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$f : [0,1] \rightarrow [0,1]$ 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?

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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 $a<b$; the case in which $b<a$ 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.

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