$f,g : X \rightarrow Y$ are continuous functions and Y is an ordered set then $\{x \in X: f(x) \leq g(x)\}$ is closed

Question

$f,g : X \rightarrow Y$ are continuous functions and Y is an ordered set then $\{x \in X: f(x) \leq g(x)\}$ is closed

I saw a proof of this by showing that its complement is open. But the way i started thinking about the problem was trying to prove that it contained all its limit points hence be closed, and using the fact that $Y$ will be Haussdorff since it will have the order topology, but i got nowhere. I was wondering if its possible to do it the way i started thinking about the problem.Thanks in advance.

Answer 1

As your set is the inverse image of $B=\{(y_1,y_2)\in Y\times Y:y_1\le y_2\}$ under the continuous map $X\to Y\times Y$ defined by $x\mapsto(f(x),g(x))$ all one needs to prove is that $B$ is closed in $Y\times Y$.

Let $(y_1,y_2)$ be in the complement of $B$, so $y_1>y_2$. If $y_1>y_0>y_2$ then $(y_1,y_2)$ is in the basic open set $(y_0,\infty)\times(-\infty,y_0)$ which is disjoint from $B$. But if there is no such $y$, use $(y_1,\infty)\times(-\infty,y_2)$ instead.