TITLE: Let $ f: ( 0, \infty) \rightarrow ( 0, \infty)$ bijective with $ f^{-1}(x) \cdot f(x) =1$.
QUESTION [8 upvotes]: Let $ f: ( 0, \infty) \rightarrow  ( 0, \infty)$ bijective  with $ f^{-1}(x) \cdot f(x) =1$.Prove that  there is an interval $ I $ such that  $f(I) $ is not an interval. I choose two elements $ x, y \in ( 0, \infty)$ such that $ f(x) <1 $  and $ f(y) >1$  and I tried to send $ x \rightarrow f(x) $  of infinetely many times to obtain that $f(I) = ( 0, \infty)$.How to try in another way?

REPLY [3 votes]: Suppose that $f(I)$ is an interval for every interval $I\subset (0,\infty)$. Then $f$ maps open intervals in open intervals. In fact, suppose that $f((a,b))=(c,d]$ (the other cases are similar), then either $f((a,f^{-1}(d)])$ or $f([f^{-1}(d),b))$ is not an interval. Hence $f^{-1}$ is a continuous function, thus by the injectivity is strictly monotone. Let say strictly increasing (the decreasing case is the same), then $f$ would be strictly increasing as well, contradicting the assumption $f(x)\cdot f^{-1}(x)=1$.