# Proving a Set in the Order Topology is Closed

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 $X$.

## 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.

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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?

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...