1. ## continuous

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

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

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

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

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

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

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