Proving a Set in the Order Topology is Closed

[Poopsilon]

Proving a Set is Closed (Topology)

Homework Statement

Let Y be an ordered set in the order topology with f,g:X\rightarrow Y continuous. Show that the set A = \{x:f(x)\leq g(x)\} is closed in X.

Homework Equations

The Attempt at a Solution

I cannot for the life of me figure this out. As far as I can tell one either needs to show the set A gathers its limit points or that it is the pre-image (under f or g) of some closed set in Y. We know the order topology is Hausdorff, so that's something. I just don't even know how to get started on this one. Hopefully someone can help, thanks.

 

[kru_]

This one is tricky.

Instead of trying to prove directly that A is closed, try to show that the complement, X-A, is open. Show the set of all x such that f(x) > g(x) to be open. You can show this by showing that every point x in X is an interior point. Just find an open set around an arbitrary x.

So let x be an arbitrary element of X such that f(x) > g(x). If there is no such x, then the set is empty, then and it is open trivially. So assume such an x exists and consider f(x) and g(x) in the order topology.

Now draw the number line to represent Y and put f(x) and g(x) in their respective places.

Then find two open sets in Y.

Does that help get you started?

 

[Poopsilon]

So going from there we use the fact that the order topology is Hausdorff and find two open disjoint sets one containing f(x) and the other containing g(x), from here I would like to map back under f^-1 and g^-1 to two open sets in X and from here take the intersection of them which would contain x and would be open, but I can't see how we know anything about these two disjoint open sets in Y...