Prime filters on a linear ordered lattice

Question

Consider a lattice algebra given by a linear order $(H,\leq)$. We say that $A\subseteq H$ is a prime filter if $\emptyset\neq A\neq H$ and

1. $x,y\in A$ implies $x\land y\in A$ (where $\land$ is the minimum),
2. $x\in A$, $x\leq y$ implies $y\in A$,
3. $x\lor y\in A$ implies $x\in A$ or $y\in A$ (where $\lor$ is the maximum).

The dual frame of $H$ is given by the set of all prime filters on $H$ ordered by $\subseteq$. I wonder, if $\leq$ is a linear order, if the prime filters are linearly ordered by $\subseteq$ as well. I tried to prove it several ways, but got nowhere. However, I also can't directly produce a counterexample.

Answer 1

In a linearly ordered lattice every proper filter is a prime filter, so it is enough to prove that the set of filters is linearly ordered too.

Now let $F$ and $G$ be filters of $H$ such that $F \nsubseteq G$, and let $a \in F \setminus G$.
Let $b \in G$, arbitrary. Then $a \leq a \vee b$, whence $a \vee b \in F$.
If $a \vee b = a$, then $b \leq a$, yielding $a \in G$, a contradiction.
Thus $a \vee b = b \in F$ and therefore $G \subseteq F$.