Problem 8 (a) in Exercises after Sec. 18 in Munkres' Topology, 2nd ed.: How to show this set is closed? Let $X$ be an arbitrary topological space, let $Y$ be an ordered set in the order topology, and let the maps $f, g \colon X \to Y$ be continuous. Then how to show that the set $S$ given by
$$S \colon= \{ \ x \in X \ \colon f(x) \leq g(x) \ \}$$ 
is closed in $X$?
I'll try to give an elaborate hint: It's very similar to the following proof

Let $h:X \rightarrow \mathbb{R}$ be defined as $h(x)=g(x)-f(x)$

$h$ is continuous, so $S=h^{-1}([0,\infty))$ is closed since $[0,\infty)$ is closed in $\mathbb{R}$.
I'll try to give an elaborate hint: It's very similar to the following proof

Let $h:X \rightarrow \mathbb{R}$ be defined as $h(x)=g(x)-f(x)$

$h$ is continuous, so $S=h^{-1}([0,\infty))$ is closed since $[0,\infty)$ is closed in $\mathbb{R}$.