TITLE: How to prove that this set is closed?
QUESTION [7 upvotes]: Suppose $Y$ be an ordered set in the order topology. Let $f, g: X \to Y$ be continuous. How to show that the set $\{x| f(x) \leq g(x)\}$ is closed?
This is a excercise from munkres. Maybe trying to show that the set $f(\{x| f(x) > g(x)\})$ is open is suffice. But I could not figure out the connection between this set and the continuity of the two functions.Could you give a hint?
Thanks.

REPLY [7 votes]: The set $$U = \{(y_1, y_2) \in Y \times Y \mid\, y_1 \leq y_2 \,\}$$ is closed in $Y \times Y\,$: 
Suppose $(z_1, z_2) \in Y \times Y\,$ is not in it, so that $z_1 > z_2$. 
If there is some $z_3 \in Y$ such that $z_1 > z_3 > z_2$, the basic open set $(z_3, \rightarrow) \times (\leftarrow, z_3)$ contains $(z_1, z_2)$ and misses $U$, otherwise 
$z_1$ and $z_2$ are neighbours and $(z_2, \rightarrow) \times (\leftarrow, z_1)$ has this property. Note that all used subsets in $Y$ are open in the order topology.
Now $f \nabla g: X \rightarrow Y \times Y$ is continuous (standard in the product topology) when $f$ and $g$ are, and your set is exactly 
$$(f \times g)^{-1}[U]$$
and thus closed as the inverse image of a closed set under a continuous function.