordered set $X$? By a complete embedding, I mean an embedding preserving arbitrary meets and joins.


The approach above does not seem to work as it might be the case that there are incomplete prime filters on the complete Heyting algebra.



Is it possible to completely embed complete Heyting Algebras into upsets of a poset?
No, not in general: for instance, the real interval $([0,1],{\le})$, or any non-atomic complete Boolean algebra, do not have such an embedding. This follows from the following characterization:



> 
> **Proposition:** Let $H$ be a complete Heyting algebra. The following are equivalent:
> 
> 
> 1. $H$ has a complete embedding into $\mathrm{Up}(X)$ for some poset $X$.
> 2. $H$ is isomorphic to $\mathrm{Up}(X)$ for some poset $X$.
> 3. Every element of $H$ is a join of a set of completely join-irreducible elements in $H$.
> 
> 
> 


**Proof sketch:**


$2\to1$ is trivial.


$3\to2$: Let $X$ be the set of all completely join-irreducible elements of $H$, ordered upside down. Then the mapping $g\colon H\to\mathrm{Up}(X)$ given by
$$g(a)=\{x\in X:x\le a\}$$
is easily checked to be an isomorphism.


$1\to3$: Let $g\colon H\to\mathrm{Up}(X)$ be a complete embedding. For any $x\in X$, the set
$$P\_x=\{a\in H:x\in g(a)\}$$
is a filter closed under arbitrary meets, i.e., principal: $P\_x=[h(x),1]$ for some $h(x)\in H$. Moreover, $P\_x$ is completely prime, hence the element $h(x)$ is completely join-irreducible.


Now, if $a,b\in H$ are such that $a\nleq b$, then there exists a completely join-irreducible element such that $u\le a$, and $u\nleq b$: indeed, $g(a)\nsubseteq g(b)$ as $g$ is an embedding, hence there is $x\in g(a)\smallsetminus g(b)$, and then $u=h(x)$ works.


This implies that any $a\in H$ is the join of all completely join-irreducible elements below it.


