Is an Order Isomorphism from (R,<) to (R,<) Always Continuous?

[TTob]

order isomorphism f:R-->R

let f is order isomorphism from (R,<) to (R,<). why f is continuous ?
so f is bijection and a<b <--> f(a)<f(b), so what ?

 

[HallsofIvy]

f maps the interval [a, a+ epsilon] one to one and onto the interval [f(a),f(a+ epsilon)]. As epsilong goes to 0, f(a+ epsilon) must go to a, hence continuity.

 

[adriank]

Another way of proving it is by showing that inverse images of open sets of R (in the order topology) are open. This is easy; testing just elements of a basis for the topology (such as the set of bounded open intervals) suffices. Given an open interval (c, d), c = f(a) and d = f(b) for some a and b, and it's easy to see that f^-1((c, d)) = (a, b), which is open.

(This works equally well for any ordered set.)