Are the connected components of a Priestley space closed?

Question

Preliminaries A Priestley space is both a poset and a topological space. The topologically connected components of the space are trivially closed. (They are just the points of the underlying set.) But a Priestley space also has connected components as a poset, i.e. the maximal connected subposets. These poset components have less global structure among each other than the topological components, i.e. we have

Any poset can be written as a disjoint union of its components.

Question I would like to know whether the quotient of a Priestley space by the equivalence relation of its poset components is always a Stone space. A necessary condition for this is that the poset components are closed sets of the topology. (I guess this condition is also sufficient, but this is not the question.) Hence my question is:

Are the connected components of the poset of a Priestley space closed sets of the topology?

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Solution attempt After trying for some time to prove this, I started to try to construct a counter-example. My strategy is to take the Stone space of a countable atom-less Boolean algebra and add a partial order relation to it. One promising idea would be to take away one point, and connect the remaining points by a fence. The fence component is not closed, because the point component is not clopen. My problem here is proving that this is really a Priestley space.

Answer 1

Let $\mathbb{N}\cup\{\infty\}$ denote the one-point compactification. Give $\mathbb{N}\cup\{\infty\}$ the partial ordering where $m>n$ if and only if $m$ is even, $n$ is odd and $|m-n|=1$. In other words, we give $\mathbb{N}$ a zig-zag ordering and we let $\infty$ be a point which is not comparable to any point of $\mathbb{N}$. Then it is easy to verify that $\mathbb{N}\cup\{\infty\}$ is a Priestley space with this partial ordering. However, the order component $\mathbb{N}$ is not closed.