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Suppose that $I\subseteq\mathbb{R}$ is nonempty. If $f:I\to\mathbb{R}$ is $1{-}1$ and continuous, then $f$ is strictly monotone on $I$.

The answer in the back of the book$^1$, which I found after writing the following proof, says this is false (no explanation). However, I cannot find an error in my proof or think of a viable counterexample. What am I overlooking?

Suppose not, and there exists some $x_1,x_2,x_3\in I$ with $x_1<x_2<x_3$ such that $f(x_1)\leq f(x_2)$ and $f(x_2)\geq f(x_3)$, or $f(x_1)\geq f(x_2)$ and $f(x_2)\leq f(x_3)$. Without loss of generality, we may assume that $f(x_1)\leq f(x_2)$ and $f(x_2)\geq f(x_3)$.

If $f(x_1)=f(x_2)$ or $f(x_2)=f(x_3)$ then $f$ would not be $1{-}1$; thus, $f(x_1)<f(x_2)$ and $f(x_2)>f(x_3)$. We are left with two possible cases.

Case 1. Suppose $f(x_1)<f(x_3)$. Then $f(x_1)<f(x_3)<f(x_2)$. By the Intermediate Value Theorem, there exists point $x_0\in (x_1, x_2)$ such that $f(x_0)=f(x_3)$ with $x_0\neq x_3$ since $x_3\notin (x_1,x_2)$, a contradiction since $f$ is $1{-}1$.

Case 2. Suppose $f(x_1)>f(x_3)$. Then $f(x_3)<f(x_1)<f(x_2)$. By the Intermediate Value Theorem, there exists point $x_0\in (x_2,x_3)$ such that $f(x_0)=f(x_1)$ with $x_0\neq x_1$ since $x_1\notin (x_2,x_3)$, a contradiction since $f$ is $1{-}1$.

If we assume that $f(x_1)\geq f(x_2)$ and $f(x_2)\leq f(x_3)$ we are lead to similar contradictions; thus, $f$ is strictly monotone. $\blacksquare$

1: An Introduction to Analysis by William R. Wade, 4th edition

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    That $I$ is an interval and not just a nonempty subset of $\mathbb R$ is taken for granted, I assume? – Hagen von Eitzen Jun 10 '13 at 21:19
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    It is true if $I$ is an interval. Is that a valid assumption? The stated question only tells us that $I$ is a subset. Could $I$ be the rationals? – Calvin Lin Jun 10 '13 at 21:21
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    @Alex As Hagen hints, the error probably is when you apply the IVT. You might not be working with an interval and so IVT might not work. – Git Gud Jun 10 '13 at 21:22

1 Answers1

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The trick in the text of this exercise that $I$ is not assumed to be an interval. Your proof is valid on an interval.

But, for example, let $I:=[0,1]\cup[2,3]$ and let $f(x):=\left\{\matrix{x,\ \text{ if }\,x\le 1\\5-x,\ \text{if }\, x\ge 2} \right.$.

Berci
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    People say that God is in the details, but sometimes I think it's the devil. At least I know it's true for intervals! Merci, Berci. –  Jun 10 '13 at 21:38