Order topology related question

Question

Say I have two topological spaces $X,Y$ s.t $Y$ has the order topology, and $f,g$ are two continuous functions $g,f:X\rightarrow Y$. I want to show that the set $O=\{x:f(x)> g(x)\}$ is open in $X$. So my attempt was defining $h=f-g$ and looking at $O=\{x:h(x)>0\}$ and proving it's open in $X$, but I'm not even sure if I'm allowed to use the concept of $0$ since I don't know anything about $Y$.

Say I can use this concept, I want to show that there exists such open $U\subseteq Y$ s.t $h^{-1}(U)=O$, and by continuity I'm done. So I thought about defining $U=Im(h)\cap V$ where $V=\{y\in Y:y>0\}$ which is simply the ray $(0,\infty)$ in $Y$ and obviously open. So $U$ is open in the subspace topology of $Im(h)$ and therefore $O$ is open in $X$.

Is my approach correct?

Answer 1

In this answer I show that $\{x\mid f(x) \le g(x)\}$ is closed when $f,g$ are continuous. Your set is its complement, thus open.

We only need the order and product topology to see this.

Answer 2

Fix $x\in O$. We will find a neighborhood $W$ of $x$ with $W\subseteq O$.

Case 1. There is some $z\in(g(x),f(x))$. Since $g(x)\in(\leftarrow,z)$ (and $(\leftarrow,z)$ is open and $g$ is continuous), there is a neighborhood $U$ of $x$ such that $g(U)\subseteq(\leftarrow,z)$. Similarly there is a neighborhood $V$ of $x$ such that $f(V)\subseteq(z,\rightarrow)$. Let $W=U\cap V$. If $y\in W$ then $g(y)<z<f(y)$ hence $y\in O$ and $W\subseteq O.$

(If you prefer to work with preimages, then fix $x\in O$ and in Case 1 fix $z\in(g(x),f(x))$ as before, and let $W=\big(g^{-1}(\leftarrow,z)\big)\cap\big(f^{-1}(z,\rightarrow)\big)$. Then $W$ is open and $x\in W\subseteq O$.)

Case 2. $(g(x),f(x))=\varnothing.$ In this case $(\leftarrow,f(x))=(\leftarrow,g(x)]$ and $(g(x),\rightarrow)=[f(x),\rightarrow)$, open sets. Let $W=\big(g^{-1}(\leftarrow,g(x)]\big)\cap\big(f^{-1}[f(x),\rightarrow)\big)$. Then $W$ is open and $x\in W\subseteq O$. (Indeed, if $y\in W$ then $g(y)\le g(x)<f(x)\le f(y).$)