# continuous open mapping of segment

$$f : [0,1] \rightarrow [0,1]$$ is a continuous open mapping. Show that it has finite number of maxima and $$\max(f(x)) = 1$$.

It is well-known that if $$g:\mathbb{R}\rightarrow\mathbb{R}$$ is continuous and open than it's monotonic (Every continuous open mapping $\mathbb{R} \to \mathbb{R}$ is monotonic). In the same way I've managed to show that $$\max(f(x)) = 1$$. But how can one prove that the number of maxima is finite?

Suppose that the set $$O=\{x\in[0,1]:f(x)=1\}$$ is an infinite set. Then it has an accumulation point $$a$$ and, since $$f$$ is continuous, $$f(a)=1$$. There is some $$\delta>0$$ such that$$|x-a|<\delta\wedge x\in[0,1]\implies|f(x)-1|<1\label{a}\tag1$$and, since $$a$$ is an accumulation point of $$O$$, $$(x-a,x+a)\cap[0,1]$$ contains some point $$b\ne a$$. I will assume that $$a; the case in which $$b is similar.
We have two possibilities here. Either $$f$$ is constant equal to $$1$$ on $$[a,b]$$ or it is not. In the first case, $$f\bigl((a,b)\bigr)=\{1\}$$, which is impossible, since $$f$$ is an open map. And if $$f$$ is not constant on $$[a,b]$$, $$f|_{[a,b]}$$ has a minimum at some point $$c\in(a,b)$$. Besides, because of \eqref{a}, $$f(c)>0$$, and then $$f\bigl((a,b)\bigr)=[c,1]$$, which is not an open subset of $$[0,1]$$. So, again we reach an impossibility.