# Prove that if $f:[a,b] \to \mathbb{R}$ is continuous, one-to-one, and satisfies $f(a) < f(b)$ then $a \leq x < y \leq b \implies f(x)<f(y)$.

So I am having a bit of trouble with the following question in the practice problem sheet for my Real Analysis course:

Prove that if $f:[a,b] \to \mathbb{R}$ is continuous, one-to-one, and satisfies $f(a) < f(b)$ then
$a \leq x < y \leq b \implies f(x)<f(y)$.

We are also provided with the following hints:

Suppose not.Then there exists $c,d \in [a,b]$ such that $c < d$ and $f(c) > f(d)$ (why?), and we must have either $f(a) < f(c)$ or $f(d) < f(b)$ (why?).

• You can assume a contrary situation and poly intermediate value theorem to contradict the fact that $f$ is one-one. This is not hard you should give it a try by yourself. – Paramanand Singh May 24 '17 at 13:44

(I fill the missing details in your question.)

Suppose not. Then there exists $c,d\in[a,b]$ such that $c<d$ and $f(c) \geq f(d)$. On the other hand, $f$ is one-to-one, hence it cannot be $f(c) = f(d)$, hence $f(c) > f(d)$.

Clearly we cannot have $f(c) \leq f(a)$ and $f(d) \geq f(b)$, for otherwise $f(d) \geq f(b) > f(a) \geq f(c)$.

Hence one of the following two cases holds:

1) $f(a) < f(c)$

2) $f(d) < f(b)$

...

Suppose that $a \leq x < y \leq b \implies f(x)<f(y)$ does not hold. Then there are $c,d \in [a,b]$ such that $c<d$ but $f(c) \ge f(d)$. We can not have that $f(c)=f(d)$ since $f$ is one-to-one and $c \ne d$. Hence $f(c) > f(d)$.

If $f(c) \le f(a)$ and $f(b) \le f(d)$, then

$f(b) \le f(d) <f(c) \le f(a)$, a contradiction, since $f(a)<f(b)$.

Thus: either $f(a)<f(c)$ or $f(d)<f(b)$ .