Let $D\subset \mathbb{R}$ and D is bounded and closed.
$$ f:D \rightarrow f(D)\in \mathbb{R}$$ is continuous and injective. Then, the inverse $ f^{-1}:f(D) \rightarrow D$ is continuous in the interval $f(D)$.
What I've got so far: Because the function is injective and is mapped onto $f(D)$ it is also bijective. Furthermore, I've proven that when it is injective, it must be either strictly increasing or strictly decreasing, therefore $f^{-1}$ is,too.
From the definition of continuity, I know:
$ \forall \varepsilon > 0 \; \; \exists \delta > 0$ such that $|f(x)-f(x_0)|< \varepsilon \; \; \forall x \in D : |x-x_0| < \delta $
My thoughts are, I need to pick some $y \in f(D)$ and use the above mentioned definition along with the strictly increasing monotonicity and the fact, that the interval is bounded and closed. I was told, this was a pretty important part, yet I am very unsure, as to what that even tells me.
Could somebody provide me with some hints, as to what useful information I can extract from the fact, that it is bounded and closed? I'd assume, I need the monotonicity mainly for the purpose of having a well defined inverse, so I'm not sure, if I still need it at this point.
Best Answer
Hint: First prove that for any closed subset $K$ of $D$, $f(K)$ is closed in $\mathbb{R}$.
To prove this you should use the fact that a subset of $\mathbb{R}$ is compact if and only if it is closed and bounded. (See: show that a subset of $\mathbb{R}$ is compact iff it is closed and bounded)