Suppose a function $f(x)$ is continuous on a closed interval $I = [a, b]$, and that $f(I) = [f(a), f(b)]$. Suppose further that as $x$ varies over $I$, $f(x)$ never repeats a value. Prove $f(x)$ is strictly increasing.
Intuitively, I can see that if $f(x)$ were not strictly increasing, it would have to repeat at one point $k \in [a,b]$ so that $f(k)$ = $f(n)$ for some $n \in [a,k)$.
How could I go about proving this formally? I suspect that it has to do with the intermediate value property of continuous functions on a closed interval, since the reverse argument holds (if $f(x)$ is strictly increasing on $[a,b]$ and has the IVP, then it is continuous on $[a,b]$.)
Thank you!
Suppose $f(x)$ is not strictly increasing.
That is, there exists $x<y$ s.t. $f(x)\geq f(y)$.
$f(a) \leq f(y)$. Therefore $f(a) \leq f(y) \leq f(x)$.
By IVT, there exists $c\in [a,x]$ s.t. $f(c) = f(y)$.