Theorem $4.17$ from Rudin Principles of Mathematical Analysis:
Theorem $4.17$
Suppose $f$ is a continuous 1-1 mapping of a compact metric space $X$ onto a metric space $Y$. Then the inverse mapping $f^{-1}$ defined on $Y$ by $$f^{-1}(f(x)) = x (x \in X)$$ is a continuous mapping of $Y$ onto $X$.
I was told that "on $X=(a,b)$ and for a real-valued function $f$ (and of course for a continuous 1-1 function $f$), the conclusion of the theorem holds (that is to say, $f^{-1}$ is continuous)".
My question is: is this true because
- $f$ is continuous and injective on $(a,b)$, hence stricly monotonous on $(a,b)$ and hence has its inverse continuous
or
- because on every $[c,d]$ included in $(a,b)$ we have, by $4.17$, $f^{-1}$ continuous on $f([c,d])$ hence for $f$ defined on $(a,b)$ we have $f^{-1}$ continuous on $f((a,b))$
Or is it true because of both above-mentioned reasons ?
PS: Is the second bullet even true?