Let $X$ and $Y$ are ordered set in the order topology. Show that if $f:X\rightarrow Y$ is order preserving and surjective, then $f$ is a homeomorphism.
i got the answer here $X$ and $Y$ are ordered set, $f:X\rightarrow Y$ is order preserving and surjective, then $f$ is a homeomorphism. .. but i didn't understand it
In an ordered space $X$, with linear order $<$, the topology is generated by two types of sets:$(\leftarrow,p)=\{x \in X: x < p\}$, where $p\in X$ can be any point, and all sets of the form $(p,\rightarrow)=\{ x \in X: x > p\}$, again with $p \in X$. Such sets are all open by definition in the order topology, and they form a subbase for it.
In the linked answer the useful fact is used that a function is continuous iff the inverse images of subbasic elements are open.