A map $\displaystyle f:\mathbb{R} \to \mathbb{R}$ is called open if $\displaystyle f(A)$ is open for every open

subset A of R. Show that every continuous open map of R into

itself is monotonic.

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- Mar 6th 2009, 07:45 AMChandru1Open set
A map $\displaystyle f:\mathbb{R} \to \mathbb{R}$ is called open if $\displaystyle f(A)$ is open for every open

subset A of R. Show that every continuous open map of R into

itself is monotonic. - Mar 6th 2009, 09:24 AMPlato
Because this function is continuous then on any interval $\displaystyle [a,b]$ it has a high point and a low point: $\displaystyle \left( {\exists \left\{ {c,d} \right\} \subseteq \left[ {a,b} \right]} \right)\left( {\forall x \in \left[ {a,b} \right]} \right)\left[ {f(d) \leqslant f(x) \leqslant f(c)} \right]$.

Suppose that $\displaystyle \left\{ {c,d} \right\} \subseteq \left( {a,b} \right)$ then $\displaystyle f\left[ {\left( {a,b} \right)} \right] = \left[ {f(d),f(c)} \right]$ which is contradictory to $\displaystyle f$ being open.

A similar argument leads to the conclusion that the maximum and minimum must happen**at the endpoints of any closed interval**.

Suppose that there are points $\displaystyle a < b < c\;. \mathrel\backepsilon .\;f(a) < f(b)\;\& \;f(c) < f(b)$.

But that would mean that the maximum on the interval $\displaystyle [a,c]$ is not at endpoint.

The same contradiction would follow if $\displaystyle a < b < c\;. \mathrel\backepsilon .\;f(a) > f(b)\;\& \;f(c) > f(b)$.

In other words $\displaystyle f$ must be monotonic.