Answer to Question #2382 in Real Analysis for Ravik
If f from R to R is continuous and open, show that f is strictly monotone.
If f is not monotone, then there are two distinct points a<b such that f(a)=f(b). Then by Weierstrass' theorem f has a maximum C and minimum c on the closed interval [a,b]. We can assume that at least one of the numbers f(C) or f(c), say f(C) differs from f(a)=f(b).
Then the image open interval f( (a,b) ) = (f(a), C] is not open, which contradicts to openness of f.
Everything works, thank you (Operator and Expert) so much! I will bookmark this page for future reference. Excellent job on the comments to by the way, now I can study and understand what is happening for the future!