Continuous Maps on Intervals

Persistent application of the Intermediate Value Theorem shows us that if $f\,:\, (0,1) \to \mathbb{R}$ is continuous, then it must be strictly monotonic.

As a flavour of the proof, suppose that $f(\tfrac13) < f(\tfrac23)$ . If $\tfrac13 < x < \tfrac23$ and $f(x) \ge f(\tfrac23)$ , then the IVT gives us $\tfrac13 \le y \le x$ with $f(y) = f(\tfrac23)$ . Thus $f(x) < f(\tfrac23)$ . Similarly $f(x) > f(\tfrac13)$ ,and hence $f(\tfrac13) < f(x) < f(\tfrac23)$ . We can similarly show that $f(x) < f(\tfrac13)$ for $x < \tfrac13$ and $f(x) > f(\tfrac23)$ for $x > \tfrac23$ . From this, we can go through a few more cases to see that $f(x) < f(y)$ whenever $0 < x < y < 1$ . Thus $f$ is strictly increasing. If $f(\tfrac13) > f(\tfrac23)$ , then $f$ will be strictly decreasing.

From this it is clear that the range of $f$ must be an open interval, and so cannot be $[0,1]$ .

$\boxed{1}$ By contradiction. If $f$ was a bijective continuous function (homeomorphism), then $f^{-1}$ would be also a homeomorphism, but $[0,1]$ is a compact set and $(0, 1)$ is not a compact sect, and compactness is a topological invariant, i.e, each homeomorphism gets a compact set from a compact set. Therefore, $f$ can't be a homeomorphism.

$\boxed{2}$ By contradiction. If $f$ was a bijective continuous function (homeomorphism), then $f^{-1}$ would be also a homeomorphism, and hence $f^{-1} ([0, 1))$ would be a connected set (Contradiction).