Assume $f$ is a continuous one-to-one function over an interval. Prove that $f$ is strictly monotone

Assume $f$ is a continuous one-to-one function over an interval. Prove that $f$ is strictly monotone.

Attempt

Since we know that $f$ is one-to-one, for every $f(x)$ there is exactly one element $x_0$ that maps to it. Thus, if $f(x) = f(y)$, then $x=y$. Now let's suppose that $f$ isn't monotone. That is, it goes from nonincreasing to nondecreasing. Thus, on some interval $[a,b]$ we must have $f'(x) < 0$ and on some other interval $[c,d]$ we must have $f'(x) > 0$. As a result we must have that $f'(x) = 0$ at some point. How do I show this contradicts the definition of $f$?

Solution 1:

It is given that $f$ is one-to-one.

Within that context, to say that $f$ is not strictly monotone is to say that one can find $a<b<c$ within the interval for which either $f(a)<f(b)>f(c)$ or $f(a)>f(b)<f(c)$.

Suppose $f(a)<f(b)>f(c)$. Then $f(b)>\max\{f(a),f(c)\}$. Since $f$ is continuous, the intermediate value theorem can be used. It tells us that there exist $x_1\in[a,b)$ and $x_2\in(b,c]$ such that $f(x_1)=f(x_2)$. But $x_1<b<x_2$ so $x_1\ne x_2$. Thus $f$ is not one-to-one. So the assumption that $f$ is not strictly monotone cannot be true.