Open set

**Chandru1** · March 6th 2009, 07:45 AM

A map $f:\mathbb{R} \to \mathbb{R}$ is called open if $f(A)$ is open for every open
subset A of R. Show that every continuous open map of R into
itself is monotonic.

**Plato** · March 6th 2009, 09:24 AM

Because this function is continuous then on any interval $[a,b]$ it has a high point and a low point: $\left( {\exists \left\{ {c,d} \right\} \subseteq \left[ {a,b} \right]} \right)\left( {\forall x \in \left[ {a,b} \right]} \right)\left[ {f(d) \leqslant f(x) \leqslant f(c)} \right]$ .
Suppose that $\left\{ {c,d} \right\} \subseteq \left( {a,b} \right)$ then $f\left[ {\left( {a,b} \right)} \right] = \left[ {f(d),f(c)} \right]$ which is contradictory to $f$ 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 $a < b < c\;. \mathrel\backepsilon .\;f(a) < f(b)\;\& \;f(c) < f(b)$ .
But that would mean that the maximum on the interval $[a,c]$ is not at endpoint.
The same contradiction would follow if $a < b < c\;. \mathrel\backepsilon .\;f(a) > f(b)\;\& \;f(c) > f(b)$ .
In other words $f$ must be monotonic.

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