Exercise 4.15

Question

Exercise 15: Call a mapping of $X$ into $Y$ {\it open} if $f(V)$ is an open set in $Y$ whenever $V$ is an open set in $X$. Prove that every
    continuous open mapping $f$ of $\mathbf R$ into $\mathbf R$ is monotonic.

analambanomenos · Accepted Answer

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By Theorem 4.16, there is an $x$ and $y$ in any closed interval $[a,b]$ such that $f(x)$ is the minimum value of $f$ on $[a,b]$ and $f(y)$
    is the maximum value. If $a<x<y<b$, then $f\bigl((a,b)\bigr)$ is the closed interval $\bigl[f(x),f(y)\bigr]$, contradicting the openness of
    $f$. Similarly, if $a=x<y<b$ or $a<x<y=b$ the image of $(a,b)$ is a half-closed interval. Hence $f$ must attain its maximum or minimum values
    at the endpoints of any closed interval. Suppose $f(a)$ is the minimum value and $f(b)$ is the maximum value of $f$ on $[a,b]$, and let
    $a<y\le b$. Then $f(y)$ would be the maximum value of $f$ on $[a,y]$, so $f(x)\le f(y)$ for $a\le x<y$, that is, $f$ is monotonically
    increasing on $[a,b]$.

Hence $f$ is monotonic on any closed interval, and if $f$ is monotonically increasing on one closed interval, then it must also be
    monotonically increasing on any larger interval. Hence $f$ is monotonic on all of $\mathbf R$.

Amit Awasthi · Answer

\begin{proof} Let $f:\mathbb{R}^1 \rightarrow \mathbb{R}^1$ be a continuous open mapping. Assume it is not monotonic. Then, there exists two cases: there exist $x_1 < x_2 < x_3$ such that $f(x_1) < f(x_2)$ and $f(x_3) < f(x_2)$, or there exist $x_1 < x_2 < x_3$ such that $f(x_1) > f(x_2)$ and $f(x_3) > f(x_2)$. \par Consider the first case. By the extreme value theorem, $f([x_1,x_3])$ has a maximum attained in this interval; since $f(x_1) < f(x_2)$ and $f(x_3) < f(x_2)$, this maximum is attained in the open interval $(x_1,x_3)$. Now assume that $f( (x_1,x_3) )$ forms an open set. However, this is a contradiction since if this set were open, we could draw a neighborhood around $f(x_2)$ that is contained in $f( (x_1,x_3) )$, but this means that there is a value of $f$ that is greater than $f(x_2)$ that is still in $f( (x_1,x_3) )$. This contradicts the fact that $f(x_2)$ was the maximum. So, the values for $f$ on the interval $(x_1,x_3)$ is not open, which contradicts that $f$ is an open mapping. \par Consider the second case. By the extreme value theorem, $f([x_1,x_3])$ has a minimum attained in this interval; since $f(x_1) > f(x_2)$ and $f(x_3) > f(x_2)$, this minimum is attained in the open interval $(x_1,x_3)$. Now assume that $f( (x_1,x_3) )$ forms an open set. However, this is a contradiction since if this set were open, we could draw a neighborhood around $f(x_2)$ that is contained in $f( (x_1,x_3) )$, but this means that there is a value of $f$ that is less than $f(x_2)$ that is still in $f( (x_1,x_3) )$. This contradicts the fact that $f(x_2)$ was the minimum. So, the values for $f$ on the interval $(x_1,x_3)$ is not open, which contradicts that $f$ is an open mapping. \par Thus, $f$ must be monotonic. \end{proof}

Exercise 4.15

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Exercise 4.15

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