Is the quotient of a complete Heyting algebra by a prime filter complete? Suppose $\mathbf A$ is a complete Heyting algebra and $F$ is a prime filter on $\mathbf A$. Do we then have that the quotient $\mathbf{A}/F$ is complete?
I have tried to prove this, but in my proof I needed that the filter $F$ fulfills
$$X\subseteq F\text{ implies }\bigwedge X\in F.$$
So one way of proving would be to show that any prime filter of a complete Heyting algebra has that property but I am not sure of that.
 A: No.  Suppose $\mathbf{A}$ is the lattice of open sets in some topological space $X$.  Then for each $x\in X$, the filter $F$ of open sets containing $x$ is prime, and the quotient $\mathbf{A}/F$ can be identified with the lattice of germs of open sets at $x$ (i.e., the colimit of the lattices of open subsets of $U$ where $U$ ranges over all neighborhoods of $x$).
In particular, let $X=\mathbb{N}\cup\{\infty\}$ and $x=\infty$.  Then the elements of $\mathbf{A}/F$ besides $1$ can be identified with the quotient of the Boolean algebra $\mathcal{P}(\mathbb{N})$ by the ideal of finite sets (an element of $\mathbf{A}/F$ besides $1$ is a germ of an open set that doesn't contain $\infty$, and that's just a subset of $\mathbb{N}$; two such subsets agree on a neighborhood of $\infty$ iff they differ by a finite set).  That is, $\mathbf{A}/F$ is the lattice $\mathcal{P}(\mathbb{N})/\mathrm{fin}$ with a greatest element added on top.  Since $\mathcal{P}(\mathbb{N})/\mathrm{fin}$ is not complete, neither is $\mathbf{A}/F$.
(Note that it is easier to come up with prime filters that are not closed under arbitrary joins--for instance, take a nonprincipal ultrafilter in $\mathcal{P}(\mathbb{N})$, or consider the case where $\mathbf{A}$ is totally ordered.  However, in those easy examples the quotient $\mathbf{A}/F$ is actually still complete!)
