Proving the Intermediate Value Theorem for One-to-One Continuous Functions

[Lee33]

Homework Statement

Let $a,b\in \mathbb{R}$, $a<b$ and let $f$ be a continuous real valued function on $[a,b]$. Prove that if $f$ is one-one then $f([a,b])$ is either $[f(a),f(b)]$ or $[f(b),f(a)].$

2. The attempt at a solution

Suppose $f(a) < f(b)$ then by IVT we have if $x\in(a,b)$ then $f(x) \in [f(a),f(b)]$. Arguing by contradiction suppose $f(x) < f(a)$ or $f(x)>f(b)$. If $f(x) <f(a)$ then by IVT $f([x,b])$ contains all points between $f(x)$ and $f(b)$ and contains $f(a)$ contradiction that $f$ is one to one.

Why does that contradict the assumption that $f$ is one to one?

 

[Dick]

Homework Statement

Let $a,b\in \mathbb{R}$, $a<b$ and let $f$ be a continuous real valued function on $[a,b]$. Prove that if $f$ is one-one then $f([a,b])$ is either $[f(a),f(b)]$ or $[f(b),f(a)].$

2. The attempt at a solution

Suppose $f(a) < f(b)$ then by IVT we have if $x\in(a,b)$ then $f(x) \in [f(a),f(b)]$. Arguing by contradiction suppose $f(x) < f(a)$ or $f(x)>f(b)$. If $f(x) <f(a)$ then by IVT $f([x,b])$ contains all points between $f(x)$ and $f(b)$ and contains $f(a)$ contradiction that $f$ is one to one.

Why does that contradict the assumption that $f$ is one to one?

Draw a picture. Suppose f(a)<=f(b) and x is in (a,b) but f(x)<f(a). Doesn't that mean that there is a y in [x,b] such that f(y)=f(a)?

 

[Lee33]

Thank you! I see my mistake in my reasoning now.

 

[Dick]

Thank you! I see my mistake in reasoning now.

I don't think there was any real mistake. Are you clear how on how f(y)=f(a) would contradict one-to-one?

 

[Lee33]

Yup, because we already know $a<x<b$ and if $f(a)\in f([x,b])$ we have that $x\le y\le b$ thus $f(y) = f(a)$ which contradicts one to one since $y\ne a$.

 

[Dick]

Yup, because we already know $a<x<b$ and if $f(a)\in f([x,b])$ we have that $x\le y\le b$ thus $f(y) = f(a)$ which contradiction one to one since $y\ne a$.

Exactly.

 

[Lee33]

Thanks again!