This is a Munkres's problem: Let Y be an ordered set with the ordered topology and f,g functions from X to Y, show that the set is closed. I can show this, for intance, when Y=R(real numbers set) but I can't prove this if Y is just an ordered set because it may have consecutive elements. Any comments will be appreciated
Let . To show W is open in X, I shall show that for every , there exists an open set containing and is contained in W.
As you said, there are two cases to consider. Let be an arbitrary point in .
1. There is no element between and such that . Then, is an open set containing and is contained in W, since f and g are continuous, Y is an ordered set in the order topology, and an intersection of open sets is open.
2. There exists element(s) between and such that . Let z be such an element . Then, is an open set containing and is contained in W.
Thus W is open in X and X-W is closed.