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

Question

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 $X$.

I am completely stumped on this problem. As far as I can tell I've either got the task of proving $A$ gathers its limit points, or showing that there is a closed set $V \subset Y$ such that $f^{-1}(V)$ or $g^{-1}(V)$ is equal to A, which would prove $A$ closed since both $f$ and $g$ are continuous.

The latter strategy seems like the more likely to succeed. Unfortunately I can't find any way of constructing this set $V$. The fact that I don't know if $f$ or $g$ are injective means I keep running into the problem of having $f^{-1}(V)$ or $g^{-1}(V)$ give me extra points not in $A$. And even if I were able to construct $V$ this wouldn't guarantee it was closed.

I suspect I may need to use the continuity of both $f$ and $g$ together in some way but I can't see how. Can anyone give me some guidance on this problem? Thanks.

Answer 1

HINT: Let $h:X\to Y:x\mapsto\max\{f(x),g(x)\}$.

1. Show that $h$ is continuous.
2. Note that $A=\{x\in X:h(x)=g(x)\}$.

Answer 2

Hint for the problem:

1. Recall that $Y$ under the order topology is Hausdorff (Exercise).

2. Show that the complement of $A$ is open. Do this by supposing that $x \notin A$. Then $g(x) < f(x)$. If this is the case, then either there exists $y$ such that $g(x) < y < f(x)$, or (exclusively) there does not exist any $y \in Y$ in between $f(x)$ and $g(x)$.

Answer 3

To show the set $A$ is closed, we only need to proof

$A^c=\{x\in X:f(x)>g(x)\}$ is open.

Proof: For any point $x \in A^c$, $f(x)>g(x)$; and the order topological space is Hausdorff, then there exist disjoint open sets $U_1$ with $f(x) \in U_1$ and $U_2$ with $g(x) \in U_2$ in $X$, such that for any point $y \in U_1$ and $z \in U_2$, $f(y)>g(z)$. Let $U= f^{-1}(U_1) \cap g^{-1}(U_2)$ is an nonempty open set in $X$: for $x \in U$. Obviously, we see $x \in U \subset A^c$, which implies the set $A^c$ is open.