order isomorphism maps $f < g$ to $Tf < Tg$? Suppose $C(X)$ is the set of functions $f:X \rightarrow \mathbb{R}$. 

A bijection $T:C(X) \rightarrow C(Y)$ is called order isomorphism if $f \leq g$ if and only if $Tf \leq Tg$. 

$f \leq g$ means that $f(x) \leq g(x)$ for all $x \in X$. Similarly, $f &lt; g $ means that $f(x) &lt; g(x)$ for all $x \in X$.

Question: If $f &lt; g$, then $Tf &lt; Tg$? 

The statement seems correct but I don't know how to prove it. 

If the statement above is not correct, then is it possible to obtain $f \geq 0 \Rightarrow Tf \geq 0$?
First of all, one usually defines $f &lt; g$ to mean $f \leq g$ and $f \neq g$. This is different from your definition, which is a little confusing. For the sake of clarity, I will write $f \prec g$ whenever $f(x) &lt; g(x)$ holds for all $x \in X$. (Now we have $f \preceq g$ if and only if $f \prec g$ or $f = g$. Note that $f \preceq g$ implies $f \leq g$, but not the other way around.)

For your first question, the answer is yes. Note that $C(X)$ is a lattice and that we have
$$ (f \wedge g)(x) = \min(f(x), g(x)),\qquad\text{for all $x\in X$}. $$
Furthermore, note that an order isomorphism between lattices automatically is a lattice isomorphism. In other words, we have $T(f \wedge g) = Tf \wedge Tg$ for all $f,g\in C(X)$. Now suppose that $f,g\in C(X)$ are given with $f \prec g$ but $Tf \not\prec Tg$, then there is some $y_0\in Y$ such that $Tf(y_0) = Tg(y_0)$ holds. We define $h \in C(Y)$ by
$$ h(y) = \begin{cases} Tf(y) + 1,&amp;\quad\text{if $y = y_0$}; \\ Tf(y),&amp;\quad\text{if $y \neq y_0$}. \end{cases} $$
Now set $h' = T^{-1}(h)$, then we have
$$ T(h' \wedge g) = h \wedge Tg = Tf. $$
Since $T$ is injective, it follows that $h' \wedge g = f$ holds. Since we have $f(x) = \min(h'(x), g(x)) &lt; g(x)$ for all $x\in X$, it follows that $h' = f$ must hold. But now we have $T(h') = h \neq Tf$, which is a contradiction.

For the second question, the answer is no. Suppose that $X = Y$ holds and consider the function $T : C(X) \to C(Y)$ given by $f \mapsto f - 1$, that is:
$$ (Tf)(x) = f(x) - 1. $$
Then clearly $T$ is an order isomorphism, but we have $T(0) \not\geq 0$.
You are correct:
$$
\def\imp{\quad\Longrightarrow\quad}
f &lt; g \imp f \le g \imp Tf \le Tg.
$$
But, if $Tf = Tg$, then $f=g$ (since $T$ is an injection), which contradicts the fact that $f &lt; g$.