topological mapping ordered sets

Question

Let $X, X'$ be two ordered sets, each topologized with the right topology ( $[x\rightarrow[$ ). How to show that a mapping $f: X \rightarrow X'$ is continuous if and only if it is order-preserving.

Answer 1

Suppose we have $x_1 \le x_2$ in $X$. Suppose $f$ is continuous.

Then $f^{-1}[f(x_1), \rightarrow)]$ is open and contains $x_1$. Hence it contains $[x_1, \rightarrow)$ (which is the minimal open set that contains $x_1$), so $$x_2 \in [x_1, \rightarrow) \subseteq f^{-1}[f(x_1), \rightarrow)]$$ and so $f(x_2) \in [f(x_1), \rightarrow)$ and so $f(x_1) \le f(x_2)$. As $x_1 \le x_2$ were arbitrary $f$ is order-preserving.

The reverse is quite clear when $f$ is moreover onto; try to show that first, and then try the general case.