Priestley space: upwards closure of closed set

Question

I am trying to prove that, in a Priestley space, the upwards closure of a closed set is closed. That is, given a Priesteley space $X$ and a closed set $F\subseteq X$ I need to show that $\uparrow F=\bigcup _{x\in F}\uparrow x$ is also closed.

I have proved this only in the case $F$ is a singleton, so this might be useful.

Thank you!

Answer 1

You can use that $F$ is closed, hence compact.

Answer 2

For all x,y, if not x <= y, then there's some clopen U(x,y)
with x in U(x,y), y not in U(x,y).

Let K be closed. up K = { x | some k in K with k <= x }.

Assume y not in up K. Then for all a in K, not a <= y. Thus
C_y = { U(a,y) : a in K } covers K.
Let Cf_y be a finite subcover of K.
Show Cf_y covers up K. Additionally,
y not in K_y = Cup Cf_y, K_y is closed and up K subset K_y

Conclude by showing
up K subset Cap{ K_y | y not in up K } subset up K,

that up K is an intersection of closed sets, hence closed.