(Solution verification request) If $Y$ is an ordered set in the order topology, $f,g\in C(X\rightarrow Y)$, show $\{x|f(x)\le g(x)\}$ is closed in $X$ Am I correct or at least on the right track? Any help is appreciated.
Proof: We will show that $\{$$\mathcal x | f(x)$ $\le$ $\mathcal g(x)$$\}$ is closed in $\mathrm X$ by showing that $\{$$\mathcal x | f(x)$ > $\mathcal g(x)$$\}$ is open in $\mathrm X$.
As $\mathrm Y$ is an ordered set, consider intervals of the form $\mathbb (a,b)$ and $\mathbb (b,c)$ where $\mathbb a < b < c$. The set $\mathrm K_1$ = $f^{-1}$ $\mathbb (b,c)$ consists of all points of $\mathrm X$ which get mapped to $\mathbb (b,c)$ by $\mathcal f$, and the set $\mathrm K_2$ = $g^{-1}$ $\mathbb (a,b)$ is all points of $\mathrm X$ which get mapped to $\mathbb (a,b)$ by $\mathcal g$. Then $\mathrm K1$ $\cap$ $\mathrm K_2$ would be the points of $\mathrm X$ where $\mathcal f(x) > g(x)$ for some subset of $\mathrm Y$.
Let $\cup$ $V_$ be the union of all sets $V_$ = $K_i$ $\cap$ $K_j$ where each $K_i$ is of the form $f^{-1}$ $\mathbb (b,c)$ and each $K_j$ is of the form $g^{-1}$ $\mathbb (a,b)$ for $\mathbb a < b < c$. Each $K_i$ and $K_j$ are open because $\mathbb (a,b)$ and $\mathbb (b,c)$ are open in the order topology and both $\mathcal f$ and $\mathcal g$ are continuous. This means that each $V_$ is open, and $\cup$ $V_$ is the arbitrary union of open sets and is thus open as well. As $\cup$ $V_$ consists of all points $\mathrm x$ ∈ $\mathrm X$ such that $\mathcal f(x) > g(x)$, we conclude that $\mathrm X$ - ($\cup$ $V_$) = $\{$$\mathcal x | f(x)$ $\le$ $\mathcal g(x)$$\}$ is closed in $\mathrm X$
 A: The $V_\alpha$ are not in general enough to cover all cases where $f(x) < g(x)$. They're only intersections of inverse images of some basic sets of $Y$. $Y$ can have more. E.g. consider what happens in a trivial case where $Y=\{0,1\}$; this has no non-empty open intervals at all....In general we can have $f(x) < g(x)$ but no element inbetween $f(x)$ and $g(x)$, so they won't be part of obvious intervals; not all ordered sets are like $\Bbb R$ or $\Bbb Q$..
It's somewhat easier to prove that $U:=\{(y_1, y_2) \in Y^2 \mid y_1 > y_2\}$ is open in $Y^2$ (in the product topology) using products of subbasic order-open sets and then note that your set equals $(f \nabla g)^{-1}[U]$ e.g. (where $f\nabla g: X \to Y^2$, given by $x \to (f(x), g(x))$, is continuous whenever $f,g$ are).
A: In my opinion, your solution is partially correct. I think the problem is $\cup V_{\alpha}$ does not cover the set $\{x| f(x) > g(x)\}$. If $f(x)$ is the immediate successor of $g(x)$, then clearly $x \notin \cup V_{\alpha}$, since you set $V_{\alpha}= f^{-1}(b,c) \cap g^{-1}(a,b)$ with $a<b<c$ (you can check this by yourself). I think you could have done it in a different way. We can prove the set $\{x| f(x) > g(x)\}$ is open:

*

*Given $x \in \{x| f(x) > g(x)\}$, since Y is ordered set in order topology, Y is Hausdorff, we can choose disjoint open intervals $(a,b)$, $(c,d)$ such that $f(x) \in (a,b)$ and $g(x) \in (c,d)$.

*Let $U= f^{-1}(a,b) \cap g^{-1}(c,d)$, then clearly U is non-empty open set. By the disjointness of $(a,b)$ and $(c,d)$ with the fact that $f(x) > g(x)$, we can conclude that $U \cap \{x| f(x) \le g(x)\}=\emptyset$. This shows that {x| f(x) > g(x)} is open,complete the proof.

Hope this helps.
