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.