1. ## continuous

Suppose $\displaystyle I$ is an interval in $\displaystyle \mathbb{R}$ and let $\displaystyle f: I \to \mathbb{R}$ be a continuous function. Prove that $\displaystyle f$ is injective $\displaystyle \Longleftrightarrow f$ is strictly monotonic.

$\displaystyle \Rightarrow$ direction: Let $\displaystyle a,b \in I$. Suppose $\displaystyle a \neq b$. Then $\displaystyle f(a) \neq f(b)$. So $\displaystyle a<b$ or $\displaystyle a > b$. And $\displaystyle f(a) < f(b)$ or $\displaystyle f(a) > f(b)$. From here, would you use the intermediate value theorem?

$\displaystyle \Leftarrow$ direction: Suppose $\displaystyle f$ is strictly monotonic. Then $\displaystyle x \neq y \implies f(x) \neq f(y)$.

2. Originally Posted by manjohn12
Suppose $\displaystyle I$ is an interval in $\displaystyle \mathbb{R}$ and let $\displaystyle f: I \to \mathbb{R}$ be a continuous function. Prove that $\displaystyle f$ is injective $\displaystyle \Longleftrightarrow f$ is strictly monotonic.

$\displaystyle \Rightarrow$ direction: Let $\displaystyle a,b \in I$. Suppose $\displaystyle a \neq b$. Then $\displaystyle f(a) \neq f(b)$. So $\displaystyle a<b$ or $\displaystyle a > b$. And $\displaystyle f(a) < f(b)$ or $\displaystyle f(a) > f(b)$. From here, would you use the intermediate value theorem?

$\displaystyle \Leftarrow$ direction: Suppose $\displaystyle f$ is strictly monotonic. Then $\displaystyle x \neq y \implies f(x) \neq f(y)$.
If $\displaystyle f$ is strictly monotonic it means if $\displaystyle a\not = b$, say $\displaystyle a<b$ WLOG then $\displaystyle f(a) < f(b)$ or $\displaystyle f(a)>f(b)$. Thus, we see $\displaystyle a\not = b \implies f(a)\not = f(b)$. This implies that $\displaystyle f$ is one-to-one.

Now we will show that if $\displaystyle f$ is one-to-one then $\displaystyle f$ is strictly monotonic. This part is more involved. The hint that will help you solve this is if: $\displaystyle a<b<c$ then $\displaystyle f(b)$ lies between $\displaystyle f(a),f(c)$. We will prove the hint and you can try finishing the proof. If $\displaystyle f(b)$ did not lie between $\displaystyle f(a),f(c)$ it means $\displaystyle f(b) > M=\text{max} \{ f(a),f(c)\}$. Pick a real number $\displaystyle r$ so that $\displaystyle f(b) > r > M$. Then by intermediate value theorem there is $\displaystyle a<x_0<b \implies f(x_0) = r \text{ and }b<x_1<c \implies f(x_1) = r$. But then $\displaystyle f(x_0) = f(x_1)$ for $\displaystyle x_0\not = x_1$. A contradiction.