Let $ f: ( 0, \infty) \rightarrow ( 0, \infty)$ bijective with $ f^{-1}(x) \cdot f(x) =1$. 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?
 A: If $f(x)$ is strictly increasing then so is $f^{-1}(x)$; but then their product would have to be strictly increasing, contradicting that it should be constant by $f^{-1}(x)\cdot f(x)=1$. In the same way $f(x)$ can't be strictly decreasing. It also can't be weakly increasing/decreasing, because it is injective.
Since $f(x)$ is not monotone either we can find $x_1<x_2<x_3$ such that $$f(x_1)<f(x_2) ~\land~ f(x_2)>f(x_3)$$ or such that $$f(x_1)>f(x_2) ~\land~ f(x_2)<f(x_3).$$ Let's assume the first case (the other is the same).
Notice we can't have $f(x_1)=f(x_3)$ (injectivity), so we can assume $f(x_1)>f(x_3)$ (the other case is the same).
             
             
             
          
Let $I=[x_2,x_3]$.
Assume that $f(I)$ was an interval. Since $f(x_2),f(x_3)\in f(I)$ then for all $y$ satisfying $f(x_2) \geq y\geq f(x_3)$ we must have $y\in f(I)$. This is true in particular for $y=f(x_1)$, since $f(x_1)<f(x_2)$ and we assumed $f(x_1)>f(x_3)$. That is,
$$f(x_1)\in f(I)=f([x_2,x_3])$$
so there is some $t\in [x_2,x_3]$ such that $f(t)=f(x_1)$, which is impossible due to injectivity, so our assumption that $f(I)$ is an interval led to a contradiction.
A: 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 $f(x)\cdot f^{-1}(x)=1$.
