I have the following proposition:
"Let $C$ be the set of continuous functions $f:[a,b] \rightarrow [a,b]$ with the sup metric. The subset of $C$ made by taking the continuous functions that are not surjective is open."
I've managed to prove this, but by using results from complete metric spaces:
- proved that $C$ with the sup metric is complete;
- proved that the subset of $C$ made by taking the continuous functions that are surjective is complete;
- proved that a complete subspace of any metric space is closed;
- finished by using that a set is open if its complement is closed.
What I'm looking for is a simpler proof, a less "complete" proof.
My idea: I thought about again proving that the subset of $C$ taking the surjective functions is closed. Maybe using that every continuous function in $f:[a,b] \rightarrow [a,b]$ is bounded. How can I go from that (if at all)?